HMMT 二月 2004 · 几何 · 第 10 题

HMMT February 2004 — Geometry — Problem 10

Right triangle XY Z has right angle at Y and XY = 228, Y Z = 2004. Angle Y is trisected, and the angle trisectors intersect XZ at P and Q so that X, P, Q, Z lie on XZ in that order. Find the value of ( P Y + Y Z )( QY + XY ). 1

解析

Right triangle XY Z has right angle at Y and XY = 228, Y Z = 2004. Angle Y is trisected, and the angle trisectors intersect XZ at P and Q so that X, P, Q, Z lie on XZ in that order. Find the value of ( P Y + Y Z )( QY + XY ). Solution: 1370736 The triangle’s area is (228 · 2004) / 2 = 228456. All the angles at Y are 30 degrees, so by the sine area formula, the areas of the three small triangles in the diagram are QY · Y Z/ 4 , P Y · QY / 4, and XY · P Y / 4, which sum to the area of the triangle. So expanding ( P Y + Y Z )( QY + XY ), we see that it equals 4 · 228456 + XY · Y Z = 6 · 228456 = 1370736 . X P Q Z Y 4