HMMT 二月 2020 · 几何 · 第 9 题
HMMT February 2020 — Geometry — Problem 9
题目详情
- Circles ω , ω , ω have centers A, B, C , respectively and are pairwise externally tangent at points
a b c
D, E, F (with D ∈ BC, E ∈ CA, F ∈ AB ). Lines BE and CF meet at T . Given that ω has ra-
a
dius 341, there exists a line
tangent to all three circles, and there exists a circle of radius 49 tangent to all three circles, compute the distance from T to.
解析
- Circles ω , ω , ω have centers A, B, C , respectively and are pairwise externally tangent at points
a b c
D, E, F (with D ∈ BC, E ∈ CA, F ∈ AB ). Lines BE and CF meet at T . Given that ω has ra-
a
dius 341, there exists a line
tangent to all three circles, and there exists a circle of radius 49 tangent to all three circles, compute the distance from T to. Proposed by: Andrew Gu Answer: 294 Solution 1: We will use the following notation: let ω be the circle of radius 49 tangent to each of ω , ω , ω . Let ω , ω , ω have radii r , r , r respectively. Let γ be the incircle of ABC , with center a b c a b c a b c I and radius r . Note that DEF is the intouch triangle of ABC and γ is orthogonal to ω , ω , ω a b c (i.e. ID , IE , IF are the common internal tangents). Since AD, BE, CF are concurrent at T , we have K = AB ∩ DE satisfies ( A, B ; F, K ) = − 1, so K is the external center of homothety of ω and ω . In a b particular, K lies on ` . Similarly, BC ∩ EF also lies on ` , so ` is the polar of T to γ . Hence IT ⊥ ` so 2 if L is the foot from I to, we have IT · IL = r . An inversion about γ preserves ω , ω , ω and sendsto the circle with diameter IT . Since inversion a b c preserves tangency, the circle with diameter IT must be ω . Therefore IT = 98 by the condition of the problem statement. Letting a , b , c be the radii of ω , ω , ω respectively and invoking Heron’s formula a b c as well as A = rs for triangle ABC , we see that γ has radius √ r r r a b c r = . r + r + r a b c We will compute this quantity using Descartes’ theorem. Note that there are two circles tangent to ω , a ω , ω , one with radius IT / 2 and one with radius ∞ . By Descartes’ circle theorem, we have (where b c k := 1 /a is the curvature) a √ 1 k + k + k + 2 k k + k k + k k = a b c a b b c a c IT / 2 and √ k + k + k − 2 k k + k k + k k = 0 , a b c a b b c c a which implies √ √ r + r + r 1 a b c = k k + k k + k k = . a b b c c a r r r 2 IT a b c 2 r Therefore r = 2 IT , which means IL = = 4 IT and T L = 3 IT = 294. IT C E I A D T F B L K Solution 2: Using the same notation as the previous solution, note that the point T can be expressed with un-normalized barycentric coordinates ( ) 1 1 1 : : r r r a b c with respect to ABC because T is the Gergonne point of triangle ABC . The distance from T tocan be expressed as a weighted average of the distances from each of the points A , B , C , which is 1 /r 1 /r 1 /r 3 a b c · r + · r + · r = . a b c 1 /r + 1 /r + 1 /r 1 /r + 1 /r + 1 /r 1 /r + 1 /r + 1 /r 1 /r + 1 /r + 1 /r a b c a b c a b c a b c Note that there are two circles tangent to ω , ω , ω , one with radius 49 and one with radius ∞ . By a b c Descartes’ circle theorem, we have (where k := 1 /r is the curvature) a a √ 1 k + k + k + 2 k k + k k + k k = a b c a b b c a c 49 and √ k + k + k − 2 k k + k k + k k = 0 , a b c a b b c c a so k + k + k = 1 / 98. The distance from T tois then 3 · 98 = 294. a b c Solution 3: As in the first solution, we deduce that IT is a diameter of ω , with T being the point on ω closest to ` . By Steiner’s porism, we can hold ω and ` fixed while making ω into a line parallel to c, resulting in the following figure: E I D T F A B L Let R be the common radius of ω and ω and r be the radius of ω . Notice that ABDE is a rectan- a b gle with center T , so R = AE = 2 IT = 4 r . The distance from T tois IL − IT = 2 R − 2 r = 6 r = 294.