problem stringlengths 10 5.15k | answer stringlengths 0 1.22k | solution stringlengths 0 11.1k | difficulty float64 0.75 2.02k | difficulty_raw listlengths 3 8 |
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A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face? | 840 | In the same ways as above, we find that there are 48 vertices. Now, notice that there are $\binom{48}{2}$ total possible ways to choose two vertices. However, we must remove the cases where the segments do not lie in the interior of the polyhedron. We get
\[\binom{48}{2}-12\binom{4}{2}-8\binom{6}{2}-6\binom{8}{2}=768\]... | 6.625 | [
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Let $w_1, w_2, \dots, w_n$ be complex numbers. A line $L$ in the complex plane is called a mean line for the points $w_1, w_2, \dots, w_n$ if $L$ contains points (complex numbers) $z_1, z_2, \dots, z_n$ such that \[\sum_{k = 1}^n (z_k - w_k) = 0.\] For the numbers $w_1 = 32 + 170i$, $w_2 = - 7 + 64i$, $w_3 = - 9 + 200i... | 163 | The mean line for $w_1, . . ., w_5$ must pass through the mean (the center of mass) of these points, which, if we graph them on the complex plane, is $(\frac{3}{5}, \frac{504i}{5})$. Since we now have two points, namely that one and $(0, 3i)$, we can simply find the slope between them, which is $\boxed{163}$ by the goo... | 6.75 | [
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Let $P$ be an interior point of triangle $ABC$ and extend lines from the vertices through $P$ to the opposite sides. Let $a$, $b$, $c$, and $d$ denote the lengths of the segments indicated in the figure. Find the product $abc$ if $a + b + c = 43$ and $d = 3$. | 441 | Let $A,B,C$ be the weights of the respective vertices. We see that the weights of the feet of the cevians are $A+B,B+C,C+A$. By mass points, we have that: \[\dfrac{a}{3}=\dfrac{B+C}{A}\] \[\dfrac{b}{3}=\dfrac{C+A}{B}\] \[\dfrac{c}{3}=\dfrac{A+B}{C}\]
If we add the equations together, we get $\frac{a+b+c}{3}=\frac{A^2B... | 6.375 | [
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Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$. | 987 | We can long divide and search for a pattern; then the remainder would be set to zero to solve for $a$. Writing out a few examples quickly shows us that the remainders after each subtraction follow the Fibonacci sequence. Carrying out this pattern, we find that the remainder is \[(F_{16}b + F_{17}a)x + F_{15}b + F_{16}a... | 6 | [
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Let $C$ be the graph of $xy = 1$, and denote by $C^*$ the reflection of $C$ in the line $y = 2x$. Let the equation of $C^*$ be written in the form
\[12x^2 + bxy + cy^2 + d = 0.\]
Find the product $bc$. | 84 | Find some simple points on the graph of C, reflect them and note down the new coordinates, and plug 'em into the given equation. After some plugging and chugging and solving the system of equations that follow, we get that $b = -7$ and $c = -12$, so $bc = 84$. The answer is $\boxed{084}$.
pi_is_3.141 | 6 | [
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In an office at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's inbox. When there is time, the secretary takes the top letter off the pile and types it. There are nine letters to be typed during the day, and the boss deliver... | 704 | Re-stating the problem for clarity, let $S$ be a set arranged in increasing order. At any time an element can be appended to the end of $S$, or the last element of $S$ can be removed. The question asks for the number of different orders in which all of the remaining elements of $S$ can be removed, given that $8$ had be... | 6.125 | [
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Compute $\sqrt{(31)(30)(29)(28)+1}$. | 869 | Note that the four numbers to multiply are symmetric with the center at $29.5$. Multiply the symmetric pairs to get $31\cdot 28=868$ and $30\cdot 29=870$. $\sqrt{868\cdot 870 + 1} = \sqrt{(869-1)(869+1) + 1} = \sqrt{869^2 - 1^2 + 1} = \sqrt{869^2} = \boxed{869}$. | 3 | [
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Ten points are marked on a circle. How many distinct convex polygons of three or more sides can be drawn using some (or all) of the ten points as vertices? | 968 | Any subset of the ten points with three or more members can be made into exactly one such polygon. Thus, we need to count the number of such subsets. There are $2^{10} = 1024$ total subsets of a ten-member set, but of these ${10 \choose 0} = 1$ have 0 members, ${10 \choose 1} = 10$ have 1 member and ${10 \choose 2} = 4... | 3.5 | [
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Suppose $n$ is a positive integer and $d$ is a single digit in base 10. Find $n$ if
$\frac{n}{810}=0.d25d25d25\ldots$ | 750 | Write out these equations:
$\frac{n}{180} = \frac{d25}{999}$
$\frac{n}{30} = \frac{d25}{37}$
$37n = 30(d25)$
Thus $n$ divides 25 and 30. The only solution for this under 1000 is $\boxed{750}$.
-jackshi2006 | 4.125 | [
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If $a<b<c<d<e$ are consecutive positive integers such that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube, what is the smallest possible value of $c$? | 675 | Let the numbers be $a,a+1,a+2,a+3,a+4.$ When then know $3a+6$ is a perfect cube and $5a+10$ is perfect cube. Since $5a+10$ is divisible by $5$ we know that $5a+10 = (5k)^3$ since otherwise we get a contradiction. This means $a = 25k^3 - 2$ in which plugging into the other expression we know $3(25k^3 - 2) + 6 = 75k^3$ i... | 5.875 | [
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When a certain biased coin is flipped five times, the probability of getting heads exactly once is not equal to $0$ and is the same as that of getting heads exactly twice. Let $\frac ij$, in lowest terms, be the probability that the coin comes up heads in exactly $3$ out of $5$ flips. Find $i+j$. | 283 | Denote the probability of getting a heads in one flip of the biased coin as $h$. Based upon the problem, note that ${5\choose1}(h)^1(1-h)^4 = {5\choose2}(h)^2(1-h)^3$. After canceling out terms, we get $1 - h = 2h$, so $h = \frac{1}{3}$. The answer we are looking for is ${5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3}\ri... | 5.875 | [
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Two skaters, Allie and Billie, are at points $A$ and $B$, respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$. At the same time Allie leaves $A$, Billie leaves $... | 160 | Label the point of intersection as $C$. Since $d = rt$, $AC = 8t$ and $BC = 7t$. According to the law of cosines,
[asy] pointpen=black; pathpen=black+linewidth(0.7); pair A=(0,0),B=(10,0),C=16*expi(pi/3); D(B--A); D(A--C); D(B--C,dashed); MP("A",A,SW);MP("B",B,SE);MP("C",C,N);MP("60^{\circ}",A+(0.3,0),NE);MP("100",(A+B... | 6 | [
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If the integer $k$ is added to each of the numbers $36$, $300$, and $596$, one obtains the squares of three consecutive terms of an arithmetic series. Find $k$. | 925 | Call the terms of the arithmetic progression $a,\ a + d,\ a + 2d$, making their squares $a^2,\ a^2 + 2ad + d^2,\ a^2 + 4ad + 4d^2$.
We know that $a^2 = 36 + k$ and $(a + d)^2 = 300 + k$, and subtracting these two we get $264 = 2ad + d^2$ (1). Similarly, using $(a + d)^2 = 300 + k$ and $(a + 2d)^2 = 596 + k$, subtracti... | 5.5 | [
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Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that \begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*} Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+8... | 334 | We let $(x_4,x_5,x_6,x_7)=(0,0,0,0)$. Thus, we have \begin{align*} x_1+4x_2+9x_3&=1,\\ 4x_1+9x_2+16x_3&=12,\\ 9x_1+16x_2+25x_3&=123.\\ \end{align*} Grinding this out, we have $(x_1,x_2,x_3)=\left(\frac{797}{4},-229,\frac{319}{4}\right)$ which gives $\boxed{334}$ as our final answer.
~Pleaseletmewin
~ MathEx | 6.125 | [
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One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that \[133^5+110^5+84^5+27^5=n^{5}.\] Find the value of $n$. | 144 | Note that $n$ is even, since the LHS consists of two odd and two even numbers. By Fermat's Little Theorem, we know $n^5\equiv n\pmod{5}.$ Hence, \[n\equiv3+0+4+2\equiv4\pmod{5}.\] Continuing, we examine the equation modulo $3,$ \[n\equiv1-1+0+0\equiv0\pmod{3}.\] Thus, $n$ is divisible by three and leaves a remainder of... | 5.75 | [
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Let $a$, $b$, $c$ be the three sides of a triangle, and let $\alpha$, $\beta$, $\gamma$, be the angles opposite them. If $a^2+b^2=1989c^2$, find
$\frac{\cot \gamma}{\cot \alpha+\cot \beta}$ | 994 | We start as in solution 1, though we'll write $A$ instead of $K$ for the area. Now we evaluate the numerator:
\[\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}\]
From the Law of Cosines and the sine area formula,
\begin{align*}\cos{\gamma}&=\frac{1988c^2}{2ab}\\ \sin{\gamma}&= \frac{2A}{ab}\\ \cot{\gamma}&= \frac{\cos ... | 6.25 | [
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A sample of 121 integers is given, each between 1 and 1000 inclusive, with repetitions allowed. The sample has a unique mode (most frequent value). Let $D$ be the difference between the mode and the arithmetic mean of the sample. What is the largest possible value of $\lfloor D\rfloor$? (For real $x$, $\lfloor x\rfloor... | 947 | With the same reasoning as Solution 1, in order to get largest possible value of D, we can construct that our set of numbers as $\underbrace{1,1,1...1,}_\text{n times}\underbrace{2,2,2...2,}_\text{n times}\underbrace{3,3,3...3,}_\text{n times}........\underbrace{1000,1000,1000....}_\text{n+1 times}$ And, we need to fin... | 6.625 | [
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Let $ABCD$ be a tetrahedron with $AB=41$, $AC=7$, $AD=18$, $BC=36$, $BD=27$, and $CD=13$, as shown in the figure. Let $d$ be the distance between the midpoints of edges $AB$ and $CD$. Find $d^{2}$. [asy] defaultpen(fontsize(10)+0.8); size(175); pair A,B,C,D,M,P,Q; C=origin; B=(8,0); D=IP(CR(C,6.5),CR(B,8)); A=(4,-3); P... | 137 | Call the midpoint of $\overline{AB}$ $M$ and the midpoint of $\overline{CD}$ $N$. $d$ is the median of triangle $\triangle CDM$. The formula for the length of a median is $m=\sqrt{\frac{2a^2+2b^2-c^2}{4}}$, where $a$, $b$, and $c$ are the side lengths of triangle, and $c$ is the side that is bisected by median $m$. The... | 6 | [
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Let $S$ be a subset of $\{1,2,3,\ldots,1989\}$ such that no two members of $S$ differ by $4$ or $7$. What is the largest number of elements $S$ can have? | 905 | We first show that we can choose at most 5 numbers from $\{1, 2, \ldots , 11\}$ such that no two numbers have a difference of $4$ or $7$. We take the smallest number to be $1$, which rules out $5,8$. Now we can take at most one from each of the pairs: $[2,9]$, $[3,7]$, $[4,11]$, $[6,10]$. Now, $1989 = 180\cdot 11 + 9$.... | 6.125 | [
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Given a positive integer $n$, it can be shown that every complex number of the form $r+si$, where $r$ and $s$ are integers, can be uniquely expressed in the base $-n+i$ using the integers $0,1,2,\ldots,n^2$ as digits. That is, the equation
$r+si=a_m(-n+i)^m+a_{m-1}(-n+i)^{m-1}+\cdots +a_1(-n+i)+a_0$
is true for a uniq... | 490 | First, we find the first three powers of $-3+i$:
$(-3+i)^1=-3+i ; (-3+i)^2=8-6i ; (-3+i)^3=-18+26i$
So we need to solve the diophantine equation $a_1-6a_2+26a_3=0 \Longrightarrow a_1-6a_2=-26a_3$.
The minimum the left hand side can go is -54, so $a_3\leq 2$, so we try cases:
Case 1: $a_3=2$
The only solution to that... | 7.25 | [
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Point $P$ is inside $\triangle ABC$. Line segments $APD$, $BPE$, and $CPF$ are drawn with $D$ on $BC$, $E$ on $AC$, and $F$ on $AB$ (see the figure below). Given that $AP=6$, $BP=9$, $PD=6$, $PE=3$, and $CF=20$, find the area of $\triangle ABC$. | 108 | Because we're given three concurrent cevians and their lengths, it seems very tempting to apply Mass points. We immediately see that $w_E = 3$, $w_B = 1$, and $w_A = w_D = 2$. Now, we recall that the masses on the three sides of the triangle must be balanced out, so $w_C = 1$ and $w_F = 3$. Thus, $CP = 15$ and $PF = 5$... | 6.375 | [
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The increasing sequence $2,3,5,6,7,10,11,\ldots$ consists of all positive integers that are neither the square nor the cube of a positive integer. Find the 500th term of this sequence. | 528 | Because there aren't that many perfect squares or cubes, let's look for the smallest perfect square greater than $500$. This happens to be $23^2=529$. Notice that there are $23$ squares and $8$ cubes less than or equal to $529$, but $1$ and $2^6$ are both squares and cubes. Thus, there are $529-23-8+2=500$ numbers in o... | 5.25 | [
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Find the value of $(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}$. | 828 | The $3/2$ power is quite irritating to work with so we look for a way to eliminate that. Notice that squaring the expression will accomplish that. Let $S$ be the sum of the given expression. \[S^2= ((52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2})^2\] \[S^2 = (52+6\sqrt{43})^{3} + (52-6\sqrt{43})^{3} - 2((52+6\sqrt{43})(52... | 6.75 | [
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Let $P_1^{}$ be a regular $r~\mbox{gon}$ and $P_2^{}$ be a regular $s~\mbox{gon}$ $(r\geq s\geq 3)$ such that each interior angle of $P_1^{}$ is $\frac{59}{58}$ as large as each interior angle of $P_2^{}$. What's the largest possible value of $s_{}^{}$? | 117 | As in above, we have $rs = 118r - 116s.$ This means that $rs + 116s - 118r = 0.$ Using SFFT we obtain $s(r+116) - 118(r+116) = -118 \cdot 116 \implies (s-118)(r+116) = -118 \cdot 116.$ Since $r+116$ is always positive, we know thta $s-118$ must be negative. Therefore the maximum value of $s$ must be $\boxed{117}$ which... | 5.875 | [
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Find the positive solution to
$\frac 1{x^2-10x-29}+\frac1{x^2-10x-45}-\frac 2{x^2-10x-69}=0$ | 13 | We could clear out the denominators by multiplying, though that would be unnecessarily tedious.
To simplify the equation, substitute $a = x^2 - 10x - 29$ (the denominator of the first fraction). We can rewrite the equation as $\frac{1}{a} + \frac{1}{a - 16} - \frac{2}{a - 40} = 0$. Multiplying out the denominators now... | 5.875 | [
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Let $n^{}_{}$ be the smallest positive integer that is a multiple of $75_{}^{}$ and has exactly $75_{}^{}$ positive integral divisors, including $1_{}^{}$ and itself. Find $\frac{n}{75}$. | 432 | The prime factorization of $75 = 3^15^2 = (2+1)(4+1)(4+1)$. For $n$ to have exactly $75$ integral divisors, we need to have $n = p_1^{e_1-1}p_2^{e_2-1}\cdots$ such that $e_1e_2 \cdots = 75$. Since $75|n$, two of the prime factors must be $3$ and $5$. To minimize $n$, we can introduce a third prime factor, $2$. Also to ... | 6.25 | [
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A biologist wants to calculate the number of fish in a lake. On May 1 she catches a random sample of 60 fish, tags them, and releases them. On September 1 she catches a random sample of 70 fish and finds that 3 of them are tagged. To calculate the number of fish in the lake on May 1, she assumes that 25% of these fish ... | 840 | First, we notice that there are 45 tags left, after 25% of the original fish have went away/died. Then, some $x$ percent of fish have been added such that $\frac{x}{x+75} = 40 \%$, or $\frac{2}{5}$. Solving for $x$, we get that $x = 50$, so the total number of fish in September is $125 \%$, or $\frac{5}{4}$ times the t... | 6.125 | [
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A triangle has vertices $P_{}^{}=(-8,5)$, $Q_{}^{}=(-15,-19)$, and $R_{}^{}=(1,-7)$. The equation of the bisector of $\angle P$ can be written in the form $ax+2y+c=0_{}^{}$. Find $a+c_{}^{}$.
[asy] import graph; pointpen=black;pathpen=black+linewidth(0.7);pen f = fontsize(10); pair P=(-8,5),Q=(-15,-19),R=(1,-7),S=(7,-1... | 89 | [asy] import graph; pointpen=black;pathpen=black+linewidth(0.7);pen f = fontsize(10); pair P=(-8,5),Q=(-15,-19),R=(1,-7),S=(7,-15),T=(-4,-17),U=IP(P--T,Q--R); MP("P",P,N,f);MP("Q",Q,W,f);MP("R",R,E,f);MP("P'",U,SE,f); D(P--Q--R--cycle);D(U);D(P--U); D((-17,0)--(4,0),Arrows(2mm));D((0,-21)--(0,7),Arrows(2mm)); [/asy]
U... | 5.25 | [
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In a shooting match, eight clay targets are arranged in two hanging columns of three targets each and one column of two targets. A marksman is to break all the targets according to the following rules:
1) The marksman first chooses a column from which a target is to be broken.
2) The marksman must then break the lowe... | 560 | Clearly, the marksman must shoot the left column three times, the middle column two times, and the right column three times.
From left to right, suppose that the columns are labeled $L,M,$ and $R,$ respectively. We consider the string $LLLMMRRR:$
Since the letter arrangements of $LLLMMRRR$ and the shooting orders hav... | 4.25 | [
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A fair coin is to be tossed $10_{}^{}$ times. Let $\frac{i}{j}^{}_{}$, in lowest terms, be the probability that heads never occur on consecutive tosses. Find $i+j_{}^{}$. | 73 | We can also split the problem into casework.
Case 1: 0 Heads
There is only one possibility.
Case 2: 1 Head
There are 10 possibilities.
Case 3: 2 Heads
There are 36 possibilities.
Case 4: 3 Heads
There are 56 possibilities.
Case 5: 4 Heads
There are 35 possibilities.
Case 6: 5 Heads
There are 6 possibilities... | 3.875 | [
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The sets $A = \{z : z^{18} = 1\}$ and $B = \{w : w^{48} = 1\}$ are both sets of complex roots of unity. The set $C = \{zw : z \in A ~ \mbox{and} ~ w \in B\}$ is also a set of complex roots of unity. How many distinct elements are in $C_{}^{}$? | 144 | The values in polar form will be $(1, 20x)$ and $(1, 7.5x)$. Multiplying these gives $(1, 27.5x)$. Then, we get $27.5$, $55$, $82.5$, $110$, $...$ up to $3960$ $(\text{lcm}(55,360)) \implies \frac{3960 \cdot 2}{55}=\boxed{144}$. | 5.5 | [
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Someone observed that $6! = 8 \cdot 9 \cdot 10$. Find the largest positive integer $n^{}_{}$ for which $n^{}_{}!$ can be expressed as the product of $n - 3_{}^{}$ consecutive positive integers. | 23 | Let the largest of the $n-3$ consecutive positive integers be $k$. Clearly $k$ cannot be less than or equal to $n$, else the product of $n-3$ consecutive positive integers will be less than $n!$.
Key observation: Now for $n$ to be maximum the smallest number (or starting number) of the $n-3$ consecutive positive integ... | 4.5 | [
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A regular 12-gon is inscribed in a circle of radius 12. The sum of the lengths of all sides and diagonals of the 12-gon can be written in the form $a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6},$ where $a^{}_{}$, $b^{}_{}$, $c^{}_{}$, and $d^{}_{}$ are positive integers. Find $a + b + c + d^{}_{}$. | 720 | Begin as in solution 2, drawing a triangle connecting the center of the 12-gon to two vertices of the 12-gon. Apply law of cosines on $\theta = {30^{\circ}, 60^{\circ}, 90^{\circ}, 120^{\circ}, 150^{\circ}, 180^{\circ}}$ to get $d^2 = 288 - 288 \cos \theta$ where $d$ is the diagonal or sidelength distance between two p... | 7.625 | [
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Let $T = \{9^k : k ~ \mbox{is an integer}, 0 \le k \le 4000\}$. Given that $9^{4000}_{}$ has 3817 digits and that its first (leftmost) digit is 9, how many elements of $T_{}^{}$ have 9 as their leftmost digit? | 184 | Since $9^{4000}$ has 3816 digits more than $9^1$, $4000 - 3816 = \boxed{184}$ numbers have 9 as their leftmost digits. | 5.5 | [
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The rectangle $ABCD^{}_{}$ below has dimensions $AB^{}_{} = 12 \sqrt{3}$ and $BC^{}_{} = 13 \sqrt{3}$. Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at $P^{}_{}$. If triangle $ABP^{}_{}$ is cut out and removed, edges $\overline{AP}$ and $\overline{BP}$ are joined, and the figure is then creased along segments... | 594 | Let $X$ be the apex of the pyramid and $M$ be the midpoint of $\overline{CD}$. We find the side lengths of $\triangle XMP$.
$MP = \frac{13\sqrt3}{2}$. $PX$ is half of $AC$, which is $\frac{\sqrt{3\cdot13^2+3\cdot12^2}}{2} = \frac{\sqrt{939}}{2}$. To find $MX$, consider right triangle $XMD$; since $XD=13\sqrt3$ and $MD=... | 6.75 | [
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Find $ax^5 + by^5$ if the real numbers $a,b,x,$ and $y$ satisfy the equations \begin{align*} ax + by &= 3, \\ ax^2 + by^2 &= 7, \\ ax^3 + by^3 &= 16, \\ ax^4 + by^4 &= 42. \end{align*} | 20 | A recurrence of the form $T_n=AT_{n-1}+BT_{n-2}$ will have the closed form $T_n=ax^n+by^n$, where $x,y$ are the values of the starting term that make the sequence geometric, and $a,b$ are the appropriately chosen constants such that those special starting terms linearly combine to form the actual starting terms.
Suppo... | 5.25 | [
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Find $x^2+y^2_{}$ if $x_{}^{}$ and $y_{}^{}$ are positive integers such that \begin{align*} xy+x+y&=71, \\ x^2y+xy^2&=880. \end{align*} | 146 | Let $a=x+y$, $b=xy$ then we get the equations \begin{align*} a+b&=71\\ ab&=880 \end{align*} After finding the prime factorization of $880=2^4\cdot5\cdot11$, it's easy to obtain the solution $(a,b)=(16,55)$. Thus \[x^2+y^2=(x+y)^2-2xy=a^2-2b=16^2-2\cdot55=\boxed{146}\] Note that if $(a,b)=(55,16)$, the answer would exce... | 4.625 | [
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Rectangle $ABCD_{}^{}$ has sides $\overline {AB}$ of length 4 and $\overline {CB}$ of length 3. Divide $\overline {AB}$ into 168 congruent segments with points $A_{}^{}=P_0, P_1, \ldots, P_{168}=B$, and divide $\overline {CB}$ into 168 congruent segments with points $C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B$. For $1_{}^{} \... | 840 | First, count the diagonal which has length $5$. For the rest of the segments, think about pairing them up so that each pair makes $5$. For example, the parallel lines closest to the diagonal would have length $\frac{167}{168}\cdot{5}$ while the parallel line closest to the corner of the rectangle would have length $\fr... | 5.25 | [
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Expanding $(1+0.2)^{1000}_{}$ by the binomial theorem and doing no further manipulation gives
${1000 \choose 0}(0.2)^0+{1000 \choose 1}(0.2)^1+{1000 \choose 2}(0.2)^2+\cdots+{1000 \choose 1000}(0.2)^{1000}$ $= A_0 + A_1 + A_2 + \cdots + A_{1000},$ where $A_k = {1000 \choose k}(0.2)^k$ for $k = 0,1,2,\ldots,1000$. For w... | 166 | We know that $A_k$ will increase as $k$ increases until certain $k=m$, where $A_0 < A_1 < A_2 < \dots < A_{m-2} < A_{m-1} < A_m$ and
$A_m > A_{m+1} > A_{m+2} > \dots > A_{1000}.$
Next, to change $A_{k-1}$ to $A_k$, we multiply $A_{k-1}$ by $\frac{1000-k+1}{5k}$. It follows that the numerator must be greater than the
... | 6 | [
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How many real numbers $x^{}_{}$ satisfy the equation $\frac{1}{5}\log_2 x = \sin (5\pi x)$? | 159 | Notice that the equation is satisfied twice for every sine period (which is $\frac{2}{5}$), except in the sole case when the two equations equate to $0$. In that case, the equation is satisfied twice but only at the one instance when $y=0$. Hence, it is double-counted in our final solution, so we have to subtract it ou... | 6 | [
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Given a rational number, write it as a fraction in lowest terms and calculate the product of the resulting numerator and denominator. For how many rational numbers between $0$ and $1$ will $20_{}^{}!$ be the resulting product? | 128 | If the fraction is in the form $\frac{a}{b}$, then $a < b$ and $\gcd(a,b) = 1$. There are 8 prime numbers less than 20 ($2, 3, 5, 7, 11, 13, 17, 19$), and each can only be a factor of one of $a$ or $b$. There are $2^8$ ways of selecting some combination of numbers for $a$; however, since $a<b$, only half of them will b... | 5.875 | [
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Suppose $r^{}_{}$ is a real number for which
$\left\lfloor r + \frac{19}{100} \right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100} \right\rfloor + \cdots + \left\lfloor r + \frac{91}{100} \right\rfloor = 546.$
Find $\lfloor 100r \rfloor$. (For real $x^{}_{}$, $\lfloor x \rflo... | 743 | There are $91 - 19 + 1 = 73$ numbers in the sequence. Since the terms of the sequence can be at most $1$ apart, all of the numbers in the sequence can take one of two possible values. Since $\frac{546}{73} = 7 R 35$, the values of each of the terms of the sequence must be either $7$ or $8$. As the remainder is $35$, $8... | 5.875 | [
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Find $A^2_{}$, where $A^{}_{}$ is the sum of the absolute values of all roots of the following equation:
$x = \sqrt{19} + \frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{x}}}}}}}}}$ | 383 | $x=\sqrt{19}+\underbrace{\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}}}}_{x}$
$x=\sqrt{19}+\frac{91}{x}$
$x^2=x\sqrt{19}+91$
$x^2-x\sqrt{19}-91 = 0$
$\left.\begin{array}{l}x_1=\frac{\sqrt{19}+\sqrt{383}}{2}\\\\x_2=\frac{\sqrt{19}-\sqrt{383}}{2}\end{array}\right\}A=|x_1|... | 5 | [
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For how many real numbers $a^{}_{}$ does the quadratic equation $x^2 + ax^{}_{} + 6a=0$ have only integer roots for $x^{}_{}$? | 10 | Let $x^2 + ax + 6a = (x - s)(x - r)$. Vieta's yields $s + r = - a, sr = 6a$. \begin{eqnarray*}sr + 6s + 6r &=& 0\\ sr + 6s + 6r + 36 &=& 36\\ (s + 6)(r + 6) &=& 36 \end{eqnarray*}
Without loss of generality let $r \le s$.
The possible values of $(r + 6,s + 6)$ are: $( - 36, - 1),( - 18, - 2),( - 12, - 3),( - 9, - 4),(... | 4.625 | [
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Suppose that $\sec x+\tan x=\frac{22}7$ and that $\csc x+\cot x=\frac mn,$ where $\frac mn$ is in lowest terms. Find $m+n^{}_{}.$ | 44 | We are given that $\frac{1+\sin x}{\cos x}=\frac{22}7\implies\frac{1+\sin x}{\cos x}\cdot\frac{1-\sin x}{1-\sin x}=\frac{1-\sin^2x}{\cos x(1-\sin x)}=\frac{\cos^2x}{\cos x(1-\sin x)}$ $=\frac{\cos x}{1-\sin x}$, or equivalently, $\cos x=\frac{7+7\sin x}{22}=\frac{22-22\sin x}7\implies\sin x=\frac{22^2-7^2}{22^2+7^2}$ $... | 6.125 | [
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Two three-letter strings, $aaa^{}_{}$ and $bbb^{}_{}$, are transmitted electronically. Each string is sent letter by letter. Due to faulty equipment, each of the six letters has a 1/3 chance of being received incorrectly, as an $a^{}_{}$ when it should have been a $b^{}_{}$, or as a $b^{}_{}$ when it should be an $a^{}... | 532 | Consider $n$ letter strings instead. If the first letters all get transmitted correctly, then the $a$ string will be first. Otherwise, the only way is for both of the first letters to be the same, and then we consider the next $n-1$ letter string following the first letter. This easily leads to a recursion: $p_n=\frac2... | 6.375 | [
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Twelve congruent disks are placed on a circle $C^{}_{}$ of radius 1 in such a way that the twelve disks cover $C^{}_{}$, no two of the disks overlap, and so that each of the twelve disks is tangent to its two neighbors. The resulting arrangement of disks is shown in the figure below. The sum of the areas of the twelve ... | 135 | We wish to find the radius of one circle, so that we can find the total area.
Notice that for them to contain the entire circle, each pair of circles must be tangent on the larger circle. Now consider two adjacent smaller circles. This means that the line connecting the radii is a segment of length $2r$ that is tangen... | 7.25 | [
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Rhombus $PQRS^{}_{}$ is inscribed in rectangle $ABCD^{}_{}$ so that vertices $P^{}_{}$, $Q^{}_{}$, $R^{}_{}$, and $S^{}_{}$ are interior points on sides $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$, respectively. It is given that $PB^{}_{}=15$, $BQ^{}_{}=20$, $PR^{}_{}=30$, and $QS^{}_{}=40$. ... | 677 | We can just use areas. Let $AP = b$ and $AS = a$. $a^2 + b^2 = 625$. Also, we can add up the areas of all 8 right triangles and let that equal the total area of the rectangle, $(a+20)(b+15)$. This gives $3a + 4b = 120$. Solving this system of equation gives $\frac{44}{5} = a$, $\frac{117}{5} = b$, from which it is stra... | 6.125 | [
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A drawer contains a mixture of red socks and blue socks, at most $1991$ in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly $\frac{1}{2}$ that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consist... | 990 | Let $r$ and $b$ denote the number of red and blue socks such that $r+b\le1991$. Then by complementary counting, the number of ways to get a red and a blue sock must be equal to $1-\frac12=\frac12=\frac{2rb}{(r+b)(r+b-1)}\implies4rb=(r+b)(r+b-1)$ $=(r+b)^2-(r+b)\implies r^2+2rb+b^2-r-b=4rb\implies r^2-2rb+b^2$ $=(r-b)^2... | 5.25 | [
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A hexagon is inscribed in a circle. Five of the sides have length $81$ and the sixth, denoted by $\overline{AB}$, has length $31$. Find the sum of the lengths of the three diagonals that can be drawn from $A_{}^{}$. | 384 | [asy]defaultpen(fontsize(9)); pair A=expi(-pi/2-acos(475/486)), B=expi(-pi/2+acos(475/486)), C=expi(-pi/2+acos(475/486)+acos(7/18)), D=expi(-pi/2+acos(475/486)+2*acos(7/18)), E=expi(-pi/2+acos(475/486)+3*acos(7/18)), F=expi(-pi/2-acos(475/486)-acos(7/18)); draw(unitcircle);draw(A--B--C--D--E--F--A);draw(A--C..A--D..A--... | 7 | [
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For positive integer $n_{}^{}$, define $S_n^{}$ to be the minimum value of the sum $\sum_{k=1}^n \sqrt{(2k-1)^2+a_k^2},$ where $a_1,a_2,\ldots,a_n^{}$ are positive real numbers whose sum is 17. There is a unique positive integer $n^{}_{}$ for which $S_n^{}$ is also an integer. Find this $n^{}_{}$. | 12 | Consider $n$ right triangles joined at their vertices, with bases $a_1,a_2,\ldots,a_n$ and heights $1,3,\ldots, 2n - 1$. The sum of their hypotenuses is the value of $S_n$. The minimum value of $S_n$, then, is the length of the straight line connecting the bottom vertex of the first right triangle and the top vertex of... | 7 | [
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Find the sum of all positive rational numbers that are less than 10 and that have denominator 30 when written in lowest terms. | 400 | Note that if $x$ is a solution, then $(300-x)$ is a solution. We know that $\phi(300) = 80.$ Therefore the answer is $\frac{80}{2} \cdot\frac{300}{30} = \boxed{400}.$
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. | 4.375 | [
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A positive integer is called ascending if, in its decimal representation, there are at least two digits and each digit is less than any digit to its right. How many ascending positive integers are there? | 502 | Note that an ascending number is exactly determined by its digits: for any set of digits (not including 0, since the only position for 0 is at the leftmost end of the number, i.e. a leading 0), there is exactly one ascending number with those digits.
So, there are nine digits that may be used: $1,2,3,4,5,6,7,8,9.$ Not... | 3.25 | [
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A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly $.500$. During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater th... | 164 | Let $n$ be the number of matches won, so that $\frac{n}{2n}=\frac{1}{2}$, and $\frac{n+3}{2n+4}>\frac{503}{1000}$.
Cross multiplying, $1000n+3000>1006n+2012$, so $n<\frac{988}{6}=164 \dfrac {4}{6}=164 \dfrac{2}{3}$. Thus, the answer is $\boxed{164}$.
The problems on this page are copyrighted by the Mathematical Assoc... | 3.25 | [
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In Pascal's Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below.
\[\begin{array}{c@{\hspace{8em}} c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{4pt}}c@{\hspace{2pt}} c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{3pt}}c@{\hspace{6pt}} ... | 62 | Call the row x, and the number from the leftmost side t. Call the first term in the ratio $N$. $N = \dbinom{x}{t}$. The next term is $N * \frac{x-t}{t+1}$, and the final term is $N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}$. Because we have the ratio, $N : N * \frac{x-t}{t+1} : N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}$ = $3:4:5... | 5.75 | [
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Let $S^{}_{}$ be the set of all rational numbers $r^{}_{}$, $0^{}_{}<r<1$, that have a repeating decimal expansion in the form $0.abcabcabc\ldots=0.\overline{abc}$, where the digits $a^{}_{}$, $b^{}_{}$, and $c^{}_{}$ are not necessarily distinct. To write the elements of $S^{}_{}$ as fractions in lowest terms, how man... | 660 | We can see that the denominator of the fraction will be 999b, where b is either 1, 1/3, 1/111, 1/9, or 1/333. This means that the numerator stays abc unless abc divides 999. Finding the numbers relatively prime to 999 gives us the numerators when the fraction cannot be simplified past 999.
$\Rightarrow 999(1 - \frac{1}... | 6.375 | [
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For how many pairs of consecutive integers in $\{1000,1001,1002,\ldots,2000\}$ is no carrying required when the two integers are added? | 156 | Consider what carrying means: If carrying is needed to add two numbers with digits $abcd$ and $efgh$, then $h+d\ge 10$ or $c+g\ge 10$ or $b+f\ge 10$. 6. Consider $c \in \{0, 1, 2, 3, 4\}$. $1abc + 1ab(c+1)$ has no carry if $a, b \in \{0, 1, 2, 3, 4\}$. This gives $5^3=125$ possible solutions.
With $c \in \{5, 6, 7, 8\... | 4.75 | [
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Faces $ABC^{}_{}$ and $BCD^{}_{}$ of tetrahedron $ABCD^{}_{}$ meet at an angle of $30^\circ$. The area of face $ABC^{}_{}$ is $120^{}_{}$, the area of face $BCD^{}_{}$ is $80^{}_{}$, and $BC=10^{}_{}$. Find the volume of the tetrahedron. | 320 | Since the area $BCD=80=\frac{1}{2}\cdot10\cdot16$, the perpendicular from $D$ to $BC$ has length $16$.
The perpendicular from $D$ to $ABC$ is $16 \cdot \sin 30^\circ=8$. Therefore, the volume is $\frac{8\cdot120}{3}=\boxed{320}$. | 5.875 | [
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For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$, define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$, whose $n^{\mbox{th}}_{}$ term is $a_{n+1}-a_n^{}$. Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$, and that $a_{19}=a_{92}^{}=0$. Find $a_1^{}$. | 819 | Note that the $\Delta$s are reminiscent of differentiation; from the condition $\Delta(\Delta{A}) = 1$, we are led to consider the differential equation \[\frac{d^2 A}{dn^2} = 1\] This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; \[a_{n} = \frac{1}{2}(n-19)(n-92)\] as we must have root... | 6.25 | [
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Trapezoid $ABCD^{}_{}$ has sides $AB=92^{}_{}$, $BC=50^{}_{}$, $CD=19^{}_{}$, and $AD=70^{}_{}$, with $AB^{}_{}$ parallel to $CD^{}_{}$. A circle with center $P^{}_{}$ on $AB^{}_{}$ is drawn tangent to $BC^{}_{}$ and $AD^{}_{}$. Given that $AP^{}_{}=\frac mn$, where $m^{}_{}$ and $n^{}_{}$ are relatively prime positive... | 164 | From $(1)$ above, $x = \frac{70r}{h}$ and $92-x = \frac{50r}{h}$. Adding these equations yields $92 = \frac{120r}{h}$. Thus, $x = \frac{70r}{h} = \frac{7}{12}\cdot\frac{120r}{h} = \frac{7}{12}\cdot92 = \frac{161}{3}$, and $m+n = \boxed{164}$.
We can use $(1)$ from Solution 1 to find that $h/r = 70/x$ and $h/r = 50/ (9... | 6.25 | [
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Consider the region $A$ in the complex plane that consists of all points $z$ such that both $\frac{z}{40}$ and $\frac{40}{\overline{z}}$ have real and imaginary parts between $0$ and $1$, inclusive. What is the integer that is nearest the area of $A$? | 572 | Let $z=a+bi \implies \frac{z}{40}=\frac{a}{40}+\frac{b}{40}i$. Since $0\leq \frac{a}{40},\frac{b}{40}\leq 1$ we have the inequality \[0\leq a,b \leq 40\]which is a square of side length $40$.
Also, $\frac{40}{\overline{z}}=\frac{40}{a-bi}=\frac{40a}{a^2+b^2}+\frac{40b}{a^2+b^2}i$ so we have $0\leq a,b \leq \frac{a^2+b... | 6.75 | [
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Lines $l_1^{}$ and $l_2^{}$ both pass through the origin and make first-quadrant angles of $\frac{\pi}{70}$ and $\frac{\pi}{54}$ radians, respectively, with the positive x-axis. For any line $l^{}_{}$, the transformation $R(l)^{}_{}$ produces another line as follows: $l^{}_{}$ is reflected in $l_1^{}$, and the resultin... | 945 | Let $l$ be a line that makes an angle of $\theta$ with the positive $x$-axis. Let $l'$ be the reflection of $l$ in $l_1$, and let $l''$ be the reflection of $l'$ in $l_2$.
The angle between $l$ and $l_1$ is $\theta - \frac{\pi}{70}$, so the angle between $l_1$ and $l'$ must also be $\theta - \frac{\pi}{70}$. Thus, $l'... | 6.625 | [
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In a game of Chomp, two players alternately take bites from a 5-by-7 grid of unit squares. To take a bite, a player chooses one of the remaining squares, then removes ("eats") all squares in the quadrant defined by the left edge (extended upward) and the lower edge (extended rightward) of the chosen square. For example... | 792 | By drawing possible examples of the subset, one can easily see that making one subset is the same as dividing the game board into two parts.
One can also see that it is the same as finding the shortest route from the upper left hand corner to the lower right hand corner; Such a route would require 5 lengths that go do... | 5.875 | [
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Triangle $ABC$ has $AB=9$ and $BC: AC=40: 41$. What's the largest area that this triangle can have? | 820 | We can start how we did above in solution 4 to get $\frac{9}{4} * \sqrt{(81x^2-1)(81-x^2)}$. Then, we can notice the inside is a quadratic in terms of $x^2$, which is $-81(x^2)^2+6562x^2-81$. This is maximized when $x^2 = \frac{3281}{81}$.If we plug it into the equation, we get $\frac{9}{4} *\frac{9}{4}*\frac{3280}{9} ... | 5.75 | [
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In triangle $ABC^{}_{}$, $A'$, $B'$, and $C'$ are on the sides $BC$, $AC^{}_{}$, and $AB^{}_{}$, respectively. Given that $AA'$, $BB'$, and $CC'$ are concurrent at the point $O^{}_{}$, and that $\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92$, find $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}$. | 94 | Using mass points, let the weights of $A$, $B$, and $C$ be $a$, $b$, and $c$ respectively.
Then, the weights of $A'$, $B'$, and $C'$ are $b+c$, $c+a$, and $a+b$ respectively.
Thus, $\frac{AO^{}_{}}{OA'} = \frac{b+c}{a}$, $\frac{BO^{}_{}}{OB'} = \frac{c+a}{b}$, and $\frac{CO^{}_{}}{OC'} = \frac{a+b}{c}$.
Therefore: $... | 6.625 | [
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Define a positive integer $n^{}_{}$ to be a factorial tail if there is some positive integer $m^{}_{}$ such that the decimal representation of $m!$ ends with exactly $n$ zeroes. How many positive integers less than $1992$ are not factorial tails? | 396 | After testing various values of $m$ in $f(m)$ of solution 1 to determine $m$ for which $f(m) = 1992$, we find that $m \in \{7980, 7981, 7982, 7983, 7984\}$. WLOG, we select $7980$. Furthermore, note that every time $k$ reaches a multiple of $25$, $k!$ will gain two or more additional factors of $5$ and will thus skip o... | 6 | [
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How many even integers between 4000 and 7000 have four different digits? | 728 | The thousands digit is $\in \{4,5,6\}$.
Case $1$: Thousands digit is even
$4, 6$, two possibilities, then there are only $\frac{10}{2} - 1 = 4$ possibilities for the units digit. This leaves $8$ possible digits for the hundreds and $7$ for the tens places, yielding a total of $2 \cdot 8 \cdot 7 \cdot 4 = 448$.
Case $... | 4.25 | [
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During a recent campaign for office, a candidate made a tour of a country which we assume lies in a plane. On the first day of the tour he went east, on the second day he went north, on the third day west, on the fourth day south, on the fifth day east, etc. If the candidate went $n^{2}_{}/2$ miles on the $n^{\mbox{th}... | 580 | On the first day, the candidate moves $[4(0) + 1]^2/2\ \text{east},\, [4(0) + 2]^2/2\ \text{north},\, [4(0) + 3]^2/2\ \text{west},\, [4(0) + 4]^2/2\ \text{south}$, and so on. The E/W displacement is thus $1^2 - 3^2 + 5^2 \ldots +37^2 - 39^2 = \left|\sum_{i=0}^9 \frac{(4i+1)^2}{2} - \sum_{i=0}^9 \frac{(4i+3)^2}{2}\right... | 6.75 | [
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The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught $n\,$ fish for various values of $n\,$.
$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & \dots & 13 & 14 & 15 \\ \hline \text{number of contestants who caught} \ n \ \text{fi... | 943 | Let $f$ be the total number of fish caught by the contestants who didn't catch $0, 1, 2, 3, 13, 14$, or $15$ fish and let $a$ be the number of contestants who didn't catch $0, 1, 2, 3, 13, 14$, or $15$ fish. From $\text{(b)}$, we know that $\frac{69+f+65+28+15}{a+31}=6\implies f=6a+9$. From $\text{(c)}$ we have $\frac{... | 6.5 | [
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How many ordered four-tuples of integers $(a,b,c,d)\,$ with $0 < a < b < c < d < 500\,$ satisfy $a + d = b + c\,$ and $bc - ad = 93\,$? | 870 | Let $b = a + m$ and $c = a + m + n$. From $a + d = b + c$, $d = b + c - a = a + 2m + n$.
Substituting $b = a + m$, $c = a + m + n$, and $d = b + c - a = a + 2m + n$ into $bc - ad = 93$, \[bc - ad = (a + m)(a + m + n) - a(a + 2m + n) = m(m + n). = 93 = 3(31)\] Hence, $(m,n) = (1,92)$ or $(3,28)$.
For $(m,n) = (1,92)$,... | 6.125 | [
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6
] |
Let $P_0(x) = x^3 + 313x^2 - 77x - 8\,$. For integers $n \ge 1\,$, define $P_n(x) = P_{n - 1}(x - n)\,$. What is the coefficient of $x\,$ in $P_{20}(x)\,$? | 763 | Notice the transformation of $P_{n-1}(x)\to P_n(x)$ adds $n$ to the roots. Thus, all these transformations will take the roots and add $1+2+\cdots+20=210$ to them. (Indeed, this is very easy to check in general.)
Let the roots be $r_1,r_2,r_3.$ Then $P_{20}(x)=(x-r_1-210)(x-r_2-210)(x-r_3-210).$ By Vieta's/expanding/c... | 6.125 | [
6,
7,
6,
6,
6,
6,
6,
6
] |
What is the smallest positive integer that can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers? | 495 | First note that the integer clearly must be divisible by $9$ and $11$ since we can use the "let the middle number be x" trick. Let the number be $99k$ for some integer $k.$ Now let the $10$ numbers be $x,x+1, \cdots x+9.$ We have $10x+45 = 99k.$ Taking mod $5$ yields $k \equiv 0 \pmod{5}.$ Since $k$ is positive, we tak... | 4.625 | [
4,
4,
5,
5,
5,
5,
4,
5
] |
Three numbers, $a_1, a_2, a_3$, are drawn randomly and without replacement from the set $\{1, 2, 3,\ldots, 1000\}$. Three other numbers, $b_1, b_2, b_3$, are then drawn randomly and without replacement from the remaining set of $997$ numbers. Let $p$ be the probability that, after suitable rotation, a brick of dimensio... | 5 | There is a total of $P(1000,6)$ possible ordered $6$-tuples $(a_1,a_2,a_3,b_1,b_2,b_3).$
There are $C(1000,6)$ possible sets $\{a_1,a_2,a_3,b_1,b_2,b_3\}.$ We have five valid cases for the increasing order of these six elements:
$aaabbb$
$aababb$
$aabbab$
$abaabb$
$ababab$
Note that the $a$'s are different from each ... | 6.25 | [
7,
6,
6,
6,
7,
6,
6,
6
] |
Let $S\,$ be a set with six elements. In how many different ways can one select two not necessarily distinct subsets of $S\,$ so that the union of the two subsets is $S\,$? The order of selection does not matter; for example, the pair of subsets $\{a, c\},\{b, c, d, e, f\}$ represents the same selection as the pair $\{... | 365 | We evaluate $f(6)$ recursively: \begin{alignat*}{6} f(0)&=1, \\ f(1)&=3f(0)-1&&=2, \\ f(2)&=3f(1)-1&&=5, \\ f(3)&=3f(2)-1&&=14, \\ f(4)&=3f(3)-1&&=41, \\ f(5)&=3f(4)-1&&=122, \\ f(6)&=3f(5)-1&&=\boxed{365}. \end{alignat*} ~MRENTHUSIASM | 5.5 | [
6,
6,
6,
5,
6,
5,
6,
4
] |
Two thousand points are given on a circle. Label one of the points $1$. From this point, count $2$ points in the clockwise direction and label this point $2$. From the point labeled $2$, count $3$ points in the clockwise direction and label this point $3$. (See figure.) Continue this process until the labels $1,2,3\dot... | 118 | The label $1993$ will occur on the $\frac12(1993)(1994) \pmod{2000}$th point around the circle. (Starting from 1) A number $n$ will only occupy the same point on the circle if $\frac12(n)(n + 1)\equiv \frac12(1993)(1994) \pmod{2000}$.
Simplifying this expression, we see that $(1993)(1994) - (n)(n + 1) = (1993 - n)(199... | 7 | [
7,
6,
7,
7,
7,
7,
8,
7
] |
Euler's formula states that for a convex polyhedron with $V$ vertices, $E$ edges, and $F$ faces, $V-E+F=2$. A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon. At each of its $V$ vertices, $T$ triangular faces and $P$ pentagonal faces meet. What is the value of $100P+10T+V$? | 250 | We know that $V-E = -30 \implies V = E-30$ based off the problem condition. Furthermore, if we draw out a few pentagons as well as triangles on each of side of the pentagons, it's clear that each vertex has 4 edges connected to it, with two triangles and two pentagons for each vertex. However, each edge is used for two... | 5.375 | [
5,
5,
6,
6,
5,
6,
6,
4
] |
Alfred and Bonnie play a game in which they take turns tossing a fair coin. The winner of a game is the first person to obtain a head. Alfred and Bonnie play this game several times with the stipulation that the loser of a game goes first in the next game. Suppose that Alfred goes first in the first game, and that the ... | 93 | The probability that the $n$th flip in each game occurs and is a head is $\frac{1}{2^n}$. The first person wins if the coin lands heads on an odd numbered flip. So, the probability of the first person winning the game is $\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\cdots = \frac{\frac{1}{2}}{1-\frac{1}{4}}=\frac{2}{3}$, and ... | 6.375 | [
6,
7,
6,
7,
7,
6,
6,
6
] |
The vertices of $\triangle ABC$ are $A = (0,0)\,$, $B = (0,420)\,$, and $C = (560,0)\,$. The six faces of a die are labeled with two $A\,$'s, two $B\,$'s, and two $C\,$'s. Point $P_1 = (k,m)\,$ is chosen in the interior of $\triangle ABC$, and points $P_2\,$, $P_3\,$, $P_4, \dots$ are generated by rolling the die repea... | 344 | If we have points $(p,q)$ and $(r,s)$ and we want to find $(u,v)$ so $(r,s)$ is the midpoint of $(u,v)$ and $(p,q)$, then $u=2r-p$ and $v=2s-q$. So we start with the point they gave us and work backwards. We make sure all the coordinates stay within the triangle. We have: $P_7=(14,92)$
$P_6=(2\cdot14-0, 2\cdot92-0)=(28... | 5.875 | [
6,
6,
6,
5,
6,
7,
5,
6
] |
Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and K... | 163 | Consider the unit cicle of radius 50. Assume that they start at points $(-50,100)$ and $(-50,-100).$ Then at time $t$, they end up at points $(-50+t,100)$ and $(-50+3t,-100).$ The equation of the line connecting these points and the equation of the circle are \begin{align}y&=-\frac{100}{t}x+200-\frac{5000}{t}\\50^2&=x^... | 5.75 | [
6,
6,
6,
6,
6,
6,
5,
5
] |
A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter ... | 448 | Note that the diagonal of the rectangle with minimum perimeter must have the diagonal along the middle segment of length 8 of the rectangle (any other inscribed rectangle can be rotated a bit, then made smaller; this one can't because then the rectangle cannot be inscribed since its longest diagonal is less than 8 in l... | 6.625 | [
6,
7,
6,
6,
7,
8,
7,
6
] |
Let $\overline{CH}$ be an altitude of $\triangle ABC$. Let $R\,$ and $S\,$ be the points where the circles inscribed in the triangles $ACH\,$ and $BCH^{}_{}$ are tangent to $\overline{CH}$. If $AB = 1995\,$, $AC = 1994\,$, and $BC = 1993\,$, then $RS\,$ can be expressed as $m/n\,$, where $m\,$ and $n\,$ are relatively ... | 997 | [asy] unitsize(48); pair A,B,C,H; A=(8,0); B=origin; C=(3,4); H=(3,0); draw(A--B--C--cycle); draw(C--H); label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,N); label("$H$",H,NE); draw(circle((2,1),1)); pair [] x=intersectionpoints(C--H,circle((2,1),1)); dot(x[0]); label("$S$",x[0],SW); draw(circle((4.29843788128,1.298... | 6.625 | [
6,
7,
7,
7,
6,
7,
6,
7
] |
The increasing sequence $3, 15, 24, 48, \ldots\,$ consists of those positive multiples of 3 that are one less than a perfect square. What is the remainder when the 1994th term of the sequence is divided by 1000? | 63 | One less than a perfect square can be represented by $n^2 - 1 = (n+1)(n-1)$. Either $n+1$ or $n-1$ must be divisible by 3. This is true when $n \equiv -1,\ 1 \equiv 2,\ 1 \pmod{3}$. Since 1994 is even, $n$ must $\equiv 1 \pmod{3}$. It will be the $\frac{1994}{2} = 997$th such term, so $n = 4 + (997-1) \cdot 3 = 2992$. ... | 5.875 | [
6,
6,
6,
5,
6,
6,
6,
6
] |
A circle with diameter $\overline{PQ}$ of length 10 is internally tangent at $P$ to a circle of radius 20. Square $ABCD$ is constructed with $A$ and $B$ on the larger circle, $\overline{CD}$ tangent at $Q$ to the smaller circle, and the smaller circle outside $ABCD$. The length of $\overline{AB}$ can be written in the ... | 312 | Call the center of the larger circle $O$. Extend the diameter $\overline{PQ}$ to the other side of the square (at point $E$), and draw $\overline{AO}$. We now have a right triangle, with hypotenuse of length $20$. Since $OQ = OP - PQ = 20 - 10 = 10$, we know that $OE = AB - OQ = AB - 10$. The other leg, $AE$, is just $... | 6.5 | [
6,
7,
6,
7,
6,
7,
7,
6
] |
The function $f_{}^{}$ has the property that, for each real number $x,\,$
$f(x)+f(x-1) = x^2.\,$
If $f(19)=94,\,$ what is the remainder when $f(94)\,$ is divided by $1000$? | 561 | \begin{align*}f(94)&=94^2-f(93)=94^2-93^2+f(92)=94^2-93^2+92^2-f(91)=\cdots \\ &= (94^2-93^2) + (92^2-91^2) +\cdots+ (22^2-21^2)+ 20^2-f(19) \\ &= 94+93+\cdots+21+400-94 \\ &= 4561 \end{align*}
So, the remainder is $\boxed{561}$. | 4.875 | [
5,
5,
5,
4,
6,
5,
4,
5
] |
Find the positive integer $n\,$ for which \[\lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor=1994\] (For real $x\,$, $\lfloor x\rfloor\,$ is the greatest integer $\le x.\,$) | 312 | Note that if $2^x \le a<2^{x+1}$ for some $x\in\mathbb{Z}$, then $\lfloor\log_2{a}\rfloor=\log_2{2^{x}}=x$.
Thus, there are $2^{x+1}-2^{x}=2^{x}$ integers $a$ such that $\lfloor\log_2{a}\rfloor=x$. So the sum of $\lfloor\log_2{a}\rfloor$ for all such $a$ is $x\cdot2^x$.
Let $k$ be the integer such that $2^k \le n<2^{... | 6.125 | [
6,
6,
6,
7,
7,
5,
6,
6
] |
Given a positive integer $n\,$, let $p(n)\,$ be the product of the non-zero digits of $n\,$. (If $n\,$ has only one digits, then $p(n)\,$ is equal to that digit.) Let
$S=p(1)+p(2)+p(3)+\cdots+p(999)$
.
What is the largest prime factor of $S\,$? | 103 | Note that $p(1)=p(11), p(2)=p(12), p(3)=p(13), \cdots p(19)=p(9)$, and $p(37)=3p(7)$. So $p(10)+p(11)+p(12)+\cdots +p(19)=46$, $p(10)+p(11)+\cdots +p(99)=46*45=2070$. We add $p(1)+p(2)+p(3)+\cdots +p(10)=45$ to get 2115. When we add a digit we multiply the sum by that digit. Thus $2115\cdot (1+1+2+3+4+5+6+7+8+9)=2115\c... | 5.875 | [
6,
6,
6,
6,
6,
6,
5,
6
] |
The graphs of the equations
$y=k, \qquad y=\sqrt{3}x+2k, \qquad y=-\sqrt{3}x+2k,$
are drawn in the coordinate plane for $k=-10,-9,-8,\ldots,9,10.\,$ These 63 lines cut part of the plane into equilateral triangles of side length $\tfrac{2}{\sqrt{3}}.\,$ How many such triangles are formed? | 660 | Picturing the diagram in your head should give you an illustration similar to the one above. The distance from parallel sides of the center hexagon is 20 units, and by extending horizontal lines to the sides of the hexagon. This shows that for every side of the hexagon there are 10 spaces. Therefore the side length of ... | 6 | [
6,
6,
7,
6,
5,
6,
6,
6
] |
For certain ordered pairs $(a,b)\,$ of real numbers, the system of equations
$ax+by=1\,$
$x^2+y^2=50\,$
has at least one solution, and each solution is an ordered pair $(x,y)\,$ of integers. How many such ordered pairs $(a,b)\,$ are there? | 72 | $x^2+y^2=50$ is the equation of a circle of radius $\sqrt{50}$, centered at the origin. The lattice points on this circle are $(\pm1,\pm7)$, $(\pm5,\pm5)$, and $(\pm7,\pm1)$.
$ax+by=1$ is the equation of a line that does not pass through the origin. (Since $(x,y)=(0,0)$ yields $a(0)+b(0)=0 \neq 1$).
So, we are looking... | 6.5 | [
6,
6,
7,
6,
6,
7,
7,
7
] |
The points $(0,0)\,$, $(a,11)\,$, and $(b,37)\,$ are the vertices of an equilateral triangle. Find the value of $ab\,$. | 315 | Consider the points on the complex plane. The point $b+37i$ is then a rotation of $60$ degrees of $a+11i$ about the origin, so:
\[(a+11i)\left(\mathrm{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.\]
Equating the real and imaginary parts, we have:
\begin{align*}b&=\frac{a}{2}-\frac{11\... | 6.125 | [
6,
6,
7,
6,
6,
6,
6,
6
] |
A solitaire game is played as follows. Six distinct pairs of matched tiles are placed in a bag. The player randomly draws tiles one at a time from the bag and retains them, except that matching tiles are put aside as soon as they appear in the player's hand. The game ends if the player ever holds three tiles, no two of... | 394 | Let $P_k$ be the probability of emptying the bag when it has $k$ pairs in it. Let's consider the possible draws for the first three cards:
Case 1. We draw a pair on the first two cards. The second card is the same as the first with probability $\frac {1}{2k - 1}$, then we have $k - 1$ pairs left. So this contributes pr... | 6.75 | [
6,
7,
6,
7,
7,
7,
7,
7
] |
In triangle $ABC,\,$ angle $C$ is a right angle and the altitude from $C\,$ meets $\overline{AB}\,$ at $D.\,$ The lengths of the sides of $\triangle ABC\,$ are integers, $BD=29^3,\,$ and $\cos B=m/n\,$, where $m\,$ and $n\,$ are relatively prime positive integers. Find $m+n.\,$ | 450 | We will solve for $\cos B$ using $\triangle CBD$, which gives us $\cos B = \frac{29^3}{BC}$. By the Pythagorean Theorem on $\triangle CBD$, we have $BC^2 - DC^2 = (BC + DC)(BC - DC) = 29^6$. Trying out factors of $29^6$, we can either guess and check or just guess to find that $BC + DC = 29^4$ and $BC - DC = 29^2$ (The... | 7 | [
6,
7,
7,
7,
7,
7,
8,
7
] |
Ninety-four bricks, each measuring $4''\times10''\times19'',$ are to be stacked one on top of another to form a tower 94 bricks tall. Each brick can be oriented so it contributes $4''\,$ or $10''\,$ or $19''\,$ to the total height of the tower. How many different tower heights can be achieved using all ninety-four of t... | 465 | Using bricks of dimensions $4''\times10''\times19''$ is comparable to using bricks of dimensions $0''\times6''\times15''$ which is comparable to using bricks of dimensions $0''\times2''\times5''$. Using 5 bricks of height $2''$ can be replaced by using 2 bricks of height $5''$ and 3 bricks of height $0''$.
It follows ... | 6 | [
5,
7,
6,
6,
6,
6,
6,
6
] |
A fenced, rectangular field measures $24$ meters by $52$ meters. An agricultural researcher has 1994 meters of fence that can be used for internal fencing to partition the field into congruent, square test plots. The entire field must be partitioned, and the sides of the squares must be parallel to the edges of the fie... | 702 | Suppose there are $n$ squares in every column of the grid, so there are $\frac{52}{24}n = \frac {13}6n$ squares in every row. Then $6|n$, and our goal is to maximize the value of $n$.
Each vertical fence has length $24$, and there are $\frac{13}{6}n - 1$ vertical fences; each horizontal fence has length $52$, and ther... | 4.75 | [
5,
4,
7,
4,
4,
4,
5,
5
] |
The equation
$x^{10}+(13x-1)^{10}=0\,$
has 10 complex roots $r_1, \overline{r_1}, r_2, \overline{r_2}, r_3, \overline{r_3}, r_4, \overline{r_4}, r_5, \overline{r_5},\,$ where the bar denotes complex conjugation. Find the value of
$\frac 1{r_1\overline{r_1}}+\frac 1{r_2\overline{r_2}}+\frac 1{r_3\overline{r_3}}+\frac 1... | 850 | Divide both sides by $x^{10}$ to get \[1 + \left(13-\dfrac{1}{x}\right)^{10}=0\]
Rearranging: \[\left(13-\dfrac{1}{x}\right)^{10} = -1\]
Thus, $13-\dfrac{1}{x} = \omega$ where $\omega = e^{i(\pi n/5+\pi/10)}$ where $n$ is an integer.
We see that $\dfrac{1}{x}=13-\omega$. Thus, \[\dfrac{1}{x\overline{x}}=(13\, -\, \o... | 6.875 | [
7,
7,
8,
6,
7,
7,
7,
6
] |
A beam of light strikes $\overline{BC}\,$ at point $C\,$ with angle of incidence $\alpha=19.94^\circ\,$ and reflects with an equal angle of reflection as shown. The light beam continues its path, reflecting off line segments $\overline{AB}\,$ and $\overline{BC}\,$ according to the rule: angle of incidence equals angle ... | 71 | At each point of reflection, we pretend instead that the light continues to travel straight. [asy] pathpen = linewidth(0.7); size(250); real alpha = 28, beta = 36; pair B = MP("B",(0,0),NW), C = MP("C",D((1,0))), A = MP("A",expi(alpha * pi/180),N); path r = C + .4 * expi(beta * pi/180) -- C - 2*expi(beta * pi/180); D(A... | 6.625 | [
6,
6,
7,
7,
6,
7,
7,
7
] |
Given a point $P^{}_{}$ on a triangular piece of paper $ABC,\,$ consider the creases that are formed in the paper when $A, B,\,$ and $C\,$ are folded onto $P.\,$ Let us call $P_{}^{}$ a fold point of $\triangle ABC\,$ if these creases, which number three unless $P^{}_{}$ is one of the vertices, do not intersect. Suppos... | 597 | Let $O_{AB}$ be the intersection of the perpendicular bisectors (in other words, the intersections of the creases) of $\overline{PA}$ and $\overline{PB}$, and so forth. Then $O_{AB}, O_{BC}, O_{CA}$ are, respectively, the circumcenters of $\triangle PAB, PBC, PCA$. According to the problem statement, the circumcenters ... | 7.625 | [
8,
8,
7,
8,
8,
7,
8,
7
] |
Square $S_{1}$ is $1\times 1.$ For $i\ge 1,$ the lengths of the sides of square $S_{i+1}$ are half the lengths of the sides of square $S_{i},$ two adjacent sides of square $S_{i}$ are perpendicular bisectors of two adjacent sides of square $S_{i+1},$ and the other two sides of square $S_{i+1},$ are the perpendicular bi... | 255 | The sum of the areas of the squares if they were not interconnected is a geometric sequence:
$1^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{16}\right)^2$
Then subtract the areas of the intersections, which is $\left(\frac{1}{4}\right)^2 + \ldots + \left(\fr... | 5 | [
5,
5,
5,
4,
5,
5,
6,
5
] |
Find the last three digits of the product of the positive roots of $\sqrt{1995}x^{\log_{1995}x}=x^2$. | 25 | Taking the $\log_{1995}$ (logarithm) of both sides and then moving to one side yields the quadratic equation $2(\log_{1995}x)^2 - 4(\log_{1995}x) + 1 = 0$. Applying the quadratic formula yields that $\log_{1995}x = 1 \pm \frac{\sqrt{2}}{2}$. Thus, the product of the two roots (both of which are positive) is $1995^{1+\s... | 6 | [
6,
6,
6,
6,
6,
6,
6,
6
] |
Starting at $(0,0),$ an object moves in the coordinate plane via a sequence of steps, each of length one. Each step is left, right, up, or down, all four equally likely. Let $p$ be the probability that the object reaches $(2,2)$ in six or fewer steps. Given that $p$ can be written in the form $m/n,$ where $m$ and $n$ a... | 67 | It takes an even number of steps for the object to reach $(2,2)$, so the number of steps the object may have taken is either $4$ or $6$.
If the object took $4$ steps, then it must have gone two steps N and two steps E, in some permutation. There are $\frac{4!}{2!2!} = 6$ ways for these four steps of occuring, and the ... | 5.375 | [
6,
5,
5,
5,
5,
5,
6,
6
] |
Circles of radius $3$ and $6$ are externally tangent to each other and are internally tangent to a circle of radius $9$. The circle of radius $9$ has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord.
[asy] pointpen = black; pathpen = black + linewidth(0.7);... | 224 | We label the points as following: the centers of the circles of radii $3,6,9$ are $O_3,O_6,O_9$ respectively, and the endpoints of the chord are $P,Q$. Let $A_3,A_6,A_9$ be the feet of the perpendiculars from $O_3,O_6,O_9$ to $\overline{PQ}$ (so $A_3,A_6$ are the points of tangency). Then we note that $\overline{O_3A_3... | 7 | [
7,
7,
7,
7,
7,
7,
7,
7
] |
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