HMMT 十一月 2016 · GEN 赛 · 第 7 题

英文解析

7. Let ABC be a triangle with AB = 13, BC = 14, CA = 15. The altitude from A intersects BC at D .
   Let ω and ω be the incircles of ABD and ACD , and let the common external tangent of ω and ω
   1 2 1 2
   (other than BC ) intersect AD at E . Compute the length of AE .
   Proposed by: Eshaan Nichani
   Answer: 7
   Solution 1:
   Let I , I be the centers of ω , ω , respectively, X , X be the tangency points of ω , ω with BC ,
   1 2 1 2 1 2 1 2
   respectively, and Y , Y be the tangency points of ω , ω with AD , respectively. Let the two common
   1 2 1 2
   external tangents of ω , ω meet at P . Note that line I I also passes through P .
   1 2 1 2
   By Heron’s formula, the area of triangle ABC is 84. Thus, AD · BC = 84, and so AD = 12. By the 1
   Pythagorean Theorem on right triangles ABD and ACD , BD = 5 and CD = 9.2
   [ ABD ]
   The inradius of ABD , r , is , where [ ABD ] is the area of ABD and s is its semiperimeter.
   ABD ABDs
   ABD is a 5-12-13 right triangle, so [ ABD ] = 30 and s = 15. Thus, r = 2. Similarly, we get aBD
   ABD ABDthat ACD ’s inradius is r = 3. I Y DX is a square, so X D = I X = r = 2, and similarly
   ACD 1 1 1 1 1 1 ABD
   X D = 3. X and X are on opposite sides of D , so X X = X D + X D = 5.
   2 1 2 1 2 1 2
   Since P lies on lines I I , X X , and I X , I X are parallel, triangles P X I and P X I are similar.
   1 2 1 2 1 1 2 2 1 1 2 2
   X I 2 P X P X P X
   1 1 1 1 1
   Thus, = = = = . Solving gives P X = 10. Letting ∠ I P X = θ , since
   1 1 1
   X I 3 P X P X + X X P X +5
   2 2 2 1 1 2 1
   X I
   1 11
   I X P is a right angle, we have tan θ = = . D and E lie on different common external tangents,
   1 1
   X P 5
   1 1
   2 tan θ 5
   so P I bisects ∠ EP D , and thus ∠ EP D = 2 θ . Thus, tan ∠ EP D = tan 2 θ = = .
   1 2
   1 − tan θ 12
   ED is perpendicular to BC , so triangle EDP is a right triangle with right angle at D . Thus, =5
   ED12
   tan ∠ EP D = . P D = P X + X D = 12, so ED = 5. AD = 12, so it follows that AE = 7 .
   1 1
   P D
   Solution 2:
   Lemma: Let Ω , Ω be two non-intersecting circles. Let ` , ` be their common external tangents, and
   1 2 A B
   ` be one of their common internal tangents. Ω intersects ` , at points A , B , respectively, and C 1 A B 1 1 Ω intersects ` , ` at points A , B . If ` intersects ` at X , and at Y , then XY = A A = B B .
   2 A B 2 2 C A B 1 2 1 2
   Proof: Let ` intersect Ω , Ω at C , C , respectively. Then, examining the tangents to Ω , Ω from
   C 1 2 1 2 1 2
   X, Y , we have A X = C X, A X = C X, B Y = C Y, B Y = C Y . Thus, 2 A A = 2 B B =
   1 1 2 2 1 1 2 2 1 2 1 2
   A A + B B = A X + A X + B Y + B Y = C X + C X + C Y + C Y = 2 XY , and the conclusion
   1 2 1 2 1 2 1 2 1 2 1 2
   follows.
   Using the notation from above, we apply the lemma to circles ω , ω , and conclude that ED = X X .
   1 2 1 2
   Then, we proceed as above to compute X X = 5 = ED . Thus, AD = 7 .
   1 2