Romania Junior TST 2021 Day 3 P2

Art of Problem Solving

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oVlad

#1 h Jun 7, 2021, 10:57 AM

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Let $O$ be the circumcenter of triangle $ABC$ and let $AD$ be the height from $A$ ( $D\in BC$ ). Let $M,N,P$ and $Q$ be the midpoints of $AB,AC,BD$ and $CD$ respectively. Let $\mathcal{C}_1$ and $\mathcal{C}_2$ be the circumcircles of triangles $AMN$ and $POQ$ . Prove that $\mathcal{C}_1\cap \mathcal{C}_2\cap AD\neq \emptyset$ .

This post has been edited 1 time. Last edited by oVlad, Jan 25, 2022, 9:59 AM

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hakN

#2 h Jun 7, 2021, 6:09 PM

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If $AB=AC$ , then statement is obvious since $O$ lies on $AD$ and both of the circles.
Let $AB\neq AC$ , let $(AMN)\cap AD = E$ . Without loss of generality assume that $AB < AC$ .
Since $OA$ is a diameter in $(AMN)$ , we have $OE\perp AD$ , implying $OE\parallel BC$ . So $PEOQ$ is a trapezoid.
We will show that $PE=OQ$ , and this will imply that $PEOQ$ is cyclic. Let $F$ be the midpoint of $BC$ .
Note that $PD = \frac{c\cdot \cos{B}}{2}$ and $FQ = \frac{a - b\cdot \cos{C}}{2}$ .
But also it is well known that $c\cdot \cos{B} + b\cdot \cos{C} = a$ , we have $PD = FQ$ , so $\triangle OFQ \cong \triangle EDP$ , implying that $OQ = PE$ , as desired.

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#3 h Jun 7, 2021, 9:04 PM

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My solution at https://stanfulger.blogspot.com/2021/06/httpsartofproblemsolvingcomcommunityc6t.html

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sunken rock

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#4 h Apr 29, 2023, 8:33 AM

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First let's prove $O\in(AMN)$ . Since $M,N$ are midpoints $OM\perp AB, ON\perp AC$ $\implies AMON$ cyclic. Let $(AMON)\cap AD=K$
Since $\angle{BAD}=\angle{OAC}$ ,and $AK\perp MN$ , $AO$ is diameter of $(AMON)$ .So $\angle{AKO}=\angle{90}$ ,which means $MN\parallel KO$ $\implies MK=ON$ .And since $KO\parallel PQ$ we need to find $KP=OQ$ . We know that $MPQN$ is a rectangle so $MP=NQ$ , and $\angle{PMK}= \angle{ONQ}$ , which implies $KP=OQ$ and we're done.

This post has been edited 1 time. Last edited by rafigamath, Oct 23, 2023, 10:47 AM
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DensSv

#5 h Mar 21, 2025, 1:30 PM

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