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Problem 481

AMC 12 late, AIME early

Geometry Difficulty 4.8 Prove it HMMT November 2018 · United States · 2018

Problem:

Let $ABCD$ be a convex quadrilateral so that all of its sides and diagonals have integer lengths. Given that $\angle ABC = \angle ADC = 90^{\circ}$ , $AB = BD$ , and $CD = 41$ , find the length of $BC$ .

Proposed by: Anders Olsen

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Solution:

Let the midpoint of $AC$ be $O$ which is the center of the circumcircle of $ABCD$ . $ADC$ is a right triangle with a leg of length $41$ , and $41^{2} = AC^{2} - AD^{2} = (AC - AD)(AC + AD)$ . As $AC, AD$ are integers and $41$ is prime, we must have $AC = 840$ , $AD = 841$ . Let $M$ be the midpoint of $AD$ . $\triangle AOM \sim \triangle ACD$ , so $BM = BO + OM = 841/2 + 41/2 = 441$ . Then $AB = \sqrt{420^{2} + 441^{2}} = 609$ (this is a $20$ - $21$ - $29$ triangle scaled up by a factor of $21$ ). Finally, $BC^{2} = AC^{2} - AB^{2}$ so $BC = \sqrt{841^{2} - 609^{2}} = 580$ .

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.