HMMT 二月 2004 · 几何 · 第 4 题
HMMT February 2004 — Geometry — Problem 4
题目详情
- P is inside rectangle ABCD . P A = 2 , P B = 3, and P C = 10. Find P D .
解析
- P is inside rectangle ABCD . P A = 2 , P B = 3, and P C = 10. Find P D . √ Solution: 95 Draw perpendiculars from P to E on AB , F on BC , G on CD , and H on DA , and let AH = BF = w , HD = F C = x , AE = DG = y , and EB = GC = z . Then 2 2 2 2 2 2 2 2 2 2 2 2 P A = w + y , P B = w + z , P C = x + z , and P D = x + y . Adding and √ 2 2 2 2 subtracting, we see that P D = P A − P B + P C = 95, so P D = 95. 1 y A E z B w w H F P x x D y z C G