HMMT 二月 2024 · 几何 · 第 2 题

HMMT February 2024 — Geometry — Problem 2

Let ABC be a triangle with ∠ BAC = 90 . Let D , E , and F be the feet of altitude, angle bisector, and median from A to BC , respectively. If DE = 3 and EF = 5 , compute the length of BC.

解析

Let ABC be a triangle with ∠ BAC = 90 . Let D , E , and F be the feet of altitude, angle bisector, and median from A to BC , respectively. If DE = 3 and EF = 5 , compute the length of BC. Proposed by: Jerry Liang Answer: 20 Solution 1: A 3 5 B C D E F Since F is the circumcenter of △ ABC , we have that AE bisects ∠ DAF . So by the angle bisector theorem, we can set AD = 3 x and AF = 5 x . Applying Pythagorean theorem to △ ADE then gives 2 2 2 (3 x ) + (5 + 3) = (5 x ) = ⇒ x = 2 . So AF = 5 x = 10 and BC = 2 AF = 20 . BA BD DA Solution 2: Let BF = F C = x . We know that △ BAD ∼ △ ACD so = = and thus AC DA DC √ √ BA BD x − 8 AB BE x − 5 = = . By Angle Bisector Theorem, we also have = = , which means that AC DC x +8 AC EC x +5 √ x − 8 x − 5 2 2 = = ⇒ ( x − 8)( x + 5) = ( x + 8)( x − 5) x + 8 x + 5 which expands to 3 2 3 2 2 x + 2 x − 55 x − 200 = x − 2 x − 55 x + 200 = ⇒ 4 x = 400 . This solves to x = 10 , and so BC = 2 x = 20 .