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Difference between revisions of "1988 USAMO Problems/Problem 4"

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Let the circumcenters of <math>\Delta IAB</math>, <math>\Delta IBC</math>, and <math>\Delta ICA</math> be <math>O_c</math>, <math>O_a</math>, and <math>O_b</math>, respectively. It then suffices to show that <math>A</math>, <math>B</math>, <math>C</math>, <math>O_a</math>, <math>O_b</math>, and <math>O_c</math> are concyclic.
 
Let the circumcenters of <math>\Delta IAB</math>, <math>\Delta IBC</math>, and <math>\Delta ICA</math> be <math>O_c</math>, <math>O_a</math>, and <math>O_b</math>, respectively. It then suffices to show that <math>A</math>, <math>B</math>, <math>C</math>, <math>O_a</math>, <math>O_b</math>, and <math>O_c</math> are concyclic.
  
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We shall prove that quadrilateral <math>ABO_aC</math> is cyclic first. Let <math>\angle BAC=\alpha</math>, <math>\angle CBA=\beta</math>, and <math>\angle ACB=\gamma</math>. Then <math>\angle ICB=\gamma/2</math> and <math>\angle IBC=\beta/2</math>. Therefore minor arc <math>\arc{BIC}</math> in the circumcircle of <math>IBC</math> has a degree measure of <math>\beta+\gamma</math>. This shows that <math>\angle CO_aB=\beta+\gamma</math>, implying that <math>\angle BAC+\angle BO_aC=\alpha+\beta+\gamma=180^{\circ}</math>. Therefore quadrilateral <math>ABO_aC</math> is cyclic.
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We shall prove that quadrilateral <math>ABO_aC</math> is cyclic first. Let <math>\angle BAC=\alpha</math>, <math>\angle CBA=\beta</math>, and <math>\angle ACB=\gamma</math>. Then <math>\angle ICB=\gamma/2</math> and <math>\angle IBC=\beta/2</math>. Therefore minor arc <math>\overarc{BIC}</math> in the circumcircle of <math>IBC</math> has a degree measure of <math>\beta+\gamma</math>. This shows that <math>\angle CO_aB=\beta+\gamma</math>, implying that <math>\angle BAC+\angle BO_aC=\alpha+\beta+\gamma=180^{\circ}</math>. Therefore quadrilateral <math>ABO_aC</math> is cyclic.
  
 
This shows that point <math>O_a</math> is on the circumcircle of <math>\Delta ABC</math>. Analagous proofs show that <math>O_b</math> and <math>O_c</math> are also on the circumcircle of <math>ABC</math>, which completes the proof. <math>\blacksquare</math>
 
This shows that point <math>O_a</math> is on the circumcircle of <math>\Delta ABC</math>. Analagous proofs show that <math>O_b</math> and <math>O_c</math> are also on the circumcircle of <math>ABC</math>, which completes the proof. <math>\blacksquare</math>

Revision as of 12:43, 9 August 2015

Problem

$\Delta ABC$ is a triangle with incenter $I$. Show that the circumcenters of $\Delta IAB$, $\Delta IBC$, and $\Delta ICA$ lie on a circle whose center is the circumcenter of $\Delta ABC$.

Solution

Let the circumcenters of $\Delta IAB$, $\Delta IBC$, and $\Delta ICA$ be $O_c$, $O_a$, and $O_b$, respectively. It then suffices to show that $A$, $B$, $C$, $O_a$, $O_b$, and $O_c$ are concyclic.

We shall prove that quadrilateral $ABO_aC$ is cyclic first. Let $\angle BAC=\alpha$, $\angle CBA=\beta$, and $\angle ACB=\gamma$. Then $\angle ICB=\gamma/2$ and $\angle IBC=\beta/2$. Therefore minor arc $\overarc{BIC}$ in the circumcircle of $IBC$ has a degree measure of $\beta+\gamma$. This shows that $\angle CO_aB=\beta+\gamma$, implying that $\angle BAC+\angle BO_aC=\alpha+\beta+\gamma=180^{\circ}$. Therefore quadrilateral $ABO_aC$ is cyclic.

This shows that point $O_a$ is on the circumcircle of $\Delta ABC$. Analagous proofs show that $O_b$ and $O_c$ are also on the circumcircle of $ABC$, which completes the proof. $\blacksquare$

Solution 2

Let $M$ denote the midpoint of arc $AC$. It is well known that $M$ is equidistant from $A$, $C$, and $I$ (to check, prove $<IAM = <AIM = \frac{<BAC + <ABC}{2}$), so that $M$ is the circumcenter of $AIC$. Similar results hold for $BIC$ and $CIA$, and hence $O_c$, $O_a$, and $O_b$ all lie on the circumcircle of $ABC$.

See Also

1988 USAMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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