ID stringlengths 8 10 | Year int64 1.98k 2.02k | Problem Number int64 1 15 | Question stringlengths 37 2.66k | Answer int64 0 997 | Part stringclasses 2 values | Solution sequencelengths 1 25 |
|---|---|---|---|---|---|---|
1983-I-1 | 1,983 | 1 | Let $x$ , $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_xw=24$ , $\log_y w = 40$ and $\log_{xyz}w=12$ . Find $\log_zw$ . | 60 | null | [
"The logarithmic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms. $x^{24}=w$, $y^{40}=w$, and $(xyz)^{12}=w$. If we now convert everything to a power of $120$, it will be easy to isolate $z$ and $w$. $x^{120}=w^5$, $y^{120}=w^3$, and $(xyz)^{120}=w^{10}$. With so... |
1983-I-2 | 1,983 | 2 | Let $f(x)=|x-p|+|x-15|+|x-p-15|$ , where $0 < p < 15$ . Determine the minimum value taken by $f(x)$ for $x$ in the interval $p \leq x\leq15$ . | 15 | null | [
"It is best to get rid of the absolute values first. Under the given circumstances, we notice that $|x-p|=x-p$, $|x-15|=15-x$, and $|x-p-15|=15+p-x$. Adding these together, we find that the sum is equal to $30-x$, which attains its minimum value (on the given interval $p \\leq x \\leq 15$) when $x=15$, giving a min... |
1983-I-3 | 1,983 | 3 | What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$ ? | 20 | null | [
"If we were to expand by squaring, we would get a quartic polynomial, which isn't always the easiest thing to deal with. Instead, we substitute $y$ for $x^2+18x+30$, so that the equation becomes $y=2\\sqrt{y+15}$. Now we can square; solving for $y$, we get $y=10$ or $y=-6$. The second root is extraneous since $2\\s... |
1983-I-4 | 1,983 | 4 | A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. [asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); // Drawing arc instead of full circle //draw(P); draw(arc(O, r, degrees(A), degrees(C))); draw(C--B--A--B); dot(A); dot(B); dot(C); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy] | 26 | null | [
"Because we are given a right angle, we look for ways to apply the Pythagorean Theorem. Let the foot of the perpendicular from $O$ to $AB$ be $D$ and let the foot of the perpendicular from $O$ to the line $BC$ be $E$. Let $OE=x$ and $OD=y$. We're trying to find $x^2+y^2$. [asy] size(150); defaultpen(linewidth(0.6)+... |
1983-I-5 | 1,983 | 5 | Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$ . What is the largest real value that $x + y$ can have? | 4 | null | [
"Solution 1 One way to solve this problem is by substitution. We have $x^2+y^2=(x+y)^2-2xy=7$ and $x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10$ Hence observe that we can write $w=x+y$ and $z=xy$. This reduces the equations to $w^2-2z=7$ and $w(7-z)=10$. Because we want the largest possible $w$, let's find an expressio... |
1983-I-6 | 1,983 | 6 | Let $a_n=6^{n}+8^{n}$ . Determine the remainder on dividing $a_{83}$ by $49$ . | 35 | null | [
"Solution 1 Firstly, we try to find a relationship between the numbers we're provided with and $49$. We notice that $49=7^2$, and both $6$ and $8$ are greater or less than $7$ by $1$. Thus, expressing the numbers in terms of $7$, we get $a_{83} = (7-1)^{83}+(7+1)^{83}$. Applying the Binomial Theorem, half of our te... |
1983-I-7 | 1,983 | 7 | Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator? | 57 | null | [
"We can use complementary counting, by finding the probability that none of the three knights are sitting next to each other and subtracting it from $1$. Imagine that the $22$ other (indistinguishable) people are already seated, and fixed into place. We will place $A$, $B$, and $C$ with and without the restriction.... |
1983-I-8 | 1,983 | 8 | What is the largest $2$ -digit prime factor of the integer $n = {200\choose 100}$ ? | 61 | null | [
"Solution 1 Expanding the binomial coefficient, we get ${200 \\choose 100}=\\frac{200!}{100!100!}$. Let the required prime be $p$; then $10 \\le p < 100$. If $p > 50$, then the factor of $p$ appears twice in the denominator. Thus, we need $p$ to appear as a factor at least three times in the numerator, so $3p<200$.... |
1983-I-9 | 1,983 | 9 | Find the minimum value of $\frac{9x^2\sin^2 x + 4}{x\sin x}$ for $0 < x < \pi$ . | 12 | null | [
"Let $y=x\\sin{x}$. We can rewrite the expression as $\\frac{9y^2+4}{y}=9y+\\frac{4}{y}$. Since $x>0$, and $\\sin{x}>0$ because $0< x<\\pi$, we have $y>0$. So we can apply AM-GM: \\[9y+\\frac{4}{y}\\ge 2\\sqrt{9y\\cdot\\frac{4}{y}}=12\\] The equality holds when $9y=\\frac{4}{y}\\Longleftrightarrow y^2=\\frac49\\Lon... |
1983-I-10 | 1,983 | 10 | The numbers $1447$ , $1005$ and $1231$ have something in common: each is a $4$ -digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there? | 432 | null | [
"Suppose that the two identical digits are both $1$. Since the thousands digit must be $1$, only one of the other three digits can be $1$. This means the possible forms for the number are $11xy,\\qquad 1x1y,\\qquad1xy1$ Because the number must have exactly two identical digits, $x\\neq y$, $x\\neq1$, and $y\\neq1$.... |
1983-I-11 | 1,983 | 11 | The solid shown has a square base of side length $s$ . The upper edge is parallel to the base and has length $2s$ . All other edges have length $s$ . Given that $s=6\sqrt{2}$ , what is the volume of the solid? [asy] import three; size(170); pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); label("A",A, S); label("B",B, S); label("C",C, S); label("D",D, S); label("E",E,N); label("F",F,N); [/asy] | 288 | null | [
"Solution 1 First, we find the height of the solid by dropping a perpendicular from the midpoint of $AD$ to $EF$. The hypotenuse of the triangle formed is the median of equilateral triangle $ADE$, and one of the legs is $3\\sqrt{2}$. We apply the Pythagorean Theorem to deduce that the height is $6$. [asy] size(180)... |
1983-I-12 | 1,983 | 12 | Diameter $AB$ of a circle has length a $2$ -digit integer (base ten). Reversing the digits gives the length of the perpendicular chord $CD$ . The distance from their intersection point $H$ to the center $O$ is a positive rational number. Determine the length of $AB$ . Pdfresizer.com-pdf-convert-aimeq12.png | 65 | null | [
"Let $AB=10x+y$ and $CD=10y+x$. It follows that $CO=\\frac{AB}{2}=\\frac{10x+y}{2}$ and $CH=\\frac{CD}{2}=\\frac{10y+x}{2}$. Scale up this triangle by 2 to ease the arithmetic. Applying the Pythagorean Theorem on $2CO$, $2OH$ and $2CH$, we deduce \\[(2OH)^2=\\left(10x+y\\right)^2-\\left(10y+x\\right)^2=99(x+y)(x-y)... |
1983-I-13 | 1,983 | 13 | For $\{1, 2, 3, \ldots, n\}$ and each of its nonempty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract succesive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+1=5$ and for $\{5\}$ it is simply $5$ . Find the sum of all such alternating sums for $n=7$ . | 448 | null | [
"Solution 1 Let $S$ be a non- empty subset of $\\{1,2,3,4,5,6\\}$. Then the alternating sum of $S$, plus the alternating sum of $S \\cup \\{7\\}$, is $7$. This is because, since $7$ is the largest element, when we take an alternating sum, each number in $S$ ends up with the opposite sign of each corresponding eleme... |
1983-I-14 | 1,983 | 14 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$ . [asy]size(160); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=intersectionpoints(C3,C1)[0]; pair R=intersectionpoints(C3,C2)[1]; draw(C1); draw(C2); draw(O2--O1); dot(O1); dot(O2); draw(Q--R); label("$Q$",Q,NW); label("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);[/asy] | 130 | null | [
"Firstly, notice that if we reflect $R$ over $P$, we get $Q$. Since we know that $R$ is on circle $B$ and $Q$ is on circle $A$, we can reflect circle $B$ over $P$ to get another circle (centered at a new point $C$, and with radius $6$) that intersects circle $A$ at $Q$. The rest is just finding lengths, as follows.... |
1983-I-15 | 1,983 | 15 | The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central angle of minor arc $AB$ is a rational number. If this number is expressed as a fraction $\frac{m}{n}$ in lowest terms, what is the product $mn$ ? [asy]size(140); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=1; pair O1=(0,0); pair A=(-0.91,-0.41); pair B=(-0.99,0.13); pair C=(0.688,0.728); pair D=(-0.25,0.97); path C1=Circle(O1,1); draw(C1); label("$A$",A,W); label("$B$",B,W); label("$C$",C,NE); label("$D$",D,N); draw(A--D); draw(B--C); pair F=intersectionpoint(A--D,B--C); add(pathticks(A--F,1,0.5,0,3.5)); add(pathticks(F--D,1,0.5,0,3.5)); [/asy] | 175 | null | [
"Solution 1 As with some of the other solutions, we analyze this with a locus—but a different one. We'll consider: given a point $P$ and a line $\\ell,$ what is the set of points $X$ such that the midpoint of $PX$ lies on line $\\ell$? The answer to this question is: a line $m$ parallel to $\\ell$, such that $m$ an... |
1984-I-1 | 1,984 | 1 | Find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ if $a_1$ , $a_2$ , $a_3\ldots$ is an arithmetic progression with common difference 1, and $a_1+a_2+a_3+\ldots+a_{98}=137$ . | 93 | null | [
"One approach to this problem is to apply the formula for the sum of an arithmetic series in order to find the value of $a_1$, then use that to calculate $a_2$ and sum another arithmetic series to get our answer. A somewhat quicker method is to do the following: for each $n \\geq 1$, we have $a_{2n - 1} = a_{2n} - ... |
1984-I-2 | 1,984 | 2 | The integer $n$ is the smallest positive multiple of $15$ such that every digit of $n$ is either $8$ or $0$ . Compute $\frac{n}{15}$ . | 592 | null | [
"Any multiple of 15 is a multiple of 5 and a multiple of 3. Any multiple of 5 ends in 0 or 5; since $n$ only contains the digits 0 and 8, the units digit of $n$ must be 0. The sum of the digits of any multiple of 3 must be divisible by 3. If $n$ has $a$ digits equal to 8, the sum of the digits of $n$ is $8a$. For t... |
1984-I-3 | 1,984 | 3 | A point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$ , the resulting smaller triangles $t_{1}$ , $t_{2}$ , and $t_{3}$ in the figure, have areas $4$ , $9$ , and $49$ , respectively. Find the area of $\triangle ABC$ . [asy] size(200); pathpen=black+linewidth(0.65);pointpen=black; pair A=(0,0),B=(12,0),C=(4,5); D(A--B--C--cycle); D(A+(B-A)*3/4--A+(C-A)*3/4); D(B+(C-B)*5/6--B+(A-B)*5/6);D(C+(B-C)*5/12--C+(A-C)*5/12); MP("A",C,N);MP("B",A,SW);MP("C",B,SE); /* sorry mixed up points according to resources diagram. */ MP("t_3",(A+B+(B-A)*3/4+(A-B)*5/6)/2+(-1,0.8),N); MP("t_2",(B+C+(B-C)*5/12+(C-B)*5/6)/2+(-0.3,0.1),WSW); MP("t_1",(A+C+(C-A)*3/4+(A-C)*5/12)/2+(0,0.15),ESE); [/asy] | 144 | null | [
"By the transversals that go through $P$, all four triangles are similar to each other by the $AA$ postulate. Also, note that the length of any one side of the larger triangle is equal to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity $K = \\dfrac{ab\\sin C}{2}... |
1984-I-4 | 1,984 | 4 | Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$ . However, if $68$ is removed, the average of the remaining numbers drops to $55$ . What is the largest number that can appear in $S$ ? | 649 | null | [
"Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$ We are given that \\begin{align*} \\frac{s+68}{n+1}&=56, \\\\ \\frac{s}{n}&=55. \\end{align*} Clearing denominators, we have \\begin{align*} s+68&=56n+56, \\\\ s&=55n. \\end{align*} Subtracting the equations, we get $68=n+56... |
1984-I-5 | 1,984 | 5 | Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$ . | 512 | null | [
"Use the change of base formula to see that $\\frac{\\log a}{\\log 8} + \\frac{2 \\log b}{\\log 4} = 5$; combine denominators to find that $\\frac{\\log ab^3}{3\\log 2} = 5$. Doing the same thing with the second equation yields that $\\frac{\\log a^3b}{3\\log 2} = 7$. This means that $\\log ab^3 = 15\\log 2 \\Longr... |
1984-I-6 | 1,984 | 6 | Three circles, each of radius $3$ , are drawn with centers at $(14, 92)$ , $(17, 76)$ , and $(19, 84)$ . A line passing through $(17,76)$ is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the absolute value of the slope of this line? | 24 | null | [
"The line passes through the center of the bottom circle; hence it is the circle's diameter and splits the circle into two equal areas. For the rest of the problem, we do not have to worry about that circle. Draw the midpoint of $\\overline{AC}$ (the centers of the other two circles), and call it $M$. If we draw th... |
1984-I-7 | 1,984 | 7 | The function f is defined on the set of integers and satisfies $f(n)= \begin{cases} n-3 & \mbox{if }n\ge 1000 \\ f(f(n+5)) & \mbox{if }n<1000 \end{cases}$ Find $f(84)$ . | 997 | null | [
"Define $f^{h} = f(f(\\cdots f(f(x))\\cdots))$, where the function $f$ is performed $h$ times. We find that $f(84) = f(f(89)) = f^2(89) = f^3(94) = \\ldots f^{y}(1004)$. $1004 = 84 + 5(y - 1) \\Longrightarrow y = 185$. So we now need to reduce $f^{185}(1004)$. Let’s write out a couple more iterations of this functi... |
1984-I-8 | 1,984 | 8 | The equation $z^6+z^3+1=0$ has complex roots with argument $\theta$ between $90^\circ$ and $180^\circ$ in the complex plane. Determine the degree measure of $\theta$ . | 160 | null | [
"We shall introduce another factor to make the equation easier to solve. If $r$ is a root of $z^6+z^3+1$, then $0=(r^3-1)(r^6+r^3+1)=r^9-1$. The polynomial $x^9-1$ has all of its roots with absolute value $1$ and argument of the form $40m^\\circ$ for integer $m$ (the ninth degree roots of unity). Now we simply need... |
1984-I-9 | 1,984 | 9 | In tetrahedron $ABCD$ , edge $AB$ has length 3 cm. The area of face $ABC$ is $15\mbox{cm}^2$ and the area of face $ABD$ is $12 \mbox { cm}^2$ . These two faces meet each other at a $30^\circ$ angle. Find the volume of the tetrahedron in $\mbox{cm}^3$ . | 20 | null | [
"[asy] /* modified version of olympiad modules */ import three; real markscalefactor = 0.03; path3 rightanglemark(triple A, triple B, triple C, real s=8) { triple P,Q,R; P=s*markscalefactor*unit(A-B)+B; R=s*markscalefactor*unit(C-B)+B; Q=P+R-B; return P--Q--R; } path3 anglemark(triple A, triple B, triple C, real t=... |
1984-I-10 | 1,984 | 10 | Mary told John her score on the American High School Mathematics Examination (AHSME), which was over $80$ . From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over $80$ , John could not have determined this. What was Mary's score? (Recall that the AHSME consists of $30$ multiple choice problems and that one's score, $s$ , is computed by the formula $s=30+4c-w$ , where $c$ is the number of correct answers and $w$ is the number of wrong answers. Students are not penalized for problems left unanswered.) | 119 | null | [
"Let Mary's score, number correct, and number wrong be $s,c,w$ respectively. Then \\begin{align*} s&=30+4c-w \\\\ &=30+4(c-1)-(w-4) \\\\ &=30+4(c+1)-(w+4). \\end{align*} Therefore, Mary could not have left at least five blank; otherwise, one more correct and four more wrong would produce the same score. Similarly, ... |
1984-I-11 | 1,984 | 11 | A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$ . | 106 | null | [
"First notice that there is no difference between the maple trees and the oak trees; we have only two types, birch trees and \"non-birch\" trees. (If you don't believe this reasoning, think about it. You could also differentiate the tall oak trees from the short oak trees, and the maple trees with many branches as ... |
1984-I-12 | 1,984 | 12 | A function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$ ? | 401 | null | [
"If $f(2+x)=f(2-x)$, then substituting $t=2+x$ gives $f(t)=f(4-t)$. Similarly, $f(t)=f(14-t)$. In particular, \\[f(t)=f(14-t)=f(14-(4-t))=f(t+10)\\] Since $0$ is a root, all multiples of $10$ are roots, and anything congruent to $4\\pmod{10}$ are also roots. To see that these may be the only integer roots, observe ... |
1984-I-13 | 1,984 | 13 | Find the value of $10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).$ | 15 | null | [
"Solution 1 We know that $\\tan(\\arctan(x)) = x$ so we can repeatedly apply the addition formula, $\\tan(x+y) = \\frac{\\tan(x)+\\tan(y)}{1-\\tan(x)\\tan(y)}$. Let $a = \\cot^{-1}(3)$, $b=\\cot^{-1}(7)$, $c=\\cot^{-1}(13)$, and $d=\\cot^{-1}(21)$. We have $\\tan(a)=\\frac{1}{3},\\quad\\tan(b)=\\frac{1}{7},\\quad\\... |
1984-I-14 | 1,984 | 14 | What is the largest even integer that cannot be written as the sum of two odd composite numbers? | 38 | null | [
"Take an even positive integer $x$. $x$ is either $0 \\bmod{6}$, $2 \\bmod{6}$, or $4 \\bmod{6}$. Notice that the numbers $9$, $15$, $21$, ... , and in general $9 + 6n$ for nonnegative $n$ are odd composites. We now have 3 cases: If $x \\ge 18$ and is $0 \\bmod{6}$, $x$ can be expressed as $9 + (9+6n)$ for some non... |
1985-I-1 | 1,985 | 1 | Let $x_1=97$ , and for $n>1$ let $x_n=\frac{n}{x_{n-1}}$ . Calculate the product $x_1x_2 \ldots x_8$ . | 384 | null | [
"Since $x_n=\\frac{n}{x_{n-1}}$, $x_n \\cdot x_{n - 1} = n$. Setting $n = 2, 4, 6$ and $8$ in this equation gives us respectively $x_1x_2 = 2$, $x_3x_4 = 4$, $x_5x_6 = 6$ and $x_7x_8 = 8$ so \\[x_1x_2x_3x_4x_5x_6x_7x_8 = 2\\cdot4\\cdot6\\cdot8 = 384 was completely unneeded!",
"Another way to do this is to realize... |
1985-I-2 | 1,985 | 2 | When a right triangle is rotated about one leg, the volume of the cone produced is $800\pi \;\textrm{cm}^3$ . When the triangle is rotated about the other leg, the volume of the cone produced is $1920\pi \;\textrm{cm}^3$ . What is the length (in cm) of the hypotenuse of the triangle? | 26 | null | [
"Let one leg of the triangle have length $a$ and let the other leg have length $b$. When we rotate around the leg of length $a$, the result is a cone of height $a$ and radius $b$, and so of volume $\\frac 13 \\pi ab^2 = 800\\pi$. Likewise, when we rotate around the leg of length $b$ we get a cone of height $b$ and ... |
1985-I-3 | 1,985 | 3 | Find $c$ if $a$ , $b$ , and $c$ are positive integers which satisfy $c=(a + bi)^3 - 107i$ , where $i^2 = -1$ . | 198 | null | [
"Expanding out both sides of the given equation we have $c + 107i = (a^3 - 3ab^2) + (3a^2b - b^3)i$. Two complex numbers are equal if and only if their real parts and imaginary parts are equal, so $c = a^3 - 3ab^2$ and $107 = 3a^2b - b^3 = (3a^2 - b^2)b$. Since $a, b$ are integers, this means $b$ is a divisor of 10... |
1985-I-4 | 1,985 | 4 | A small square is constructed inside a square of area $1$ by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices, as shown in the figure. Find the value of $n$ if the the area of the small square is exactly $\frac1{1985}$ . AIME 1985 Problem 4.png | 32 | null | [
"The lines passing through $A$ and $C$ divide the square into three parts, two right triangles and a parallelogram. Using the smaller side of the parallelogram, $1/n$, as the base, where the height is 1, we find that the area of the parallelogram is $A = \\frac{1}{n}$. By the Pythagorean Theorem, the longer base of... |
1985-I-5 | 1,985 | 5 | A sequence of integers $a_1, a_2, a_3, \ldots$ is chosen so that $a_n = a_{n - 1} - a_{n - 2}$ for each $n \ge 3$ . What is the sum of the first $2001$ terms of this sequence if the sum of the first $1492$ terms is $1985$ , and the sum of the first $1985$ terms is $1492$ ? | 986 | null | [
"The problem gives us a sequence defined by a recursion, so let's calculate a few values to get a feel for how it acts. We aren't given initial values, so let $a_1 = a$ and $a_2 = b$. Then $a_3 = b - a$, $a_4 = (b - a) - b = -a$, $a_5 = -a - (b - a) = -b$, $a_6 = -b - (-a) = a - b$, $a_7 = (a - b) - (-b) = a$ and $... |
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