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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 1a (Pythagorean theorem)

Let $O$ be the center of the circle, $2a$ be the side length of square $ABCD$, and $2b$ be the side length of square $EFGH$. By symmetry, the horizontal and vertical displacements of $C$ from $O$ are both $\frac{2a}{2} = a$, so by the Pythagorean theorem, the radius of the circle is $OC = \sqrt{a^2+a^2} = a\sqrt{2}$.

[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 let $I$ be the midpoint of $\overline{GH}$, giving $\overline{GI} = \frac{2b}{2} = b$, and let $J$ be the point where $\overline{OI}$ intersects $\overline{CD}$. We observe that since a diameter bisects a chord perpendicular to it, $\overline{GH}$ must be perpendicular to the diameter passing through $I$. This means that triangle $OGI$ has a right angle at $I$, and that $\overline{OJ}$ and $\overline{JI}$ are both parallel to $\overline{BC}$ and $\overline{FG}$. As the horizontal displacement of $C$ from $O$ is $a$ (from above), it follows that $\overline{OJ} = a$, and hence \[\overline{OI} = \overline{OJ}+\overline{JI} = a+\overline{FG} = a+2b,\] so by the Pythagorean Theorem again,

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

Since lengths must be positive, we clearly cannot have $a+b = 0$, so the solution must instead be $a = 5b$. Since the ratio of the areas of two similar figures is the square of the ratio of their corresponding side lengths, we deduce that the required ratio is $\left(\frac{1}{5}\right)^2 = \frac{1}{25}$, and so the answer is $10 \cdot 25 + 1 = \boxed{251}$.

Solution 1b (alternative finish for Solution 1a)

After establishing that $a^2-4ab-5b^2 = 0$, another way to proceed is to observe that what we actually need to calculate is $\frac{b}{a}$. Accordingly, we divide both sides of this equation by $a^2$ (or equivalently, choose units such that the area of square $ABCD$ is $1$, without loss of generality), giving \[1-4\left(\frac{b}{a}\right)-5\left(\frac{b}{a}\right)^2 = 0,\] which is a quadratic in precisely the variable $\frac{b}{a}$. Thus, by solving it, we immediately obtain $\frac{b}{a} = \frac{1}{5}$ or $\frac{b}{a} = -1$, and as in Solution 1a, the negative root is obviously extraneous. Hence the required ratio of areas is $\left(\frac{1}{5}\right)^2 = \frac{1}{25}$, and so the answer is $10 \cdot 25 + 1 = \boxed{251}$.

Solution 2 (coordinates)

Let $A$ be the top-left vertex of square $ABCD$, and label the rest of the vertices in alphabetical order going clockwise from $A$. Let $D$ have coordinates $(0,0)$ and the side length of square $ABCD$ be $a$. Let $\overline{DF} = b$, the midpoint of $\overline{EF}$ be $J$, and $\overline{PQ}$ be the diameter of the circle that passes through $J$. Since a diameter bisects a chord perpendicular to it, we deduce that $\overline{CJ} = \overline{DJ}$, while $\overline{JE} = \overline{JF}$ by the definition of the midpoint, so $\overline{CE} = \overline{CJ}+\overline{JE} = \overline{DJ}+\overline{JF} = \overline{DF} = b$. It follows that \[2b = \overline{CE}+\overline{DF} = \left(\overline{CF}+\overline{FE}\right)+\left(\overline{DE}+\overline{FE}\right) = \left(\overline{CF}+\overline{DE}\right)+2\overline{FE} = \left(\overline{CD}-\overline{FE}\right)+2\overline{FE} = \overline{CD}+\overline{FE} = a+\overline{FE},\] so the side length of square $EFGH$ is $\overline{FE} = 2b-a$, and as $F$ has coordinates $(b,0)$, $G$ therefore has coordinates $(b,2b-a)$.

Now, by symmetry, 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 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 $b = \overline{DF} < \overline{DC} = a$, we cannot have $a = b$, so the solution must instead be $b = \frac{2}{5}a$. Thus the required 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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