Let $ABC$ be an acute triangle, where $AB$ is the smallest side and let $D$ be the midpoint of $AB$. Let $P$ be a point in the interior of the triangle $ABC$ such that $\angle CAP = \angle CBP = \angle ACB$. From the point $P$, we draw perpendicular lines on $BC$ and $AC$ where the intersection point with $BC$ is $M$, and with $AC$ is $N$ . Through the point $M$ we draw a line parallel to $AC$, and through $N$ parallel to $BC$. These lines intercept at the point $K$. Prove that $D$ is the center of the circumscribed circle for the triangle $MNK$.
Problem
Source: 2021 3nd Final Mathematical Cup Junior Division P2 FMC
Tags: geometry, Circumcenter, equal angles
15.10.2021 21:08
Let $X=AP\cap BC$ , $Y=BP\cap CA$ , $R=AP\cap NK$ and $S=BP\cap MK$. Claim 1. $P$ is the orthocenter of $\triangle MNK$. $$ \left\{\begin{array}{lll} PM\perp BC \Longrightarrow PM\perp NK \\ PN\perp CA \Longrightarrow PN\perp MK \\ \end{array} \right. $$Hence Claim 1 is proved. Claim 2 Quadrilateral $DRPS$ is a parallelogram. From the conditions of the heading, it holds $$ \left\{\begin{array}{lll} AX=XC \Longrightarrow AR=RN \\ BY=YC \Longrightarrow BS=SM \\ \end{array} \right. $$Therefore $R$ and $S$ are the circumcenters of the right triangles $\triangle ANP$ and $\triangle BMP$, respectively. This implies that $D,R,S$ are the midpoints of the sides of triangle $\triangle ABP$. Therefore, $$ DR=SP \quad ; \quad DS=RP \quad \Longrightarrow \quad \text{Claim 2 is proved.} $$ Claim 3. Triangles $\triangle DRN$ and $\triangle MPN$ are similar. Let $\measuredangle ACB=\gamma$. Then, some easy angle chasing shows $$ \left\{\begin{array}{lll} \measuredangle AYP=2\gamma \Longrightarrow \measuredangle YPN=90^o-2\gamma \\ \measuredangle PXB=2\gamma \Longrightarrow \measuredangle MPX=90^o-2\gamma \\ MPNC \quad \text{cyclic} \quad \Longrightarrow \measuredangle MPN=180^o-\gamma \\ \measuredangle XPY=3\gamma \Longrightarrow \measuredangle RPS=3\gamma \Longrightarrow \measuredangle DRP=180^o-3\gamma \\ \measuredangle RAN=\gamma=\measuredangle ANR \Longrightarrow \measuredangle PRN=2\gamma \\ \end{array} \right. $$Therefore, $$ \measuredangle DRN=\measuredangle MPN=180^o-\gamma \qquad (\star) $$Again, easy angle chasing shows that triangles $\triangle MSP$ and $\triangle PRN$ are both isosceles and similar. Consequently, $$ \frac{SP}{MP}=\frac{RN}{PN} $$But in light of Claim 2, this implies $$ \frac{DR}{MP}=\frac{RN}{PN} \Longrightarrow \frac{DR}{RN}=\frac{MP}{PN} \qquad (\star\star) $$Clearly, $(\star)$ and $(\star\star)$ prove Claim 3. Claim 4. Triangles $\triangle MSP$ and $\triangle MPN$ are similar. Angle chasing shows $$ \measuredangle MSP=\measuredangle DRN $$Also, $$ \left\{\begin{array}{lll} MS=SP=DR \\ RN=PR=SD \\ \end{array} \right. \Longrightarrow \triangle MSP \quad \text{and} \quad \triangle DRN \quad \text{are congruent.} $$Together with Claim 3, this last observation proves Claim 4. In light of Claim 3 and Claim 4, it holds $$ \measuredangle RND=\measuredangle PNM \quad ; \quad \measuredangle MNP=\measuredangle DMS $$Thus $D$ is the isogonal conjugate of $P$ with respect to $\triangle KNM$. But, in light of Claim 1, and since the isogonal conjugate of the orthocenter is always the circumcenter, the problem is solved.
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16.10.2021 09:11
is a little bit different! Best regards, sunken rock
16.10.2021 12:42
Let $A', B'$ be the reflections of $A,B$ over $N,M$, respectively. Note that $\triangle PBB'\sim\triangle PA'A$, thus by spiral sim, $AB'\cap A'B$ lies on $(PA'A)$ and $(PBB')$, and, $\triangle PAB'\cong\triangle PA'B$. Thus, $\measuredangle (AB',A'B)=\measuredangle AA'P+\measuredangle PB'P=2\measuredangle ACB$, and, $AB'=A'B$. Note that $DM$ and $DN$ are the midsegments of $\triangle ABB'$ and $\triangle ABA'$, respectively. Therefore, $DM=\tfrac{1}{2}AB'=\tfrac{1}{2}A'B=DN$. Also, $\measuredangle MDN=\measuredangle (AB',A'B)=2\measuredangle ABC=2\measuredangle MKN$, where the last equality holds as $CMKN$ is a parallelogram. We conclude that $D$ is the circumcenter of $\triangle MNK$.