O1A = O2A => ABC Isosceles

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O1A = O2A => ABC Isosceles

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Source: 2002 All-Russian MO, Grade 9, Problem 2

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#1 h Jan 2, 2012, 7:26 AM • 2 Y

Y by HamstPan38825, Adventure10

Point $A$ lies on one ray and points $B,C$ lie on the other ray of an angle with the vertex at $O$ such that $B$ lies between $O$ and $C$ . Let $O_1$ be the incenter of $\triangle OAB$ and $O_2$ be the center of the excircle of $\triangle OAC$ touching side $AC$ . Prove that if $O_1A = O_2A$ , then the triangle $ABC$ is isosceles.

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#2 h Jan 2, 2012, 10:54 AM • 2 Y

Y by Adventure10, Mango247

Awesome problem!

Draw $E$ the $A$ -excenter of $\triangle ABC$ and observe that $EO_1O_2A$ is cyclic as by well-known angle formulas we have
$\angle BO_1A = 90^\circ + \frac12 \angle AOB$ and $\angle AO_2C = 90^\circ - \frac12 \angle AOB$ (and $O_1B \cap O_2C = E$ ).

Since $O_1A = O_2A$ , line $EA$ is angle bisector in $\triangle EBC$ and thus $\triangle ABC$ is symmetric about it's angle bisector $EA$ . Hence it is isosceles.

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#3 h Dec 3, 2012, 4:41 PM • 2 Y

Y by Adventure10, Mango247

it's even easier by cyclic quads

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#4 h Apr 7, 2017, 11:17 PM • 1 Y

Y by Adventure10

Note that $O,O_1,O_2$ are collinear because they all lie on the angle bisector of $\angle AOC.$

Because $\angle ACB + \angle ABC = 2\angle O_2AC + 2\angle O_1AB,$ it suffices to show that $\angle ACB = \angle O_2AC + \angle O_1AB.$

Note that $\angle O_2AC + \angle OO_2A = \angle ACB + \angle O_2OC.$ Then $\angle ACB = \angle O_2AC + \angle OO_2A - \angle O_2OC.$ It suffices to show that $\angle OO_2A - \angle O_2OC=\angle O_1AB.$ Since $\angle OO_2A = \angle AO_1O_2$ and $\angle O_2OC = \angle O_1OA$ and $\angle O_1AB = \angle O_1AO,$ this becomes $\angle AO_1O_2 = \angle O_1OA + \angle O_1AO$ which is true.

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