ADH' = APF iff ABP = CBM

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ADH' = APF iff ABP = CBM

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Source: DeuX Mathematics Olympiad 2020 Shortlist G2 (Level I Problem 2)

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#1 h Jul 12, 2020, 6:00 PM • 1 Y

Y by Mango247

Let $ABC$ be an acute triangle such that $D,E,F$ lies on $BC, CA, AB$ respectively and $AD,BE,CF$ be its altitudes. Let $P$ be the common point of $EF$ with the circumcircle of $\triangle ABC$ , with $P$ on the minor arc of $AC$ . Define $H' \not= B$ as the common point of $BE$ with the circumcircle of $\triangle ABC$ . Prove that $\angle ADH' = \angle APF$ if and only if $\angle ABP = \angle CBM$ where $M$ denotes the midpoint of $AC$ .

Proposed by Orestis Lignos, Greece

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#2 h Jul 12, 2020, 6:02 PM • 5 Y

Y by Aryan-23, amar_04, RudraRockstar, adityaguharoy, parmenides51

My problem!

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#3 h Jul 12, 2020, 9:39 PM • 1 Y

Y by Muaaz.SY

Suppose first that $BP$ is $B-$ symmedian on $\triangle ABC$ . So $ABCP$ is harmonic, and let $G$ be the intersection of the tangents to $(ABC)$ at $A, C$ and $BP$ and $I=AC\cap BP$ . Notice that $\angle AEF=\angle ABC=\angle GAC$ , so $EF\|AG$ . Since $ABCP$ harmonic we have $(A, P; I, G)=-1 \implies A(F, P; E, G)=-1$ , but $EF\|AG$ , hence $E$ is midpoint of $FP$ . Notice now that $\triangle BFE\sim \triangle DHE$ , indeed $\angle BFE=\angle DHE=180^{\circ}-\angle C$ and $\angle FBE=\angle HDE$ . Then $\frac{EF}{EB}=\frac{EP}{EB}=\frac{EH}{ED}=\frac{EH'}{ED}$ , also easy angle chasing leads to $\angle H'ED=\angle PEB$ . With this we notice that there is a clear spiral similarity centered at $E$ that sends $H'D \mapsto PB$ ans it is known that $J=H'D\cap PB$ lies in both $(EH'P), (EDB)$ so $EH'PJ, AEJDB$ are cyclic. Finally $\angle ADH'=\angle ADJ=\angle ABJ=\angle ACP=\angle PAG=\angle APF$ .

Now, suppose $\angle ADH' = \angle APF$ , let $G$ be the intersection of the tangent to $(ABC)$ at $A$ and $BP$ , $I=AC\cap BP$ and $J=H'D\cap PB$ , notice that $\angle AEF=\angle ABC=\angle GAC$ , so $EF\|AG$ , then $\angle APF=\angle GAP=\angle PCA=\angle PBA=\angle H'DA$ , so $AEJDB$ is cyclic. Also $\angle H'EP=\angle HEF =\angle BED=\angle BJD=\angle H'JP$ and $EH'PJ$ is cyclic as well. With this we notice that there is a clear spiral similarity centered at $E$ that sends $H'D \mapsto PB$ , then $\frac{EH'}{ED}=\frac{EP}{EB}$ . Notice now that $\triangle BFE\sim \triangle DHE$ , indeed $\angle BFE=\angle DHE=180^{\circ}-\angle C$ and $\angle FBE=\angle HDE$ , hence $\frac{EH}{ED}=\frac{EF}{EB}=\frac{EH'}{ED}=\frac{EP}{EB} \implies EF=EP$ Now just see $A(F, P; E, G)=-1 \implies (A, P; I, G)=-1$ and $ABCP$ harmonic with $BP$ the $B-$ symmedian of $\triangle ABC$

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#4 h Jul 13, 2020, 4:53 AM • 2 Y

Y by mkomisarova, Muaaz.SY

Part 1: $\angle CBM = \angle ABP$ implies that $\angle ADH'=\angle APF$
Clearly $BP$ is symedian, therefore $(P,B;A,C)=-1$ . Note that:
$-1 = (P,B;A,C) \overset{H'}{=} (PH' \cap AC, E;A,C)$ On another hand:
$(FD \cap AC, E;A,C) =-1$ We conclude that $PH', AC, FD$ are concurrent. Denote this point of concurrency by $X$ . Then from PoP follows:
$XF \cdot XD = XA \cdot XC = XH' \cdot XP$ This implies that $DFH'P$ is cyclic quadrilateral. Therefore $\angle FDH'= \angle FPH'$ . It also well - known that $AC$ is perpendicular bisector of $HH'$ , therefore:
$\angle H'PA= \angle ACH'= \angle FCA = \angle FDA$ As a result:
$\angle ADH' = \angle FDH' - \angle FDA = \angle FPH' - \angle H'PA = \angle APF$
as desired.

Part 2: $\angle ADH'=\angle APF$ implies $\angle CBM = \angle ABP$
It also well - known that $AC$ is perpendicular bisector of $HH'$ , therefore:
$\angle H'PA= \angle ACH'= \angle FCA = \angle FDA$ As a result:
$\angle FDH' =\angle ADH' + \angle FDA = \angle APF + \angle H'PA =\angle FPH'$ This implies that $DFH'P$ is cyclic quadrilateral. Now by radical axis theorem on $\odot(ABC), \odot (DFH'P), \odot( ACDF)$ , we have that $AC,PH', FD$ are concurrent. Denote point of concurrency by $X$ . Then:
$-1=(X,E;A,C) \overset{H'}{=} (P,B;A,C)$ We conclude that $BP$ is symmedian and as a result $\angle CBM = \angle ABP$ as desired.

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