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

Revision as of 18:10, 18 July 2016

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

Solution 1

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 $ABC$ .

@@ Line 3: / Line 3: @@
 ==Solution==
+===Solution 1===
 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.
@@ Line 9: / Line 10: @@
 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>
-==Solution 2==
+===Solution 2===
 Let <math>M</math> denote the midpoint of arc <math>AC</math>. It is well known that <math>M</math> is equidistant from <math>A</math>, <math>C</math>, and <math>I</math> (to check, prove <math><IAM = <AIM = \frac{<BAC + <ABC}{2}</math>), so that <math>M</math> is the circumcenter of <math>AIC</math>. Similar results hold for <math>BIC</math> and <math>CIA</math>, and hence <math>O_c</math>, <math>O_a</math>, and <math>O_b</math> all lie on the circumcircle of <math>ABC</math>.

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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