HMMT 二月 2011 · TEAM1 赛 · 第 12 题

英文解析

12. [ 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 asolution. 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 themidpoint of AH . Let I be the incenter of triangle ABC . Let ω be the incircle of triangle ABC . Let γbe 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 , ifthe circumcenter of triangle AEF lies on the incircle of triangle ABC , then ω and η intersect in twopoints: M and D . Since ω and η are tangent by Feuerbach’s Theorem, it follows that ω and η arecoincident, in which case triangle ABC is equilateral.
    FγMAωIHEη
    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 followsthat I and H are coincident, in which case triangle ABC is equilateral. Since it is clear that if ABCis 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 oftriangle ABC , and let α be the A -mixtilinear incircle. Let α be tangent to AB at P , to AC at Q , andto Ω at R . Since AB 6 = AC , lines P Q and BC are not parallel. Let their intersection be Z , and letthe bisectors of ∠ A , ∠ B , and ∠ C intersect Ω at T , U , and V , respectively. Let the point diametricallyopposite 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
    AΩVS
    F γ
    HUPM
    QKIEω
    B Y D C Zη
    Let lines AR and P Q intersect at K . Since P A and QA are tangent to α at P and Q , respectively, RKTαRis 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 bisectorof ∠ 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, BSand CS are tangent to the circumcircle of triangle BIC at B and C , respectively. It follows that IS isthe 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 harmonicconjugates with respect to B and C , as desired.
    Team Round A
    ΩAVSγ
    PUMF
    EHIXηωQK
    B Y D C Zα
    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 ,TRthen M is the Feuerbach point. Let X be the center of the homothety of positive magnitude mappingω to Ω. Since the center of the homothety of positive magnitude ω to η is M and the center of thehomothety 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 sametheorem 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 arecoincident, and, therefore, that H and K are coincident. We may conclude that ∠ AIH is right, whichimplies the desired result because γ is the circle on diameter AH . (Do think carefully about why thisargument 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 isperpendicular 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 thehomothety 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 Athat H is the center of the homothety of positive magnitude mapping Ω to η . Since the center of thehomothety of positive magnitude mapping ω to η is the Feuerbach point, it follows by the same theoremthat the Feuerbach point lies on line AR as well. Since AR intersects η at M and D , the Feuerbachpoint 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 todetermine how to proceed. Let ABC be a triangle, and let E and F be the feet of the altitudes from Band C , respectively. If A is not a right angle, prove that the circumcenter of triangle AEF lies on theincircle of triangle ABC if and only if the incenter of triangle ABC lies on the circumcircle of triangle
    AEF .
    Last Writes [110]