HMMT 十一月 2020 · 冲刺赛 · 第 29 题

HMMT November 2020 — Guts Round — Problem 29

[15] In acute triangle ABC , let H be the orthocenter and D the foot of the altitude from A . The circumcircle of triangle BHC intersects AC at E 6 = C , and AB at F 6 = B . If BD = 3, CD = 7, and AH 5 a = , the area of triangle AEF can be expressed as , where a, b are relatively prime positive integers. HD 7 b Compute 100 a + b .

解析

[15] In acute triangle ABC , let H be the orthocenter and D the foot of the altitude from A . The circumcircle of triangle BHC intersects AC at E 6 = C , and AB at F 6 = B . If BD = 3, CD = 7, AH 5 a and = , the area of triangle AEF can be expressed as , where a, b are relatively prime positive HD 7 b integers. Compute 100 a + b . Proposed by: Andrew Yao Answer: 12017 Solution: A E H F B C D P Q Let AH intersect the circumcircle of 4 ABC again at P , and the circumcircle of 4 BHC again at Q . Because ∠ BHC = 180 − ∠ A = ∠ BP C , P is the reflection of H over D . Thus, we know that 12 HD P D = HD . From power of a point and AD = , 7 2 12 HD BD · CD = AD · P D = . 7 7 5 From this, HD = and AH = . Furthermore, because 4 BHC is the reflection of 4 BP C over 2 2 BC , the circumcircle of 4 BHC is the reflection of the circumcircle of 4 ABC over BC . Then, AQ = 2 AD = 12. Applying Power of a Point, AC · AE = AB · AF = AH · AQ = 30 . √ √ √ √ 6 85 We can compute AC = 85 and AB = 3 5, which means that AE = and AF = 2 5. Also, 17 BC · AD [ ABC ] = = 30. Therefore, 2 AE · AF 4 120 [ AEF ] = · [ ABC ] = · 30 = . AC · AB 17 17