# 2016 USAJMO Problems/Problem 1

## Problem

The isosceles triangle $\triangle ABC$, with $AB=AC$, is inscribed in the circle $\omega$. Let $P$ be a variable point on the arc $\stackrel{\frown}{BC}$ that does not contain $A$, and let $I_B$ and $I_C$ denote the incenters of triangles $\triangle ABP$ and $\triangle ACP$, respectively.

Prove that as $P$ varies, the circumcircle of triangle $\triangle PI_BI_C$ passes through a fixed point.

## Solution 1 $[asy] size(8cm); pair A = dir(90); pair B = dir(-10); pair C = dir(190); pair P = dir(-70); pair U = incenter(A,B,P); pair V = incenter(A,C,P); pair M = dir(-90); draw(circle((0,0),1)); dot("A", A, dir(A)); dot("B", B, dir(B)); dot("C", C, dir(C)); dot("P", P, dir(P)); dot("I_B", U, NE); dot("I_C", V, NW); dot("M", M, dir(M)); draw(A--B--C--A); draw(circumcircle(P,U,V)); [/asy]$

We claim that $M$ (midpoint of arc $BC$) is the fixed point. We would like to show that $M$, $P$, $I_B$, $I_C$ are cyclic.

We extend $PI_B$ to intersect $\omega$ again at R. We extend $PI_C$ to intersect $\omega$ again at S.

We invert around a circle centered at $P$ with radius $1$ (for convenience). (I will denote X' as the reflection of X for all the points) The problem then becomes: Prove $I_B'$, $I_C'$, and $M'$ are collinear.

Now we look at triangle $\triangle PR'S'$. We apply Menelaus (the version where all three points lie outside the triangle). It suffices to show that $$\dfrac{PI_B'}{I_B'R'} \cdot \dfrac{R'M'}{M'S'} \cdot \dfrac{S'I_C'}{I_C'P} = 1$$

By inversion, we know $PX' = \dfrac{1}{PX}$ for any point $X$ and $X'Y' = \dfrac{XY}{PX \cdot PY}$ for any points $X$ and $Y$.

Plugging this into our Menelaus equation we obtain that it suffices to show $$\dfrac{\dfrac{1}{PI_B}}{\dfrac{RI_B}{PI_B \cdot PR}} \cdot \dfrac{\dfrac{RM}{PR \cdot PM}}{\dfrac{SM}{PS \cdot PM}} \cdot \dfrac{\dfrac{SI_C}{PI_C \cdot PS}}{\dfrac{1}{PI_C}} = 1$$ We cancel out the like terms and rewrite. It suffices to show $$\dfrac{RM \cdot SI_C}{SM \cdot RI_B} = 1$$ We know that $AM$ is the diameter of $\omega$ because $\triangle ABC$ is isosceles and $AM$ is the angle bisector. We also know $\angle RMA = \dfrac{\angle ACB}{2} = \dfrac{\angle ABC}{2} = \angle SMA$ so $R$ and $S$ are symmetric with respect to $AM$ so $RM = SM$.

Thus, it suffices to show $\dfrac{SI_C}{RI_B} = 1$. This is obvious because $RI_B = RA = SA = SI_C$. Therefore we are done. $\blacksquare$

## Solution 2

We will use complex numbers as mentioned here. Set the circumcircle of $\triangle ABC$ to be the unit circle. Let $$A=1, B=w^2,C=\frac{1}{w^2}, P=p^2,$$ such that $$I_B=w+wp-p, I_C=p-\frac{p}{w}+\frac{1}{w}.$$ We claim that the circumcircle of $\triangle PI_BI_C$ passes through $M=-1.$ This is true if $$k=\frac{(I_B-M)(I_C-P)}{(I_B-P)(I_C-M)}=\frac{(w+wp-p+1)(p+\frac{1}{w}-\frac{p}{w}-p^2)}{(w+wp-p-p^2)(p+\frac{1}{w}-\frac{p}{w}+1)}=\frac{(wp+1)(1-p)}{(p+1)(w-p)}$$ is real. Now observe that $$\overline{k}=\frac{(1+\frac{1}{wp})(1-\frac{1}{p})}{(\frac{1}{p}+1)(\frac{1}{w}-\frac{1}{p})}=k,$$ so $k$ is real and we are done. $\blacksquare$

## Solution 3 $[asy] size(8cm); pair A = dir(90); pair B = dir(-10); pair C = dir(190); pair P = dir(-70); pair U = incenter(A,B,P); pair V = incenter(A,C,P); pair M = dir(-90); pair D = dir(40); pair E = dir(140); draw(circle((0,0),1)); dot("A", A, dir(A)); dot("B", B, dir(B)); dot("C", C, dir(C)); dot("D", D, dir(D)); dot("E", E, dir(E)); dot("P", P, dir(P)); dot("I_B", U, NE); dot("I_C", V, NW); dot("M", M, dir(M)); draw(A--B--C--A--P--B ^^ P--C ^^ E--D--P--E--M--V ^^ D--M--U--V); [/asy]$

Let $M$ be the midpoint of arc $BC$. Let $D$ be the midpoint of arc $AB$. Let $E$ be the midpoint of arc $AC$. Then, $P, I_B$, and $D$ are collinear and $P, I_C$, and $E$ are collinear.

We'll prove $MPI_B I_C$ is cyclic. (Intuition: we'll show that $M$ is the Miquel's point of quadrilateral $DE I_C I_B$. $D$ is the center of the circle $A I_B B$ (the $P-$ excenter of $PAB$ is also on the same circle). Therefore $D I_B = DB$. Similarly $E I_C = EC$. Since $AB=AC$, $DB=EC$. Therefore $D I_B = E I_C$. Obviously $ME = MD$ and $\angle MEI_C = \angle MEP = \angle MDP = \angle MDI_B$. Thus by SAS, $\triangle MEI_C \cong \triangle MDI_B$.

Hence $\angle I_B M I_C = \angle DME = \angle DPE = \angle I_B P I_C$, so $MPI_B I_C$ is cyclic and we are done.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. ## See also

 2016 USAJMO (Problems • Resources) First Problem Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 All USAJMO Problems and Solutions
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