HMMT 二月 2022 · 团队赛 · 第 9 题
HMMT February 2022 — Team Round — Problem 9
题目详情
- [55] Let Γ and Γ be two circles externally tangent to each other at N that are both internally tangent 1 2 to Γ at points U and V , respectively. A common external tangent of Γ and Γ is tangent to Γ and 1 2 1 Γ at P and Q , respectively, and intersects Γ at points X and Y . Let M be the midpoint of the arc 2 d XY that does not contain U and V . Let Z be on Γ such M Z ⊥ N Z , and suppose the circumcircles of QV Z and P U Z intersect at T ̸ = Z . Find, with proof, the value of T U + T V , in terms of R , r , and 1 r , the radii of Γ, Γ , and Γ , respectively. 2 1 2
解析
- [55] Let Γ and Γ be two circles externally tangent to each other at N that are both internally tangent 1 2 to Γ at points U and V , respectively. A common external tangent of Γ and Γ is tangent to Γ and 1 2 1 Γ at P and Q , respectively, and intersects Γ at points X and Y . Let M be the midpoint of the arc 2 d XY that does not contain U and V . Let Z be on Γ such M Z ⊥ N Z , and suppose the circumcircles of QV Z and P U Z intersect at T ̸ = Z . Find, with proof, the value of T U + T V , in terms of R , r , and 1 r , the radii of Γ, Γ , and Γ , respectively. 2 1 2 Proposed by: Akash Das √ ( Rr + Rr − 2 r r )2 r r 1 2 1 2 1 2 √ Answer: | r − r | ( R − r )( R − r ) 1 2 1 2 Solution: By Archimedes lemma, we have M, Q, V are collinear and M, P, U are collinear as well. Note that inversion at M with radius M X shows that P QU V is cyclic. Thus, we have M P · M U = M Q · M V , so M lies on the radical axis of ( P U Z ) and ( QV Z ), thus T must lie on the line M Z . Thus, we have 2 M Z · M T = M Q · M V = M N , which implies triangles M ZN and M N T are similar. Thus, we have N T ⊥ M N . However, since the line through O and O passes through N and is perpendicular to 1 2 2 2 M N , we have T lies on line O O . Additionally, since M Z · M T = M N = M X , inversion at M 1 2 with radius M X swaps Z and T , and since ( M XY ) maps to line XY , this means T also lies on XY . Therefore, T is the intersection of P Q and O O , and thus by Monge’s Theorem, we must have T lies 1 2 on U V . Now, to finish, we will consider triangle OU V . Since O O T is a line that cuts this triangle, by 1 2 Menelaus, we have OO U T V O 1 2 · · = 1 . O U V T O O 1 2 Using the values of the radii, this simplifies to R − r U T r U T r ( R − r ) 1 2 1 2 · · = 1 = ⇒ = . r V T R − r V T r ( R − r ) 1 2 2 1 Now, note that 2 2 4 r r 1 2 T U · T V = T P · T Q = . 2 ( r − r ) 1 2 Now, let T U = r ( R − r ) k and T U = r ( R − r ) k . Plugging this into the above equation gives 1 2 2 1 2 4( r r ) 1 2 2 r r ( R − r )( R − r ) k = . 1 2 1 2 2 ( r − r ) 1 2 Solving gives √ 2 r r 1 2 p k = . | r − r | ( R − r )( R − r ) 1 2 1 2 To finish, note that √ 2( Rr + Rr − 2 r r ) r r 1 2 1 2 1 2 T U + T V = k ( Rr + Rr − 2 r r ) = p . 1 2 1 2 | r − r | ( R − r )( R − r ) 1 2 1 2