Easy complex bash

Let $X\in AB$ such that $HX\parallel BC$ and $M$ midpoint of $AC$ Notice that \[ \angle HDA = \angle MOA =\angle CBA = \angle HXA \]meaning that $HDXA$ is cyclic. Now, $\angle AHX = 90$ so $P$ is the center of $(HDXA)$ i.e. the midpoint of $AX$. The rest is also angle chasing, \[ \angle DPB = 2\angle DAB = \angle DAB +\angle ABO = 180 -\angle AOB \] which implies $OBPD$ is cyclic.

AlanLG wrote: Let $ABC$ an acutangle triangle with orthocenter $H$ and circumcenter $O$. Let $D$ the intersection of $AO$ and $BH$. Let $P$ be the point on $AB$ such that $PH=PD$. Prove that the points $B, D, O$ and $P$ lie on a circle. Let $(BOD)$ meet $AB$ again at $P$. We show $PH=PD$. Indeed, $\angle APD = \angle AOB = 2\angle C$ and $\angle PAD = 90^{\circ}-\angle C$, so $\triangle PAD$ is isosceles. Now $\angle AHD = 180^{\circ}-\angle AHB = \angle C = \tfrac{1}{2}\angle PAD$ hence $P$ is the circumcentre of $\triangle AHD$ so $PH=PD$.

At first I didn't see that just a simple angle chase does the job, so here is a bit more complicated approach (by constructing one additional point). Redefine $P$ as the circumcenter of $\triangle ADH$. We will prove that $P \in AB$. Firstly, notice that this is sufficient, as then $\angle PDA= \angle PAD =\angle ABO$, so $BODP$ is cyclic. Now, add the reflection $B'$ of $B$ in $AC$. Since $\angle BAO =\angle B'AO, \angle BAH = \angle OAC$, we get that $\angle DAH = \angle B'AC$. In addition, $B'$ lies on the circumcircle of $ABC$ and $AB=AB'$, so $\angle ACB' = \angle ACB = \angle AHD$, so $\triangle ADH \sim \triangle AB'C$ (this is actually spiral similarity). But $P$ and $O$ are corresponding circumcenters, so $\angle PAH =\angle OAC =\angle BAH$ and thus $P \in AB$, done.

Just let $P'$ be the circumcenter of $AHD$, then $P'$ is in the perpendicular bisector of $HD$ and $\angle P'AD=90^\circ -\angle DHA=\angle HAC=90^\circ -\angle ACB=\angle BAO=\angle BAD$, so $P'\in AB$. Hence $P\equiv P'$, so from $\angle PAD=90^\circ -\angle ACB$ and $PA=PD$ we find $\angle DPA=2\angle ACB=\angle AOB=\angle DOB$, so $BPDO$ is cyclic.

Alternatively, let $H'$ be the reflection of $H$ wrt $AB$, then $H'\in (ABC)$ and $\angle AH'H=\angle AH'C=\angle ABC$, but $\angle ADH=\angle ADB=180^\circ -\angle HBA-\angle BAD=180^\circ -(90^\circ -\angle BAC)-(90^\circ -\angle ACB)=\angle BAC+\angle ACB=180^\circ -\angle ABC$. Hence $ADHH'$ is cyclic. Now, $P$ is clearly the circumcenter of $HH'D$, so it is the circumcenter of $ADH$, and we are done in the same way as above.

Let $P'$ be the circumcenter of $ADH$, then angle chasing gives $\angle PAH = 90^{\circ} - \angle B = \angle BAH$ whereupon $A, P', B$ are collinear and thus $P = P'$. Then \[\angle ODP = \angle ADP = \angle PAD = \angle BAO = \angle OBA = \angle OBP\] Which proves $ODBP$ is cyclic as desired.

This problem was proposed by Colombia. It was created by one of my students Mateo Lizcano (16 years old) with a bit of my help. I’m quite proud of him. Hopefully you enjoy the problem as much as I do.

$WLOG$ let's suppose that $AB < AC$. We say that $HC$ cut to $\odot (ABC)$ again at $H'$. It's well know that $H'$ is the reflection of $H$ wrt $AB$, so $PH' = PH = PD$. By angle chasing $\angle PHH' = 2\angle B - 90^{\circ} \Rightarrow \angle H'PB = 180^{\circ} - 2\angle B$. Hence $\angle H'OB = 2 (90^{\circ} - \angle B) = 180^{\circ} - 2\angle B = \angle H'PB \Rightarrow H'POB$ is cyclic. Finally $\angle BH'O = 90^{\circ} - \angle H'AB = 90^{\circ} - (90^{\circ} - \angle B) = \angle B = \angle BDO$ (since $\angle DAC = \angle OAC = 90^{\circ} - \angle B$), therefore $B$, $D$, $O$, $P$ and $H'$ lie on a circle $\blacksquare$

JuanDelPan wrote: Easy complex bash barycentric coordinates works as well (maybe it is quicker?)

Let $E$ be the point on $\overline{AB}$ such that $EH \parallel BC$. We first have the following claim. Claim : Quadrilateral $ADHE$ is cyclic. Proof : Clearly, \begin{align*} \measuredangle HDA &= \measuredangle BDA \\ &= \measuredangle DBA + \measuredangle BAD \\ &= (90+ \measuredangle CAB) + (90 + \measuredangle BCA)\\ &= \measuredangle CBA \\ &= \measuredangle HEA \end{align*}which implies that the points $A$ , $D$ , $H$ and $E$ lie on a common circle, say $\omega$. Now clearly, $\measuredangle AHE = 90 ^\circ$ (due to the parallel line definition) and thus $AE$ must be the diameter of $\omega$. But then, note that the midpoint of $AE$ must be the circumcenter of $\omega$ and thus, this point lies on the perpendicular bisector of $DH$. But, by the given condition, $P$ is the intersection of $\overline{AB}$ and the perpendicular bisector of $DH$ which implies that $P$ must indeed be, the midpoint of $AE$. Thus, \[\measuredangle DPB = \measuredangle DPE = 2\measuredangle DAP= 2\measuredangle OAB = \measuredangle AOB = \measuredangle DOB\]which finishes. Remark : One can also easily find other nice things hiding inside this configuration. For example, the circles $(BDO)$, $(ABC)$ and $\omega$ pass through a common point.

It suffices to show that $P$ is the circumcenter of $\triangle ADH$, as then $\angle APD=2\angle AHD=2\angle C=\angle AOB$. From here, it is more natural to think in terms of $\triangle ADH$. Letting $S$ be the circumcenter of $\triangle ADH$, we see that $$\angle SAH=\angle HDA-90=\angle DAC,$$so $AS$ is isogonal to $AC$ in $\triangle AHD$ so $S$ lies on $AB$ as desired.

Let $E$ be the reflection of $H$ wrt $AB$. Then $E\in (ABC)$, $PE=PH=PD$ and $\angle DBP=\angle HBP=\angle PBE$, so $BEPD$ is cyclic. Hence it suffices to prove that $DOBE$ is cyclic. We know that $\angle DBE=2\angle HBA=180^\circ -2\angle BAC$, so it suffices to prove that $\angle AOE=180^\circ -2\angle BAC$. From $OA=OE$ we get that it suffices to prove that $\angle EAO=\angle BAC$, i.e. that $\angle EAB=\angle OAC$. This is true because$$\angle EAB=\angle BAH=90^\circ -\angle CBA=\angle OAC.$$Done.

Trivial? maybe $$\measuredangle CAK =\measuredangle BAD=\measuredangle EBC $$$$\measuredangle ABE=90^{\circ}-\measuredangle DAH-\measuredangle EBK $$$$\measuredangle DPH=2\measuredangle DAH $$Lemma: $PH=PD=PA$ $$\measuredangle PAD=\measuredangle ADP=\measuredangle OBA$$hence we are done

JuanDelPan wrote: Easy complex bash Average complex basher lovers be like

KHOMNYO2 wrote: JuanDelPan wrote: Easy complex bash Average complex basher lovers be like Average MIT student: