HMMT 十一月 2016 · GEN 赛 · 第 7 题
HMMT November 2016 — GEN Round — Problem 7
题目详情
- 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 .
解析
- 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
1
By Heron’s formula, the area of triangle ABC is 84. Thus, AD · BC = 84, and so AD = 12. By the
2
Pythagorean Theorem on right triangles ABD and ACD , BD = 5 and CD = 9.
[ ABD ]
The inradius of ABD , r , is , where [ ABD ] is the area of ABD and s is its semiperimeter.
ABD ABD
s
ABD
ABD is a 5-12-13 right triangle, so [ ABD ] = 30 and s = 15. Thus, r = 2. Similarly, we get
ABD ABD
that 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
1 1
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
5
ED is perpendicular to BC , so triangle EDP is a right triangle with right angle at D . Thus, =
12
ED
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 , andat 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