Four points on AP

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Four points on AP

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Source: Turkey National Mathematical Olympiad 2020 P2

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#1 h Mar 8, 2021, 12:17 PM • 2 Y

Y by Mango247, Mango247

Let $P$ be an interior point of acute triangle $\Delta ABC$ , which is different from the orthocenter. Let $D$ and $E$ be the feet of altitudes from $A$ to $BP$ and $CP$ , and let $F$ and $G$ be the feet of the altitudes from $P$ to sides $AB$ and $AC$ . Denote by $X$ the midpoint of $[AP]$ , and let the second intersection of the circumcircles of triangles $\Delta DFX$ and $\Delta EGX$ lie on $BC$ . Prove that $AP$ is perpendicular to $BC$ or $\angle PBA = \angle PCA$ .

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mehmetakifyildiz

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mehmetakifyildiz

#2 h Mar 9, 2021, 11:17 AM • 3 Y

Y by CinarArslan, mijail, ehuseyinyigit

Official Solution: Let $Y$ , $Z$ , $K$ , $L$ , $M$ be midpoints of $[BP]$ , $[CP]$ , $[BC]$ , $[CA]$ , $[AB]$ , respectively. By considering the nine point circles of the triangles $PAB$ and $PAC$ , it is easy to see that $(DFX)$ and $(XYM)$ are the same, $(EGX)$ and $(XZL)$ are the same. Let $T$ be the second intersection point of $(XYM)$ and $(XZL)$ . By using the parallel lines, simple angle chasing gives: $\widehat{YTZ}=\widehat{YTX}+\widehat{ZTX}=180^\circ-\widehat{YMX}+180^\circ-\widehat{ZLX}=360^\circ-\widehat{XPY}-\widehat{XPZ}=\widehat{BPC}=\widehat{YKZ}$ This implies $T\in (YKZ)$ , so we can conclude that $T$ lies on the nine point circle of the triangle $PBC$ . Since $T$ also lies on $BC$ , there are two possibilities: $T=K$ or $PT\perp BC$ .

(i) If $T=K$ , angle chasing using parallel lines gives $\widehat{PBA}=\widehat{ABC}-\widehat{PBC}=\widehat{LKC}-\widehat{ZKC}=\widehat{LKZ}$ . Hence, by considering the circle $(XLZK)$ we get $\widehat{PCA}=\widehat{LCZ}=\widehat{LXZ}=\widehat{LKZ}$ , which implies $\widehat{PBA}=\widehat{PCA}$ .

(ii) If $PT\perp BC$ , we have $|YB|=|YT|$ . Then, we get $|MB|=|MT|$ by using $\widehat{MTY}=\widehat{MXY}=\widehat{MBY}$ . As a result, $|MA|=|MB|=|MT|$ implies $AT\perp BC$ and the result follows.

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#3 h Mar 9, 2021, 12:16 PM • 5 Y

Y by mijail, joseph02, Mango247, pokpokben, ehuseyinyigit

Case 1: $DE\parallel FG$
Since $DEFG$ is cyclic, we get $EF=DG$ so $\angle EPF=\angle DPG$ . $\Rightarrow \angle DPF=\angle EPG\Rightarrow \angle PFB+\angle PBF=\angle PGC+\angle PCG\Rightarrow \angle PBF=\angle PCG\Rightarrow \angle PBA = \angle PCA$ .
Case 2: $DE\nparallel FG$
Let's say $DE\cap FG=N, EF\cap DG=M, EG\cap DF=K$ and let the second intersection of the circumcircles of triangles $\Delta DFX$ and $\Delta EGX$ be $H$ . When we apply Pascal's Theorem to the hexagon $GAFEPD$ , we get $M$ lies on $BC$ . By using a known lemma we can see that $H$ lies on $NM$ . But we know $H, M \in BC$ . Thus, $N$ lies on $BC$ too. Desargues theorem (for the triangles $EGC$ and $DFB$ ) indicates $ED,GF,CB$ are concurrent $\Leftrightarrow K,P,A$ are collinear. Since $ED,GF,CB$ intersect at $N$ , we get $K,P,A$ are collinear. Now, look at the radical axis of the circumcircles of triangles $\Delta DFX$ and $\Delta EGX$ . We know the radical axis is $XH$ . Since $EDGF$ is cyclic, $FK.KD=EK.KG$ . So the point $K$ lies on the radical axis. Thus, $X,K,H$ are collinear $\Rightarrow A,P,H$ are collinear ( $A,X,K,P$ are collinear). By Brocard Theorem we get, $XK\perp MN\Rightarrow AP\perp BC$ .

Remark 1
Remark 2

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#4 h Mar 19, 2021, 2:52 PM • 1 Y

Y by ehuseyinyigit

mehmetakifyildiz wrote:

so we can conclude that $T$ lies on the nine point circle of the triangle $PBC$ .

It is indeed, the result of the definition of the Poncelet Point on the quadrilateral $ABCP$

This post has been edited 1 time. Last edited by Pqrq, Mar 19, 2021, 2:54 PM
Reason: Added the quadrilateral

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#5 h Dec 6, 2022, 2:19 PM

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BarisKoyuncu wrote:

By using a known lemma we can see that $H$ lies on $NM$ .

Can you tell the lemma?

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