PUMaC 2016 · 几何（A 组） · 第 8 题

PUMaC 2016 — Geometry (Division A) — Problem 8

Let 4 ABC have side lengths AB = 4 , BC = 6 , CA = 5. Let M be the midpoint of BC and ◦ let P be the point on the circumcircle of 4 ABC such that ∠ M P A = 90 . Let D be the foot of the altitude from B to AC , and let E be the foot of the altitude from C to AB . Let P D and P E intersect line BC at X and Y , respectively. Compute the square of the area of 4 AXY . 1

解析

Let H be the orthocenter of ABC ; since the reflection H of H over BC lies on ( ABC ), ′′ ′′ the reflection H of H over M lies on ( ABC ). Since AH is a diameter of ( ABC ), it fol- lows that M, H, P are collinear. We then have ADEHP concyclic. Let F be the foot of ′ the altitude from A and M the reflection of M over F . Angle chasing, ∠ CDX, BEY = ′ ′ ∠ AEP, ADP = ∠ AHP = ∠ F HM = ∠ M HF . Then, ∠ XDH, Y EH = 90 − ∠ M HF = ′ ′ ′ ′ 180 − ∠ HM X, HM Y , so DHM X, EHM Y are concyclic. By power of a point: ′ ′ BM · BX = BH · BD = BE · BA CM · CY = CH · CE = CD · CA By the Pythagorean theorem, we find that: 27 3 9 9 ′ ′ BE = , BM = CD = , CM = 8 2 2 2 2 √ 5 7 This gives us BX = 9 , CY = 5, and XY = 9 + 5 − 6 = 8. A quick calculation gives AF = , 4 √ so the area of 4 AXY is 5 7. Thus, the square of the area is 175 . Problem written by Bill Huang. If you believe that any of these answers is incorrect, or that a problem had multiple reason- able interpretations or was incorrectly stated, you may appeal at tinyurl.com/pumacappeals . All appeals must be in by 1 PM to be considered. 3