\documentclass[12pt]{article} \addtolength{\voffset}{-.5in} \addtolength{\textheight}{1in} \addtolength{\hoffset}{-.5in} \addtolength{\textwidth}{1in} \begin{document} \pagestyle{empty} \begin{center} Western Mass.\ ARML Summer Homework \#2 SOLUTIONS / ANSWERS \end{center} \begin{enumerate} \item[] {\bf The mandatory questions:} \item[21.] The numbers $a$ and $b$ are both perfect squares, and are both represented by four-digit decimal numerals. The digits of these two numerals are the same but in the reverse order. If the square root of $a$ divides the square root of $b$, find $a$. {\bf Answer:} The divisibility says that $\frac{\sqrt{b}}{\sqrt{a}}=k$, for some $k$, so $b=k^2a$. Since $a$ and $b$ are both 4-digit squares, $a$ must be at least 1000 ($32^2=1024$, actually) and $b$ can be at most 9999 ($99^2=9801$, actually). Since the biggest possible ratio is just over 9, $k^2$ can only be 4 or 9 (or 1, if the square is a palindrome. There are no palindromic 4-digit perfect squares, but I don't have a nice proof other than checking them all). Say $k^2=4$. We have $b=4a$, which means the last digit of $b$ is even, which means the first digit of $a$ is even. (Remember, the first digit of $a$ is the last digit of $b$.) If $a$ were 4xxx, 6xxx, or 8xxx, $4a$ would be more than four digits. So the first digit of $a$ must be 2. So the last digit of $b$ is 2. But $b$ is a perfect square, and no perfect square ends in 2. ($0^2\to0$, $1^2\to1$, $2^2\to4$, $3^2\to9$, $4^2\to6$, $5^2\to5$, $6^2\to6$, $7^2\to9$, $8^2\to4$, $9^2\to1$). This is a contradiction, so $k^2 \neq 4$. Say $k^2=9$. We have $b=9a$. $a$ must be at least $1024=32^2$. Since the biggest four-digit square is 9801, $a$ can be at most $9801/9=1089$. Well, there it is. You could also say (here I'm reusing variables) $$9000d+900c+90b+9a = 1000a+100b+10c+d$$ $$8999d+890c-10b-991a = 0$$ Using modular arithmetic, we see $$8999d+890c-10b-991a \equiv 0 \pmod{10}$$ $$9d-a \equiv 0 \pmod{10}$$ $$9d \equiv a \pmod{10}$$ So either $a=d=0$ or $d=1$ and $a=9$. These are four digit numbers, so choose the latter. Then we have $$9000+900c+90b+81 = 9000 + 100b + 10c + 1$$ $$890c-10b+80=0$$ $$89c-b+8=0$$ Since $b$ and $c$ are digits, if $c>0$ we can't make this true. So $c=0$. Then $-b+8=0$, so $b=8$. This gives you the same answer, \fbox{1089.} \newpage \item[5.] The $n$th term, $a_n$, of a sequence of numbers is given by $a_1=1$ and $a_n=a_{n-1}+2n$ for $n>1$. Write an equation expressing $a_n$ as a polynomial in $n$. {\bf Answer:} Under a time constraint, the first thing to do is to find some terms of the sequence. You pretty quickly find that the terms are 1, 5, 11, 19, 29, 41, 55, 71, \ldots. Use what you learned from Dr.~Little in May about finite differences. First, you need to start at the zeroth term. Since the differences you see are 4, 6, 8, 10, etc., the previous difference was 2. So the zeroth term is $1-2=-1$. Set up a table of successive differences: \begin{tabular}{ccccccccccccccccccc} $-1$ && 1 && 5 && 11 && 19 && 29 && 41 && 55 && 71 & \\ & 2 && 4 && 6 && 8 && 10 && 12 && 14 && 16 && \\ && 2 && 2 && 2 && 2 && 2 && 2 && 2 &&&& \\ \end{tabular} Here's the powerful result: a closed form for the $n$th term is $$-1{n \choose 0} + 2{n \choose 1} + 2{n \choose 2}$$ $$-1 + 2n + 2\frac{n(n-1)}{2}$$ $$-1 + 2n + n^2-n$$ $$\fbox{$n^2+n-1$}$$ If you don't remember, the $-1$, 2, and 2 come from the first diagonal in your table of differences. \newpage \item[20.] Two cardboard rectangles each have dimensions 2 and 8. They are placed on a table so that one pair of diagonally opposite corners coincides, but the other pair of diagonally opposite corners does not coincide. Find the area of the region in which the rectangles overlap. {\bf Answer:} If you draw a diagram, you should see pretty quickly that you have a rhombus. Let the length of a side of the rhombus be $a$, and what's left over is obviously $8-a$. Because you have overlapping {\em rectangles\/}, you now have four right triangles, with legs 2 and $8-a$, and hypotenuse $a$. So $$2^2 + (8-a)^2 = a^2$$ $$4 + 64 - 16a + a^2 = a^2$$ $$68 = 16a$$ So $a = \frac{68}{16} = \frac{17}{4}$ and the remaining piece is $\frac{15}{4}$. {\bf Method 1:} The area of the shaded region is the area of both rectangles, minus the four triangles, divided by 2. $$A = \frac{2 \cdot 16 - 4\left(\frac{1}{2} \cdot 2 \cdot \frac{15}{4}\right)}{2}$$ $$A = \frac{32 - 15}{2}$$ $$A = \fbox{$\frac{17}{2}$}$$ {\bf Method 2:} Use the formula for area of a rhombus. Draw both diagonals of the rhombus. You get four congruent right triangles. The long diagonal of the rhombus is $\sqrt{4+64}=\sqrt{68}=2\sqrt{17}$, so the longer leg of these triangles is $\sqrt{17}$. The hypotenuse, as above, is $\frac{17}{4}$. The other leg of each triangle, $x$, satisfies $$\sqrt{17}^2 + x^2 = \left(\frac{17}{4}\right)^2$$ $$17 + x^2 = \frac{17^2}{16}$$ $$x^2 = \frac{17^2}{16}-\frac{17\cdot16}{16}$$ $$x^2 = \frac{17}{16}$$ $$x = \frac{\sqrt{17}}{4}$$ So the area of the rhombus is $$A=4 \cdot \frac{1}{2} \cdot \sqrt{17} \cdot \frac{\sqrt{17}}{4}$$ $$A=\frac{1}{2} \cdot 17$$ $$A = \fbox{$\frac{17}{2}$}$$ \newpage \item[15.] Find the sum of the seventeenth powers of the 17 roots of the equation $x^{17} - 3x + 1 = 0$. {\bf Answer:} The first thing you should see is that the sum of the 17 roots is 0, because the coefficient of the $x^{16}$ term is 0. If the 17 roots are $x_1$, $x_2$, \ldots, $x_{17}$, we want to know what $x_1^{17} + x_2^{17} + \cdots + x_{17}^{17}$ is. But notice that $x^{17}=3x-1$. So, for example, $x_1^{17}=3x_1-1$. So\ldots $$x_1^{17} + x_2^{17} + \cdots + x_{17}^{17}$$ $$(3x_1 - 1) + (3x_2 - 1) + \cdots + (3x_{17}-1)$$ $$3(x_1 + x_2 + \cdots + x_{17}) - 17$$ But the sum of the roots is zero: $$3(0)-17$$ $$\fbox{$-17$}$$ \newpage \item[29.] One hundred pennies are arranged in seven stacks, of which no two stacks contain the same number of pennies. A student counts the number of pennies in each stack and takes 50 pennies in such a way as to disturb the fewest number of stacks. He ends up taking pennies from $N$ stacks. For all such arrangements of pennies, what is the largest possible value of $N$ that will be necessary? {\bf Answer:} The least disturbance is caused by taking all pennies from the biggest pile, then the next biggest, and so on. So to force the disturbance of the most pennies, you need the largest stacks to be as small as possible, and you need the smallest stacks to be as large as possible, to compensate. $100/7=14.2\ldots$, so you should center your stacks around 14. $11 + 12 + 13 + 14 + 15 + 16 + 17 = 98$. Maybe make the 16 an 18. Then your three biggest stacks add up to $15+17+18=50$ exactly. So the largest possible value of $N$ is at least 3. Could it be four? To be four, your three largest piles must sum to 49 or less (forcing you to touch a fourth pile) and your four smallest piles must sum to 51 or more. Call your smallest large pile $a$ and your largest small pile $b$. $a>b$ by definition ($a$ is one of the three largest piles, $b$ one of the four smallest, piles can't have the same number of pennies). Consider your smallest large pile, $a$. If it is 16, our large piles must be at least $16+17+18=51$, too big. So our smallest large pile is 15 or less, to keep us under the limit of 49. Consider your largest small pile, $b$. If it is 14, our small piles can be at most $11+12+13+14=50$, too small. So our largest small pile is 15 or more, to keep us over the limit of 51. This gives us $15 \geq a>b \geq 15$, a contradiction. So \fbox{$N=3$} is the maximum value of $N$. \newpage \item[] {\bf The optional questions.} \item[1.] Instead of finding twice the square of a number, a student found twice the square root of that number and got 10 as her answer. What is the correct answer? $\left(\mbox{1250}\right)$ \item[2.] Diameter $\overline{AB}$ of a circle whose center is $O$ is extended past $A$ to point $P$, and tangent segment $\overline{PT}$ is drawn. On $\overline{AO}$ as diameter, a small semicircle is drawn on the same side of $\overline{AB}$ as point $T$. If $\overline{OT}$ intersects the smaller semicircle at $X$ and $m\angle TPB=40^\circ$, find the degree measure of arc $OX$. $\left(\mbox{$80^\circ$}\right)$ \item[3.] In triangle $ABC$, $AC=6$, $BC=8$, and $AB=10$. Squares $ACXY$ and $BCWZ$ are drawn exterior to the triangle (the triangle has no interior point in common with either of the squares). Find the distance between the centers of the two squares. $\left(\mbox{$7\sqrt{2}$}\right)$ \item[4.] If $a=\log_{10}(\sqrt{13}+\sqrt{3})$, express in terms of $a$ the value of $\log_{10}(\sqrt{13}-\sqrt{3})$. $\left(\mbox{$1-a$}\right)$ \item[6.] In parallelogram $PQRS$, point $X$ is on $\overline{PQ}$ and $PX:PQ=1:3$. Point $Y$ is on $\overline{PS}$ and $PY:PS=1:4$. The line $XY$ intersects diagonal $\overline{PR}$ in $Z$. Find the numerical value of the ratio $PZ:PR$. $\left(\mbox{$\frac{1}{7}$}\right)$ \item[7.] The number $N$ has three digits when written in base ten notation. Its cube root is the sum of its three digits. Find $N$. $\left(\mbox{512}\right)$ \item[8.] A train leaves the Jerome Avenue subway terminal every 12 minutes. These trains travel down the Jerome branch line, then down the Lexington Avenue trunk line. A train leaves the White Plains Road terminal every 6 minutes. These trains travel down the White Plains branch line, then also join the Lexington Avenue trunk line. All the trains are scheduled so that there are constant intervals between them as they run down Lexington Avenue. How many minutes are there in this interval? $\left(\mbox{4 or 4 minutes}\right)$ \item[9.] Find the largest integer smaller than $\log_4 9 + \log_9 28$. $\left(\mbox{3}\right)$ \item[10.] In equilateral triangle $ABC$, $AB=12$. One vertex of a square is at the midpoint of $\overline{AB}$, and the two {\em adjacent\/} vertices are on the other two sides of the triangle. The length of a side of the square may be expressed as $p\sqrt{2} + q\sqrt{6}$ where $p$ and $q$ are rational numbers. Find the ordered pair $(p,q)$. $\left(\mbox{$(9,-3)$}\right)$ \item[11.] If $0 < x < \frac{\pi}{4}$ and $\cos x + \sin x = \frac{5}{4}$, find the numerical value of $\cos x - \sin x$. $\left(\mbox{$\frac{\sqrt{7}}{4}$}\right)$ \item[12.] Find the least positive remainder when the integer $2^0 + 0^2 + 2^1 + 1^2 + 2^2 + 2^2 + 2^3 + 3^2 + 2^4 + 4^2 + \cdots + 2^{100} + 100^2$ is divided by 8. $\left(\mbox{5}\right)$ \item[13.] The number $N$ is represented by a three-digit base ten numeral. $N$ is equal to the cube of its units digit, and is also equal to the square of a two-digit numeral formed by its other two digits. Find $N$. $\left(\mbox{729}\right)$ \item[14.] In triangle $ACB$, $AC=7$ and $BC=5$. Squares $ACXY$ and $BCWZ$ are drawn exterior to the triangle (the triangle has no interior point in common with either square). Find the numerical value of $AB^2 + XW^2$. $\left(\mbox{148}\right)$ \item[16.] A train leaves Main Street for Times Square every 15 minutes, starting at 8 A.M. A train also leaves Times Square for Main Street on the same schedule. All trains run at a constant speed along the same tracks, and the trip either way takes 90 minutes. Including the trains it meets at each terminal, how many trains will the one leaving Main Street at noon encounter on its trip to Times Square? $\left(\mbox{13}\right)$ \item[17.] Altitude $\overline{CD}$ to hypotenuse $\overline{AB}$ of right triangle $ABC$ is a diameter of circle $O$. This circle intersects $\overline{AC}$ in $E$ and $\overline{BC}$ in $F$. If $AC=9$ and $BC=12$, find $EF$. $\left(\mbox{$\frac{36}{5}$}\right)$ \item[18.] Find the degree-measure of all angles $x$ such that $0^\circ \leq x \leq 180^\circ$ and $$\cos^6 x - \sin^6 x + \frac{(\sin^2 2x)(\cos 2x)}{4}=0.$$ $\left(\mbox{45, 135}\right)$ \item[19.] $AB$ and $CA$ are decimal numerals, and $A$, $B$, $C$ are distinct digits. If four times $AB$ equals $CA$, find the ordered triple $(A,B,C)$. $\left(\mbox{$(2,3,9)$}\right)$ \item[22.] In trapezoid $ABCD$, the ratio of the base $AB$ to base $CD$ is $2:3$. Diagonals $\overline{AC}$ and $\overline{BD}$ intersect in $X$, and the line through $X$ parallel to $\overline{AB}$ intersects $\overline{AD}$ at $P$. Find the ratio of the area of triangle $PAX$ to the area of triangle $ABD$. $\left(\mbox{$6:25$}\right)$ \item[23.] From a point $P$ on the hyperbola with equation $x^2-y^2=9-4y$, tangents are drawn to the circle centered at the origin with radius 1. What is the minimal possible length of these tangents? $\left(\mbox{$\sqrt{6}$}\right)$ \item[24.] Walking along a bus line, a student found that a bus caught up with her (going in the same direction) every 12 minutes and a bus passed her (going in the opposite direction) every 4 minutes. The student travels at a constant rate, and all buses travel at the same constant rate as one another. The buses leave their terminals at equally spaced intervals of time. How long, in minutes, is each such interval? $\left(\mbox{6 minutes}\right)$ \item[25.] A boy has as many brothers as he has sisters. His sister has twice as many brothers as she has sisters. How many children are in this family? $\left(\mbox{7}\right)$ \item[26.] Radii $\overline{OA}$ and $\overline{OB}$ of circle $O$ meet at an angle of $120^\circ$. A square has one vertex on $\overline{OA}$, another on $\overline{OB}$, and two more on minor arc $AB$. If $OA=13$, the area of the square can be written as $p+q\sqrt{3}$, where $p$ and $q$ are rational. Find the ordered pair $(p,q)$. $\left(\mbox{$(156,-39)$}\right)$ \item[27.] If $r_1$, $r_2$, \ldots, $r_7$ are the roots of $2x^7 - x^6 - 5x^5 + 17x^4 - 419x^2 - 372x - 10 = 0$, find $(1+r_1)(1+r_2)\cdots(1+r_7)$. $\left(\mbox{19}\right)$ \item[28.] Find all ordered pairs $(x,y)$ of {\em positive\/} integers satisfying $$x=\frac{6-x}{y^2-x}.$$ $\left(\mbox{$(2,2)$, $(3,2)$}\right)$ \item[30.] A right angle has its vertex at the centroid of an equilateral triangle of side 1 unit. Find the maximum possible area interior to both the angle and the triangle. $\left(\mbox{$\frac{1}{3(1+\sqrt{3})}$ or $\frac{\sqrt{3}-1}{6}$ or equivalent}\right)$ \end{enumerate} \end{document}