2007 AIME II Problems/Problem 3

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

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 $EF^2 = (17\sqrt{2})^2 = 578$ .

2007 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2007 AIME II Problems/Problem 3

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