PUMaC 2011 · 几何（A 组） · 第 8 题
PUMaC 2011 — Geometry (Division A) — Problem 8
题目详情
- [ 8 ] Let ABC be a triangle with ∠ BAC = 60 , BA = 2, and CA = 3. A point M is located inside ABC such that M B = 1 and M C = 2. A semicircle tangent to M B and M C has its center O on BC . Let P be the intersection of the angle bisector of ∠ BAC and the perpendicular bisector of AC . If the ratio OP/M O is a/b , where a and b are positive integers and gcd( a, b ) = 1, find a + b . 1
解析
- Calculating side BC using the Theorem of Cosines, we get that BC = 7. Then, calculating ◦ ∠ BM C using the Theorem of Cosines in triangle BM C , we get that ∠ BM C = 120 . Now, we reflect triangle BM C over line BC , and let D be the reflection of M . Note that quadrilateral BDCM is a kite with the circle inscribed in it. Now, consider quadrilateral ABDC . It has ◦ opposite angles adding to 180 , so it is an inscribed quadrilateral. Since AB = DC = 2, it 6 is an isosceles trapezoid. Moreover, AB + DC = 4 = BD + AC implies that ABDC is a circumscribed quadrilateral. Note that because the perpendicular bisector of AC is the line of symmetry of ABDC , it passes through the incenter of ABDC . Also, the angle bisector of ∠ BAC passes through the incenter of ABDC . Thus, the point P in the problem is actually the incenter of ABDC . Note that points O and P both lie on the angle bisector of ∠ BDC , so D, O, P are collinear. Moreover, by symmetry, DO = OM , so DO/DP = M O/ ( M O + OP ). Thus, ( M O + OP ) /M O = DP/DO , so OP/M O = DP/DO − 1. By the homothety centered at point D , DP/DO is the same as the ratio of the radii of the two circles. To find the ◦ ◦ ◦ radius of the larger circle, we consider the 30 -60 -90 right triangle with hypotenuse AP . √ From this, the larger radius can immediately be seen to be 3 / 2. To find the radius of the smaller circle, consider the area S of triangle BM C . If r is the radius of the smaller 1 3 circle, then S = (1 · r + 2 · r ) = r . On the other hand, the area of triangle BM C is 2 2 √ 1 1 ◦ S = BM · M C sin( ∠ BM C ) = · 1 · 2 · sin(120 ) = 3 / 2. Equating the two expressions 2 2 √ √ √ for S , we get that r = 3 / 3. Thus, DP/DO = ( 3 / 2) / ( 3 / 3) = 3 / 2. Thus, OP/M O = DP/DO − 1 = 3 / 2 − 1 = 1 / 2. Thus, our answer is 1 + 2 = 3 . Figure 10: Problem 8 diagram. 7