Find < BAC given MB = OI

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Find < BAC given MB = OI

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Source: Baltic Way 2017 Problem 13

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#1 h Nov 11, 2017, 2:14 PM • 1 Y

Y by Adventure10

Let $ABC$ be a triangle in which $\angle ABC = 60^{\circ}$ . Let $I$ and $O$ be the incentre and circumcentre of $ABC$ , respectively. Let $M$ be the midpoint of the arc $BC$ of the circumcircle of $ABC$ , which does not contain the point $A$ . Determine $\angle BAC$ given that $MB = OI$ .

This post has been edited 6 times. Last edited by math163, Aug 15, 2018, 6:36 PM

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#2 h Nov 11, 2017, 2:32 PM • 1 Y

Y by Adventure10

Denote $\angle BAC = \alpha$ . Using that $A, C, O, I$ are cyclic, we get $MB= 2R \cdot \sin \frac{\alpha}{2}$ and $OI = 2R \cdot \sin |\frac{\alpha}{2} - 30|$ , so $\alpha=30$ will be our answer. (I think)

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#3 h Nov 11, 2017, 4:09 PM • 2 Y

Y by AlastorMoody, Adventure10

This problem was proposed by Burii.

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#4 h Nov 11, 2017, 9:12 PM • 2 Y

Y by Adventure10, Mango247

$\angle AOC=2\angle B= 120, \angle AIC=90+\angle B/2=120$ so $AOIC$ is cyclic and so $\angle AIO=\angle ACO=30$
$\angle IBM=\frac{ \angle A+\angle B}{2}= 90-\frac{\angle C}{2}= 180-(90+\frac{\angle C}{2})=180-\angle AIB=\angle MIB$ so $MB=IM=OI$
So $\angle IMO=15$ and so $\angle MOB=30$
But $\angle A=\angle MOB=30$

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#5 h Apr 25, 2018, 7:11 AM • 2 Y

Y by Adventure10, Mango247

RagvaloD wrote:

$\angle AOC=2\angle B= 120, \angle AIC=90+\angle B/2=120$ so $AOIC$ is cyclic and so $\angle AIO=\angle ACO=30$
$\angle IBM=\frac{ \angle A+\angle B}{2}= 90-\frac{\angle C}{2}= 180-(90+\frac{\angle C}{2})=180-\angle AIB=\angle MIB$ so $MB=IM=OI$
So $\angle IMO=15$ and so $\angle MOB=30$
But $\angle A=\angle MOB=30$

For the thought of the pre last line: you don't know that points A, O and B are colinear! You have to prove that!

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#6 h Apr 30, 2018, 8:40 AM • 1 Y

Y by Adventure10

MI=MC (fact 5)
<MIC=60, so MIC - equilateral and CI=MC
AOIC - cyclic (<AOC = <AIC = 120)
so <OIC = <BMC (180 - <BAC)
Given OI=BM, triangle BMC congruent with triangle OIC.
So OC = BC. BOC - equilateral (BO=OC). O lay on AB.
So <BAC = 30.

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#7 h Aug 13, 2021, 9:20 PM

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Solution 1.[Trigonometry.]
Let $x=\angle BAC$ . As $MB=OI$ , we get that $OM$ is tangent to $(AOI)$ , hence $r^2=OM^2=MI\cdot MA=MC\cdot MA$ .

By Law of Sines, $\frac{MC}{\sin{\frac{x}{2}}}=\frac{MA}{\sin{120^\circ-\frac{x}{2}}}=\frac{AC}{\sin{60^\circ}}=\frac{2\sqrt{3}}{3}AC$ and $AC=2r\cdot \sin{60^\circ}=\sqrt{3}r,$ hence we get that
$\begin{align*} 4\cdot \sin{(120^\circ-\frac{x}{2})}\cdot \sin{\frac{x}{2}}&=1\\ 4\cdot (\frac{\sin{\frac{x}{2}}}{2}+\frac{\sqrt{3}}{2}\cos{\frac{x}{2}})\cdot \sin{\frac{x}{2}}&=1\\ \sqrt{3}\sin{x}+2\sin^2{\frac{x}{2}} &=1\\ \sqrt{3}\sin{x}+\frac{\sin{x} \cdot \sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}&=1\\ \sqrt{3}+\frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}&=\frac{1}{\sin{x}}\\ \sqrt{3}+\frac{\sin{x}}{2\cos^2{\frac{x}{2}}}&=\frac{1}{\sin{x}}\\ \sqrt{3}+\frac{\sin{x}}{\cos{x}+1}&=\frac{1}{\sin{x}}\\ \sqrt{3}(\cos{x}+1)\sin{x}+\sin^2{x}&=\cos{x}+1\\ \sqrt{3}\sin{x}+\sin^2{x}-1&=\cos{x}-\sqrt{3}\cos{x}\sin{x}\\ (\sqrt{3}\sin{x}+\sin^2{x}-1)^2&=(1-\sqrt{3}\sin{x})^2(1-\sin^2{x})\\ 4\sin^4{x}&=\sin^2{x}\\ \sin{x}&=\frac{1}{2}\\ x&=30^{\circ}.\\ \end{align*}$
Solution 2.[Synthetic.]
As $\angle ABC=60^\circ$ , we get that $\angle CMI=60^\circ$ and as $MB=MI=MC$ , we get that $IMC$ is equilateral. Also problem statement yields that $I$ is the circumcentre of $\triangle OMC$ . Let $\angle BAM=a$ , then $\angle OIM=180^\circ-2a$ , hence $\angle OCM=90^\circ-a$ , however on the other hand, $\angle OCM=\angle OCB+\angle BCM=60^\circ+a$ , therefore $a=15^\circ$ . We conclude that $\angle BAC=30^\circ$ .

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#8 h Apr 7, 2025, 7:55 AM

Y by

Alternative solution.

Let $D=MO \cap AB$ . Some angle chase gives that $ADOIC$ is cyclic. $MI=MB=OI$ is equivalent to $MD=DA$ . Since $MO=OA$ we can conclude that $O=D$ . And answer follows trivially.

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