PUMaC 2011 · 几何（A 组） · 第 7 题
PUMaC 2011 — Geometry (Division A) — Problem 7
题目详情
- [ 7 ] Let ABC be a triangle with AB = 2 , BC = 5 , AC = 4. Let M be the projection of C onto the external angle bisector at vertex B . Similarly, let N be the projection of B onto the external angle bisector at vertex C . If the ratio of the area of quadrilateral BCN M to the area of triangle ABC is a/b , where a and b are positive integers and gcd( a, b ) = 1, find a + b . ◦
解析
- First Solution : Extend BM and CN to meet at the excenter O . Let [ P P ...P ] denote the 1 2 n area of polygon P P ...P . Since quadrilateral BM N C is cyclic, we have that triangle OM N 1 2 n 2 2 is similar to triangle OCB . Thus, we have that [ OM N ] / [ OCB ] = ( ON/OB ) = cos ( ∠ O ). We have ◦ ∠ O = 180 − ( ∠ CBO + ∠ BCO ) ◦ ◦ ◦ = 180 − [(180 − ∠ ABC ) / 2 + (180 − ∠ ACB ) / 2] = ( ∠ ABC + ∠ ACB ) / 2 ◦ = (180 − ∠ A ) / 2 ◦ = 90 − ∠ A/ 2 . 2 1 2 2 ◦ Thus, cos ( ∠ O ) = cos (90 − ∠ A/ 2) = sin ( ∠ A/ 2) = (1 − cos ∠ A ). By the cosine theorem 2 2 2 2 for triangle ABC , we have cos ∠ A = ( AB + AC − BC ) / (2 · AB · AC ) = (4 + 16 − 25) / 16 = 2 1 1 5 21 − 5 / 16. Thus, [ OM N ] / [ OCB ] = cos ∠ O = (1 − cos ∠ A ) = (1 + ) = . It follows that 2 2 16 32 21 11 [ BM N C ] / [ OCB ] = 1 − [ OM N ] / [ OCB ] = 1 − = . Thus, 32 32 1 · CB · r [ BM N C ] [ BM N C ] [ OCB ] 11 A 2 = · = · , [ ABC ] [ OCB ] [ ABC ] 32 [ ABC ] where r is the altitude from O of triangle OBC , which is the exradius corresponding to A . A This exradius is r = [ ABC ] / ( s − a ), where s is the semiperimeter of triangle ABC , and A a = BC . Thus, r / [ ABC ] = 1 / ( s − a ) = 1 / [(2 + 4 − 5) / 2] = 2. Finally, we find that A 1 · CB · r [ BM N C ] 11 11 r 11 55 A A 2 = · = · 5 · = · 5 · 2 = . [ ABC ] 32 [ ABC ] 64 [ ABC ] 64 32 Thus, the answer is 55 + 32 = 87 . 5 Figure 9: Problem 7 diagram. ′ ′ Second Solution : Extend CM to meet AB at B and BN to meet AC at C . Then, ′ ′ B M = M C and C N = N B . Let P, Q, R be the midpoints of AB, AC, BC . Then, we see ′ that M, R, Q all lie on the midline of triangle AB C . Similarly, P, R, N all lie on the midline ′ of triangle ABC . Observe that [ P BR ] = [ P RQ ] = [ RQC ] = 1 / 4[ ABC ]. We have that [ BCN M ] [ BM R ] [ M RN ] [ RN C ] = + + [ ABC ] [ ABC ] [ ABC ] [ ABC ] 1 [ BM R ] 1 [ M RN ] 1 [ RN C ] = + + 4 [ RQC ] 4 [ P QR ] 4 [ P RB ] ( ) 1 RB · RM RN · RM RN · RC = + + 4 RQ · RC RP · RQ RP · RB ( ) 2 2 2 1 BC BC BC = + + 4 AB · BC AB · AC AC · BC 1 AC + BC + AB 2 = BC · 4 AB · BC · AC 1 AC + BC + AB = BC · 4 AB · AC 1 11 55 = · 5 · = , 4 2 · 4 32 so the answer is 55 + 32 = 87 . √