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

Latest revision as of 18:25, 14 October 2021

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 $\angle IAM = \angle AIM = \frac{\angle BAC + \angle ABC}{2}$ ), 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.

Solution 4

Let the centers be $T, R, S$ . We want to show that $ABC$ and $TRS$ have the same circumcircle. By Fact 5 we know that $CATB$ lie on a circle and similarly with the others. Thus the two triangles have the same circumcircle. ~coolmath_2018

@@ 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.
-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.
+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>
+===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>\angle IAM = \angle AIM = \frac{\angle BAC + \angle 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.
+==Solution 4==
+Let the centers be <math>T, R, S</math>. We want to show that <math>ABC</math> and <math>TRS</math> have the same circumcircle. By Fact 5 we know that <math>CATB</math> lie on a circle and similarly with the others. Thus the two triangles have the same circumcircle.
+~coolmath_2018
 ==See Also==
 {{USAMO box|year=1988|num-b=3|num-a=5}}
+{{MAA Notice}}
 [[Category:Olympiad Geometry Problems]]

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

Page

Toolbox

Search