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a469991347d8b43b6bc84d80f5bc9385 | https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_14 | For any real number a and positive integer k, define
${a \choose k} = \frac{a(a-1)(a-2)\cdots(a-(k-1))}{k(k-1)(k-2)\cdots(2)(1)}$
What is
${-\frac{1}{2} \choose 100} \div {\frac{1}{2} \choose 100}$
$\textbf{(A)}\ -199\qquad \textbf{(B)}\ -197\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ 197\qquad \textbf{(E)}\ 199$ | We expand both the numerator and the denominator.
\begin{align*} \binom{-\frac{1}{2}}{100}\div\binom{\frac{1}{2}}{100} &= \frac{ \dfrac{ (-\frac{1}{2}) (-\frac{1}{2} - 1) (-\frac{1}{2} - 2) \cdots (-\frac{1}{2} - (100 - 1)) }{\cancel{(100)(99)\cdots(1)}} }{ \dfrac{ (\frac{1}{2}) (\frac{1}{2} - 1) (\frac{1}{2} - 2) \cdots (\frac{1}{2} - (100 - 1)) }{\cancel{(100)(99)\cdots(1)}} } \\ &= \frac{ (-\frac{1}{2}) (-\frac{1}{2} - 1) (-\frac{1}{2} - 2) \cdots (-\frac{1}{2} - 99) }{ (\frac{1}{2}) (\frac{1}{2} - 1) (\frac{1}{2} - 2) \cdots (\frac{1}{2} - 99) } \end{align*}
Now, note that $-\frac{1}{2}-1=\frac{1}{2}-2$ $-\frac{1}{2}-2=\frac{1}{2}-3$ , etc.; in essence, $-\frac{1}{2}-n=\frac{1}{2}-(n+1)$ . We can then simplify the numerator and cancel like terms.
\begin{align*} \frac{ (-\frac{1}{2}) (-\frac{1}{2} - 1) (-\frac{1}{2} - 2) \cdots (-\frac{1}{2} - 99) }{ (\frac{1}{2}) (\frac{1}{2} - 1) (\frac{1}{2} - 2) \cdots (\frac{1}{2} - 99) } &= \frac{ \cancel{(\frac{1}{2} - 1)} \cancel{(\frac{1}{2} - 2)} \cancel{(\frac{1}{2} - 3)} \cdots (\frac{1}{2} - 100) }{ (\frac{1}{2}) \cancel{(\frac{1}{2} - 1)} \cancel{(\frac{1}{2} - 2)} \cdots \cancel{(\frac{1}{2} - 99)} } \\ &= \frac{\frac{1}{2}-100}{\frac{1}{2}} \\ &= \frac{-\frac{199}{2}}{\frac{1}{2}} \\ &= \boxed{199.} | A | 199. |
c5e9e853d0614dbddab465fa09aa18f8 | https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_15 | If $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^3 + bx^2 + 1$ , then $b$ is
$\textbf{(A)}\ -2\qquad \textbf{(B)}\ -1\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2$ | Using polynomial division, we find that the remainder is $(2a+b)x+(a+b+1)$ , so for the condition to hold, we need this remainder to be $0$ . This gives $2a+b=0$ and $a+b+1=0$ , so $b=-2a$ and $a-2a+1=0 \implies a=1 \implies b=-2$ , which is $\boxed{2}.$ | A | 2 |
2ea06b38262305decf5689cdba769f7a | https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_21 | The complex number $z$ satisfies $z + |z| = 2 + 8i$ . What is $|z|^{2}$ ? Note: if $z = a + bi$ , then $|z| = \sqrt{a^{2} + b^{2}}$
$\textbf{(A)}\ 68\qquad \textbf{(B)}\ 100\qquad \textbf{(C)}\ 169\qquad \textbf{(D)}\ 208\qquad \textbf{(E)}\ 289$ | Let the complex number $z$ equal $a+bi$ . Then the preceding equation can be expressed as \[a+bi+\sqrt{a^2+b^2} = 2+8i\] Because $a$ and $b$ must both be real numbers, we immediately have that $bi = 8i$ , giving $b = 8$ . Plugging this in back to our equation gives us $a+\sqrt{a^2+64} = 2$ .
Rearranging this into $2-a = \sqrt{a^2+64}$ , we can square each side of the equation resulting in \[4-4a+a^2 = a^2+64\] Further simplification will yield $60 = -4a$ meaning that $-15 = a$ . Knowing both $a$ and $b$ , we can plug them in into $a^2+b^2$ . Our final answer is $\boxed{289}$ | null | 289 |
2c739cb9458a31d1caefe2c78d074712 | https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_22 | For how many integers $x$ does a triangle with side lengths $10, 24$ and $x$ have all its angles acute?
$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ \text{more than } 7$ | We first notice that the sides $10$ and $24$ , can be part of $2$ different right triangles, one with sides $10,24,26$ , and the other with a leg
somewhere between $21$ and $22$ . We now notice that if $x$ is less than or equal to $21$ , one of the angles is obtuse, and that the same is the same for
any value of $x$ above $26$ . Thus the only integer values of $x$ that fit the conditions, are $x=22, 23, 24, \text{and }25.$ So, the answer is $\boxed{4}$ | A | 4 |
520afaab734b0734020e16a8f542df03 | https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_23 | The six edges of a tetrahedron $ABCD$ measure $7, 13, 18, 27, 36$ and $41$ units. If the length of edge $AB$ is $41$ , then the length of edge $CD$ is
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 13\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$ | By the triangle inequality in $\triangle ABC$ , we find that $BC$ and $CA$ must sum to greater than $41$ , so they must be (in some order) $7$ and $36$ $13$ and $36$ $18$ and $27$ $18$ and $36$ , or $27$ and $36$ . We try $7$ and $36$ , and now by the triangle inequality in $\triangle ABD$ , we must use the remaining numbers $13$ $18$ , and $27$ to get a sum greater than $41$ , so the only possibility is $18$ and $27$ . This works as we can put $BC = 36$ $AC = 7$ $AD = 18$ $BD = 27$ $CD = 13$ , so that $\triangle ADC$ and $\triangle BDC$ also satisfy the triangle inequality. Hence we have found a solution that works, and it can be verified that the other possibilities don't work, though as this is a multiple-choice competition, you probably wouldn't do that in order to save time. In any case, the answer is $CD = 13$ , which is $\boxed{13}$ | B | 13 |
aa23d696340e88a23264f40fe1a40dac | https://artofproblemsolving.com/wiki/index.php/1987_AHSME_Problems/Problem_3 | How many primes less than $100$ have $7$ as the ones digit? (Assume the usual base ten representation)
$\text{(A)} \ 4 \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 6 \qquad \text{(D)} \ 7 \qquad \text{(E)} \ 8$ | List out all numbers that have 7 as the ones digit less than 100: ${7, 17, 27, 37, 47, 57, 67, 77, 87, 97}$ . Only $7, 17,37, 47,67,$ and $97$ are prime. Thus, it is $\boxed{6}$ . -slackroadia | C | 6 |
6d79b91c0e42633be0d13f7c7aef166e | https://artofproblemsolving.com/wiki/index.php/1987_AHSME_Problems/Problem_5 | A student recorded the exact percentage frequency distribution for a set of measurements, as shown below.
However, the student neglected to indicate $N$ , the total number of measurements. What is the smallest possible value of $N$
\[\begin{tabular}{c c}\text{measured value}&\text{percent frequency}\\ \hline 0 & 12.5\\ 1 & 0\\ 2 & 50\\ 3 & 25\\ 4 & 12.5\\ \hline\ & 100\\ \end{tabular}\]
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 16 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 50$ | Note that $12.5\% = \frac{1}{8}$ $25\% = \frac{1}{4}$ , and $50\% = \frac{1}{2}$ . Thus, since the frequencies must be integers, $N$ must be divisible by $2$ $4$ , and $8$ (so that $\frac{N}{8}$ etc. are integers), or in other words, $N$ is divisible by $8$ . Thus the smallest possible value of $N$ is the smallest positive multiple of $8$ , which is $8$ itself, or $\boxed{8}$ | B | 8 |
31644055f1991fb77cc28aaa5783ddf0 | https://artofproblemsolving.com/wiki/index.php/1987_AHSME_Problems/Problem_10 | How many ordered triples $(a, b, c)$ of non-zero real numbers have the property that each number is the product of the other two?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$ | We have $ab = c$ $bc = a$ , and $ca = b$ , so multiplying these three equations together gives $a^{2}b^{2}c^{2} = abc \implies abc(abc-1)=0$ , and as $a$ $b$ , and $c$ are all non-zero, we cannot have $abc = 0$ , so we must have $abc = 1$ . Now substituting $bc = a$ gives $a(bc) = 1 \implies a^2 = 1 \implies a = \pm 1$ . If $a = 1$ , then the system becomes $b = c, bc = 1, c = b$ , so either $b = c = 1$ or $b = c = -1$ , giving $2$ solutions. If $a = -1$ , the system becomes $-b = c, bc = -1, -c = b$ , so $-b = c = 1$ or $b = -c = 1$ , giving another $2$ solutions. Thus the total number of solutions is $2 + 2 = 4$ , which is answer $\boxed{4}$ | D | 4 |
06a2d2882af2052ed5e85e4d99c347f5 | https://artofproblemsolving.com/wiki/index.php/1987_AHSME_Problems/Problem_13 | A long piece of paper $5$ cm wide is made into a roll for cash registers by wrapping it $600$ times around a cardboard tube of diameter $2$ cm,
forming a roll $10$ cm in diameter. Approximate the length of the paper in meters.
(Pretend the paper forms $600$ concentric circles with diameters evenly spaced from $2$ cm to $10$ cm.)
$\textbf{(A)}\ 36\pi \qquad \textbf{(B)}\ 45\pi \qquad \textbf{(C)}\ 60\pi \qquad \textbf{(D)}\ 72\pi \qquad \textbf{(E)}\ 90\pi$ | Notice (by imagining unfolding the roll), that the length of the paper is equal to the sum of the circumferences of the concentric circles, which is $\pi$ times the sum of the diameters. Now the, the diameters form an arithmetic series with first term $2$ , last term $10$ , and $600$ terms in total, so using the formula $\frac{1}{2}n(a+l)$ , the sum is $300 \times 12 = 3600$ , so the length is $3600\pi$ centimetres, or $36\pi$ metres, which is answer $\boxed{36}$ | A | 36 |
f535efcec2d8685809e74b07aac3f5b2 | https://artofproblemsolving.com/wiki/index.php/1987_AHSME_Problems/Problem_15 | If $(x, y)$ is a solution to the system $xy=6$ and $x^2y+xy^2+x+y=63$ ,
find $x^2+y^2$
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ \frac{1173}{32} \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 69 \qquad \textbf{(E)}\ 81$ | First note that $x^2y+xy^2+x+y= (xy+1)(x+y)$ . Substituting $6$ for $xy$ gives $7(x+y)= 63$ , giving a result of $x+y=9$ . Squaring this equation and subtracting by $12$ , gives us $x^2+y^2= \boxed{69}$ | null | 69 |
d7e538ca9a6e61760b52fce513cab657 | https://artofproblemsolving.com/wiki/index.php/1987_AHSME_Problems/Problem_16 | A cryptographer devises the following method for encoding positive integers. First, the integer is expressed in base $5$ .
Second, a 1-to-1 correspondence is established between the digits that appear in the expressions in base $5$ and the elements of the set $\{V, W, X, Y, Z\}$ . Using this correspondence, the cryptographer finds that three consecutive integers in increasing
order are coded as $VYZ, VYX, VVW$ , respectively. What is the base- $10$ expression for the integer coded as $XYZ$
$\textbf{(A)}\ 48 \qquad \textbf{(B)}\ 71 \qquad \textbf{(C)}\ 82 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 113$ | Since $VYX + 1 = VVW$ , i.e. adding $1$ causes the "fives" digit to change, we must have $X = 4$ and $W = 0$ . Now since $VYZ + 1 = VYX$ , we have $X = Z + 1 \implies Z = 4 - 1 = 3$ . Finally, note that in $VYX + 1 = VVW$ , adding $1$ will cause the "fives" digit to change by $1$ if it changes at all, so $V = Y + 1$ , and thus since $1$ and $2$ are the only digits left (we already know which letters are assigned to $0$ $3$ , and $4$ ), we must have $V = 2$ and $Y = 1$ . Thus $XYZ = 413_{5} = 4 \cdot 5^{2} + 1 \cdot 5 + 3 = 100 + 5 + 3 = 108$ , which is answer $\boxed{108}$ | D | 108 |
f1302daa2efd364bdbd0d75bd7638353 | https://artofproblemsolving.com/wiki/index.php/1987_AHSME_Problems/Problem_19 | Which of the following is closest to $\sqrt{65}-\sqrt{63}$
$\textbf{(A)}\ .12 \qquad \textbf{(B)}\ .13 \qquad \textbf{(C)}\ .14 \qquad \textbf{(D)}\ .15 \qquad \textbf{(E)}\ .16$ | We have $\sqrt{65} > 8 > 7.5$ . Also $7.5^2 = (7 + 0.5)^2 = 7^2 + 2 \cdot 7 \cdot 0.5 + 0.5^2 = 49 + 7 + 0.25 = 56.25 < 63$ , so $\sqrt{63} > 7.5$ . Thus $\sqrt{65} + \sqrt{63} > 7.5 + 7.5 = 15$ . Now notice that $\sqrt{65} - \sqrt{63} = \frac{(\sqrt{65} - \sqrt{63})(\sqrt{65} + \sqrt{63})}{\sqrt{65} + \sqrt{63}} = \frac{2}{\sqrt{65} + \sqrt{63}}$ , so $\sqrt{65} - \sqrt{63} < \frac{2}{15} = 0.1333333...$ , so the answer must be $A$ or $B$ . To determine which, we write $\sqrt{65} - \sqrt{63} > 0.125 \iff 65 - 2\sqrt{65 \cdot 63} + 63 > 0.015625 \iff 128 - 0.015625 > 2\sqrt{4095} \iff \sqrt{4095} < 64 - 0.0078125 \iff 4095 < 4096 - 128 \cdot 0.0078125 + 0.0078125^2 = 4096 - 1 + 0.0078125^2$ which is true. Hence as the expression is greater than $0.125$ , and less than or equal to $0.13$ (since we showed it is certainly less than $0.1333333...$ ), it is closest to $0.13$ , which is answer $\boxed{.13}$ | B | .13 |
912ba7d671f7eb6bdd0ab5f45d6b876b | https://artofproblemsolving.com/wiki/index.php/1987_AHSME_Problems/Problem_21 | There are two natural ways to inscribe a square in a given isosceles right triangle.
If it is done as in Figure 1 below, then one finds that the area of the square is $441 \text{cm}^2$ .
What is the area (in $\text{cm}^2$ ) of the square inscribed in the same $\triangle ABC$ as shown in Figure 2 below?
[asy] draw((0,0)--(10,0)--(0,10)--cycle); draw((-25,0)--(-15,0)--(-25,10)--cycle); draw((-20,0)--(-20,5)--(-25,5)); draw((6.5,3.25)--(3.25,0)--(0,3.25)--(3.25,6.5)); label("A", (-25,10), W); label("B", (-25,0), W); label("C", (-15,0), E); label("Figure 1", (-20, -5)); label("Figure 2", (5, -5)); label("A", (0,10), W); label("B", (0,0), W); label("C", (10,0), E); [/asy]
$\textbf{(A)}\ 378 \qquad \textbf{(B)}\ 392 \qquad \textbf{(C)}\ 400 \qquad \textbf{(D)}\ 441 \qquad \textbf{(E)}\ 484$ | We are given that the area of the inscribed square is $441$ , so the side length of that square is $21$ . Since the square divides the $45-45-90$ larger triangle into 2 smaller congruent $45-45-90$ , then the legs of the larger isosceles right triangle ( $BC$ and $AB$ ) are equal to $42$ [asy] draw((0,0)--(10,0)--(0,10)--cycle); draw((6.5,3.25)--(3.25,0)--(0,3.25)--(3.25,6.5)); label("A", (0,10), W); label("B", (0,0), W); label("C", (10,0), E); label("S", (25/3,11/6), E); label("S", (11/6,25/3), E); label("S", (5,5), NE); [/asy]
We now have that $3S=42\sqrt{2}$ , so $S=14\sqrt{2}$ . But we want the area of the square which is $S^2=(14\sqrt{2})^2= \boxed{392}$ | B | 392 |
b4f61da8d1b0a8b21f8e78f2d4432d04 | https://artofproblemsolving.com/wiki/index.php/1987_AHSME_Problems/Problem_27 | A cube of cheese $C=\{(x, y, z)| 0 \le x, y, z \le 1\}$ is cut along the planes $x=y, y=z$ and $z=x$ . How many pieces are there?
(No cheese is moved until all three cuts are made.)
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$ | The cut $x = y$ separates the cube into points with $x < y$ and points with $x > y$ , and analogous results apply for the other cuts. Thus, which piece a particular point is in depends only on the relative sizes of its coordinates $x$ $y$ , and $z$ - for example, all points with the ordering $x < y < z$ are in the same piece. Thus, as there are $3! = 6$ possible orderings, there are $6$ pieces, which is answer $\boxed{6}$ | B | 6 |
83fe6d16c245668809d32e340e906833 | https://artofproblemsolving.com/wiki/index.php/1987_AHSME_Problems/Problem_28 | Let $a, b, c, d$ be real numbers. Suppose that all the roots of $z^4+az^3+bz^2+cz+d=0$ are complex numbers lying on a circle
in the complex plane centered at $0+0i$ and having radius $1$ . The sum of the reciprocals of the roots is necessarily
$\textbf{(A)}\ a \qquad \textbf{(B)}\ b \qquad \textbf{(C)}\ c \qquad \textbf{(D)}\ -a \qquad \textbf{(E)}\ -b$ | Let's denote the roots of the polynomial as $z_1, z_2, z_3, z_4$ . We know that the magnitudes of these 4 roots are 1 as given in the problem statement. Therefore, we have $z_1 \overline{z_1}, z_2 \overline{z_2}, z_3 \overline{z_3}, z_4 \overline{z_4} = 1$ . We want to find $\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} + \frac{1}{z_4}$ which is $\overline{z_1} + \overline{z_2} + \overline{z_3} + \overline{z_4}$ . Remember that these are the roots of polynomials. Whenever complex numbers are the roots of a polynomial with real coefficients, we know they come in complex-conjugate pairs. Therefore, $\overline{z_1} + \overline{z_2} + \overline{z_3} + \overline{z_4} = z_1 + z_2 + z_3 + z_4$ . However, by Vieta's, $z_1 + z_2 + z_3 + z_4 = -a$ . Thus, the answer is $\boxed{01}$ | D | 01 |
b7ce4b62669a04adc8072a6b3d3bcfe9 | https://artofproblemsolving.com/wiki/index.php/1987_AHSME_Problems/Problem_29 | Consider the sequence of numbers defined recursively by $t_1=1$ and for $n>1$ by $t_n=1+t_{(n/2)}$ when $n$ is even
and by $t_n=\frac{1}{t_{(n-1)}}$ when $n$ is odd. Given that $t_n=\frac{19}{87}$ , the sum of the digits of $n$ is
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 17 \qquad \textbf{(C)}\ 19 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 23$ | If $n$ is even, then $t_{(n/2)}$ would be negative, which is not possible. Therefore, $n$ is odd. With this function, backwards thinking is the key. If $t_x < 1$ , then $x$ is odd, and $t_{(x-1)} = \frac{1}{t_{x}}$ . Otherwise, you keep on subtracting 1 and halving x until $t_\frac{x}{2^{n}} < 1$ .
We can use this logic to go backwards until we reach $t_1 = 1$ , like so:
$t_n=\frac{19}{87}\\\\t_{n-1} = \frac{87}{19}\\\\t_{\frac{n-1}{2}} = \frac{68}{19}\\\\t_{\frac{n-1}{4}} = \frac{49}{19}\\\\t_{\frac{n-1}{8}} = \frac{30}{19}\\\\t_{\frac{n-1}{16}} = \frac{11}{19}\\\\t_{\frac{n-1}{16} - 1} = \frac{19}{11}\\\\t_{\frac{\frac{n-1}{16} - 1}{2}} = \frac{8}{11}\\\\t_{\frac{\frac{n-1}{16} - 1}{2} - 1} = \frac{11}{8}\\\\t_{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2}} = \frac{3}{8}\\\\t_{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1} = \frac{8}{3}\\\\t_{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{2}} = \frac{5}{3}\\\\t_{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4}} = \frac{2}{3}\\\\t_{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1} = \frac{3}{2}\\\\t_{\frac{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2}} = \frac{1}{2}\\\\t_{\frac{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2} - 1} = 2\\\\t_{\frac{\frac{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2} - 1}{2}} = t_1 = 1 \Rightarrow \frac{\frac{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2} - 1}{2} = 1 \Rightarrow n = 1905$ , so the answer is $\boxed{15}$ | A | 15 |
cce247b6e17026f847c9516f4eda5e05 | https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_1 | $[x-(y-z)] - [(x-y) - z] =$
$\textbf{(A)}\ 2y \qquad \textbf{(B)}\ 2z \qquad \textbf{(C)}\ -2y \qquad \textbf{(D)}\ -2z \qquad \textbf{(E)}\ 0$ | The expression becomes $(x-y+z)-(x-y-z) = x-y+z-x+y+z = 2z$ , which is $\boxed{2}$ | B | 2 |
079160ce21ee2ddf3e55fa26693ea9de | https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_3 | $\triangle ABC$ has a right angle at $C$ and $\angle A = 20^\circ$ . If $BD$ $D$ in $\overline{AC}$ ) is the bisector of $\angle ABC$ , then $\angle BDC =$
$\textbf{(A)}\ 40^\circ \qquad \textbf{(B)}\ 45^\circ \qquad \textbf{(C)}\ 50^\circ \qquad \textbf{(D)}\ 55^\circ\qquad \textbf{(E)}\ 60^\circ$ | Since $\angle C = 90^{\circ}$ and $\angle A = 20^{\circ}$ , we have $\angle ABC = 70^{\circ}$ . Thus $\angle DBC = 35^{\circ}$ . It follows that $\angle BDC = 90^{\circ} - 35^{\circ} = 55^{\circ}$ , which is $\boxed{55}$ | D | 55 |
f9e07f1421840ec4a959fc4e509afa45 | https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_6 | Using a table of a certain height, two identical blocks of wood are placed as shown in Figure 1. Length $r$ is found to be $32$ inches. After rearranging the blocks as in Figure 2, length $s$ is found to be $28$ inches. How high is the table?
[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13pt)); path table = origin--(1,0)--(1,6)--(6,6)--(6,0)--(7,0)--(7,7)--(0,7)--cycle; path block = origin--(3,0)--(3,1.5)--(0,1.5)--cycle; path rotblock = origin--(1.5,0)--(1.5,3)--(0,3)--cycle; draw(table^^shift((14,0))*table); filldraw(shift((7,0))*block^^shift((5.5,7))*rotblock^^shift((21,0))*rotblock^^shift((18,7))*block,gray); draw((7.25,1.75)--(8.5,3.5)--(8.5,8)--(7.25,9.75),Arrows(size=5)); draw((21.25,3.25)--(22,3.5)--(22,8)--(21.25,8.25),Arrows(size=5)); unfill((8,5)--(8,6.5)--(9,6.5)--(9,5)--cycle); unfill((21.5,5)--(21.5,6.5)--(23,6.5)--(23,5)--cycle); label("$r$",(8.5,5.75)); label("$s$",(22,5.75)); [/asy]
$\textbf{(A) }28\text{ inches}\qquad\textbf{(B) }29\text{ inches}\qquad\textbf{(C) }30\text{ inches}\qquad\textbf{(D) }31\text{ inches}\qquad\textbf{(E) }32\text{ inches}$ | Let $h$ $l$ , and $w$ represent the height of the table and the length and width of the wood blocks, respectively, in inches. From Figure 1, we have $l+h-w=32$ , and from Figure 2, $w+h-l=28$ . Adding the equations gives $2h=60 \implies h=30$ , which is $\boxed{30}$ | C | 30 |
6211a3a685e67f63d18c76d380e02dd3 | https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_8 | The population of the United States in $1980$ was $226,504,825$ . The area of the country is $3,615,122$ square miles. There are $(5280)^{2}$ square feet in one square mile. Which number below best approximates the average number of square feet per person?
$\textbf{(A)}\ 5,000\qquad \textbf{(B)}\ 10,000\qquad \textbf{(C)}\ 50,000\qquad \textbf{(D)}\ 100,000\qquad \textbf{(E)}\ 500,000$ | With about $230$ million people and under $4$ million square miles, there are about $60$ people per square mile. Since a square mile is about $(5000 \ \text{ft})^{2} = 25$ million square feet, that gives approximately $\frac{25}{60}$ of a million square feet per person. $\frac{25}{60}$ is approximately half, so the answer is approximately half a million, which is $\boxed{500,000}$ | E | 500,000 |
3c7360f29bcaaefb51ce04731504aa8f | https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_11 | In $\triangle ABC, AB = 13, BC = 14$ and $CA = 15$ . Also, $M$ is the midpoint of side $AB$ and $H$ is the foot of the altitude from $A$ to $BC$ .
The length of $HM$ is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair H=origin, A=(0,6), B=(-4,0), C=(5,0), M=B+3.6*dir(B--A); draw(B--C--A--B^^M--H--A^^rightanglemark(A,H,C)); label("A", A, NE); label("B", B, W); label("C", C, E); label("H", H, S); label("M", M, dir(M)); [/asy]
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 6.5\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 7.5\qquad \textbf{(E)}\ 8$ | Warning: this solution is very intensive in calculation. Please do NOT try this on the test!
Let's start by finding $AH$ . By Heron's Formula, $s=\frac{13+14+15}{2}=21, [ABC]=\sqrt{21*(21-13)(21-14)(21-15)}=84$ . Using the area formula $A=0.5bh$ $AH=12$ . Now using the Pythagorean Theorem, $BH=5, HC=9$
Now $AM=MB=6.5$ . Using Stewart's Theorem on $\triangle{ABH}$ , letting $HM=x$
$13*6.5*6.5+13x^2=12*6.5*12+5*6.5*5$ (remember that Stewart's Theorem is $man+dad=bmb+cnc$ ).
Thus $x=6.5$ or $x=-6.5$ (reject this solution since $x$ is positive). Thus $HM=6.5$ . Select $\boxed{6.5}$ | B | 6.5 |
59c4d4b616eb41a19777ed109fcdd8b2 | https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_13 | A parabola $y = ax^{2} + bx + c$ has vertex $(4,2)$ . If $(2,0)$ is on the parabola, then $abc$ equals
$\textbf{(A)}\ -12\qquad \textbf{(B)}\ -6\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 12$ | Consider the quadratic in completed square form: it must be $y=a(x-4)^{2}+2$ . Now substitute $x=2$ and $y=0$ to give $a=-\frac{1}{2}$ . Now expanding gives $y=-\frac{1}{2}x^{2}+4x-6$ , so the product is $-\frac{1}{2} \cdot 4 \cdot -6 = 3 \cdot 4 = 12$ , which is $\boxed{12}$ | E | 12 |
69e011d833240d68e11b47de6c9cc41e | https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_16 | In $\triangle ABC, AB = 8, BC = 7, CA = 6$ and side $BC$ is extended, as shown in the figure, to a point $P$ so that $\triangle PAB$ is similar to $\triangle PCA$ . The length of $PC$ is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, P=(1.5,5), B=(8,0), C=P+2.5*dir(P--B); draw(A--P--C--A--B--C); label("A", A, W); label("B", B, E); label("C", C, NE); label("P", P, NW); label("6", 3*dir(A--C), SE); label("7", B+3*dir(B--C), NE); label("8", (4,0), S); [/asy]
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 9\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 11$ | Since we are given that $\triangle{PAB}\sim\triangle{PCA}$ , we have $\frac{PC}{PA}=\frac{6}{8}=\frac{PA}{PC+7}$
Solving for $PA$ in $\frac{PC}{PA}=\frac{6}{8}=\frac{3}{4}$ gives us $PA=\frac{4PC}{3}$
We also have $\frac{PA}{PC+7}=\frac{3}{4}$ . Substituting $PA$ in for our expression yields $\frac{\frac{4PC}{3}}{PC+7}=\frac{3}{4}$
Which we can further simplify to $\frac{16PC}{3}=3PC+21$
$\frac{7PC}{3}=21$
$PC=9\implies\boxed{9}$ | C | 9 |
d1611c6387c0878330ebd2ed29862c2d | https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_17 | A drawer in a darkened room contains $100$ red socks, $80$ green socks, $60$ blue socks and $40$ black socks.
A youngster selects socks one at a time from the drawer but is unable to see the color of the socks drawn.
What is the smallest number of socks that must be selected to guarantee that the selection contains at least $10$ pairs?
(A pair of socks is two socks of the same color. No sock may be counted in more than one pair.)
$\textbf{(A)}\ 21\qquad \textbf{(B)}\ 23\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 30\qquad \textbf{(E)}\ 50$ | Solution by e_power_pi_times_i
Suppose that you wish to draw one pair of socks from the drawer. Then you would pick $5$ socks (one of each kind, plus one). Notice that in the worst possible situation, you will continue to draw the same sock, until you get $10$ pairs. This is because drawing the same sock results in a pair every $2$ of that sock, whereas drawing another sock creates another pair. Thus the answer is $5+2\cdot(10-1) = \boxed{23}$ | B | 23 |
63f16be931365d641e9cf8cd43ffc5ff | https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_19 | A park is in the shape of a regular hexagon $2$ km on a side. Starting at a corner,
Alice walks along the perimeter of the park for a distance of $5$ km.
How many kilometers is she from her starting point?
$\textbf{(A)}\ \sqrt{13}\qquad \textbf{(B)}\ \sqrt{14}\qquad \textbf{(C)}\ \sqrt{15}\qquad \textbf{(D)}\ \sqrt{16}\qquad \textbf{(E)}\ \sqrt{17}$ | We imagine this problem on a coordinate plane and let Alice's starting position be the origin. We see that she will travel along two edges and then go halfway along a third. Therefore, her new $x$ -coordinate will be $1 + 2 + \frac{1}{2} = \frac{7}{2}$ because she travels along a distance of $2 \cdot \frac{1}{2} = 1$ km because of the side relationships of an equilateral triangle, then $2$ km because the line is parallel to the $x$ -axis, and the remaining distance is $\frac{1}{2}$ km because she went halfway along and because of the logic for the first part of her route. For her $y$ -coordinate, we can use similar logic to find that the coordinate is $\sqrt{3} + 0 - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$ . Therefore, her distance is \[\sqrt{\left(\frac{7}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{49}{4} + \frac{3}{4}} = \sqrt{\frac{52}{4}} = \sqrt{13},\] giving an answer of $\boxed{13}$ | A | 13 |
1ecd9902d8ac932cdc5d9e001c5da32b | https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_26 | It is desired to construct a right triangle in the coordinate plane so that its legs are parallel
to the $x$ and $y$ axes and so that the medians to the midpoints of the legs lie on the lines $y = 3x + 1$ and $y = mx + 2$ . The number of different constants $m$ for which such a triangle exists is
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ \text{more than 3}$ | In any right triangle with legs parallel to the axes, one median to the midpoint of a leg has slope $4$ times that of the other. This can easily be shown with coordinates: any triangle of this sort may be labelled with right angle at $P(a,b)$ , other vertices $Q(a,b+2c)$ and $R(a-2d,b)$ , and thus midpoints $(a,b+c)$ and $(a-d,b)$ , so that the slopes are $\frac{c}{2d}$ and $\frac{2c}{d} = 4(\frac{c}{2d})$ , thus showing that one is $4$ times the other as required.
Thus in our problem, $m$ is either $3 \times 4 = 12$ or $3 \div 4 = \frac{3}{4}$ . In fact, both are possible, and each for infinitely many triangles. We shall show this for $m=12$ , and the argument is analogous for $m=\frac{3}{4}$ . Take any right triangle with legs parallel to the axes and a hypotenuse with slope $12 \div 2 = 6$ , e.g. the triangle with vertices $(0,0)$ $(1,0)$ , and $(1,6)$ . Then quick calculations show that the medians to the legs have slopes $12$ and $3$ . Now translate the triangle (without rotating it) so that its medians intersect at the point where the lines $y=12x+2$ and $y=3x+1$ intersect. This forces the medians to lie on these lines (since their slopes are determined, and now we force them to go through a particular point; a slope and a point uniquely determine a line). Finally, for any central dilation of this triangle (a larger or smaller triangle with the same centroid and sides parallel to this one's sides), the medians will still lie on these lines, showing the "infinitely many" part of the result.
Hence, to sum up, $m$ can in fact be both $12$ or $\frac{3}{4}$ , which is exactly $2$ values, i.e. $\boxed{2}$ | C | 2 |
0b2eb9231c3697342227866440d0fa74 | https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_28 | $ABCDE$ is a regular pentagon. $AP, AQ$ and $AR$ are the perpendiculars dropped from $A$ onto $CD, CB$ extended and $DE$ extended,
respectively. Let $O$ be the center of the pentagon. If $OP = 1$ , then $AO + AQ + AR$ equals
[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair O=origin, A=2*dir(90), B=2*dir(18), C=2*dir(306), D=2*dir(234), E=2*dir(162), P=(C+D)/2, Q=C+3.10*dir(C--B), R=D+3.10*dir(D--E), S=C+4.0*dir(C--B), T=D+4.0*dir(D--E); draw(A--B--C--D--E--A^^E--T^^B--S^^R--A--Q^^A--P^^rightanglemark(A,Q,S)^^rightanglemark(A,R,T)); dot(O); label("$O$",O,dir(B)); label("$1$",(O+P)/2,W); label("$A$",A,dir(A)); label("$B$",B,dir(B)); label("$C$",C,dir(C)); label("$D$",D,dir(D)); label("$E$",E,dir(E)); label("$P$",P,dir(P)); label("$Q$",Q,dir(Q)); label("$R$",R,dir(R)); [/asy]
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 1 + \sqrt{5}\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 2 + \sqrt{5}\qquad \textbf{(E)}\ 5$ | To solve the problem, we compute the area of regular pentagon $ABCDE$ in two different ways. First, we can divide regular pentagon $ABCDE$ into five congruent triangles. [asy] unitsize(2 cm); pair A, B, C, D, E, O, P, Q, R; A = dir(90); B = dir(90 - 360/5); C = dir(90 - 2*360/5); D = dir(90 - 3*360/5); E = dir(90 - 4*360/5); O = (0,0); P = (C + D)/2; Q = (A + reflect(B,C)*(A))/2; R = (A + reflect(D,E)*(A))/2; draw((2*R - E)--D--C--(2*Q - B)); draw(A--P); draw(A--Q); draw(A--R); draw(B--A--E); draw((O--B),dashed); draw((O--C),dashed); draw((O--D),dashed); draw((O--E),dashed); label("$A$", A, N); label("$B$", B, dir(0)); label("$C$", C, SE); label("$D$", D, SW); label("$E$", E, W); dot("$O$", O, NE); label("$P$", P, S); label("$Q$", Q, dir(0)); label("$R$", R, W); label("$1$", (O + P)/2, dir(0)); [/asy]
If $s$ is the side length of the regular pentagon, then each of the triangles $AOB$ $BOC$ $COD$ $DOE$ , and $EOA$ has base $s$ and height 1, so the area of regular pentagon $ABCDE$ is $5s/2$
Next, we divide regular pentagon $ABCDE$ into triangles $ABC$ $ACD$ , and $ADE$
[asy] unitsize(2 cm); pair A, B, C, D, E, O, P, Q, R; A = dir(90); B = dir(90 - 360/5); C = dir(90 - 2*360/5); D = dir(90 - 3*360/5); E = dir(90 - 4*360/5); O = (0,0); P = (C + D)/2; Q = (A + reflect(B,C)*(A))/2; R = (A + reflect(D,E)*(A))/2; draw((2*R - E)--D--C--(2*Q - B)); draw(A--P); draw(A--Q); draw(A--R); draw(B--A--E); draw(A--C,dashed); draw(A--D,dashed); label("$A$", A, N); label("$B$", B, dir(0)); label("$C$", C, SE); label("$D$", D, SW); label("$E$", E, W); dot("$O$", O, dir(0)); label("$P$", P, S); label("$Q$", Q, dir(0)); label("$R$", R, W); label("$1$", (O + P)/2, dir(0)); [/asy] Triangle $ACD$ has base $s$ and height $AP = AO + 1$ . Triangle $ABC$ has base $s$ and height $AQ$ . Triangle $ADE$ has base $s$ and height $AR$ . Therefore, the area of regular pentagon $ABCDE$ is also \[\frac{s}{2} (AO + AQ + AR + 1).\] Hence, \[\frac{s}{2} (AO + AQ + AR + 1) = \frac{5s}{2},\] which means $AO + AQ + AR + 1 = 5$ , or $AO + AQ + AR = \boxed{4}$ | C | 4 |
0b2eb9231c3697342227866440d0fa74 | https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_28 | $ABCDE$ is a regular pentagon. $AP, AQ$ and $AR$ are the perpendiculars dropped from $A$ onto $CD, CB$ extended and $DE$ extended,
respectively. Let $O$ be the center of the pentagon. If $OP = 1$ , then $AO + AQ + AR$ equals
[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair O=origin, A=2*dir(90), B=2*dir(18), C=2*dir(306), D=2*dir(234), E=2*dir(162), P=(C+D)/2, Q=C+3.10*dir(C--B), R=D+3.10*dir(D--E), S=C+4.0*dir(C--B), T=D+4.0*dir(D--E); draw(A--B--C--D--E--A^^E--T^^B--S^^R--A--Q^^A--P^^rightanglemark(A,Q,S)^^rightanglemark(A,R,T)); dot(O); label("$O$",O,dir(B)); label("$1$",(O+P)/2,W); label("$A$",A,dir(A)); label("$B$",B,dir(B)); label("$C$",C,dir(C)); label("$D$",D,dir(D)); label("$E$",E,dir(E)); label("$P$",P,dir(P)); label("$Q$",Q,dir(Q)); label("$R$",R,dir(R)); [/asy]
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 1 + \sqrt{5}\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 2 + \sqrt{5}\qquad \textbf{(E)}\ 5$ | Now, we know that angle $D$ has measure $\frac{180 \cdot 3}{5} = 108$ . Since \[\sin 54 = \frac{OP}{DO} = \frac{1}{DO}, DO = \frac{1}{\sin 54}\] \[\tan 54 = \frac{OP}{DP} = \frac{1}{DP}, DP = \frac{1}{\tan 54}\] Therefore, $AB = 2DP = \frac{2}{\tan 54}$ \[\sin 72 = \frac{AQ}{AB} = AQ \tan 54 \cdot \frac{1}{2}, AQ = \frac{2 \sin 72}{\tan 54}\] Therefore, $AO + AQ + AR = AO + 2AQ = \frac{1}{\sin 54}+\frac{4 \sin 72}{\tan 54} = \frac{1}{\sin 54} + 8 \sin 36 \cos 54 = \frac{1}{\cos 36} + 8-8\cos^2(36)$ . Recalling that $\cos 36 = \frac{1 + \sqrt{5}}{4}$ gives a final answer of $\boxed{4}$ | null | 4 |
fa4d5d94e691035b1033a49de7c01179 | https://artofproblemsolving.com/wiki/index.php/1986_AHSME_Problems/Problem_30 | The number of real solutions $(x,y,z,w)$ of the simultaneous equations $2y = x + \frac{17}{x}, 2z = y + \frac{17}{y}, 2w = z + \frac{17}{z}, 2x = w + \frac{17}{w}$ is
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 16$ | Consider the cases $x>0$ and $x<0$ , and also note that by AM-GM, for any positive number $a$ , we have $a+\frac{17}{a} \geq 2\sqrt{17}$ , with equality only if $a = \sqrt{17}$ . Thus, if $x>0$ , considering each equation in turn, we get that $y \geq \sqrt{17}, z \geq \sqrt{17}, w \geq \sqrt{17}$ , and finally $x \geq \sqrt{17}$
Now suppose $x > \sqrt{17}$ . Then $y - \sqrt{17} = \frac{x^{2}+17}{2x} - \sqrt{17} = (\frac{x-\sqrt{17}}{2x})(x-\sqrt{17}) < \frac{1}{2}(x-\sqrt{17})$ , so that $x > y$ . Similarly, we can get $y > z$ $z > w$ , and $w > x$ , and combining these gives $x > x$ , an obvious contradiction.
Thus we must have $x \geq \sqrt{17}$ , but $x \ngtr \sqrt{17}$ , so if $x > 0$ , the only possibility is $x = \sqrt{17}$ , and analogously from the other equations we get $x = y = z = w = \sqrt{17}$ ; indeed, by substituting, we verify that this works.
As for the other case, $x < 0$ , notice that $(x,y,z,w)$ is a solution if and only if $(-x,-y,-z,-w)$ is a solution, since this just negates both sides of each equation and so they are equivalent. Thus the only other solution is $x = y = z = w = -\sqrt{17}$ , so that we have $2$ solutions in total, and therefore the answer is $\boxed{2}$ | B | 2 |
946190fc6df9267f2a72339f366722cc | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_1 | If $2x+1=8$ , then $4x+1=$
$\mathrm{(A)\ } 15 \qquad \mathrm{(B) \ }16 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ }19$ | We have \begin{align*}2x+1 = 8 &\iff 2x = 7 \\ &\iff x = \frac{7}{2},\end{align*} so \begin{align*}4x+1 &= 4\left(\frac{7}{2}\right)+1 \\ &= 2(7)+1 \\ &= \boxed{15} | A | 15 |
946190fc6df9267f2a72339f366722cc | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_1 | If $2x+1=8$ , then $4x+1=$
$\mathrm{(A)\ } 15 \qquad \mathrm{(B) \ }16 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ }19$ | From $2x = 7$ (as above), we can directly compute \begin{align*}4x &= 2(2x) \\ &= 2(7) \\ &= 14,\end{align*} so $4x+1 = 14+1 = \boxed{15}$ | A | 15 |
f239d60051a396ab9fe7257e29758e13 | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_3 | In right $\triangle ABC$ with legs $5$ and $12$ , arcs of circles are drawn, one with center $A$ and radius $12$ , the other with center $B$ and radius $5$ . They intersect the hypotenuse in $M$ and $N$ . Then $MN$ has length
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(12,7), C=(12,0), M=12*dir(A--B), N=B+B.y*dir(B--A); real r=degrees(B); draw(A--B--C--cycle^^Arc(A,12,0,r)^^Arc(B,B.y,180+r,270)); pair point=incenter(A,B,C); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, dir(point--M)); label("$N$", N, dir(point--N)); label("$12$", (6,0), S); label("$5$", (12,3.5), E);[/asy]
$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }\frac{13}{5} \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 4 \qquad \mathrm{(E) \ }\frac{24}{5}$ | Firstly, the Pythagorean theorem gives \begin{align*}AB &=\sqrt{AC^2+BC^2} \\ &= \sqrt{12^2+5^2} \\ & =\sqrt{144+25} \\ &=\sqrt{169} \\ &= 13.\end{align*} Also, $AM = AC = 12$ and $BN = BC = 5$ since they are both radii of the respective circles. Thus $MB = AB-AM = 13-12 = 1$ , and so $MN = BN-BM = 5-1 = \boxed{4}$ | D | 4 |
703b2cf2cd75454caa41f0e1f5e9d786 | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_4 | A large bag of coins contains pennies, dimes and quarters. There are twice as many dimes as pennies and three times as many quarters as dimes. An amount of money which could be in the bag is
$\mathrm{(A)\ } $306 \qquad \mathrm{(B) \ } $333 \qquad \mathrm{(C)\ } $342 \qquad \mathrm{(D) \ } $348 \qquad \mathrm{(E) \ } $360$ | If there are $x$ pennies in the bag, then there are $2x$ dimes and $3(2x) = 6x$ quarters. Since pennies are $$0.01$ , dimes are $$0.10$ , and quarters are $$0.25$ , the total amount of money in the bag is \[$ \left(0.01x+(0.10)(2x)+(0.25)(6x)\right) = $1.71x.\] Therefore, the possible amounts of money are precisely the integer multiples of $$1.71$
Since the answer choices are all integer numbers of dollars, we multiply by $100$ to deduce that the answer must be an integer multiple of $$171$ . The only such multiple among the answer choices is $$(2 \cdot 171) = \boxed{342}$ | C | 342 |
9f67b27afa2af214d71857a03b788c03 | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_10 | An arbitrary circle can intersect the graph of $y = \sin x$ in
$\mathrm{(A) \ } \text{at most }2\text{ points} \qquad \mathrm{(B) \ }\text{at most }4\text{ points} \qquad \mathrm{(C) \ } \text{at most }6\text{ points} \qquad \mathrm{(D) \ } \text{at most }8\text{ points}$ $\mathrm{(E) \ }\text{more than }16\text{ points}$ | Consider a circle whose center lies on the positive $y$ -axis and which passes through the origin. As the radius of this circle becomes arbitrarily large, its curvature near the $x$ -axis becomes almost flat, and so it can intersect the curve $y = \sin x$ arbitrarily many times (since the $x$ -axis itself intersects the curve infinitely many times). Hence, in particular, we can choose a radius sufficiently large that the circle intersects the curve at $\boxed{16}$ | E | 16 |
ed6079924f782d79093a4963997c4d01 | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_11 | How many distinguishable rearrangements of the letters in $CONTEST$ have both the vowels first? (For instance, $OETCNST$ is one such arrangement, but $OTETSNC$ is not.)
$\mathrm{(A)\ } 60 \qquad \mathrm{(B) \ }120 \qquad \mathrm{(C) \ } 240 \qquad \mathrm{(D) \ } 720 \qquad \mathrm{(E) \ }2520$ | We consider the vowels and consonants separately. There are $2$ vowels ( $O$ and $E$ ), giving $2! = 2$ choices for the first two letters; similarly, there are $5$ consonants ( $C$ $N$ $S$ , and two $T$ s), which would give $5! = 120$ possible choices for letters $3$ to $7$ , except that since the two $T$ s are indistinguishable, this actually counts each order exactly twice. Therefore the number of possible orderings of the consonants is $\frac{120}{2} = 60$ , giving a total of $2 \cdot 60 = \boxed{120}$ possible rearrangements. | B | 120 |
fcdb5418e7126817fa20c3ff71d47363 | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_13 | Pegs are put in a board $1$ unit apart both horizontally and vertically. A rubber band is stretched over $4$ pegs as shown in the figure, forming a quadrilateral. Its area in square units is
[asy] int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<4; j=j+1) { dot((i,j)); }} draw((0,1)--(1,3)--(4,1)--(3,0)--cycle, linewidth(0.7));[/asy]
$\mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }4.5 \qquad \mathrm{(C) \ } 5 \qquad \mathrm{(D) \ } 5.5 \qquad \mathrm{(E) \ }6$ | [asy] int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<4; j=j+1) { dot((i,j)); }} draw((0,1)--(1,3)--(4,1)--(3,0)--cycle, linewidth(0.7)); draw((0,0)--(4,0)--(4,3)--(0,3)--cycle); label("$A$",(0,3),NW); label("$B$",(4,3),NE); label("$C$",(4,0),SE); label("$D$",(0,0),SW); label("$E$",(1,3),N); label("$F$",(4,1),E); label("$G$",(3,0),S); label("$H$",(0,1),W); [/asy] We draw in the rectangle bounding the given quadrilateral and label the points as shown. The area of rectangle $ABCD$ is $(3)(4) = 12$ , while the areas of the triangles $AEH$ $EBF$ $FCG$ , and $GDH$ are, respectively, \begin{align*}&\text{area of } \triangle AEH = \frac{1}{2}(1)(2) = 1, \\ &\text{area of } \triangle EBF = \frac{1}{2}(3)(2) = 3, \\ &\text{area of } \triangle FCG = \frac{1}{2}(1)(1) = \frac{1}{2}, \text{ and} \\ &\text{area of } \triangle GDH = \frac{1}{2}(3)(1) = \frac{3}{2}.\end{align*} Hence the area of the given quadrilateral $EFGH$ is $12-\left(1+3+\frac{1}{2}+\frac{3}{2}\right) = \boxed{6}$ | E | 6 |
fcdb5418e7126817fa20c3ff71d47363 | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_13 | Pegs are put in a board $1$ unit apart both horizontally and vertically. A rubber band is stretched over $4$ pegs as shown in the figure, forming a quadrilateral. Its area in square units is
[asy] int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<4; j=j+1) { dot((i,j)); }} draw((0,1)--(1,3)--(4,1)--(3,0)--cycle, linewidth(0.7));[/asy]
$\mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }4.5 \qquad \mathrm{(C) \ } 5 \qquad \mathrm{(D) \ } 5.5 \qquad \mathrm{(E) \ }6$ | The number of lattice points (i.e. pegs on the board) strictly inside the quadrilateral is $5$ and the number of lattice points on its boundary is $4$ . Therefore, by Pick's theorem , its area is $5+\frac{4}{2}-1 = \boxed{6}$ | E | 6 |
aa61b7c2e2799f0152a1e6e43666f9d5 | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_14 | Exactly three of the interior angles of a convex polygon are obtuse. What is the maximum number of sides of such a polygon?
$\mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }5 \qquad \mathrm{(C) \ } 6 \qquad \mathrm{(D) \ } 7 \qquad \mathrm{(E) \ }8$ | Suppose that such a polygon has $n$ sides. Let the three obtuse angle measures, in degrees, be $o_1$ $o_2$ , and $o_3$ and the $(n-3)$ acute angle measures, again in degrees, be $a_1,a_2,a_3, \dotsc a_{n-3}$
Since $90 < o_i < 180$ for each $i$ , we have \[3(90) = 270 < o_1+o_2+o_3 < 3(180) = 540,\] and similarly, since $0 < a_i < 90$ for each $i$ \[0 < a_1+a_2+a_3+\dotsb+a_{n-3} < 90(n-3) = 90n-270.\] It follows that \[270+0 < o_1+o_2+o_3+a_1+a_2+a_3+\dotsb+a_{n-3} < 540+90n-270,\] and recalling that the sum of the interior angle measures of an $n$ -gon is $180(n-2) = 180n-360$ , this reduces to $270 < 180n-360 < 90n+270$ . Hence \[\frac{540}{180} < n < \frac{270+360}{90} \iff 3 < n < 7,\] so an upper bound is $n \leq 6$ , and it is easy to check that this bound can be attained by e.g. a convex hexagon with a right angle, $2$ acute angles, and $3$ obtuse angles, as shown below:
HrKCfF2ETNSQSy3uGlkg hexagonsurvey1.gif
Accordingly, the maximum possible number of sides of such a polygon is $\boxed{6}$ | C | 6 |
47b6b2ea9b79f40d1489fdff76bd370d | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_16 | If $\usepackage{gensymb} A = 20 \degree$ and $\usepackage{gensymb} B = 25 \degree$ , then the value of $\left(1+\tan A\right)\left(1+\tan B\right)$ is
$\mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \ }\text{none of these}$ | Noting that $\usepackage{gensymb} 25 \degree + 20 \degree = 45 \degree$ , we apply the angle sum formula \[\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B},\] giving \begin{align*}1 &= \tan 45^{\circ} \\ &= \tan(A+B) \\ &= \frac{\tan A+\tan B}{1-\tan A\tan B},\end{align*} so \[\tan A + \tan B = 1-\tan A\tan B.\] Hence \begin{align*}(1+\tan A)(1+\tan B) &= 1+\tan A+\tan B+\tan A\tan B \\ &= 1+\left(1-\tan A\tan B\right)+\tan A\tan B \\ &= \boxed{2} | B | 2 |
47b6b2ea9b79f40d1489fdff76bd370d | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_16 | If $\usepackage{gensymb} A = 20 \degree$ and $\usepackage{gensymb} B = 25 \degree$ , then the value of $\left(1+\tan A\right)\left(1+\tan B\right)$ is
$\mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \ }\text{none of these}$ | Expanding in terms of sines and cosines, we obtain \begin{align*}\left(1+\tan A\right)\left(1+\tan B\right) &= \left(\frac{\cos A}{\cos A}+\frac{\sin A}{\cos A}\right)\left(\frac{\cos B}{\cos B}+\frac{\sin B}{\cos B}\right) \\ &=\frac{\left(\sin A+\cos A\right)\left(\sin B+\cos B\right)}{\cos A\cos B} \\ &= \frac{\sin A\sin B+\cos A\cos B+\sin A\cos B+\sin B\cos A}{\cos A\cos B}.\end{align*}
Recalling the angle sum identities \begin{align*}&\cos(A-B) = \sin A\sin B+\cos A\cos B \text{ and} \\ &\sin(A+B)=\sin A\cos B+\sin B\cos A,\end{align*} this reduces to \[\frac{\cos(A-B)+\sin(A+B)}{\cos A\cos B}.\]
Now, using the product-to-sum formula \[\cos A\cos B = \frac{1}{2}\left(\cos(A-B)+\cos(A+B)\right),\] we can simplify the denominator, yielding \[\left(1+\tan A\right)\left(1+\tan B\right) = \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}\left(\cos(A-B)+\cos(A+B)\right)}.\]
Finally, since $\usepackage{gensymb} A+B = 45 \degree$ , we have $\sin(A+B) = \cos(A+B)$ , so \begin{align*}\left(1+\tan A\right)\left(1+\tan B\right) &= \frac{\cos(A-B)+\sin(A+B)}{\frac{1}{2}\left(\cos(A-B)+\sin(A+B)\right)} \\ &= \frac{1}{\left(\frac{1}{2}\right)} \\ &= \boxed{2} | B | 2 |
47b6b2ea9b79f40d1489fdff76bd370d | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_16 | If $\usepackage{gensymb} A = 20 \degree$ and $\usepackage{gensymb} B = 25 \degree$ , then the value of $\left(1+\tan A\right)\left(1+\tan B\right)$ is
$\mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \ }\text{none of these}$ | As in Solution 2, we rewrite the expression as \[\frac{\cos 20^{\circ} \cos 25^{\circ} + \cos 20^{\circ} \sin 25^{\circ} + \sin 20^{\circ} \cos 25^{\circ} + \sin 20^{\circ} \sin 25^{\circ}}{\cos 20^{\circ} \cos 25^{\circ}},\] and hence as \[1 + \frac{\cos 20^{\circ} \sin 25^{\circ} + \sin 20^{\circ} \cos 25^{\circ} + \sin 20^{\circ} \sin 25^{\circ}}{\cos 20^{\circ} \cos 25^{\circ}}.\] Using the angle sum identities \begin{align*}&\cos(A-B) = \sin A\sin B+\cos A\cos B \text{ and} \\ &\sin(A+B)=\sin A\cos B+\sin B\cos A,\end{align*} we obtain \begin{align*}&\cos 20^{\circ} \sin 25^{\circ} + \sin 20^{\circ} \cos 25^{\circ} = \sin\left(20^{\circ}+25^{\circ}\right) = \sin 45^{\circ} \text{ and} \\ &\cos 20^{\circ} \cos 25^{\circ} - \sin 20^{\circ} \sin 25^{\circ} = \cos\left(20^{\circ}+25^{\circ}\right) = \cos 45^{\circ}.\end{align*} Therefore the expression becomes \begin{align*}1+\frac{\sin 45^{\circ} + \sin 20^{\circ} \sin 25^{\circ}}{\cos 45^{\circ} + \sin 20^{\circ} \sin 25^{\circ}} &= 1+1 \qquad \text{(since } \sin 45^{\circ} = \cos 45^{\circ}\text{)} \\ &= \boxed{2} | B | 2 |
47b6b2ea9b79f40d1489fdff76bd370d | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_16 | If $\usepackage{gensymb} A = 20 \degree$ and $\usepackage{gensymb} B = 25 \degree$ , then the value of $\left(1+\tan A\right)\left(1+\tan B\right)$ is
$\mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \ }\text{none of these}$ | As in Solutions 2 and 3, the expression becomes \[\left(\frac{\cos A+\sin A}{\cos A}\right)\left(\frac{\cos B+\sin B}{\cos B}\right).\]
Now, using the identity $\cos^2 A + \sin^2 A = 1$ and the double-angle identity $\sin(2A) = 2\sin A\cos A$ , we observe that \begin{align*}\left(\cos A + \sin A\right)^2 &= \cos^2 A + \sin^2 A + 2\sin A \cos A \\ &= 1 + 2\sin A \cos A \\ &= 1 + \sin(2A).\end{align*}
Since $A$ and $B$ are acute, we have $\sin A,\cos A,\sin B,\cos B > 0$ , so $\cos A + \sin A > 0$ and $\cos B + \sin B > 0$ . Hence, taking the positive square root of both sides in the above identity, the expression becomes \[\left(\frac{\sqrt{1+\sin(2A)}}{\cos A}\right)\left(\frac{\sqrt{1+\sin(2B)}}{\cos B}\right).\] Recalling the further identity $\sin A = \cos\left(90^{\circ}-A\right)$ , together with the half-angle identity \[\cos\left(\frac{A}{2}\right) = \sqrt{\frac{1+\cos A}{2}} \qquad \text{for } 0^{\circ} \leq A \leq 180^{\circ},\] we finally obtain \begin{align*}\left(1+\tan A\right)\left(1+\tan B\right) &= 2\left(\frac{\sqrt{\frac{1+\cos\left(90^{\circ}-2A\right)}{2}}}{\cos A}\right)\left(\frac{\sqrt{\frac{1+\cos\left(90^{\circ}-2B\right)}{2}}}{\cos B}\right) \\ &= 2\left(\frac{\cos\left(45^{\circ}-A\right)}{\cos A}\right)\left(\frac{\cos\left(45^{\circ}-B\right)}{\cos B}\right) \\ &= 2\left(\frac{\cos 25^{\circ} \cos 20^{\circ}}{\cos 20^{\circ} \cos 25^{\circ}}\right) = \boxed{2} | B | 2 |
fbcb58faade3a1b557d11604f9111a82 | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_17 | Diagonal $DB$ of rectangle $ABCD$ is divided into three segments of length $1$ by parallel lines $L$ and $L'$ that pass through $A$ and $C$ and are perpendicular to $DB$ . The area of $ABCD$ , rounded to the one decimal place, is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); real x=sqrt(6), y=sqrt(3), a=0.4; pair D=origin, A=(0,y), B=(x,y), C=(x,0), E=foot(C,B,D), F=foot(A,B,D); real r=degrees(B); pair M1=F+3*dir(r)*dir(90), M2=F+3*dir(r)*dir(-90), N1=E+3*dir(r)*dir(90), N2=E+3*dir(r)*dir(-90); markscalefactor=0.02; draw(B--C--D--A--B--D^^M1--M2^^N1--N2^^rightanglemark(A,F,B)^^rightanglemark(N1,E,B)); pair W=A+a*dir(135), X=B+a*dir(45), Y=C+a*dir(-45), Z=D+a*dir(-135); label("A", A, NE); label("B", B, NE); label("C", C, dir(0)); label("D", D, dir(180)); label("$L$", (x/2,0), SW); label("$L^\prime$", C, SW); label("1", D--F, NW); label("1", F--E, SE); label("1", E--B, SE); clip(W--X--Y--Z--cycle);[/asy]
$\mathrm{(A)\ } 4.1 \qquad \mathrm{(B) \ }4.2 \qquad \mathrm{(C) \ } 4.3 \qquad \mathrm{(D) \ } 4.4 \qquad \mathrm{(E) \ }4.5$ | Let $E$ be the point of intersection of $L$ and $\overline{BD}$ . Then, because $AE$ is the altitude to the hypotenuse of right triangle $ABD$ , triangles $ADE$ and $BAE$ are similar, giving \[\frac{AE}{BE} = \frac{ED}{EA},\] and so \begin{align*}AE &= \sqrt{BE \cdot ED} \\ &= \sqrt{(1+1)(1)} \\ &= \sqrt{2}.\end{align*} Thus, taking $BD$ and $AE$ as the base and perpendicular height, respectively, of triangle $ABD$ , we may compute its area as $\frac{1}{2}(3)\left(\sqrt{2}\right) = \frac{3\sqrt{2}}{2}$ . By symmetry, the area of the entire rectangle $ABCD$ is \[2\left(\frac{3\sqrt{2}}{2}\right) = 3\sqrt{2} \approx (3)(1.4) = \boxed{4.2}.\] | B | 4.2 |
a4dbd0265af755fd8d023ca3e0d6da4b | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_18 | Six bags of marbles contain $18, 19, 21, 23, 25$ and $34$ marbles, respectively. One bag contains chipped marbles only. The other $5$ bags contain no chipped marbles. Jane takes three of the bags and George takes two of the others. Only the bag of chipped marbles remains. If Jane gets twice as many marbles as George, how many chipped marbles are there?
$\mathrm{(A)\ } 18 \qquad \mathrm{(B) \ }19 \qquad \mathrm{(C) \ } 21 \qquad \mathrm{(D) \ } 23 \qquad \mathrm{(E) \ }25$ | Let George's bags contain a total of $x$ marbles, so Jane's bag contains $2x$ marbles. This means the total number of non-chipped marbles is $3x \equiv 0 \pmod{3}$ , while the total number of marbles is $18+19+21+23+25+34 = 140 \equiv 2 \pmod{3}$ , so the number of chipped marbles must also be congruent to $2-0 \equiv 2 \pmod{3}$
The answer choices are congruent modulo 3 to $0$ $1$ $0$ $2$ , and $1$ respectively, so the only possible number of chipped marbles among these is $23$ . Indeed, if Jane takes the bags containing $19$ $25$ , and $34$ marbles and George takes the remaining bags containing $18$ and $21$ marbles, then Jane will have a total of $19+25+34 = 78$ marbles, which is twice as many as George's $18+21 = 39$ marbles, as desired. Thus the answer is precisely $\boxed{23}$ | D | 23 |
48ce81497345bf462f880b4b26338090 | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_19 | Consider the graphs of $y = Ax^2$ and $y^2+3 = x^2+4y$ , where $A$ is a positive constant and $x$ and $y$ are real variables. In how many points do the two graphs intersect?
$\mathrm{(A) \ }\text{exactly }4 \qquad \mathrm{(B) \ }\text{exactly }2 \qquad$
$\mathrm{(C) \ }\text{at least }1,\text{ but the number varies for different positive values of }A \qquad$
$\mathrm{(D) \ }0\text{ for at least one positive value of }A \qquad \mathrm{(E) \ }\text{none of these}$ | Substituting $y = Ax^2$ into the equation $y^2+3 = x^2+4y$ gives \begin{align*}\left(Ax^2\right)+3 = x^2+4\left(Ax^2\right) &\iff A^2x^4+3 = x^2+4Ax^2 \\ &\iff A^2x^4-\left(4A+1\right)x^2+3 = 0 \\ &\iff x^2 = \frac{4A+1 \pm \sqrt{4A^2+8A+1}}{2A^2} \\ &\text{(using the quadratic formula)}.\end{align*} Now observe that since $A$ is positive, $4A^2+8A+1$ is also positive, so the square root will always give two distinct real values. Moreover, \[\left(4A+1\right)^2 = 16A^2+8A+1 > 4A^2+8A+1,\] so $4A+1-\sqrt{4A^2+8A+1} > 0$ , meaning that both solutions for $x^2$ are positive. Hence both solutions will give $2$ distinct values of $x$ (the positive and negative square roots), and each of these will correspond to a distinct point of intersection of the graphs, so there are $2 \cdot 2 = \boxed{4}$ points of intersection. | A | 4 |
48ce81497345bf462f880b4b26338090 | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_19 | Consider the graphs of $y = Ax^2$ and $y^2+3 = x^2+4y$ , where $A$ is a positive constant and $x$ and $y$ are real variables. In how many points do the two graphs intersect?
$\mathrm{(A) \ }\text{exactly }4 \qquad \mathrm{(B) \ }\text{exactly }2 \qquad$
$\mathrm{(C) \ }\text{at least }1,\text{ but the number varies for different positive values of }A \qquad$
$\mathrm{(D) \ }0\text{ for at least one positive value of }A \qquad \mathrm{(E) \ }\text{none of these}$ | Firstly, note that $y = Ax^2$ is an upward-facing parabola (since $A > 0$ ) whose vertex is at the origin. We now manipulate the equation of the second graph as follows: \begin{align*}y^2+3 = x^2+4y &\iff y^2-4y+3-x^2 = 0 \\ &\iff y^2-4y+4-x^2 = 1 \\ &\iff \frac{(y-2)^2}{1}-\frac{(x-0)^2}{1} = 1,\end{align*} showing that it is a vertical (upward- and downward-opening) hyperbola with center $(0,2)$ and asymptotes $y=x+1$ and $y=-x+1$ . It therefore remains to consider graphically where the parabola will intersect the hyperbola.
On the lower branch of the hyperbola, the maximum point is $(0,2-1) = (0,1)$ , which is above the vertex of the parabola. Therefore, by continuity and the symmetry of both the parabola and the hyperbola in the $y$ -axis, there are always exactly $2$ intersection points here.
For the top branch, as it approaches the asymptote $y = x+1$ , its slope also approaches that of this asymptote, which is $1$ . However, for any upward-opening parabola, the slope approaches infinity as $x$ does, so no matter how small $A$ is (i.e. how 'flat' the parabola is), the parabola will eventually overtake the hyperbola, giving a point of intersection with positive $x$ -coordinate. As above, symmetry gives another point of intersection with negative $x$ -coordinate, so that there are $2$ intersection points with this branch too.
Thus there are a total of $2+2 = \boxed{4}$ intersection points. | A | 4 |
8ead3ae86b4c76afadfb899c862d82b3 | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_20 | A wooden cube with edge length $n$ units (where $n$ is an integer $>2$ ) is painted black all over. By slices parallel to its faces, the cube is cut into $n^3$ smaller cubes each of unit edge length. If the number of smaller cubes with just one face painted black is equal to the number of smaller cubes completely free of paint, what is $n$
$\mathrm{(A)\ } 5 \qquad \mathrm{(B) \ }6 \qquad \mathrm{(C) \ } 7 \qquad \mathrm{(D) \ } 8 \qquad \mathrm{(E) \ }\text{none of these}$ | Observe that if we remove the outer layer of unit cubes from the entire cube, what remains is a smaller cube of side length $(n-2)$ , which contains all of the unpainted cubes and no others. This shows that there are exactly $(n-2)^3$ unpainted cubes. Similarly, taking one face of the cube and removing the outer edge leaves a square of side length $(n-2)$ containing all of the cubes on that face with exactly one face painted. Making the same argument for the other $5$ faces as well, we deduce that there are a total of $6(n-2)^2$ cubes with only one face painted.
Accordingly, we require \begin{align*}(n-2)^3 = 6(n-2)^2 &\iff n-2 = 6 \qquad \text{(as } n > 2\text{, so } n-2 \neq 0\text{)} \\ &\iff n = \boxed{8} | D | 8 |
0810b041ce99d05627cba68ac8441bcd | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_21 | How many integers $x$ satisfy the equation \[\left(x^2-x-1\right)^{x+2} = 1?\]
$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ } 5 \qquad \mathrm{(E) \ }\text{none of these}$ | We recall that for real numbers $a$ and $b$ , there are exactly $3$ ways in which we can have $a^b = 1$ , namely $a = 1$ $b = 0$ and $a \neq 0$ ; or $a = -1$ and $b$ is an even integer.
The first case therefore gives \begin{align*}x^2-x-1 = 1 &\iff x^2-x-2 = 0 \\&\iff (x-2)(x+1) = 0 \\&\iff x = 2 \text{ or } x = -1.\end{align*}
Similarly, the second case gives $x+2 = 0$ , i.e. $x = -2$ , and this indeed gives $x^2-x-1 = 4+2-1 = 5 \neq 0$ , so $x = -2$ is a further valid solution.
Lastly, for the third case, we have \begin{align*}x^2-x-1 = -1 &\iff x^2-x = 0 \\&\iff x(x-1) = 0 \\&\iff x = 0 \text{ or } x = 1,\end{align*} but $x = 1$ would give $x+2 = 3$ , which is odd, whereas $x = 0$ gives $x+2 = 2$ , which is even. Therefore, this case gives only one further solution, namely $x = 0$
Accordingly, the possible values of $x = -2$ $-1$ $0$ , or $2$ , yielding a total of $\boxed{4}$ solutions. | C | 4 |
1341db1361e33aa1f76ece7fbd0787fc | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_22 | In a circle with center $O$ $AD$ is a diameter, $ABC$ is a chord, $BO = 5$ and $\angle ABO = \ \stackrel{\frown}{CD} \ = 60^{\circ}$ . Then the length of $BC$ is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair O=origin, A=dir(35), C=dir(155), D=dir(215), B=intersectionpoint(dir(125)--O, A--C); draw(C--A--D^^B--O^^Circle(O,1)); pair point=O; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$O$", O, dir(305)); label("$5$", B--O, dir(O--B)*dir(90)); label("$60^\circ$", dir(185), dir(185)); label("$60^\circ$", B+0.05*dir(-25), dir(-25));[/asy]
$\mathrm{(A)\ } 3 \qquad \mathrm{(B) \ }3+\sqrt{3} \qquad \mathrm{(C) \ } 5-\frac{\sqrt{3}}{2} \qquad \mathrm{(D) \ } 5 \qquad \mathrm{(E) \ }\text{none of the above}$ | Since $\angle CAD$ is an angle inscribed in a $60{^\circ}$ arc, we obtain $\angle CAD =\frac{60^{\circ}}{2} = 30^{\circ}$ , so $\triangle ABO$ is a $30^{\circ}$ $60^{\circ}$ $90^{\circ}$ right triangle. This gives $AO = BO\sqrt{3} = 5\sqrt{3}$ and $AB = 2BO = 10$ , and now since $AD$ is a diameter, $AD = 2AO = 10\sqrt{3}$ . The fact that $AD$ is a diameter also means that $\angle ACD = 90^{\circ}$ , so $\triangle ACD$ is again a $30^{\circ}$ $60^{\circ}$ $90^{\circ}$ right triangle, yielding $CD = \frac{1}{2}AD = 5\sqrt{3}$ and $AC = \frac{\sqrt{3}}{2}AD = 15$ , which finally gives $BC = AC-AB = 15-10 = \boxed{5}$ | D | 5 |
4c37ac143fb3b14ea34f1e4055e7bfc0 | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_24 | A non-zero digit is chosen in such a way that the probability of choosing digit $d$ is $\log_{10}{(d+1)}-\log_{10}{d}$ . The probability that the digit $2$ is chosen is exactly $1/2$ the probability that the digit chosen is in the set
$\mathrm{(A)\ } \{2,3\} \qquad \mathrm{(B) \ }\{3,4\} \qquad \mathrm{(C) \ } \{4,5,6,7,8\} \qquad \mathrm{(D) \ } \{5,6,7,8,9\} \qquad \mathrm{(E) \ }\{4,5,6,7,8,9\}$ | We have $\log_{10}{(d+1)}-\log_{10}{d} = \log_{10}{\left(\frac{d+1}{d}\right)}$ , so the probability of choosing $2$ is $\log_{10}{\left(\frac{3}{2}\right)}$ . The probability that the digit chosen is in the set must therefore be \begin{align*}2\log_{10}{\left(\frac{3}{2}\right)} = &\log_{10}{\left(\left(\frac{3}{2}\right)^2\right)} \\ = &\log_{10}{\left(\frac{9}{4}\right)} \\ = &\log_{10}{9}-\log_{10}{4} \\ = &\left(\log_{10}{9}-\log_{10}{8}\right)+\left(\log_{10}{8}-\log_{10}{7}\right)+\left(\log_{10}{7}-\log_{10}{6}\right) \\ &+\left(\log_{10}{6}-\log_{10}{5}\right)+\left(\log_{10}{5}-\log_{10}{4}\right),\end{align*} which, by definition, is the probability that the digit chosen is in the set $\boxed{4,5,6,7,8}$ | C | 4,5,6,7,8 |
2dbf62edb02592b62c9b2644bf2357aa | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_25 | The volume of a certain rectangular solid is $8$ cm , its total surface area is $32$ cm , and its three dimensions are in geometric progression. The sums of the lengths in cm of all the edges of this solid is
$\mathrm{(A)\ } 28 \qquad \mathrm{(B) \ }32 \qquad \mathrm{(C) \ } 36 \qquad \mathrm{(D) \ } 40 \qquad \mathrm{(E) \ }44$ | As the dimensions are in geometric progression, let them be $\frac{b}{r}$ $b$ , and $br$ cm, so the volume is $\left(\frac{b}{r}\right)(b)(br) = b^3$ , giving $b^3 = 8$ and thus $b = 2$ . The surface area condition now yields \begin{align*}2\left(\frac{2}{r}\right)(2)+2(2)(2r)+2(2r)\left(\frac{2}{r}\right) = 32 &\iff \frac{8}{r}+8+8r = 32 \\ &\iff r+\frac{1}{r} = 3 \\ &\iff r^2-3r+1 = 0 \\ &\iff r = \frac{3 \pm \sqrt{5}}{2}.\end{align*}
Since \[\frac{3-\sqrt{5}}{2} = \frac{1}{\left(\frac{3+\sqrt{5}}{2}\right)},\] the two possible values of $r$ do not actually give different dimensions, but merely determine whether they are in increasing or decreasing order. Therefore, without loss of generality, we take $r = \frac{3+\sqrt{5}}{2}$ , giving the dimensions (in cm) as \begin{align*}&\frac{2}{\left(\frac{3+\sqrt{5}}{2}\right)}, 2, \text{ and } 2\left(\frac{3+\sqrt{5}}{2}\right) \\ &= \frac{4}{3+\sqrt{5}}, 2, \text{ and } 3+\sqrt{5} \\ &= 3-\sqrt{5}, 2, \text{ and } 3+\sqrt{5}.\end{align*}
As there are $4$ edges with each of these distinct lengths, it follows that the sum of all the edge lengths (in cm) is \begin{align*}4\left(3-\sqrt{5}+2+3+\sqrt{5}\right) &= 4(8) \\ &= \boxed{32} | B | 32 |
2dbf62edb02592b62c9b2644bf2357aa | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_25 | The volume of a certain rectangular solid is $8$ cm , its total surface area is $32$ cm , and its three dimensions are in geometric progression. The sums of the lengths in cm of all the edges of this solid is
$\mathrm{(A)\ } 28 \qquad \mathrm{(B) \ }32 \qquad \mathrm{(C) \ } 36 \qquad \mathrm{(D) \ } 40 \qquad \mathrm{(E) \ }44$ | Similarly to in Solution 1, we let the dimensions (in cm) be $b$ $br$ , and $br^2$ , so that the volume condition gives $8 = b^3r^3 = (br)^3$ and thus $br = 2$ . The surface area condition now becomes \begin{align*}2(b)(br)+2(br)\left(br^2\right)+2\left(br^2\right)(b) = 32 &\iff b^2r+b^2r^2+b^2r^3 = 16 \\&\iff br\left(b+br+br^2\right) = 16,\end{align*} so substituting $br = 2$ from above immediately gives \[b+br+br^2 = \frac{16}{2} = 8,\] and hence, without needing to actually compute the dimensions, we deduce that the sum of the edge lengths (in cm) is \[4b + 4br + 4br^2 = 4(8) = \boxed{32}.\] | B | 32 |
bf511a0ce89a202915da4c5a340d3f13 | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_27 | Consider a sequence $x_1,x_2,x_3,\dotsc$ defined by: \begin{align*}&x_1 = \sqrt[3]{3}, \\ &x_2 = \left(\sqrt[3]{3}\right)^{\sqrt[3]{3}},\end{align*} and in general \[x_n = \left(x_{n-1}\right)^{\sqrt[3]{3}} \text{ for } n > 1.\]
What is the smallest value of $n$ for which $x_n$ is an integer?
$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ } 9 \qquad \mathrm{(E) \ }27$ | Firstly, we will show by induction that \[x_n = \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\right)}.\] For the base case, we indeed have \begin{align*}x_1 &= \sqrt[3]{3} \\ &= \left(\sqrt[3]{3}\right)^1 \\ &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^0\right)},\end{align*} and for the inductive step, if our claim is true for $x_n$ , then \begin{align*}x_{n+1} &= \left(x_n\right)^{\sqrt[3]{3}} \\ &= \left(\left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\right)}\right)^{\sqrt[3]{3}} \\ &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\cdot\sqrt[3]{3}\right)} \\ &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^n\right)},\end{align*} which completes the proof.
We now rewrite our formula for $x_n$ as follows: \begin{align*}x_n &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\right)} \\ &= \left(\sqrt[3]{3}\right)^{\left(3^{\frac{n-1}{3}}\right)} \\ &=3^{\left(\frac{1}{3} \cdot 3^{\frac{n-1}{3}}\right)} \\ &= 3^{\left(3^{\left(\frac{n-1}{3}-1\right)}\right)} \\ &= 3^{\left(3^{\left(\frac{n-4}{3}\right)}\right)},\end{align*} and as $3$ is not a perfect power, we deduce that $x_n$ is an integer if and only if the exponent, $3^{\left(\frac{n-4}{3}\right)}$ , is itself an integer. By precisely the same argument, this reduces to $\frac{n-4}{3}$ being an integer, so the smallest possible (positive) value of $n$ is $\boxed{4}$ | C | 4 |
c35762325b03e77657cf1cd5d03e0810 | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_28 | In $\triangle ABC$ , we have $\angle C = 3\angle A$ $a = 27$ and $c = 48$ . What is $b$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(14,0), C=(10,6); draw(A--B--C--cycle); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$a$", B--C, dir(B--C)*dir(-90)); label("$b$", A--C, dir(C--A)*dir(-90)); label("$c$", A--B, dir(A--B)*dir(-90));[/asy]
$\mathrm{(A)\ } 33 \qquad \mathrm{(B) \ }35 \qquad \mathrm{(C) \ } 37 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ }\text{not uniquely determined}$ | Let $\angle A = x^{\circ}$ , so $\angle C = 3x^{\circ}$ , and thus $\angle B = \left(180-4x\right)^{\circ}$ . Now let $D$ be a point on side $AB$ such that $\angle ACD = x^{\circ}$ , so $\angle BCD = 3x^{\circ}-x^{\circ} = 2x^{\circ}$ , which gives \[\angle CDB = 180^{\circ}-2x^{\circ}-\left(180-4x\right)^{\circ} = 2x^{\circ},\] meaning that $\triangle CDB$ and $\triangle CDA$ are both isosceles, with $BC = BD$ and $AD = CD$ . In particular, $BD = BC = 27$ and $CD = AD = AB-BD = 48-27 = 21$ . Hence by Stewart's theorem on triangle $ABC$ \begin{align*}&BD \cdot AB \cdot AD + CD^2 \cdot AB = AC^2 \cdot BD + BC^2 \cdot AD \\ &\iff 27 \cdot 48 \cdot 21 + 21^2 \cdot 48 = AC^2 \cdot 27 + 27^2 \cdot 21 \\ &\iff AC^2 = \frac{27(21)(48-27) + 21^2 \cdot 48}{27} \\ &\iff AC^2 = \frac{21^2(27+48)}{27} \\ &\iff AC^2 = \frac{21^2 \cdot 25}{9} \\ &\iff AC = \frac{21 \cdot 5}{3} \qquad \text{(as } AC > 0\text{)} \\ &\iff AC = \boxed{35} | B | 35 |
c35762325b03e77657cf1cd5d03e0810 | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_28 | In $\triangle ABC$ , we have $\angle C = 3\angle A$ $a = 27$ and $c = 48$ . What is $b$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(14,0), C=(10,6); draw(A--B--C--cycle); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$a$", B--C, dir(B--C)*dir(-90)); label("$b$", A--C, dir(C--A)*dir(-90)); label("$c$", A--B, dir(A--B)*dir(-90));[/asy]
$\mathrm{(A)\ } 33 \qquad \mathrm{(B) \ }35 \qquad \mathrm{(C) \ } 37 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ }\text{not uniquely determined}$ | We apply the law of sines in the form \[\frac{\sin(A)}{a} = \frac{\sin(C)}{c},\] yielding \[\frac{\sin(A)}{27} = \frac{\sin(3A)}{48} \iff 9\sin(3A) = 16\sin(A).\]
Now, the angle sum and double angle identities give \begin{align*}\sin(3A) &= \sin(2A+A) \\ &= \sin(2A)\cos(A)+\cos(2A)\sin(A) \\ &= \left(2\sin(A)\cos(A)\right)\cos(A)+\left(\cos^2(A)-\sin^2(A)\right)\sin(A) \\ &= 2\sin(A)\cos^2(A)+\sin(A)\cos^2(A)-\sin^3(A) \\ &= 3\sin(A)\left(1-\sin^2(A)\right)-\sin^3(A) \\ &\text{(using the further identity } \cos^2(\theta)+\sin^2(\theta) = 1\text{)} \\ &= 3\sin(A)-4\sin^3(A).\end{align*}
Thus our equation becomes \begin{align*}9\left(3\sin(A)-4\sin^3(A)\right) = 16\sin(A) &\iff 27\sin(A)-36\sin^3(A) = 16\sin(A) \\ &\iff 36\sin^3(A) = 11\sin(A) \\ &\iff \sin(A) = 0 \text{ or } \pm\frac{\sqrt{11}}{6}.\end{align*} Notice, however, that we must have $0^{\circ} < A < 45^{\circ}$ , the latter because otherwise $A+3A \geq 180^{\circ}$ , which would contradict the fact that $A$ and $3A$ are angles in a (non-degenerate) triangle. This means $\sin(A) > 0$ , so the only valid solution is \[\sin(A) = \frac{\sqrt{11}}{6},\] and the fact that $A$ is acute also means $\cos(A) > 0$ , so we deduce \begin{align*}\cos(A) &= \sqrt{1-\left(\frac{\sqrt{11}}{6}\right)^2} \\ &= \sqrt{1-\frac{11}{36}} \\ &=\frac{5}{6}.\end{align*} Accordingly, using the double angle identities again, \begin{align*}\sin(4A) &= \sin(2 \cdot 2A) \\ &= 2\sin(2A)\cos(2A) \\ &= 2\left(2\sin(A)\cos(A)\right)\left(\cos^2(A)-\sin^2(A)\right) \\ & =2\left(2 \cdot \frac{\sqrt{11}}{6} \cdot\frac{5}{6}\right)\left(\left(\frac{5}{6}\right)^2-\left(\frac{\sqrt{11}}{6}\right)^2\right) \\ &= \frac{5\sqrt{11}}{9} \cdot \frac{25-11}{36} \\ &= \frac{5\sqrt{11}}{9} \cdot \frac{7}{18} \\ &= \frac{35\sqrt{11}}{162}.\end{align*} Finally, the law of sines now gives \begin{align*}\frac{\sin(A)}{27} &= \frac{\sin(B)}{b} \\ &= \frac{\sin(180^{\circ}-3A-A)}{b} \\ &= \frac{\sin(4A)}{b} \qquad \text{(using the identity } \sin\left(180^{\circ}-\theta\right) = \sin(\theta)\text{)},\end{align*} so, substituting the above results, \[\frac{\left(\frac{\sqrt{11}}{6}\right)}{27} = \frac{\left(\frac{35\sqrt{11}}{162}\right)}{b} \iff b = \frac{6 \cdot 27 \cdot 35}{162} = \boxed{35}.\] | B | 35 |
a972e3763af786fb3e374208b63c8fe1 | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_29 | In their base $10$ representations, the integer $a$ consists of a sequence of $1985$ eights and the integer $b$ consists of a sequence of $1985$ fives. What is the sum of the digits of the base $10$ representation of the integer $9ab$
$\mathrm{(A)\ } 15880 \qquad \mathrm{(B) \ }17856 \qquad \mathrm{(C) \ } 17865 \qquad \mathrm{(D) \ } 17874 \qquad \mathrm{(E) \ }19851$ | By the formula for the sum of a geometric series, \begin{align*}a &= 8 \cdot 10^0 + 8 \cdot 10^1 + \dotsb + 8 \cdot 10^{1984} \\ &= \frac{8\left(10^{1985}-1\right)}{10-1} \\ &= \frac{8\left(10^{1985}-1\right)}{9},\end{align*} and similarly \[b = \frac{5\left(10^{1985}-1\right)}{9},\] so \begin{align*}9ab &= 9\cdot\frac{8\left(10^{1985}-1\right)}{9}\cdot\frac{5\left(10^{1985}-1\right)}{9} \\ &= \frac{40\left(10^{1985}-1\right)^2}{9} \\ &= \frac{40\left(10^{3970}-2 \cdot 10^{1985}+1\right)}{9} \\ &= \frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9}.\end{align*}
We now compute the decimal expansion of this expression. Firstly, $10^{3971} = 100 \dotsb 0$ , with $1$ one and $3971$ zeroes, and $2 \cdot 10^{1986} = 200 \dotsb 0$ , with $1$ two and $1986$ zeroes. Subtracting therefore gives \[10^{3971}-2 \cdot 10^{1986} = 99 \dotsb 9800 \dotsb 0,\] where there are $3971-1986-1 = 1984$ nines followed by $1$ eight and then $1986$ zeroes. Adding $10$ transforms this to $99 \dotsb 9800 \dotsb 010$ , now with $1984$ nines followed by $1$ eight, $1984$ zeroes, $1$ one, and a final zero.
Using long division, and noting that $80 = 8 \cdot 9 + 8$ and $81 = 9 \cdot 9$ , it follows that \[\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 11 \dotsb 1088 \dotsb 890,\] with $1984$ ones, $1$ zero, then $1984$ eights, $1$ nine, and a final zero. Lastly, using long multiplication and noting that $9 \cdot 4 = 36$ $8 \cdot 4 = 32$ , and $8 \cdot 4 + 3 = 35$ , we obtain \[\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 44 \dotsb 4355 \dotsb 560,\] where there are $1984$ fours, $1$ three, $1984$ fives, $1$ six, and a final zero, so the sum of the digits is \begin{align*}1984 \cdot 4 + 3 + 1984 \cdot 5 + 6 + 0 &= 1984 \cdot 9 + 9 \\ &= 1985 \cdot 9 \\ &= \boxed{17865} | C | 17865 |
3adb034c822193b2d331eefa1e337cf3 | https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_30 | Let $\left\lfloor x\right\rfloor$ be the greatest integer less than or equal to $x$ . Then the number of real solutions to $4x^2-40\left\lfloor x\right\rfloor+51 = 0$ is
$\mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ } 3 \qquad \mathrm{(E) \ }4$ | We rearrange the equation as $4x^2 = 40\left\lfloor x\right\rfloor-51$ , where the right-hand side is now clearly an integer, meaning that $4x^2 = n$ for some non-negative integer $n$ . Therefore, in the case where $x \geq 0$ , substituting $x = \frac{\sqrt{n}}{2}$ gives \[40\left\lfloor\frac{\sqrt{n}}{2}\right\rfloor-51 = n.\] To proceed, let $a$ be the unique non-negative integer such that $a \leq \frac{\sqrt{n}}{2} < a+1$ , so that \begin{align*}&\left\lfloor \frac{\sqrt{n}}{2}\right\rfloor = a, \text{ and} \\ &4a^2 \leq n < 4a^2+8a+4,\end{align*} and our equation reduces to \[40a-51 = n.\]
The above inequalities therefore become \[4a^2 \leq 40a-51 < 4a^2+8a+4 \iff 4a^2-40a+51 < 0 \text{ and } 4a^2-32a+55 > 0,\] where the first inequality can now be rewritten as $(2a-10)^2 \leq 49$ , i.e. $\left\lvert 2a-10\right\rvert \leq 7$ . Since $(2a-10)$ is even for all integers $a$ , we must in fact have \begin{align*}\left\lvert 2a-10\right\rvert \leq 6 &\iff \left\lvert a-5\right\rvert \leq 3 \\ &\iff 2 \leq a \leq 8.\end{align*} The second inequality similarly simplifies to $(2a-8)^2 > 9$ , i.e. $\left\lvert 2a-8\right\rvert > 3$ . As $(2a-8)$ is even, this is equivalent to \begin{align*}\left\lvert 2a-8 \right\rvert \geq 4 &\iff \left\lvert a-4\right\rvert \geq 2 \\ &\iff a \geq 6 \text{ or } a \leq 2,\end{align*} so the values of $a$ satisfying both inequalities are $2$ $6$ $7$ , and $8$ . Since $n = 40a-51$ , each of these distinct values of $a$ gives a distinct solution for $n$ , and thus for $x = \frac{\sqrt{n}}{2}$ , giving a total of $4$ solutions in the $x \geq 0$ case.
As $4$ is already the largest of the answer choices, this suffices to show that the answer is $\text{(E)}$ , but for completeness, we will show that the $x < 0$ case indeed gives no other solutions. If $x = -\frac{\sqrt{n}}{2}$ (and so $n > 0$ ), we require \[40\left\lfloor -\frac{\sqrt{n}}{2}\right\rfloor-51 = n,\] and recalling that $\left\lfloor -x\right\rfloor = -\left\lceil x\right\rceil$ for all $x$ , this equation can be rewritten as \[-40\left\lceil \frac{\sqrt{n}}{2}\right\rceil-51 = n.\] Since $n$ is positive, the least possible value of $\left\lceil \frac{\sqrt{n}}{2}\right\rceil$ is $1$ , but this means \begin{align*}n &= -40\left\lceil\frac{\sqrt{n}}{2}\right\rceil-51 \\ &\leq -40 \cdot 1 - 51 \\ &= -91,\end{align*} which is a contradiction. Therefore the $x < 0$ case indeed gives no further solutions, confirming that the total number of solutions is precisely $\boxed{4}$ | E | 4 |
cea2c787e19d31bdb6d2d7b09169d7c9 | https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_4 | A rectangle intersects a circle as shown: $AB=4$ $BC=5$ , and $DE=3$ . Then $EF$ equals:
[asy]defaultpen(linewidth(0.7)+fontsize(10)); pair D=origin, E=(3,0), F=(10,0), G=(12,0), H=(12,1), A=(0,1), B=(4,1), C=(9,1), O=circumcenter(B,C,F); draw(D--G--H--A--cycle); draw(Circle(O, abs(O-C))); label("$A$", A, NW); label("$B$", B, NW); label("$C$", C, NE); label("$D$", D, SW); label("$E$", E, SE); label("$F$", F, SW); label("4", (2,0.85), N); label("3", D--E, S); label("5", (6.5,0.85), N); [/asy] $\mathbf{(A)}\; 6\qquad \mathbf{(B)}\; 7\qquad \mathbf{(C)}\; \frac{20}3\qquad \mathbf{(D)}\; 8\qquad \mathbf{(E)}\; 9$ | [asy] defaultpen(linewidth(0.7)+fontsize(10)); pair D=origin, E=(3,0), F=(10,0), X=(12,0), Y=(12,1), A=(0,1), B=(4,1), C=(9,1), O=circumcenter(B,C,F), G=foot(E,A,C), H=foot(B,D,F), I=foot(C,D,F); draw(D--X--Y--A--cycle); draw(Circle(O, abs(O-C))); label("$A$", A, NW); label("$B$", B, N); label("$C$", C, NE); label("$D$", D, SW); label("$E$", E, S); label("$F$", F, S); label("$G$", G, N); label("$H$", H, S); label("$I$", I, S); label("4", (2,0.85), N); label("3", D--E, S); label("5", (6.5,0.85), N); draw(E--G^^H--B^^I--C, linetype("4 4")); [/asy]
Draw $BE$ and $CF$ , forming a trapezoid . Since it's cyclic, this trapezoid must be isosceles . Also, drop altitudes from $E$ to $AC$ $B$ to $DF$ , and $C$ to $DF$ , and let the feet of these altitudes be $G$ $H$ , and $I$ respectively. $AGED$ is a rectangle since it has $4$ right angles . Therefore, $AG=DE=3$ , and $GB=4-3=1$ . By the same logic, $GBHE$ is also a rectangle, and $EH=GB=1$ $BH=CI$ since they're both altitudes to a trapezoid, and $BE=CF$ since the trapezoid is isosceles. Therefore, $\triangle BHE \cong \triangle CIF$ by HL congruence , so $IF=EH=1$ . Also, $BCIH$ is a rectangle from $4$ right angles, and $HI=BC=5$ . Therefore, $EF=EH+HI+IF=1+5+1=\boxed{7}$ | null | 7 |
cf75b54d7cc422b2d8c96e8c9a04e3b1 | https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_5 | The largest integer $n$ for which $n^{200}<5^{300}$ is
$\mathrm{(A) \ }8 \qquad \mathrm{(B) \ }9 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ }11 \qquad \mathrm{(E) \ } 12$ | Since both sides are positive, we can take the $100th$ root of both sides to find the largest integer $n$ such that $n^2<5^3$ . Fortunately, this is simple to evaluate: $5^3=125$ , and the largest square less than $125$ is $11^2=121$ , so the largest $n$ is $11, \boxed{11}$ | D | 11 |
e1d9eb2a70f2a18bc6d7215cb43a741e | https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_9 | The number of digits in $4^{16}5^{25}$ (when written in the usual base $10$ form) is
$\mathrm{(A) \ }31 \qquad \mathrm{(B) \ }30 \qquad \mathrm{(C) \ } 29 \qquad \mathrm{(D) \ }28 \qquad \mathrm{(E) \ } 27$ | We can rewrite this as $2^{32}5^{25}$ . We can also combine some of the factors to introduce factors of $10$ , whose digit count is simple to evaluate because it simply adds $0$ s. Thus, we have $2^{32}5^{25}=2^72^{25}5^{25}=2^710^{25}$ . We can see that this final number is $2^7$ with $25$ $0$ s annexed onto it. $2^7=128$ , which has $3$ digits, so the entire number has $25+3=28$ digits, $\boxed{28}$ | D | 28 |
50dff3bdea77a525bad9b8089c72449a | https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_12 | If the sequence $\{a_n\}$ is defined by
$a_1=2$
$a_{n+1}=a_n+2n$
where $n\geq1$
Then $a_{100}$ equals
$\mathrm{(A) \ }9900 \qquad \mathrm{(B) \ }9902 \qquad \mathrm{(C) \ } 9904 \qquad \mathrm{(D) \ }10100 \qquad \mathrm{(E) \ } 10102$ | We begin to evaluate the first couple of terms of the sequence, hoping to find a pattern: $2, 4, 8, 14, 22, ....$ . We notice that the difference between succesive terms of the sequence are $2, 4, 6, 8, ....$ , a clear pattern. We can see that this pattern continues infinitely because of the recursive definition: each term is the previous term plus the next even number. Therefore, since the differences of consecutive terms form an arithmetic sequence, then the terms satisfy a quadratic , specifically, the one that contains the points $(1, 2), (2, 4),$ , and $(3, 8)$ . Let the quadratic be $f(x)=ax^2+bx+c$ , so:
$a+b+c=2$ (1)
$4a+2b+c=4$ (2)
$9a+3b+c=8$ (3)
Subtracting (1) from (2) and (2) from (3) yields the two-variable system of equations
$3a+b=2$ (4)
$5a+b=4$ (5)
We can subtract (4) from (5) to find that $2a=2$ , so $a=1$ . Substituting this back in yields $b=-1$ , and substituting these back into one of the original equations yields $c=2$ , so the closed form for the terms is
$f(x)=x^2-x+2$ , or
$a_n=n^2-n+2$
Substituting in $n=100$ yields $a_{100}=100^2-100+2=9902, \boxed{9902}$ | B | 9902 |
4771de0ba88ce6e9dcfcee73027e2798 | https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_15 | If $\sin{2x}\sin{3x}=\cos{2x}\cos{3x}$ , then one value for $x$ is
$\mathrm{(A) \ }18^\circ \qquad \mathrm{(B) \ }30^\circ \qquad \mathrm{(C) \ } 36^\circ \qquad \mathrm{(D) \ }45^\circ \qquad \mathrm{(E) \ } 60^\circ$ | We divide both sides of the equation by $\cos{2x}\times\cos{3x}$ to get $\frac{\sin{2x}\times\sin{3x}}{\cos{2x}\times\cos{3x}}=1$ , or $\tan{2x}\times\tan{3x}=1$
This looks a lot like the formula relating the slopes of two perpendicular lines , which is $m_1\times m_2=-1$ , where $m_1$ and $m_2$ are the slopes . It's made even more relatable by the fact that the tangent of an angle can be defined by the slope of the line that makes that angle with the x-axis
We can make this look even more like the slope formula by multiplying both sides by $-1$
$\tan{2x}\times-\tan{3x}=-1$ , and using the trigonometric identity $-\tan{x}=\tan{-x}$ , we have $\tan{2x}\times\tan{-3x}=-1$
Now it's time for a diagram:
[asy] unitsize(2.54cm); draw(unitcircle); draw((0,-1.25)--(0,1.25)); draw((-1.25,0)--(1.25,0)); draw((0,0)--(cos(2pi/10),sin(2pi/10))); draw((0,0)--(cos(-3pi/10),sin(-3pi/10))); label("$\tan{-3x}$",(cos(-3pi/10),sin(-3pi/10)),SE); label("$\tan{2x}$",(cos(2pi/10),sin(2pi/10)),NE); label("$2x$",(.125,.03),ENE); label("$3x$",(.125,-.06),ESE); [/asy] Since the product of the two slopes, $\tan{2x}$ and $\tan{-3x}$ , is $-1$ , the lines are perpendicular, and the angle between them is $\frac{\pi}{2}$ . The angle between them is also $2x+3x=5x$ , so $5x=\frac{\pi}{2}$ and $x=\frac{\pi}{10}$ , or $18^\circ, \boxed{18}$ | A | 18 |
4771de0ba88ce6e9dcfcee73027e2798 | https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_15 | If $\sin{2x}\sin{3x}=\cos{2x}\cos{3x}$ , then one value for $x$ is
$\mathrm{(A) \ }18^\circ \qquad \mathrm{(B) \ }30^\circ \qquad \mathrm{(C) \ } 36^\circ \qquad \mathrm{(D) \ }45^\circ \qquad \mathrm{(E) \ } 60^\circ$ | Start by subtracting $\sin{2x}\sin{3x}$ from both sides to get $\cos{2x}\cos{3x}-\sin{2x}\sin{3x}=0$ . We recognize that this is of the form $\cos{a}\cos{b}-\sin{a}\sin{b}=\cos{(a+b)}$ , so $\cos{2x}\cos{3x}-\sin{2x}\sin{3x}=\cos{(2x+3x)}=0$ $\cos{90^\circ}=0$ , so $x=\boxed{18}$ | null | 18 |
6ac2e0451e25ffe6f8a2fa90dbb4d0d7 | https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_16 | The function $f(x)$ satisfies $f(2+x)=f(2-x)$ for all real numbers $x$ . If the equation $f(x)=0$ has exactly four distinct real roots , then the sum of these roots is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$ | Let one of the roots be $r_1$ . Also, define $x$ such that $2+x=r_1$ . Thus, we have $f(2+x)=f(r_1)=0$ and $f(2+x)=f(2-x)$ . Therefore, we have $f(2-x)=0$ , and $2-x$ is also a root. Let this root be $r_2$ . The sum $r_1+r_2=2+x+2-x=4$ . Similarly, we can let $r_3$ be a root and define $y$ such that $2+y=r_3$ , and we will find $2-y$ is also a root, say, $r_4$ , so $r_3+r_4=2+y+2-y=4$ . Therefore, $r_1+r_2+r_3+r_4=4+4=8, \boxed{8}$ | E | 8 |
4ec6e1f809c38848d51a8d2e6bc6f5bb | https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_20 | The number of the distinct solutions to the equation
$|x-|2x+1||=3$ is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ } 4$ | We can create a tree of possibilities, progressively eliminating the absolute value signs by creating different cases.
[asy] unitsize(2cm); draw((0,0)--(2,-2)); draw((0,0)--(-2,-2)); label("$|x-|2x+1||=3$",(0,0),N); label("$x-|2x+1|=3$",(-2,-2),S); label("$x-|2x+1|=-3$",(2,-2),S); label("$|2x+1|=x-3$",(-2,-2.25),S); label("$|2x+1|=x+3$",(2,-2.25),S); draw((2,-2.5)--(3,-3.5)); draw((2,-2.5)--(1,-3.5)); draw((-2,-2.5)--(-3,-3.5)); draw((-2,-2.5)--(-1,-3.5)); label("$2x+1=x-3$",(-3,-3.5),S); label("$x=-2$",(-3,-3.75),S); label("$2x+1=-x+3$",(-1,-3.5),S); label("$x=\frac{2}{3}$",(-1,-3.75),S); label("$2x+1=x+3$",(1,-3.5),S); label("$x=2$",(1,-3.75),S); label("$2x+1=-x-3$",(3,-3.5),S); label("$x=-\frac{4}{3}$",(3,-3.75),S); [/asy]
So we have $4$ possible solutions: $-2, \frac{2}{3}, 2,$ and $-\frac{4}{3}$ . Checking for extraneous solutions, we find that the only ones that work are $2$ and $-\frac{4}{3}$ , so there are $2$ solutions, $\boxed{2}$ | C | 2 |
fee812e4b2750384d07cdfd809a432d8 | https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_21 | The number of triples $(a, b, c)$ of positive integers which satisfy the simultaneous equations
$ab+bc=44$
$ac+bc=23$
is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ } 4$ | We can factor the second equation to get $c(a+b)=23$ , so we see that $c$ must be a factor of $23$ , and since this is prime $c=1$ or $c=23$ . However, if $c=23$ , then $a+b=1$ , which is impossible for the field of positive integers. Therefore, $c=1$ for all possible solutions. Substituting this into the original equations gives
$ab+b=44$
and
$a+b=23$
From the second equation, $a=23-b$ , and substituting this into the first equation yields $b(23-b)+b=44$ , or $b^2-24b+44=0$ . Factoring this gives $(b-2)(b-22)=0$ , so $b=2$ or $b=22$ . Both of these yield integer solutions for $a$ , giving $a=21$ or $a=1$ , respectively.
Therefore, the only solutions are $(21, 2, 1)$ and $(1, 22, 1)$ , yielding $2$ solutions, $\boxed{2}$ | C | 2 |
f4d442979ab01843891ec66821be6fe1 | https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_24 | If $a$ and $b$ are positive real numbers and each of the equations $x^2+ax+2b=0$ and $x^2+2bx+a=0$ has real roots , then the smallest possible value of $a+b$ is
$\mathrm{(A) \ }2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ } 6$ | Since both of the equations have real roots, both of their discriminants are nonnegative. Therefore, we have
$a^2-4(1)(2b)=a^2-8b\geq0\implies a^2\geq8b$ from the first equation, and
$(2b)^2-4(1)(a)=4b^2-4a\geq0\implies 4b^2\geq4a\implies b^2\geq a$ from the second.
We can square the second equation to get $b^4\geq a^2$ , and combining this with the first one gives $b^4\geq a^2\geq8b$ , so $b^4\geq8b$ . We can divide both sides by $b$ , since it is positive, and take the cubed root of that to get $b\geq2$ . Therefore, we have $a^2\geq8b\geq8(2)=16$ , and since $a$ is positive, we can take the square root of this to get $a\geq4$ . Therefore, $a+b\geq 2+4=6$ , and the smallest possible value is $6, \boxed{6}$ | E | 6 |
94e7660723d0b0dc2f3ea1257db7b7dd | https://artofproblemsolving.com/wiki/index.php/1984_AHSME_Problems/Problem_28 | The number of distinct pairs of integers $(x, y)$ such that $0<x<y$ and $\sqrt{1984}=\sqrt{x}+\sqrt{y}$ is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ }4 \qquad \mathrm{(E) \ } 7$ | We can simplify $\sqrt{1984}$ to $8\sqrt{31}$ . Therefore, the only solutions are $a\sqrt{31}+b\sqrt{31}$ such that $a+b=8$ and $0<a<b$ . The only solutions to this are $a=1, b=7; a=2, b=6; a=3, b=5$ . Each of these gives distinct pairs of $(x, y)$ , so there are $3$ pairs, $\boxed{3}$ | C | 3 |
de6c02c18fdf08d05670bfb4fc279b4f | https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_1 | If $x \neq 0, \frac x{2} = y^2$ and $\frac{x}{4} = 4y$ , then $x$ equals
$\textbf{(A)}\ 8\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 32\qquad \textbf{(D)}\ 64\qquad \textbf{(E)}\ 128$ | From $\frac{x}{4} = 4y$ , we get $x=16y$ . Plugging in the other equation, $\frac{16y}{2} = y^2$ , so $y^2-8y=0$ . Factoring, we get $y(y-8)=0$ , so the solutions are $0$ and $8$ . Since $x \neq 0$ , we also have $y \neq 0$ , so $y=8$ . Hence $x=16\cdot 8 = \boxed{128}$ | E | 128 |
840945b88bb41cdb9ea316ff449c9e82 | https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_2 | Point $P$ is outside circle $C$ on the plane. At most how many points on $C$ are $3$ cm from $P$
$\textbf{(A)} \ 1 \qquad \textbf{(B)} \ 2 \qquad \textbf{(C)} \ 3 \qquad \textbf{(D)} \ 4 \qquad \textbf{(E)} \ 8$ | The points $3$ cm away from $P$ can be represented as a circle centered at $P$ with radius $3$ cm. The maximum number of intersection points of two circles is $\boxed{2}$ | B | 2 |
4ca30b9244315b04f730e9086717bb9e | https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_3 | Three primes $p,q$ , and $r$ satisfy $p+q = r$ and $1 < p < q$ . Then $p$ equals
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 13\qquad \textbf{(E)}\ 17$ | We are given that $p,q$ and $r$ are primes. In order for $p$ and $q$ to sum to another prime, either $p$ or $q$ has to be even, because the sum of two odd numbers would be even, and the only even prime is $2$ (but $p + q = 2$ would have, as the only solution in positive integers, $p = q = 1$ , and $1$ is not prime). Thus, with one of either $p$ or $q$ being even, either $p$ or $q$ must be $2$ , and as $p < q$ , we deduce $p = 2$ (as $2$ is the smallest prime). This means the answer is $\boxed{2}$ | A | 2 |
ca93406464596042e414eae604545a75 | https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_6 | When $x^5, x+\frac{1}{x}$ and $1+\frac{2}{x} + \frac{3}{x^2}$ are multiplied, the product is a polynomial of degree.
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8$ | We have $x^5\left(x+\frac{1}{x}\right)\left(1+\frac{2}{x}+\frac{3}{x^2}\right) = (x^6+x^4)\left(1+\frac{2}{x}+\frac{3}{x^2}\right) = x^6 + \text{lower order terms}$ , where we know that the $x^6$ will not get cancelled out by e.g. a $-x^6$ term since all the terms inside the brackets are positive. Thus the degree is $6$ , which is choice $\boxed{6}$ | C | 6 |
8488c8a18ff4afd817d08c31b05a1785 | https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_7 | Alice sells an item at $$10$ less than the list price and receives $10\%$ of her selling price as her commission.
Bob sells the same item at $$20$ less than the list price and receives $20\%$ of his selling price as his commission.
If they both get the same commission, then the list price is
$\textbf{(A) } $20\qquad \textbf{(B) } $30\qquad \textbf{(C) } $50\qquad \textbf{(D) } $70\qquad \textbf{(E) } $100$ | If $x$ is the list price, then $10\%(x-10)=20\%(x-20)$ . Solving this equation gives $x=30$ , so the answer is $\boxed{30}$ | B | 30 |
fa10c775b938ce788a4fb3ce6ffd14ee | https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_10 | Segment $AB$ is both a diameter of a circle of radius $1$ and a side of an equilateral triangle $ABC$ .
The circle also intersects $AC$ and $BC$ at points $D$ and $E$ , respectively. The length of $AE$ is
$\textbf{(A)} \ \frac{3}{2} \qquad \textbf{(B)} \ \frac{5}{3} \qquad \textbf{(C)} \ \frac{\sqrt 3}{2} \qquad \textbf{(D)}\ \sqrt{3}\qquad \textbf{(E)}\ \frac{2+\sqrt 3}{2}$ | Note that since $AB$ is a diameter, $\angle AEB = 90^{\circ}$ , which means $AB$ is an altitude of equilateral triangle $ABC$ . It follows that $\triangle ABE$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, and so $AE = AB \cdot \frac{\sqrt{3}}{2} = (2 \cdot 1) (\frac{\sqrt{3}}{2}) = \boxed{3}$ | D | 3 |
cfe5109498856e4c3edf2a82c79da365 | https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_14 | The units digit of $3^{1001} 7^{1002} 13^{1003}$ is
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9$ | First, we notice that $3^0$ is congruent to $1 \ \text{(mod 10)}$ $3^1$ is $3 \ \text{(mod 10)}$ $3^2$ is $9 \ \text{(mod 10)}$ $3^3$ is $7 \ \text{(mod 10)}$ $3^4$ is $1 \ \text{(mod 10)}$ , and so on. This turns out to be a cycle repeating every $4$ terms, so $3^{1001}$ is congruent to $3 \ \text{(mod 10)}$
The number $7$ has a similar cycle, going $1, 7, 9, 3, 1, ...$ Hence we have that $7^{1002}$ is congruent to $9 \ \text{(mod 10)}$ . Finally, $13^{1003}$ is congruent to $3^{1003} \equiv 7 \ \text{(mod 10)}$ . Thus the required units digit is $3\cdot 9\cdot 7 \equiv 9 \ \text{(mod 10)}$ , so the answer is $\boxed{9}$ | E | 9 |
d2d2cdceafd7b08bb5b04dde1346c585 | https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_16 | Let $x = .123456789101112....998999$ , where the digits are obtained by writing the integers $1$ through $999$ in order.
The $1983$ rd digit to the right of the decimal point is
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8$ | We consider the first $1983$ digits, letting the $1983$ rd digit be $z$ . We can break the string of digits into three segments: let $A$ denote $123456789$ (the $1$ -digit numbers), let $B$ denote $1011...9899$ (the $2$ -digit numbers), and let $C$ denote $100101...z$ (the $3$ -digit numbers). Clearly there are $9$ digits in $A$ ; in $B$ , there are $99-10+1 = 90$ numbers, so $90 \cdot 2 = 180$ digits. This leaves $1983 - 9 - 180 = 1794$ digits in $C$ . Notice that $1794 = 3 \cdot 598$ with no remainder, so $C$ consists of precisely the first $598$ $3$ -digit numbers. Since the first $3$ -digit number is $100$ , the $598$ th is $100 + 598 - 1 = 697$ , so as $z$ is the last digit, the answer is $\boxed{7}$ | D | 7 |
b97b359d888113183004c79fa886dc0a | https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_19 | Point $D$ is on side $CB$ of triangle $ABC$ . If $\angle{CAD} = \angle{DAB} = 60^\circ\mbox{, }AC = 3\mbox{ and }AB = 6$ ,
then the length of $AD$ is
$\textbf{(A)} \ 2 \qquad \textbf{(B)} \ 2.5 \qquad \textbf{(C)} \ 3 \qquad \textbf{(D)} \ 3.5 \qquad \textbf{(E)} \ 4$ | Let $AD = y$ . Since $AD$ bisects $\angle{BAC}$ , the Angle Bisector Theorem gives $\frac{DB}{CD} = \frac{AB}{AC} = 2$ , so let $CD = x$ and $DB = 2x$ . Applying the Law of Cosines to $\triangle CAD$ gives $x^2 = 3^2 + y^2 - 3y$ , and to $\triangle DAB$ gives $(2x)^2 = 6^2 + y^2 - 6y$ . Subtracting $4$ times the first equation from the second equation therefore yields $0 = 6y - 3y^2 \Rightarrow y(y-2) = 0$ , so $y$ is $0$ or $2$ . But since $y \neq 0$ $y$ is the length of a side of a triangle), $y$ must be $2$ , so the answer is $\boxed{2}$ | A | 2 |
2a5295c8921ce89775694278a3489866 | https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_20 | If $\tan{\alpha}$ and $\tan{\beta}$ are the roots of $x^2 - px + q = 0$ , and $\cot{\alpha}$ and $\cot{\beta}$ are the roots of $x^2 - rx + s = 0$ , then $rs$ is necessarily
$\textbf{(A)} \ pq \qquad \textbf{(B)} \ \frac{1}{pq} \qquad \textbf{(C)} \ \frac{p}{q^2} \qquad \textbf{(D)}\ \frac{q}{p^2}\qquad \textbf{(E)}\ \frac{p}{q}$ | By Vieta's Formulae, we have $\tan(\alpha)\tan(\beta)=q$ and $\cot(\alpha)\cot(\beta)=s$ . Recalling that $\cot\theta=\frac{1}{\tan\theta}$ , we have $\frac{1}{\tan(\alpha)\tan(\beta)}=\frac{1}{q}=s$
Also by Vieta's Formulae, we have $\tan(\alpha)+\tan(\beta)=p$ and $\cot(\alpha)+\cot(\beta)=r$ , and again using $\cot\theta=\frac{1}{\tan\theta}$ , we have $\tan(\alpha)+\tan(\beta)=r(\tan(\alpha)\tan(\beta))$ . Using $\tan(\alpha)\tan(\beta)=q$ and $\tan(\alpha)+\tan(\beta)=p$ , we therefore deduce that $r=\frac{p}{q}$ , which yields $rs = \frac{1}{q}\cdot\frac{p}{q}=\frac{p}{q^2}$
Thus, the answer is $\boxed{2}$ | C | 2 |
6166b6385f646970dcffe3c7d432cb6f | https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_23 | In the adjoining figure the five circles are tangent to one another consecutively and to the lines $L_1$ and $L_2$ .
If the radius of the largest circle is $18$ and that of the smallest one is $8$ , then the radius of the middle circle is
[asy] size(250);defaultpen(linewidth(0.7)); real alpha=5.797939254, x=71.191836; int i; for(i=0; i<5; i=i+1) { real r=8*(sqrt(6)/2)^i; draw(Circle((x+r)*dir(alpha), r)); x=x+2r; } real x=71.191836+40+20*sqrt(6), r=18; pair A=tangent(origin, (x+r)*dir(alpha), r, 1), B=tangent(origin, (x+r)*dir(alpha), r, 2); pair A1=300*dir(origin--A), B1=300*dir(origin--B); draw(B1--origin--A1); pair X=(69,-5), X1=reflect(origin, (x+r)*dir(alpha))*X, Y=(200,-5), Y1=reflect(origin, (x+r)*dir(alpha))*Y, Z=(130,0), Z1=reflect(origin, (x+r)*dir(alpha))*Z; clip(X--Y--Y1--X1--cycle); label("$L_2$", Z, S); label("$L_1$", Z1, dir(2*alpha)*dir(90));[/asy]
$\textbf{(A)} \ 12 \qquad \textbf{(B)} \ 12.5 \qquad \textbf{(C)} \ 13 \qquad \textbf{(D)} \ 13.5 \qquad \textbf{(E)} \ 14$ | Pdfresizer.com-pdf-convert-q23.png
Consider three consecutive circles, as shown in the diagram above; observe that their centres $P$ $Q$ , and $R$ are collinear by symmetry. Let $A$ $B$ , and $C$ be the points of tangency, and let $PS$ and $QT$ be segments parallel to the upper tangent (i.e. $L_1$ ), as also shown. Since $PQ$ is parallel to $QR$ (the three points are collinear), $PS$ is parallel to $QT$ (as both are parallel to $L_1$ ), and $SQ$ is parallel to $TR$ (as both are perpendicular to $L_1$ , due to the tangent being perpendicular to the radius), we have $\triangle PQS \sim \triangle QRT$
Now, if we let $x, y$ , and $z$ be the radii of the three circles (from smallest to largest), then $QS = y - x$ and $RT = z - y$ . Thus, from the similarity that we just proved, $\frac{QS}{PQ} = \frac{RT}{QR} \Rightarrow \frac{y-x}{x+y} = \frac{z-y}{y+z}$ (where e.g. $PQ = x + y$ because of collinearity). This equation reduces to $y^2 = zx$ , i.e. $\frac{y}{x} = \frac{z}{y}$ , so the ratio of consecutive radii is constant, forming a geometric sequence. In this case, as the first radius is $8$ and, four radii later, the radius is $18$ , this constant ratio is $\sqrt[\leftroot{-2}\uproot{2}{4}]{\frac{18}{8}}$ . Therefore the middle radius is $8 \cdot {\left(\sqrt[\leftroot{-2}\uproot{2}{4}]{\frac{18}{8}}\right)}^{2} = 8 \sqrt{\frac{18}{8}} = \sqrt{18 \cdot 8} = \sqrt{144} = 12$ , which is choice $\boxed{12}$ | A | 12 |
fb0f32b07366d2edb43ea0587465cfc4 | https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_25 | If $60^a=3$ and $60^b=5$ , then $12^{(1-a-b)/\left(2\left(1-b\right)\right)}$ is
$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2\sqrt{3}\qquad$ | We have that $12=\frac{60}{5}$ . We can substitute our value for 5, to get \[12=\frac{60}{60^b}=60\cdot 60^{-b}=60^{1-b}.\] Hence \[12^{(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-b)(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-a-b)/2}.\] Since $4=\frac{60}{3\cdot 5}$ , we have \[4=\frac{60}{60^a 60^b}=60^{1-a-b}.\] Therefore, we have \[60^{(1-a-b)/2}=4^{1/2}=\boxed{2}\] | B | 2 |
41bc894e67aa749d61e0d4f05a39b1d9 | https://artofproblemsolving.com/wiki/index.php/1983_AHSME_Problems/Problem_30 | Distinct points $A$ and $B$ are on a semicircle with diameter $MN$ and center $C$ .
The point $P$ is on $CN$ and $\angle CAP = \angle CBP = 10^{\circ}$ . If $\stackrel{\frown}{MA} = 40^{\circ}$ , then $\stackrel{\frown}{BN}$ equals
[asy] import geometry; import graph; unitsize(2 cm); pair A, B, C, M, N, P; M = (-1,0); N = (1,0); C = (0,0); A = dir(140); B = dir(20); P = extension(A, A + rotate(10)*(C - A), B, B + rotate(10)*(C - B)); draw(M--N); draw(arc(C,1,0,180)); draw(A--C--B); draw(A--P--B); label("$A$", A, NW); label("$B$", B, E); label("$C$", C, S); label("$M$", M, SW); label("$N$", N, SE); label("$P$", P, S); [/asy]
$\textbf{(A)}\ 10^{\circ}\qquad \textbf{(B)}\ 15^{\circ}\qquad \textbf{(C)}\ 20^{\circ}\qquad \textbf{(D)}\ 25^{\circ}\qquad \textbf{(E)}\ 30^{\circ}$ | Since $\angle CAP = \angle CBP = 10^\circ$ , quadrilateral $ABPC$ is cyclic (as shown below) by the converse of the theorem "angles inscribed in the same arc are equal".
[asy] import geometry; import graph; unitsize(2 cm); pair A, B, C, M, N, P; M = (-1,0); N = (1,0); C = (0,0); A = dir(140); B = dir(20); P = extension(A, A + rotate(10)*(C - A), B, B + rotate(10)*(C - B)); draw(M--N); draw(arc(C,1,0,180)); draw(A--C--B); draw(A--P--B); draw(A--B); draw(circumcircle(A,B,C),dashed); label("$A$", A, NW); label("$B$", B, E); label("$C$", C, S); label("$M$", M, SW); label("$N$", N, SE); label("$P$", P, S); [/asy]
Since $\angle ACM = 40^\circ$ $\angle ACP = 140^\circ$ , so, using the fact that opposite angles in a cyclic quadrilateral sum to $180^{\circ}$ , we have $\angle ABP = 40^\circ$ . Hence $\angle ABC = \angle ABP - \angle CBP = 40^ \circ - 10^\circ = 30^\circ$
Since $CA = CB$ , triangle $ABC$ is isosceles, with $\angle BAC = \angle ABC = 30^\circ$ . Now, $\angle BAP = \angle BAC - \angle CAP = 30^\circ - 10^\circ = 20^\circ$ . Finally, again using the fact that angles inscribed in the same arc are equal, we have $\angle BCP = \angle BAP = \boxed{20}$ | C | 20 |
19f68cfe88880c261ec9cb0e7281f752 | https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_3 | Evaluate $(x^x)^{(x^x)}$ at $x = 2$
$\text{(A)} \ 16 \qquad \text{(B)} \ 64 \qquad \text{(C)} \ 256 \qquad \text{(D)} \ 1024 \qquad \text{(E)} \ 65,536$ | Plugging in $2$ as $x$ gives $4^4$ , which is merely $\boxed{256}$ | C | 256 |
97eb86105e8fe2f9417558d44eb32e97 | https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_6 | The sum of all but one of the interior angles of a convex polygon equals $2570^\circ$ . The remaining angle is
$\text{(A)} \ 90^\circ \qquad \text{(B)} \ 105^\circ \qquad \text{(C)} \ 120^\circ \qquad \text{(D)}\ 130^\circ\qquad \text{(E)}\ 144^\circ$ | Note that the sum of the interior angles of a convex polygon of $n$ sides is $180(n-2)^\circ$ , and each interior angle belongs to $[0, 180^\circ)$ . Therefore, we must have $n - 2 = \lfloor \frac{2570^\circ}{180^\circ} \rfloor = 15$ . Then the missing angle must be $180*15^\circ - 2570^\circ = 130^\circ$ , so our answer is $\boxed{130}$ and we are done. | D | 130 |
bd3e05d8d5f6a7e86eade46b32f6e50c | https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_14 | In the adjoining figure, points $B$ and $C$ lie on line segment $AD$ , and $AB, BC$ , and $CD$ are diameters of circle $O, N$ , and $P$ , respectively. Circles $O, N$ , and $P$ all have radius $15$ and the line $AG$ is tangent to circle $P$ at $G$ . If $AG$ intersects circle $N$ at points $E$ and $F$ , then chord $EF$ has length
[asy] size(250); defaultpen(fontsize(10)); pair A=origin, O=(1,0), B=(2,0), N=(3,0), C=(4,0), P=(5,0), D=(6,0), G=tangent(A,P,1,2), E=intersectionpoints(A--G, Circle(N,1))[0], F=intersectionpoints(A--G, Circle(N,1))[1]; draw(Circle(O,1)^^Circle(N,1)^^Circle(P,1)^^G--A--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F^^G^^O^^N^^P); label("$A$", A, W); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, dir(0)); label("$P$", P, S); label("$N$", N, S); label("$O$", O, S); label("$E$", E, dir(120)); label("$F$", F, NE); label("$G$", G, dir(100));[/asy] | Drop a perpendicular line from $N$ to $AG$ at point $H$ $AN=45$ , and since $\triangle{AGP}$ is similar to $\triangle{AHN}$ $NH=9$ $NE=NF=15$ so by the Pythagorean Theorem, $EH=HF=12$ . Thus $EF=\boxed{24.}$ | A | 24. |
0b0642b3aaca2ea5abd8285c37ce751c | https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_17 | How many real numbers $x$ satisfy the equation $3^{2x+2}-3^{x+3}-3^{x}+3=0$
$\text {(A)} 0 \qquad \text {(B)} 1 \qquad \text {(C)} 2 \qquad \text {(D)} 3 \qquad \text {(E)} 4$ | Let $a = 3^x$ . Then the preceding equation can be expressed as the quadratic, \[9a^2-28a+3 = 0\] Solving the quadratic yields the roots $3$ and $1/9$ . Setting these equal to $3^x$ , we can immediately see that there are $\boxed{2}$ real values of $x$ that satisfy the equation. | null | 2 |
8611fd5f02d14032145981c16a9466db | https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_19 | Let $f(x)=|x-2|+|x-4|-|2x-6|$ for $2 \leq x\leq 8$ . The sum of the largest and smallest values of $f(x)$ is
$\textbf {(A)}\ 1 \qquad \textbf {(B)}\ 2 \qquad \textbf {(C)}\ 4 \qquad \textbf {(D)}\ 6 \qquad \textbf {(E)}\ \text{none of these}$ | Note that at $x=2,3,4,$ one of the three absolute values is equal to $0.$
Without using absolute values, we rewrite $f(x)$ as a piecewise function: \[f(x) = \begin{cases} (x-2)+(4-x)-(6-2x) & \mathrm{if} \ 2\leq x<3 \\ (x-2)+(4-x)-(2x-6) & \mathrm{if} \ 3\leq x<4 \\ (x-2)+(x-4)-(2x-6) & \mathrm{if} \ 4\leq x\leq8 \end{cases},\] which simplifies to \[f(x) = \begin{cases} 2x-4 & \mathrm{if} \ 2\leq x<3 \\ -2x+8 & \mathrm{if} \ 3\leq x<4 \\ 0 & \mathrm{if} \ 4\leq x\leq8 \end{cases}.\] The graph of $y=f(x)$ is shown below. [asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -2; int xMax = 10; int yMin = -2; int yMax = 4; //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (int i = yMin+1; i < yMax; ++i) { draw((-3/16,i)--(3/16,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (int i = xMin+1; i < xMax; ++i) { draw((i,-3/16)--(i,3/16), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); pair A[]; A[0] = (2,0); A[1] = (3,2); A[2] = (4,0); A[3] = (8,0); draw(A[0]--A[1]--A[2]--A[3],red+linewidth(1.5)); for(int i = 0; i <= 3; ++i) { dot(A[i],red+linewidth(4.5)); } label("$(2,0)$",A[0],(0,-1.5),UnFill); label("$(3,2)$",A[1],(0,1.5),UnFill); label("$(4,0)$",A[2],(0,-1.5),UnFill); label("$(8,0)$",A[3],(0,-1.5),UnFill); [/asy] The largest value of $f(x)$ is $2,$ and the smallest value of $f(x)$ is $0.$ So, their sum is $\boxed{2}.$ | B | 2 |
502056a3d6d31cbb6f1fdfd9f60609b8 | https://artofproblemsolving.com/wiki/index.php/1982_AHSME_Problems/Problem_30 | Find the units digit of the decimal expansion of \[\left(15 + \sqrt{220}\right)^{19} + \left(15 + \sqrt{220}\right)^{82}.\]
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ \text{none of these}$ | Let $A=15+\sqrt{220}$ and $B=15-\sqrt{220}.$ Note that $A^{19}+B^{19}$ and $A^{82}+B^{82}$ are both integers: When we expand (Binomial Theorem) and combine like terms for each expression, the rational terms are added and the irrational terms are canceled.
We have \begin{align*} A^{19}+B^{19} &= \left[\binom{19}{0}15^{19}\sqrt{220}^0+\binom{19}{1}15^{18}\sqrt{220}^1+\cdots+\binom{19}{19}15^0\sqrt{220}^{19}\right] + \left[\binom{19}{0}15^{19}\sqrt{220}^0-\binom{19}{1}15^{18}\sqrt{220}^1+\cdots-\binom{19}{19}15^0\sqrt{220}^{19}\right] \\ &= 2\left[\binom{19}{0}15^{19}\sqrt{220}^0+\binom{19}{2}15^{17}\sqrt{220}^2+\cdots+\binom{19}{18}15^1\sqrt{220}^{18}\right] \\ &= 2\left[\binom{19}{0}15^{19}+\binom{19}{2}15^{17}220+\cdots+\binom{19}{18}15^1 220^9\right]. \end{align*} Similarly, we have \[A^{82}+B^{82}=2\left[\binom{82}{0}15^{82}+\binom{82}{2}15^{80}220+\cdots+\binom{82}{82}220^{41}\right].\] We add the two equations and take the sum modulo $10:$ \begin{align*} \left(A^{19}+A^{82}\right)+\left(B^{19}+B^{82}\right) &= 2\Biggl[\binom{19}{0}15^{19}+\phantom{ }\underbrace{\binom{19}{2}15^{17}220+\cdots+\binom{19}{18}15^1 220^9}_{0\pmod{10}}\phantom{ }\Biggr]+2\Biggl[\binom{82}{0}15^{82}+\phantom{ }\underbrace{\binom{82}{2}15^{80}220+\cdots+\binom{82}{82}220^{41}}_{0\pmod{10}}\phantom{ }\Biggr] \\ &\equiv 2\left[\binom{19}{0}15^{19}\right]+2\left[\binom{82}{0}15^{82}\right] \\ &\equiv 2\left[5\right]+2\left[5\right] \\ &\equiv 0\pmod{10}. \end{align*} It is clear that $0<B^{82}<B^{19}<B<0.5,$ from which $0<B^{19}+B^{82}<1.$ We conclude that the units digit of the decimal expansion of $B^{19}+B^{82}$ is $0.$ Since the units digit of the decimal expansion of $\left(A^{19}+A^{82}\right)+\left(B^{19}+B^{82}\right)$ is $0,$ the units digit of the decimal expansion of $A^{19}+A^{82}$ is $\boxed{9}.$ | D | 9 |
bf30aeb9182527c276b5dcd1462cb7d6 | https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_1 | If $\sqrt{x+2}=2$ , then $(x+2)^2$ equals:
$\textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 16$ | If we square both sides of the $\sqrt{x+2} = 2$ , we will get $x+2 = 4$ , if we square that again, we get $(x+2)^2 = \boxed{16}$ | E | 16 |
bf30aeb9182527c276b5dcd1462cb7d6 | https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_1 | If $\sqrt{x+2}=2$ , then $(x+2)^2$ equals:
$\textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 16$ | We can immediately get that $x = 2$ , after we square $(2+2)$ , we get $\boxed{16}$ | E | 16 |
99f33cd31bd92cc737c7dcf863e0be76 | https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_2 | Point $E$ is on side $AB$ of square $ABCD$ . If $EB$ has length one and $EC$ has length two, then the area of the square is
$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ \sqrt{5}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 2\sqrt{3}\qquad\textbf{(E)}\ 5$ | Note that $\triangle BCE$ is a right triangle. Thus, we do Pythagorean theorem to find that side $BC=\sqrt{3}$ . Since this is the side length of the square, the area of $ABCD$ is $\boxed{3}$ | C | 3 |
5c4b22d002492d9c6a21e723095b6db1 | https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_14 | In a geometric sequence of real numbers, the sum of the first $2$ terms is $7$ , and the sum of the first $6$ terms is $91$ . The sum of the first $4$ terms is
$\textbf{(A)}\ 28\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 84$ | Denote the sum of the first $2$ terms as $x$ . Since we know that the sum of the first $6$ terms is $91$ which is $7 \cdot 13$ , we have $x$ $xy$ $xy^2$ $13x$ because it is a geometric series. We can quickly see that $y$ $3$ , and therefore, the sum of the first $4$ terms is $4x = 4 \cdot 7 = \boxed{28}$ | A | 28 |
d811c376510d1b6747b8d555dd531334 | https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_20 | A ray of light originates from point $A$ and travels in a plane, being reflected $n$ times between lines $AD$ and $CD$ before striking a point $B$ (which may be on $AD$ or $CD$ ) perpendicularly and retracing its path back to $A$ (At each point of reflection the light makes two equal angles as indicated in the adjoining figure. The figure shows the light path for $n=3$ ). If $\measuredangle CDA=8^\circ$ , what is the largest value $n$ can have?
[asy] unitsize(1.5cm); pair D=origin, A=(-6,0), C=6*dir(160), E=3.2*dir(160), F=(-2.1,0), G=1.5*dir(160), B=(-1.4095,0); draw((-6.5,0)--D--C,black); draw(A--E--F--G--B,black); dotfactor=4; dot("$A$",A,S); dot("$C$",C,N); dot("$R_1$",E,N); dot("$R_2$",F,S); dot("$R_3$",G,N); dot("$B$",B,S); markscalefactor=0.015; draw(rightanglemark(G,B,D)); draw(anglemark(C,E,A,12)); draw(anglemark(F,E,G,12)); draw(anglemark(E,F,A)); draw(anglemark(E,F,A,12)); draw(anglemark(B,F,G)); draw(anglemark(B,F,G,12)); draw(anglemark(E,G,F)); draw(anglemark(E,G,F,12)); draw(anglemark(E,G,F,16)); draw(anglemark(B,G,D)); draw(anglemark(B,G,D,12)); draw(anglemark(B,G,D,16)); [/asy]
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 38\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ \text{There is no largest value.}$ | Notice that when we start, we want the smallest angle possible of reflection. The ideal reflection would be $0$ , but that would be impossible. Therefore we start by working backwards. Since angle $CDA$ is $8$ , the reflection would give us a triangle with angles $16, 90$ , and $74$ . Then, when we reflect again, we will have $180 - 74 - 74$ $32$ . Since the other side of the reflection when we had the $82$ degrees had carried over to the other side, we have a $32-82-66$ triangle.
Notice that we keep decreasing by increments of $8$ . This is because the starting angle was $8$ and since we always have to decrease $8$ every time and that every triangle has every increasing angles of $8$ , we must decrease by $8$ every time. This is the most optimal path of the light beam.
The pattern of light will be $82-74-66-58-50-42-34-26-18-10$ . When we get to the angle of $2$ degrees, we have reached angle $A$ . Therefore, we don't count the $2$ , so our total number of reflections between $CD$ and $AD$ is $\boxed{10}$ | B | 10 |
bbdfd88ba8ef5847e595d4a243ca8771 | https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_21 | In a triangle with sides of lengths $a$ $b$ , and $c$ $(a+b+c)(a+b-c) = 3ab$ . The measure of the angle opposite the side length $c$ is
$\textbf{(A)}\ 15^\circ\qquad\textbf{(B)}\ 30^\circ\qquad\textbf{(C)}\ 45^\circ\qquad\textbf{(D)}\ 60^\circ\qquad\textbf{(E)}\ 150^\circ$ | \[(a+b+c)(a+b-c)=3ab\] \[a^2+2ab+b^2-c^2=3ab\] \[a^2+b^2-c^2=ab\] \[c^2=a^2+b^2-ab\] This looks a lot like Law of Cosines, which is $c^2=a^2+b^2-2ab\cos{c}$ \[c^2=a^2+b^2-ab=a^2+b^2-2ab\cos{c}\] \[ab=2ab\cos{c}\] \[\frac{1}{2}=\cos{c}\] $\cos{c}$ is $\frac{1}{2}$ , so the angle opposite side $c$ is $\boxed{60}$ | null | 60 |
6438c2fc0e50cdc28db0d26bcb78da47 | https://artofproblemsolving.com/wiki/index.php/1981_AHSME_Problems/Problem_24 | If $\theta$ is a constant such that $0 < \theta < \pi$ and $x + \dfrac{1}{x} = 2\cos{\theta}$ , then for each positive integer $n$ $x^n + \dfrac{1}{x^n}$ equals
$\textbf{(A)}\ 2\cos\theta\qquad \textbf{(B)}\ 2^n\cos\theta\qquad \textbf{(C)}\ 2\cos^n\theta\qquad \textbf{(D)}\ 2\cos n\theta\qquad \textbf{(E)}\ 2^n\cos^n\theta$ | Multiply both sides by $x$ and rearrange to $x^2-2x\cos(\theta)+1=0$ . Using the quadratic equation, we can solve for $x$ . After some simplifying:
\[x=\cos(\theta) + \sqrt{\cos^2(\theta)-1}\] \[x=\cos(\theta) + \sqrt{(-1)(\sin^2(\theta))}\] \[x=\cos(\theta) + i\sin(\theta)\]
Substituting this expression in to the desired $x^n + \dfrac{1}{x^n}$ gives:
\[(\cos(\theta) + i\sin(\theta))^n + (\cos(\theta) + i\sin(\theta))^{-n}\]
Using DeMoivre's Theorem:
\[=\cos(n\theta) + i\sin(n\theta) + \cos(-n\theta) + i\sin(-n\theta)\]
Because $\cos$ is even and $\sin$ is odd:
\[=\cos(n\theta) + i\sin(n\theta) + \cos(n\theta) - i\sin(n\theta)\]
$=\boxed{2},$ (Error compiling LaTeX. Unknown error_msg) | null | 2 |
f9822f0b2f1f60cc3f95e3a9ca606e07 | https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_1 | The largest whole number such that seven times the number is less than 100 is
$\text{(A)} \ 12 \qquad \text{(B)} \ 13 \qquad \text{(C)} \ 14 \qquad \text{(D)} \ 15 \qquad \text{(E)} \ 16$ | We want to find the smallest integer $x$ so that $7x < 100$ . Dividing by 7 gets $x < 14\dfrac{2}{7}$ , so the answer is 14. $\boxed{14}$ | C | 14 |
4f5326001de581a8e805978aa6fc941d | https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_2 | The degree of $(x^2+1)^4 (x^3+1)^3$ as a polynomial in $x$ is
$\text{(A)} \ 5 \qquad \text{(B)} \ 7 \qquad \text{(C)} \ 12 \qquad \text{(D)} \ 17 \qquad \text{(E)} \ 72$ | It becomes $(x^{8}+...)(x^{9}+...)$ with 8 being the degree of the first factor and 9 being the degree of the second factor, making the degree of the whole thing 17, or $\boxed{17}$ | D | 17 |
534b490bf8359c75c79ceb56a4fb65f9 | https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_7 | Sides $AB,BC,CD$ and $DA$ of convex polygon $ABCD$ have lengths 3, 4, 12, and 13, respectively, and $\angle CBA$ is a right angle. The area of the quadrilateral is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); real r=degrees((12,5)), s=degrees((3,4)); pair D=origin, A=(13,0), C=D+12*dir(r), B=A+3*dir(180-(90-r+s)); draw(A--B--C--D--cycle); markscalefactor=0.05; draw(rightanglemark(A,B,C)); pair point=incenter(A,C,D); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$3$", A--B, dir(A--B)*dir(-90)); label("$4$", B--C, dir(B--C)*dir(-90)); label("$12$", C--D, dir(C--D)*dir(-90)); label("$13$", D--A, dir(D--A)*dir(-90));[/asy]
$\text{(A)} \ 32 \qquad \text{(B)} \ 36 \qquad \text{(C)} \ 39 \qquad \text{(D)} \ 42 \qquad \text{(E)} \ 48$ | Connect C and A, and we have a 3-4-5 right triangle and 5-12-13 right triangle. The area of both is $\frac{3\cdot4}{2}+\frac{5\cdot12}{2}=36\Rightarrow\boxed{36}$ | B | 36 |
2773b0e23731e1e2ca86cbf0a58e1fa0 | https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_11 | If the sum of the first $10$ terms and the sum of the first $100$ terms of a given arithmetic progression are $100$ and $10$ ,
respectively, then the sum of first $110$ terms is:
$\text{(A)} \ 90 \qquad \text{(B)} \ -90 \qquad \text{(C)} \ 110 \qquad \text{(D)} \ -110 \qquad \text{(E)} \ -100$ | Let $a$ be the first term of the sequence and let $d$ be the common difference of the sequence.
Sum of the first 10 terms: $\frac{10}{2}(2a+9d)=100 \Longleftrightarrow 2a+9d=20$ Sum of the first 100 terms: $\frac{100}{2}(2a+99d)=10 \Longleftrightarrow 2a+99d=\frac{1}{5}$
Solving the system, we get $d=-\frac{11}{50}$ $a=\frac{1099}{100}$ . The sum of the first 110 terms is $\frac{110}{2}(2a+109d)=55(-2)=-110$
Therefore, $\boxed{110}$ | D | 110 |
f910ed299c54967d0d25f09721d1fa01 | https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_12 | The equations of $L_1$ and $L_2$ are $y=mx$ and $y=nx$ , respectively. Suppose $L_1$ makes twice as large of an angle with the horizontal (measured counterclockwise from the positive x-axis ) as does $L_2$ , and that $L_1$ has 4 times the slope of $L_2$ . If $L_1$ is not horizontal, then $mn$ is
$\text{(A)} \ \frac{\sqrt{2}}{2} \qquad \text{(B)} \ -\frac{\sqrt{2}}{2} \qquad \text{(C)} \ 2 \qquad \text{(D)} \ -2 \qquad \text{(E)} \ \text{not uniquely determined}$ | Solution by e_power_pi_times_i
$4n = m$ , as stated in the question. In the line $L_1$ , draw a triangle with the coordinates $(0,0)$ $(1,0)$ , and $(1,m)$ . Then $m = \tan(\theta_1)$ . Similarly, $n = \tan(\theta_2)$ . Since $4n = m$ and $\theta_1 = 2\theta_2$ $\tan(2\theta_2) = 4\tan(\theta_2)$ . Using the angle addition formula for tangents, $\dfrac{2\tan(\theta_2)}{1-\tan^2(\theta_2)} = 4\tan(\theta_2)$ . Solving, we have $\tan(\theta_2) = 0, \dfrac{\sqrt{2}}{2}$ . But line $L_1$ is not horizontal, so therefore $(m,n) = (2\sqrt{2},\dfrac{\sqrt{2}}{2})$ . Looking at the answer choices, it seems the answer is $(2\sqrt{2})(\dfrac{\sqrt{2}}{2}) = \boxed{2}$ | C | 2 |
43cdf6da266d36776c56a16b8e296659 | https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_16 | Four of the eight vertices of a cube are the vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron.
$\text{(A)} \ \sqrt 2 \qquad \text{(B)} \ \sqrt 3 \qquad \text{(C)} \ \sqrt{\frac{3}{2}} \qquad \text{(D)} \ \frac{2}{\sqrt{3}} \qquad \text{(E)} \ 2$ | We assume the side length of the cube is $1$ . The side length of the tetrahedron is $\sqrt2$ , so the surface area is $4\times\frac{2\sqrt3}{4}=2\sqrt3$ . The surface area of the cube is $6\times1\times1=6$ , so the ratio of the surface area of the cube to the surface area of the tetrahedron is $\frac{6}{2\sqrt3}=\boxed{3}$ | null | 3 |