HMMT 十一月 2010 · 冲刺赛 · 第 31 题

HMMT November 2010 — Guts Round — Problem 31

[ 20 ] What is the perimeter of the triangle formed by the points of tangency of the incircle of a 5-7-8 triangle with its sides? a b

解析

[ 20 ] What is the perimeter of the triangle formed by the points of tangency of the incircle of a 5-7-8 triangle with its sides? Guts Round √ 9 21 Answer: + 3 Let 4 ABC be a triangle with sides a = 7, b = 5, and c = 8. Let the incircle of 7 4 ABC be tangent to sides BC , CA , and AB at points D , E , and F . By the law of cosines (using the 2 2 2 b + c − a form cos( A ) = ), we have 2 bc 2 2 2 8 + 5 − 7 1 cos( A ) = = 2(5)(8) 2 2 2 2 8 + 7 − 5 11 cos( B ) = = 2(7)(8) 14 2 2 2 5 + 7 − 8 1 cos( C ) = = 2(7)(5) 7 Now we observe that AEF , BDF , and CDE are all isosceles. Let us call the lengths of the legs of these triangles s , t , and u , respectively. Then we know that s + t = 8, t + u = 7, and u + s = 5, so s = 3, t = 5, and u = 2. Our final observation is that an isosceles angle with legs of length l and whose non-equal angle is θ √ has a base of length l 2(1 − cos( θ )). This can be proven using the law of cosines or the Pythagorean theorem. Using this, we can calculate that √ DE = 2 2(1 − cos( C )) √ 12 = 2 7 √ EF = 3 2(1 − cos( A )) = 3 √ F D = 5 2(1 − cos( B )) √ 3 = 5 , 7 and then √ √ 12 3 DE + EF + F D = 2 + 3 + 5 7 7 √ 3 = 3 + 9 7 √ 21 = 3 + 9 . 7 a b