2001 AIME II Problems/Problem 6

Problem

Square $ABCD$ is inscribed in a circle. Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. If the area of square $ABCD$ is $1$ , then the area of square $EFGH$ can be expressed as $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m < n$ . Find $10n + m$ .

Solution 1(Pythagorean Theorem)

Let $O$ be the center of the circle, and $2a$ be the side length of $ABCD$ , $2b$ be the side length of $EFGH$ . By the Pythagorean Theorem, the radius of $\odot O = OC = a\sqrt{2}$ .

$[asy] size(150); pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("4 4") + blue + linewidth(0.7); pair C=(1,1), D=(1,-1), B=(-1,1), A=(-1,-1), E= (1, -0.2), F=(1, 0.2), G=(1.4, 0.2), H=(1.4, -0.2); D(MP("A",A)--MP("B",B,N)--MP("C",C,N)--MP("D",D)--cycle); D(MP("E",E,SW)--MP("F",F,NW)--MP("G",G,NE)--MP("H",H,SE)--cycle); D(CP(D(MP("O",(0,0))), A)); D((0,0) -- (2^.5, 0), d); D((0,0) -- G -- (G.x,0), d); [/asy]$

Now consider right triangle $OGI$ , where $I$ is the midpoint of $\overline{GH}$ . Then, by the Pythagorean Theorem,

$\begin{align*} OG^2 = 2a^2 &= OI^2 + GI^2 = (a+2b)^2 + b^2 \\ 0 &= a^2 - 4ab - 5b^2 = (a - 5b)(a + b) \end{align*}$

Thus $a = 5b$ (since lengths are positive, we discard the other root). The ratio of the areas of two similar figures is the square of the ratio of their corresponding side lengths, so $\frac{[EFGH]}{[ABCD]} = \left(\frac 15\right)^2 = \frac{1}{25}$ , and the answer is $10n + m = \boxed{251}$ .

Another way to proceed from $0 = a^2 - 4ab - 5b^2$ is to note that $\frac{b}{a}$ is the quantity we need; thus, we divide by $a^2$ to get

$0 = 1 - 4\left(\frac{b}{a}\right) - 5\left(\frac{b}{a}\right)^2$ This is a quadratic in $\frac{b}{a}$ , and solving it gives $\frac{b}{a} = \frac{1}{5},-1$ . The negative solution is extraneous, and so the ratio of the areas is $\left(\frac{1}{5}\right)^2 = \frac{1}{25}$ and the answer is $10\cdot 25 + 1 = \boxed{251}$ .

Solution 2 (Coordinates)

Let point $A$ be the top-left corner of square $ABCD$ and the rest of the vertices be arranged, in alphabetical order, in a clockwise arrangement from there. Let $D$ have coordinates $(0,0)$ and the side length of square $ABCD$ be $a$ . Let $DF$ = $b$ and diameter $HI$ go through $J$ the midpoint of $EF$ . Since a diameter always bisects a chord perpendicular to it, $DJ$ = $JC$ and since $F$ and $E$ must be symmetric around the diameter, $FJ = JE$ and it follows that $DF = EC = b.$ Hence $FE$ the side of square $EFGH$ has length $a - 2b$ . $F$ has coordinates $(b,0)$ and $G$ has coordinates $(b, 2b - a).$ We know that point $G$ must be on the circle $O$ - hence it must satisfy the circle equation. Since the center of the circle is at the center of the square $(a/2, a/2)$ and has radius $a *$ $\sqrt{2} / 2$ , half the diagonal of the square, $(x - a/2)^2 + (y - a/2)^2 = 1/2a^2$ follows as the circle equation. Then substituting coordinates of $G$ into the equation, $(b - a/2)^2 + (2b - a - a/2)^2 = a^2/2$ . Simplifying and factoring, we get $2a^2-7ab+5b^2 = (2a-5b)(a-b) = 0.$ Since $a = b$ would imply $m = n$ , and $m < n$ in the problem, we must use the other factor. We get $b = 2/5a$ , meaning the ratio of areas $((a-2b)/a)^2$ = $(1/5)^2$ = $1/25$ = $m/n.$ Then $10n + m = 25 * 10 + 1 = \boxed{251}$ .

2001 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2001 AIME II Problems/Problem 6

Contents

Problem

Solution 1(Pythagorean Theorem)

Solution 2 (Coordinates)

See also