HMMT 二月 2020 · 几何 · 第 7 题
HMMT February 2020 — Geometry — Problem 7
题目详情
- Let Γ be a circle, and ω and ω be two non-intersecting circles inside Γ that are internally tangent to 1 2 Γ at X and X , respectively. Let one of the common internal tangents of ω and ω touch ω and ω 1 2 1 2 1 2 at T and T , respectively, while intersecting Γ at two points A and B . Given that 2 X T = X T and 1 2 1 1 2 2 that ω , ω , and Γ have radii 2, 3, and 12, respectively, compute the length of AB . 1 2
解析
- Let Γ be a circle, and ω and ω be two non-intersecting circles inside Γ that are internally tangent to 1 2 Γ at X and X , respectively. Let one of the common internal tangents of ω and ω touch ω and ω 1 2 1 2 1 2 at T and T , respectively, while intersecting Γ at two points A and B . Given that 2 X T = X T and 1 2 1 1 2 2 that ω , ω , and Γ have radii 2, 3, and 12, respectively, compute the length of AB . 1 2 Proposed by: James Lin √ 96 10 Answer: 13 Solution 1: Let ω , ω , Γ have centers O , O , O and radii r , r , R respectively. Let d be the 1 2 1 2 1 2 distance from O to AB (signed so that it is positive if O and O are on the same side of AB ). 1 X 2 O 2 T M 1 A B T 2 X 1 O 1 O Note that OO = R − r , i i O T − OM r − d 1 1 1 cos ∠ T O O = = , 1 1 OO R − r 1 1 O T + OM r + d 2 2 2 cos ∠ T O O = = . 2 2 OO R − r 1 2 Then √ X T = r 2 − 2 cos ∠ X O T 1 1 1 1 1 1 √ = r 2 + 2 cos ∠ T O O i 1 1 √ r − d 1 = r 2 + 2 1 R − r 1 √ R − d = r 2 . 1 R − r 1 Likewise, √ R + d X T = r 2 . 2 2 2 R − r 2 From 2 X T = X T we have 1 1 2 2 ( ) ( ) R − d R + d 2 2 2 2 8 r = 4 X T = X T = 2 r . 1 2 1 1 2 2 R − r R − r 1 2 √ √ 36 96 10 2 2 Plugging in r = 2, r = 3, R = 12 and solving yields d = . Hence AB = 2 R − d = . 1 2 13 13 Solution 2: We borrow the notation from the previous solution. Let X T and X T intersect Γ again 1 1 2 2 at M and M . Note that, if we orient AB to be horizontal, then the circles ω and ω are on opposite 1 2 1 2 R sides of AB . In addition, for i ∈ { 1 , 2 } there exist homotheties centered at X with ratio which send i r i ω to Γ. Since T and T are points of tangencies and thus top/bottom points, we see that M and M i 1 2 1 2 are the top and bottom points of Γ, and so M M is a diameter perpendicular to AB . 1 2 M 1 X 2 T 1 M A B T 2 X 1 M 2 Now, note that through power of a point and the aforementioned homotheties, ( ) ( ) R R 2 2 P ( M , ω ) = M T · M X = X T − 1 = 30 X T , 1 1 1 1 1 1 1 1 1 1 r r 1 1 2 and similarly P ( M , ω ) = 12 X T . (Here P is the power of a point with respect to a circle). Then 2 2 2 2 2 P ( M , ω ) 30 X T 30 5 1 1 1 1 = = = . 2 2 P ( M , ω ) 12 X T 12(2) 8 2 2 2 2 Let M be the midpoint of AB , and suppose M M = R + d (here d may be negative). Noting that 1 M and M are arc bisectors, we have ∠ AX M = ∠ T AM , so 4 M AT ∼ 4 M X A , meaning 1 2 1 1 1 1 1 1 1 1 2 2 that M A = M T · M X = P ( M , ω ). Similarly, 4 M AT ∼ 4 M X A , so M A = P ( M , ω ). 1 1 1 1 1 1 1 2 2 2 2 2 2 2 Therefore, 2 2 2 2 2 P ( M , ω ) M A ( R − d ) + ( R + d ) 2 R + 2 Rd R + d 5 1 1 1 = = = = = , 2 2 2 2 2 P ( M , ω ) M A ( R − d ) + ( R − d ) 2 R − 2 Rd R − d 8 2 2 2 √ √ √ ( ) 2 3 3 8 R 10 96 10 giving d = − R . Finally, we compute AB = 2 R 1 − = = . 13 13 13 13