Talk:2021 AIME I Problems/Problem 9

Based on Solution 1:

Note: Instead of solving the system of equations (1)(2) which can be time consuming, by noting that $\triangle ACF \sim \triangle ABG$ by AA, we could find out $\frac{AB}{AG} = \frac{AC}{AF}$ , which gives $AC = \frac{9}{5}x$ . We also know that $EB = \sqrt{x^2 - 15^2}$ by Pythagorean Theorem on $\triangle ABE$ . $BC = AD = \frac{6}{5}x$ . Then using Pythagorean Theorem on $\triangle ACE$ we obtain: $AC^2 = (EB+BC)^2 + AE^2$ substituting, we get: $\frac{81}{25}x^2 = (\sqrt{x^2 -225}+\frac{6}{5}x)^2+225 \iff x = 3\sqrt{x^2 - 15^2}$ Finally, we solve to obtain $x = \frac{45\sqrt{2}}{4}$ .

~Chupdogs

However, I don't think $\triangle ACF \sim \triangle ABG$ is convincing. How do you find that by AA?

One right angle is shared and $\angle ABD = \angle BDC = \angle ACD$ by alternate angles theorem (for first part) and symmetry for the second part as its an isosceles trapezoid ( $BD$ and $AC$ are both diagonals and the angles it makes with the base edge is the same)

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Talk:2021 AIME I Problems/Problem 9