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2001 AIME II Problems/Problem 6

Revision as of 13:12, 3 June 2025 by Sevenoptimus (talk | contribs) (Improved Solution 2)

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 $10\cdot 25 + 1 = \boxed{251}$.

Remark: The division by $a^2$ is equivalent to simply setting the original area of square $ABCD$ to 1.

Solution 2 (Coordinates)

Let point $A$ be the top-left corner of square $ABCD$, and the rest of the vertices be labelled clockwise from $A$ in alphabetical order. Let $D$ have coordinates $(0,0)$ and the side length of square $ABCD$ be $a$. Let $DF = b$, and diameter $HI$ pass through $J$, the midpoint of $EF$. Since a diameter always bisects a chord perpendicular to it, we deduce that $DJ$ = $JC$, while as $F$ must be the reflection of $E$ in the diameter, we also have $FJ = JE$, which gives $DF = DJ+FJ = JC+JE = EC = b$. Hence the side length of square $EFGH$ is $FE = a - 2b$, so since $F$ has coordinates $(b,0)$, $G$ has coordinates $(b,2b-a)$.

Now, the center of the circle is the same as the center of the square, i.e. $\left(\frac{a}{2},\frac{a}{2}\right)$, and so its radius is half of the square's diagonal, i.e. $\frac{a\sqrt{2}}{2}$. This means the equation of the circle $O$ is \[\left(x-\frac{a}{2}\right)^2 + \left(y-\frac{a}{2}\right)^2 = \left(\frac{a\sqrt{2}}{2}\right)^2 = \frac{a^2}{2},\] and as $G$ lies on the circle, its coordinates must satisfy this equation, yielding \[\left(b-\frac{a}{2}\right)^2 + \left(\left(2b-a\right)-\frac{a}{2}\right)^2 = \frac{a^2}{2}.\] Upon simplifying, this becomes $2a^2-7ab+5b^2 = 0$, which factors as $(2a-5b)(a-b) = 0$. Recalling that $b = DF < DC = a$, we cannot have $a = b$, so the solution must instead be given by the other factor, i.e. $b = \frac{2}{5}a$. Thus the required the ratio of areas is \[\left(\frac{a-2b}{a}\right)^2 = \left(1-2\cdot\frac{b}{a}\right)^2 = \left(1-2\cdot\frac{2}{5}\right)^2 = \left(\frac{1}{5}\right)^2 = \frac{1}{25},\] so the answer is $10 \cdot 25 + 1 = \boxed{251}$.

See also

2001 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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