HMMT 十一月 2021 · 冲刺赛 · 第 33 题
HMMT November 2021 — Guts Round — Problem 33
题目详情
- [17] Point P lies inside equilateral triangle ABC so that ∠ BP C = 120 and AP 2 = BP + CP . AB √ a b can be written as , where a, b, c are integers, c is positive, b is square-free, and gcd( a, c ) = 1. Find c 100 a + 10 b + c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2021, November 13, 2021 — GUTS ROUND Organization Team Team ID#
解析
- [17] Point P lies inside equilateral triangle ABC so that ∠ BP C = 120 and AP 2 = BP + CP . AB √ a b can be written as , where a, b, c are integers, c is positive, b is square-free, and gcd( a, c ) = 1. Find c 100 a + 10 b + c . Proposed by: Joseph Heerens Answer: 255 Solution: Let O be the center of ABC . First, we draw in the circumcircle of ABC and the circumcircle of BOC , labeled ω and ω , respectively. Note that ω is the reflection of ω over BC and that P 1 2 1 2 lies on ω . Now, let P be the second intersection of ray CP with ω . Additionally, label the second 2 C 1 ′ intersections of ray AP with ω and ω be M and X , respectively. Lastly, let A be the diametrically 1 2 opposite point from A on ω . 1 ′ ′ We first note that A is the center of ω . Thus, A lies on the perpendicular bisector of segment P X . 2 ′ But since AA is a diameter of ω , this also means that the midpoint of P X lies on ω . This implies 1 1 that M is the midpoint of P X . ◦ ◦ From a simple angle chase, we have ∠ P P B = 180 − ∠ BP C = 60 . Also, ∠ BP C = ∠ BAC = 60 . C C Therefore, we find that triangle BP P is equilateral with side length BP . C 2 2 Now we begin computations. By Law of Cosines in triangle BP C , we see that BP + CP + BP · CP = 2 2 BC = AB . However, we can rewrite this as 2 2 2 2 2 AB = BP + CP + BP · CP = ( BP + CP ) − BP · CP = 2 · AP − BP · CP. AP To find an equation for , it suffices to simplify the expression BP · CP. Since BP P is equilateral, C AB we can proceed through Power of a Point. By looking at ω , we see that 1 1 BP · CP = P P · CP = AP · P M = · AP · AX. C 2 Then, from Power of a Point on ω , we see that 2 ( ) 1 1 1 1 1 2 2 2 · AP · AX = · AP · ( AX − AP ) = · AP · AX − · AP = AB − AP . 2 2 2 2 2 ( ) 1 2 2 Combining everything, we find that BP · CP = AB − AP which means that 2 √ ( ) 1 5 3 AP 15 2 2 2 2 2 2 AB = 2 · AP − AB − AP = ⇒ AB = AP = ⇒ = . 2 2 2 AB 5