Midpoints and angles

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Source: Japan Olympiad Finals 2014, #1

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202 posts

#1 h May 17, 2014, 7:33 AM • 2 Y

Y by Adventure10, Mango247

Let $O$ be the circumcenter of triangle $ABC$ , and let $l$ be the line passing through the midpoint of segment $BC$ which is also perpendicular to the bisector of angle $\angle BAC$ . Suppose that the midpoint of segment $AO$ lies on $l$ . Find $\angle BAC$ .

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81 posts

mefeke

#2 h May 17, 2014, 8:00 AM • 4 Y

Y by mathematiculperson, thunderz28, Adventure10, Mango247

Let the angle bisector of $\angle BAC$ meet the circumcenter at $N$ .
Let $AA'$ be a diameter of the circumcenter.
Let $M$ be the midpoint of $BC$ .
Let $X$ be the the midpoint of $AN$ . $OX \perp AN$ .
Let $Y$ be the midpoint of $AO$ . $MY \perp AN$ is given. Let $Z$ be the intersection of $MY$ and $AN$ .

In $\triangle AXO$ , $YZ \parallel OX$ and $AY=OY$ . So $AZ=ZX = AN/4$ .
Clearly, $O, M, N$ are collinear.
In $\triangle ZMN$ , $OX \parallel MZ$ and $XN=2\cdot XZ$ . So $2\cdot ON = OM = OC \Rightarrow \angle MCO = 30^\circ$ . Thus $\angle BAC = 120^\circ$ .

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111 posts

#3 h May 17, 2014, 9:21 AM • 2 Y

Y by Adventure10, Mango247

let $X$ on $OM$ such that $XM=OM$
now it is easy to see $AXCO$ is cycle.

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314 posts

#4 h May 20, 2014, 7:20 PM • 2 Y

Y by Adventure10, Mango247

Let the midpoint of BC be M and G be the intersection of OM and circle of ABC(A and G on diferent sides of BC),and PG be the diameter(G,O,M and P are collinear),so AP is perpendicular to AG and so is MN,so MN is parallel to AP =>OM=PM,and now easy calculation of angles gives <BAC=120,and we are finished

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46 posts

#5 h May 24, 2014, 9:29 AM • 3 Y

Y by mathematiculperson, Adventure10, Mango247

Let $w$ is circumcircle of triangle $ABC$ . $D$ lies on $w$ as well as $AD$ is bisector of $\angle BAC$ . Suppose that $\angle BAD=\angle CAD=\frac{x} {2}$ and $\angle DCA=y$ . $E$ is intersection $DO$ and $BC$ ; $G$ and $F$ intersections $OA$ and $AD$ with the line $l$ , respectively. From theorem Menelaus in triangle $AOD$ : $\frac{OG} {AG} \frac{AF} {DF} \frac{DE} {OE}=1$ . From theorem sinus we easily obtain $cos(\frac{x} {2}+y)+2cos^2\frac{x} {2}=0$ $(I)$ . It is shown $cos(180^0-\frac{x} {2}+y)= \frac{OE} {OC}= \frac{OE} {R}$ in triangle $OEC$ . From angles we can easily $OG=OE$ . $\Rightarrow$ $cos( 180^0-\frac{x} {2}+y)= \frac{OE} {R}= -\frac{1} {2}$ $\Rightarrow$ $cos^2 \frac{x} {2}=0$ . $\Rightarrow$ $\frac{x} {2}=60^0$ $\Rightarrow$ $\angle BAC=120^0$ . Done

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sayantanchakraborty

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sayantanchakraborty

#6 h Oct 2, 2014, 11:01 AM • 2 Y

Y by mathematiculperson, Adventure10

Let the angle bisector of $A$ meet $l$ at $J$ and $AO$ meet $l$ at $X$ .Also let $N$ be the midpoint of $BC$ .Then by easy angle chasing we get $\angle{XNO}=\angle{JNO}=\angle{JMN}=\angle{AMB}=\frac{A}{2}+C$ and $\angle{NXO}=\angle{AXJ}=90^{\circ}-\angle{JAX}=90^{\circ}-\frac{B}{2}+\frac{C}{2}=\frac{A}{2}+C$ .Thus $ON=OX=\frac{R}{2} \implies Rsin(A-90)=\frac{R}{2} \implies A=120^{\circ}$ .

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595 posts

#7 h Jan 20, 2015, 6:46 AM • 3 Y

Y by mathematiculperson, Adventure10, Mango247

Let $l$ intesect $AO$ and the angle bisector of $A$ at $X,Y$ resp.Let the midpt of $BC$ be $D$
Draw the altitude through $A$ which meets $l$ at $Z$ .We know that $AY$ bisects $\angle{ZAO}$ , and further since $AY$ is perpendicular to $l$ , we have $AZ=AX=XO$ .Also $AZ$ is parallel to $OD$ and we get that $AZX$ and $OXD$ are congruent and isosceles.So $OX=OD=1/2R=1/2OC$
So $\angle {DOC}=60$ and ultimately $A=120$

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SergeyKrakowska

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SergeyKrakowska

#8 h Jun 4, 2017, 12:12 PM • 2 Y

Y by Adventure10, Mango247

Let $J$ and $K$ be the intersection of $l$ with $AO$ and the bisector of $\angle BAC$ respectively. Let $M$ be the midpoint of $BC$ . Let the bisector of $\angle BAC$ meet circumcenter at $N$ . As $\triangle OAN$ is isosceles, then we have $\angle OAN = \angle ONA$ . Then, $\angle KMN = 90^{\circ} - \angle ONA$ . We also have $\angle OJM = \angle AJK = 90^{\circ} - \angle OAN$ . So, $\angle KMN = \angle OJM$ and we have $\triangle OJM$ isosceles. Since $AO = OC = 2 \cdot OM$ so, $\angle OCM = 30^{\circ}$ . SInce $\triangle BOC$ is isosceles, we have $\angle BOC = 120^{\circ}$ and thus $\angle BAC = 120^{\circ}$ .

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3042 posts

#9 h Jan 16, 2018, 7:45 AM • 3 Y

Y by Maths_Guy, Adventure10, Mango247

Let $AD$ bisect $\angle BAC$ and $D$ lies on $(ABC)$ . Let $l\cap AD=X, l\cap AO = Z$ and $OY\perp AD$ with $Y$ on $AD$ . Let $M$ be the midpoint of $BC$ . Let $R$ be the radius of $(ABC)$ . We have $AZ = ZO$ which gives $OX = XY$ . Also, $OY\perp AD$ gives $AY=YD$ . So, we have $2=\frac{YD}{YX}=\frac{DO}{OM}=\frac{R}{OM}.$ So, $OM = \frac{R}{2}$ . Let ray $OM$ meet the circle again at $A'$ . So, $OBA'C$ is a rhombus. So, $180^{\circ}-\frac{\angle BOC}{2}=180^{\circ}-\angle BDC=\angle BAC = \angle BA'C = \angle BOC.$ So, $\angle BAC = \angle BOC = 120^{\circ}$ .

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643 posts

#10 h Jan 16, 2018, 3:28 PM • 2 Y

Y by Maths_Guy, Adventure10

syk0526 wrote:

Let $O$ be the circumcenter of triangle $ABC$ , and let $l$ be the line passing through the midpoint of segment $BC$ which is also perpendicular to the bisector of angle $\angle BAC$ . Suppose that the midpoint of segment $AO$ lies on $l$ . Find $\angle BAC$ .

Solution.

Let $O'$ be the reflection of $O$ upon $\overline{BC},$ $X=AO\cap l,$ $M$ be the midpoint of $\overline{BC},$ $M_a$ be the midpoint of arc $\overarc{BC}.$ Beacuse $\overline{AX}=\overline{XO},$ so $l\parallel AO'.$ Therefore, $\angle O'AM_a=\angle M_aAC+\angle O'AC=\frac{\angle A}{2}+\left(90^{\circ}-\frac{\angle A}{2}\right)=90^{\circ}\implies O'\in\odot(ABC).$ Thus, it follows that,
$\begin{align*}\angle A&=\angle BO'C\\&=\angle BOC\\&=2\cdot\angle BM_aC\\&=360^{\circ}-2\cdot\angle A\end{align*}$ $\implies \boxed{\angle A=120^{\circ}}\qquad\blacksquare$
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This post has been edited 1 time. Last edited by ayan.nmath, Jan 16, 2018, 3:35 PM
Reason: asymtote edit

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#11 h Oct 13, 2019, 6:29 PM • 2 Y

Y by Adventure10, Mango247

here's an easy solution using complex numbers:
let $a=1 , B=b^2 , C=c^2$
let $n=\frac{b^2 + c^2 }{2} , m = 1/2 , k=-bc$
the problem equivalent to : $NM$ is perpindacular to $AK$
so
$\frac{n-m}{ \overline{n}- \overline{m}}=-\frac{a-k}{ \overline{a}- \overline{k}} \implies\ b^2 + c^2=bc$
thus $\vec{OB}+\vec{OC}=\vec{OL}$ where L is the midpoint of arc $BAC$
so $OBLC$ is a parallelogram then
$\angle CBO=\angle BCL \implies a-90=90-a/2 \ implies \angle BAC=120$

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