HMMT, 2013

Geometry Difficulty 5.6 AIME, harder Prove it United States

Problem:

Pentagon $A B C D E$ is given with the following conditions:
(a) $\angle C B D + \angle D A E = \angle B A D = 45^{\circ}, \angle B C D + \angle D E A = 300^{\circ}$
(b) $\frac{B A}{D A} = \frac{2 \sqrt{2}}{3},\ C D = \frac{7 \sqrt{5}}{3}$ , and $D E = \frac{15 \sqrt{2}}{4}$
(c) $A D^{2} \cdot B C = A B \cdot A E \cdot B D$

Compute $B D$ .

Solution

Solution:

Answer: $\sqrt{\sqrt{39}}$ As a preliminary, we may compute that by the law of cosines, the ratio $\frac{A D}{B D} = \frac{3}{\sqrt{5}}$ . Now, construct the point $P$ in triangle $A B D$ such that $\triangle A P B \sim \triangle A E D$ . Observe that $\frac{A P}{A D} = \frac{A E \cdot A B}{A D \cdot A D} = \frac{B C}{B D}$ (where we have used first the similarity and then condition 3). Furthermore, $\angle C B D = \angle D A B - \angle D A E = \angle D A B - \angle P A B = \angle P A D$ so by SAS, we have that $\triangle C B D \sim \triangle P A D$ .

Therefore, by the similar triangles, we may compute $P B = D E \cdot \frac{A B}{A D} = 5$ and $P D = C D \cdot \frac{A D}{B D} = 7$ . Furthermore, $\angle B P D = 360 - \angle B P A - \angle D P A = 360 - \angle B C D - \angle D E A = 60$ and therefore, by the law of cosines, we have that $B D = \sqrt{39}$ .

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