HMMT 二月 2011 · TEAM1 赛 · 第 12 题
HMMT February 2011 — TEAM1 Round — Problem 12
题目详情
- [ 70 ] Let ABC be a triangle, and let E and F be the feet of the altitudes from B and C , respectively. If A is not a right angle, prove that the circumcenter of triangle AEF lies on the incircle of triangle ABC if and only if the incenter of triangle ABC lies on the circumcircle of triangle AEF . Let ABC be a triangle, and let E and F be the feet of the altitudes from B and C , respectively. If A is not a right angle, prove that the circumcenter of triangle AEF lies on the incircle of triangle ABC if and only if the incenter of triangle ABC lies on the circumcircle of triangle AEF . Last Writes [110]
解析
- [ 70 ] Let ABC be a triangle, and let E and F be the feet of the altitudes from B and C , respectively. If A is not a right angle, prove that the circumcenter of triangle AEF lies on the incircle of triangle ABC if and only if the incenter of triangle ABC lies on the circumcircle of triangle AEF . Team Round A Solution: This problem is arguably the most difficult among all those appearing in the 2011 Harvard-MIT Mathematics Tournament. Do not feel badly if your team wasted time in a vain attempt to find a solution. It was intended by the author as a serious test for serious geometers. Let D be the foot of the altitude from A . Let H be the orthocenter of triangle ABC . Let M be the midpoint of AH . Let I be the incenter of triangle ABC . Let ω be the incircle of triangle ABC . Let γ be the circumcircle of AEF . Let η be the nine-point circle of triangle ABC . We first dispense with the case in which AB = AC . Since M is the circumcenter of triangle AEF , if the circumcenter of triangle AEF lies on the incircle of triangle ABC , then ω and η intersect in two points: M and D . Since ω and η are tangent by Feuerbach’s Theorem, it follows that ω and η are coincident, in which case triangle ABC is equilateral. A M γ F E H I ω η B D C If the incenter of triangle ABC lies on the circumcircle of triangle AEF , then A , I , and H lie on γ . Since no circle may intersect a line in three distinct points and A is distinct from I and H , it follows that I and H are coincident, in which case triangle ABC is equilateral. Since it is clear that if ABC is equilateral, M lies on ω and I lies on γ , this completes the proof in the case that AB = AC . Throughout the remainder of this solution, we assume that AB 6 = AC . Let Ω be the circumcircle of triangle ABC , and let α be the A -mixtilinear incircle. Let α be tangent to AB at P , to AC at Q , and to Ω at R . Since AB 6 = AC , lines P Q and BC are not parallel. Let their intersection be Z , and let the bisectors of ∠ A , ∠ B , and ∠ C intersect Ω at T , U , and V , respectively. Let the point diametrically opposite T on Ω be S . Let lines RS and BC intersect at Y . We introduce a lemma here, namely that I lies on P Q and that Y and Z are harmonic conjugates with respect to B and C . Team Round A S V Ω A F γ M P U H E I K Q ω B Y D C Z η R α T Let lines AR and P Q intersect at K . Since P A and QA are tangent to α at P and Q , respectively, RK is the symmedian from R in triangle P QR . By Pascal’s Theorem applied to cyclic hexagon ACV RU B , points Q , I , and P are collinear. Since AP = AQ and AI bisects ∠ P AQ , I is the midpoint of P Q . Hence RI is the median from R in triangle P QR . It follows that ∠ ICQ = ∠ P RA = ∠ IRQ and ∠ IBP = ∠ QRA = ∠ IRP . Hence quadrilaterals RBP I and RCQI are cyclic, so RI is the bisector of ∠ BRC . It follows also that ∠ BIP = ∠ BCI , so line P Q is tangent to the circumcircle of triangle BIC at I . Since T is the circumcenter of triangle BIC and S is diametrically opposite T on Ω, BS and CS are tangent to the circumcircle of triangle BIC at B and C , respectively. It follows that IS is the symmedian from I in triangle BIC ; since RI is the bisector of ∠ BRC , R , I , and S are collinear. Hence RS is the polar of Z with respect to the circumcircle of triangle BIC , so Y and Z are harmonic conjugates with respect to B and C , as desired. Team Round A S V A Ω γ F M U P X I H E K Q ω η B Y D C Z α R T We now take up the main problem. We first prove that I lies on γ if M lies on ω . If M lies on ω , then M is the Feuerbach point. Let X be the center of the homothety of positive magnitude mapping ω to Ω. Since the center of the homothety of positive magnitude ω to η is M and the center of the homothety of positive magnitude mapping η to Ω is H , it follows by Monge’s Circle Theorem that X lies on line M H . Note that R is the center of the homothety of positive magnitude mapping α to Ω. Since the center of homothety of positive magnitude mapping α to ω is A , it follows by the same theorem that R lies on line AX , so points A , M , H , and R are collinear. Note that triangle BRC is the reflection of triangle BHC about line BC . Since line RY bisects angle BRC , it follows that line HY bisects ∠ BHC . Let the line through H parallel to P Q intersect BC ′ at Z . Since ∠ HBA = ∠ HCA , the internal bisectors of ∠ BAC and ∠ BHC are parallel. Since P Q is ′ ′ perpendicular to AI and BZ is parallel to P Q , it follows that BZ is the external bisector of ∠ BHC . ′ ′ Hence Y and Z are harmonic conjugates with respect to B and C . It follows that Z and Z are coincident, and, therefore, that H and K are coincident. We may conclude that ∠ AIH is right, which implies the desired result because γ is the circle on diameter AH . (Do think carefully about why this argument fails if AB = AC ). Conversely, if I lies on γ , then ∠ AIH is right, so points P , H , I , and Q are collinear. Since P Q is perpendicular to AI and the interal bisectors of ∠ BAC and ∠ BHC are parallel, it follows that P Q ′ ′ is the external bisector of ∠ BHC . Let the internal bisector of ∠ BHC intersect BC at Y . Then Y ′ ′ and Z are harmonic conjugates with respect to B and C , so Y and Y are coincident. Let R be the ′ ′ ′ reflection of H about line BC . Clearly R Y is the bisector of ∠ BR C , so R , Y , and S are collinear. ′ ′ Since R , Y , and S are collinear and R also lies on Ω, it follows that R and R are coincident. Since ′ HR is perpendicular to BC , it follows that points A , H , and R are collinear. ′ Let X be the center of the homothety of positive magnitude mapping ω to Ω. Since the center of the homothety of positive magnitude mapping ω to α is A , and the center of the homothety of positive ′ magnitude mapping α to Ω is R , it follows by Monge’s Circle Theorem that X lies on line AR . Note Team Round A that H is the center of the homothety of positive magnitude mapping Ω to η . Since the center of the homothety of positive magnitude mapping ω to η is the Feuerbach point, it follows by the same theorem that the Feuerbach point lies on line AR as well. Since AR intersects η at M and D , the Feuerbach point may be either M or D . However, D may not be the Feuerbach point, else the the altitude from A and the bisector of ∠ A in triangle ABC would be coincident, contradicting the assumption that AB 6 = AC . We may conclude that M is the Feuerbach point, so, in particular, M lies on ω , as desired. This completes the proof. Remark: more computationally intensive solutions are also possible; in particular, it may be observed ′ ′ that the condition IJ = AM is equivalent to the condition M D/DA = IJ/JA , where J denotes ′ the point where ω is tangent to BC and A denotes the midpoint of BC . It is left to the reader to determine how to proceed. Let ABC be a triangle, and let E and F be the feet of the altitudes from B and C , respectively. If A is not a right angle, prove that the circumcenter of triangle AEF lies on the incircle of triangle ABC if and only if the incenter of triangle ABC lies on the circumcircle of triangle AEF . Last Writes [110]