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

(Created page with "== Problem == The isosceles triangle <math>\triangle ABC</math>, with <math>AB=AC</math>, is inscribed in the circle <math>\omega</math>. Let <math>P</math> be a variable poi...")
 
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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>
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{{MAA Notice}}
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==See also==
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{{USAJMO newbox|year=2016|beforetext=|before=First Problem|num-a=2}}

Revision as of 15:33, 21 April 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

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. $\blacksquare$

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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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