Problem:
Let ABC be a triangle with incenter I, incircle γ and circumcircle Γ. Let M,N,P be the midpoints of sides BC,CA,AB and let E,F be the tangency points of γ with CA and AB, respectively. Let U,V be the intersections of line EF with line MN and line MP, respectively, and let X be the midpoint of arc BAC of Γ.
(1) Prove that I lies on ray CV.
(2) Prove that line XI bisects UV.
Solution:
Set ∠ABC=2y and ∠BCA=2z. First, we start with a known fact that I lies on ray CV. Let V1​ be the foot of the perpendicular from B to ray CI. Then in right triangle BV1​C,V1​M=MB=MC and ∠MV1​C=∠MCV1​=z=∠V1​CA, implying that MV1​​∥CA; in particular, V1​ lies on line MP. Because ∠BV1​I=∠BFI=90∘,BIFV1​ is cyclic, from which it follows that ∠V1​FB=∠V1​IB=y+z=∠AEF=∠AFE; in particular, V1​ lies on EF. Because V1​ lies on both line MP and line EF,V=V1​ and V lies on line CI. Likewise we can prove that U lies on line BI.
Rays BI and CI intersect again at Y and Z. Note that ∠UVC=∠EVC=∠AEV− ∠ECV=∠AEF−∠ECV=y. Because BCYZ is cyclic, we have ∠YZC=∠YBC= y. Therefore, UV∥YZ. It suffices to show that IX bisects segment YZ, which is clearly true because IYXZ is a parallelogram. (Indeed, ∠YZX=XAY=∠XBC−∠YBC= y+z−y=z=∠ZYB, from which it follows that ZX∥IY. Likewise, we can show that IZ∥XY.)
OR
First, note that U and V lie on the bisectors BI and CI, respectively. Indeed, let D be the tangency point of γ with BC and let U′ be the intersection of BI with EF. Note that triangles BFU′ and BDU′ are congruent (by SAS), so ∠BU′F=∠BU′D. In addition, the pencil (U′F,U′B,U′D,U′C) is harmonic; thus, it follows that U′B⊥U′C, so, in particular, U′M=MB, which gives ∠MU′B=∠MBU′=21​∠B=∠ABU′; thus, MU′∥AB; hence U′=U, which proves the claim that U lies on BI. Similarly, we get that V is on CI. Also, remember the perpendicularities IB⊥CU and IC⊥VB, which we will use soon.
Next, note that the lines XB and XC are tangent to the circumcircle of triangle IBC; indeed, observe that
∠XBI​=∠ABI−∠ABX=21​∠B−(∠BCX−∠C)=21​∠B−21​(180∘−∠A)+∠C=21​∠C=∠BCI​
Similarly, ∠XCI=∠IBC. This means that X is the intersection of the tangents at B and C to the circumcircle of IBC; hence, IX is the I-symmedian of triangle IBC.
But we proved before that U and V are on IB and IC, respectively and that IB⊥CU and IC⊥VB. In other words, we showed that U and V are the feet of the altitudes from C and B in triangle IBC - so, in particular, we have that BCUV is cyclic and that UV is an antiparallel to BC in triangle IBC. This yields the conclusion, since we know that the I-symmedian of IBC is the locus of the midpoints of the antiparallels to BC in triangle IBC; hence we showed that IX bisects UV, as claimed.
This problem and and the second solution were suggested by Titu Andreescu and Cosmin Pohoata. The first solution was suggested by Zuming Feng.
The problems on this page are the property of the MAA's American Mathematics Competitions