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Geometry Difficulty 4.3 AIME Find the answer United States

In square $ABCD$ , points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$ , respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$ , with $BR = 6$ and $PR = 7$ . What is the area of the square?

Pick one

Solution

Because $\angle RBC$ is complementary to both $\angle RCB$ and $\angle PBA$ , those two angles are congruent. Therefore $\triangle BAP \cong \triangle CBQ$ by ASA, so $CQ = BP = 13$ . Let $d = CR$ ; then $QR = 13 - d$ , so the Altitude-to-Hypotenuse Theorem yields $6^2 = d(13 - d)$ , which has solutions $d = 4$ and $d = 9$ . Because $QR < RC$ , in fact $d = 9$ . It follows that the area of the square is
$BC^2 = BR^2 + RC^2 = 6^2 + 9^2 = 117.$

Let $c = BQ$ and $s = AB$ . Because $\triangle BRQ$ is similar to $\triangle BAP$ , it follows that $\frac{6}{c} = \frac{s}{13}$ , so $cs = 78$ . As before, $CQ = BP = 13$ , so by the Pythagorean Theorem, $s^2 + c^2 = 169$ . Then $(s+c)^2 = 169 + 2 \cdot 78 = 325$ , so $s + c = 5\sqrt{13}$ . Similarly $(s-c)^2 = 169 - 2 \cdot 78 = 13$ , so $s - c = \sqrt{13}$ . Solving this system of equations yields $s = 3\sqrt{13}$ , and the area of the square is $s^2 = 117$ .

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