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 $\mathcal{C}_1\cap \mathcal{C}_2\cap AD\neq \emptyset$.
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Tags: geometry, romania, Romanian TST
hakN
07.06.2021 21:09
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.
sunken rock
08.06.2021 00:04
My solution at https://stanfulger.blogspot.com/2021/06/httpsartofproblemsolvingcomcommunityc6t.html Best regards, sunken rock
rafigamath
29.04.2023 11:33
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.