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

Revision as of 19:06, 19 February 2018

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 $ABC$ .

Solution 3

Extend $CI$ to point $L$ on $(ABC)$ . By The Incenter-Excenter Lemma, B, I, A are all concyclic. Thus, L is the circumcenter of triangle $IAB$ . In other words, $L=O_c$ , so $O_c$ is on $(ABC)$ . Similarly, we can show that $O_a$ and $O_b$ are on $(ABC)$ , and thus, $A,B,C,O_a,O_b,O_c$ are all concyclic. It follows that the circumcenters are equal.

@@ Line 12: / Line 12: @@
 ===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>.
+===Solution 3===
+Extend <math>CI</math> to point <math>L</math> on <math>(ABC)</math>. By The Incenter-Excenter Lemma, B, I, A are all concyclic. Thus, L is the circumcenter of triangle <math>IAB</math>. In other words, <math>L=O_c</math>, so <math>O_c</math> is on <math>(ABC)</math>. Similarly, we can show that <math>O_a</math> and <math>O_b</math> are on <math>(ABC)</math>, and thus, <math>A,B,C,O_a,O_b,O_c</math> are all concyclic. It follows that the circumcenters are equal.
 ==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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