AIME 2001 II · 第 6 题

Solution 1

Begin by drawing another square congruent to $EFGH$ but on side $AB$ instead of side $CD$. Call this square $IJKL$, such that $I$ lies on $\overline{AB}$ and $\overline{IJ}$ lies on line $FG$. Let the side length of $EFGH$ and $IJKL$ be $s$ and the side length of $ABCD$ be $1$.

By Power of a Point on point $F$, we find that

$$FC \cdot FD = FG \cdot FJ$$ $$\frac{1-s}{2} \cdot \frac{1+s}{2} = s(1+s)$$ Expanding and simplifying we get the quadratic $5s^2 +4s-1 = 0$. Factoring, we get $(5s-1)(s+1) = 0$. Therefore, $s= \frac{1}{5}$ and the ratio of the squares' areas is $\frac{1}{25}$. Our answer is therefore $10 \cdot 25 + 1 = \boxed{251}$.

~lprado

Solution 2

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 $OC = \sqrt{a^2+a^2} = a\sqrt{2}$.

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,

$$\begin{aligned} \overline{OG}^2 = \overline{OI}^2 + \overline{GI}^2 &\iff \left(a\sqrt{2}\right)^2 = (a+2b)^2 + b^2 \\ &\iff 2a^2 = a^2+4ab+4b^2+b^2 \\ &\iff a^2-4ab-5b^2 = 0 \\ &\iff (a-5b)(a+b) = 0 \end{aligned}$$ 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 $10 \cdot 25 + 1 = \boxed{251}$.

Alternative Finish for Solution 1:

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 3 (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 $10 \cdot 25 + 1 = \boxed{251}$.