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Difference between revisions of "2016 USAJMO Problems/Problem 1"

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(→‎Solution 3)
 
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Let <math>M</math> be the midpoint of arc <math>BC</math>. Let <math>D</math> be the midpoint of arc <math>AB</math>. Let <math>E</math> be the midpoint of arc <math>AC</math>.
 
Let <math>M</math> be the midpoint of arc <math>BC</math>. Let <math>D</math> be the midpoint of arc <math>AB</math>. Let <math>E</math> be the midpoint of arc <math>AC</math>.
−
Then <math>P I_B D</math> collinear. And <math>P I_C E</math> collinear.
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Then, <math>P, I_B</math>, and <math>D</math> are collinear and <math>P, I_C</math>, and <math>E</math> are collinear.
  
 
We'll prove <math>MPI_B I_C</math> is cyclic. (Intuition: we'll show that <math>M</math> is the Miquel's point of quadrilateral <math>DE I_C I_B</math>.
 
We'll prove <math>MPI_B I_C</math> is cyclic. (Intuition: we'll show that <math>M</math> is the Miquel's point of quadrilateral <math>DE I_C I_B</math>.

Latest revision as of 14:15, 28 June 2020

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. AMC logo.png

See also

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