HMMT 二月 2004 · 几何 · 第 10 题
HMMT February 2004 — Geometry — Problem 10
题目详情
英文原题
- Right triangle XY Z has right angle at Y and XY = 228, Y Z = 2004. Angle Y istrisected, 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
英文解析
- Right triangle XY Z has right angle at Y and XY = 228, Y Z = 2004. Angle Y istrisected, 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. Soexpanding ( P Y + Y Z )( QY + XY ), we see that it equals
4 · 228456 + XY · Y Z = 6 · 228456 = 1370736 .
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