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Difference between revisions of "2016 USAJMO Problems/Problem 1"
(Added Solution 1 and moved previous solution to solution 2.)
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Prove that as <math>P</math> varies, the circumcircle of triangle <math>\triangle PI_BI_C</math> passes through a fixed point.
 
Prove that as <math>P</math> varies, the circumcircle of triangle <math>\triangle PI_BI_C</math> passes through a fixed point.
  
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== Solution ==
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==Solution 1==
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We claim that <math>M</math> (midpoint of arc <math>BC</math>) is the fixed point.
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We would like to show that <math>M</math>, <math>P</math>, <math>I_B</math>, <math>I_C</math> are cyclic.
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We extend <math>PI_B</math> to intersect <math>\omega</math> again at R.
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We extend <math>PI_C</math> to intersect <math>\omega</math> again at S.
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We invert around a circle centered at <math>P</math> with radius <math>1</math> (for convenience).
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(I will denote X' as the reflection of X for all the points)
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The problem then becomes: Prove <math>I_B'</math>, <math>I_C'</math>, and <math>M'</math> are collinear.
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Now we look at triangle <math>\triangle P'R'S'</math>. We apply Menelaus (the version where all three points lie outside the triangle).
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It suffices to show that
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<cmath>\dfrac{P'I_B'}{I_B'R'} \cdot \dfrac{R'M'}{M'S'} \cdot \dfrac{S'I_C'}{I_C'P'} = 1</cmath>
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By inversion, we know <math>P'X' = \dfrac{1}{PX}</math> for any point <math>X</math> and <math>X'Y' = \dfrac{XY}{PX \cdot PY}</math> for any points <math>X</math> and <math>Y</math>.
 +
 
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Plugging this into our Menelaus equation we obtain that it suffices to show
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<cmath>\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</cmath>
 +
We cancel out the like terms and rewrite. It suffices to show
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<cmath>\dfrac{RM \cdot SI_C}{SM \cdot RI_B} = 1</cmath>
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We know that <math>AM</math> is the diameter of <math>\omega</math> because <math>\triangle ABC</math> is isosceles and <math>AM</math> is the angle bisector. We also know <math>\angle RMA = \dfrac{\angle ACB}{2} = \dfrac{\angle ABC}{2} = \angle SMA</math> so <math>R</math> and <math>S</math> are symmetric with respect to <math>AM</math> so <math>RM = SM</math>.
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Thus, it suffices to show <math>\dfrac{SI_C}{RI_B} = 1</math>.
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This is obvious because <math>RI_B = RA = SA = SI_C</math>.
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Therefore we are done. <math>\Box</math>
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== Solution 2==
  
 
We will use complex numbers, with the circumcircle of <math>\triangle ABC</math> as the unit circle. Let <cmath>A=1, B=w^2,C=\frac{1}{w^2}, P=p^2,</cmath> such that <cmath>I_B=w+wp-p, I_C=p-\frac{p}{w}+\frac{1}{w}.</cmath> We claim that the circumcircle of <math>\triangle PI_BI_C</math> passes through <math>M=-1.</math> This is true iff <cmath>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)}</cmath> is real. This is true iff <math>k=\overline{k}.</math> We can compute <cmath>\overline{k}=\frac{(1+\frac{1}{wp})(1-\frac{1}{p})}{(\frac{1}{p}+1)(\frac{1}{w}-\frac{1}{p})}=k,</cmath> so we are done. <math>\blacksquare</math>
 
We will use complex numbers, with the circumcircle of <math>\triangle ABC</math> as the unit circle. Let <cmath>A=1, B=w^2,C=\frac{1}{w^2}, P=p^2,</cmath> such that <cmath>I_B=w+wp-p, I_C=p-\frac{p}{w}+\frac{1}{w}.</cmath> We claim that the circumcircle of <math>\triangle PI_BI_C</math> passes through <math>M=-1.</math> This is true iff <cmath>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)}</cmath> is real. This is true iff <math>k=\overline{k}.</math> We can compute <cmath>\overline{k}=\frac{(1+\frac{1}{wp})(1-\frac{1}{p})}{(\frac{1}{p}+1)(\frac{1}{w}-\frac{1}{p})}=k,</cmath> so we are done. <math>\blacksquare</math>

Revision as of 19:46, 24 June 2016

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

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

By inversion, we know $P'X' = \dfrac{1}{PX}$ 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. $\Box$

Solution 2

We will use complex numbers, with the circumcircle of $\triangle ABC$ as 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 iff \[k=\frac{(I_B-M)(I_C-P)}{(I_B-P)(I_C-M)}=\frac{(w+wp-p+1)(p+\frac{1}{w}-\frac{p}{w}-p^2)}{(w+wp-p-p^2)(p+\frac{1}{w}-\frac{p}{w}+1)}=\frac{(wp+1)(1-p)}{(p+1)(w-p)}\] is real. This is true iff $k=\overline{k}.$ We can compute \[\overline{k}=\frac{(1+\frac{1}{wp})(1-\frac{1}{p})}{(\frac{1}{p}+1)(\frac{1}{w}-\frac{1}{p})}=k,\] so we are done. $\blacksquare$

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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