2007 AIME II Problems/Problem 3

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Problem

Square $ABCD$ has side length $13$ , and points $E$ and $F$ are exterior to the square such that $BE=DF=5$ and $AE=CF=12$ . Find $\displaystyle EF^{2}$ .

Solution

Solution 1

Extend $\overline{AE}, \overline{DF}$ and $\overline{BE}, \overline{CF}$ to their points of intersection. Since $\triangle ABE \cong \triangle CDF$ and are both $5-12-13$ right triangles, we can come to the conclusion that the two new triangles are also congruent to these two (use ASA, as we know all the sides are $13$ and the angles are mostly complementary). Thus, we create a square with sides $5 + 12 = 17$ .

$\overline{EF}$ is the diagonal of the square, with length $17\sqrt{2}$ ; the answer is $EF^2 = (17\sqrt{2})^2 = 578$ .

Solution 2

A slightly more analytic/brute-force approach:

Drop perpendiculars from $E$ and $F$ to $I$ and $J$ , respectively; construct right triangle $EKF$ with right angle at K and $EK || BC$ . Since $2[CDF]=DF*CF=CD*JF$ , we have $JF=5\times12/13 = \frac{60}{13}$ . Similarly, $EI=\frac{60}{13}$ . Since $\triangle DJF \sim \triangle DFC$ , we have $DJ=\frac{5JF}{12}=\frac{25}{13}$ .

Now, we see that $FK=DC-(DJ+IB)=DC-2DJ=13-\frac{50}{13}=\frac{119}{13}$ . Also, $EK=BC+(JF+IE)=BC+2JF=13+\frac{120}{13}=\frac{289}{13}$ . By the Pythagorean Theorem, we have $EF=\sqrt{\left(\frac{289}{13}\right)^2+\left(\frac{119}{13} \right)^2}=\frac{\sqrt{(17^2)(17^2+7^2)}}{13}$ $=\frac{17\sqrt{338}}{13}=\frac{17(13\sqrt{2})}{13}=17\sqrt{2}$ . Therefore, $EF^2=(17\sqrt{2})^2=578$ .

2007 AIME II (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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