61plus wrote: Given circle $O$ with radius $R$, the inscribed triangle $ABC$ is an acute scalene triangle, where $AB$ is the largest side. $AH_A, BH_B,CH_C$ are heights on $BC,CA,AB$. Let $D$ be the symmetric point of $H_A$ with respect to $H_BH_C$, $E$ be the symmetric point of $H_B$ with respect to $H_AH_C$. $P$ is the intersection of $AD,[color=\#FF0000]B[/color]E$, $H$ is the orthocentre of $\triangle ABC$. Prove: $OP\cdot OH$ is fixed, and find this value in terms of $R$. Best regards, sunken rock

We will first show that $O,H,P$ are collinear. Note that since $\angle DH_CA=\angle AH_C-\angle DH_CH_B=\angle H_AH_CB-\angle H_AH_CE=\angle BH_CE$, and $\frac {AH_C}{DH_C}=\frac {AH_C}{H_AH_C}=\frac {H_BH_C}{BH_C}=\frac {EH_C}{BH_C}$ therefore $\triangle DAH_C\sim \triangle BEH_C\Rightarrow P,E,A,H_C$ are concyclic$\Rightarrow BE*BP=BH_C*BA=BH*BH_B\Rightarrow P,E,H_B,H$ are concyclic$\Rightarrow \angle PHH_B=\angle PEH_B$. Hence we just have to prove $\angle FHH_B=\angle BEH_B$. Construct the nine point center $F$ of $ABC$, which we know lies on the midpoint of $HO$ and is also the circumcenter of $H_AH_BH_C$. Since $\angle H_CFH_B=90-\angle H_CH_AH_B=\angle H_AH_BE\Rightarrow \angle FH_BH=\angle EH_BH$ and $\triangle H_BFH_C\sim \triangle H_BH_AE$, therefore it now just suffices to show $\triangle H_BHF\sim \triangle H_BEB$$\iff HH_B*BH_B=FH_B*EH_B=H_AH_B*H_CH_B$ which is true since $\triangle HH_CH_B\sim \triangle H_ABH_B$. Now since $BO\perp H_AH_C$, therefore $EH_B\parallel BO\Rightarrow \angle BHO=\angle PEH_B=\angle PBO\Rightarrow OH*OP=\boxed {R^2}$

It is easy to observe that $\angle AH_{C}D=\angle EH_{C}B$ and $\Delta AH_{C}H_{B} \sim \Delta H_{A}H_{C}B \Rightarrow \frac{AH_{C}}{H_{C}H_{B}}=\frac{H_{A}H_{C}}{BH_{C}}\Rightarrow \frac{AH_{C}}{H_{C}E}=\frac{DH_{C}}{BH_{C}}\Rightarrow \Delta AH_{C}D=\Delta EH_{C}B\Rightarrow AH_{C}EP$ and $PDH_{C}B$ are cyclic $\Rightarrow BE.BP=BH_{C}.BA=BH.BH_{B} \Rightarrow H_{B}{HEP}$ is cyclic .Similarly $H_{A}HDP$ is cyclic. Observe that $OA||DH_{A} \Rightarrow \angle C-\angle B=\angle OAH=\angle HH_{A}D=\angle HPD$ .Similarly $\angle HPE=\angle C-\angle A$. Now extend $OH$ and let $L$ be the point on ray $OH$ and $R^2=OH.OL$ .Now it easily follows that $ALPB$ is on a circle and $ALPH$ are on a circle $\Rightarrow AHPB$ is on circle $\Rightarrow \angle AHB=\angle APB$ Which results $\angle C =90$ #.$\Rightarrow L\equiv P$ .The rest follows easily

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Let $ P^* $ be the image of $ H $ under the inversion WRT $ \odot (O). $ Consider the inversion with center $ A $ with factor $ AH \cdot AH_A $ and denote $ Z' $ as the image of $ Z. $ Since $H_A' $ $ = $ $ H, $ $ H_B' $ $ = $ $ C, $ $ H_C' $ $ = $ $ B, $ so the image of $ H_BH_C $ is $ \odot (ABC), $ hence $ D'=P^* $ $ \Longrightarrow $ $ AD $ passes through $ P^*. $ Similarly, we can prove $ P^* $ $ \in $ $ BE, $ so $ P^* $ $ \equiv $ $ P. $ i.e. $ O, $ $ P, $ $ H $ are collinear and $ OP $ $ \cdot $ $ OH $ $ = $ $ {R}^2 $ Remark: $ P $ is $ X_{484} $ (First Evans perspector) of $ \triangle H_AH_BH_C $

My solution:It is easy to prove that triangles $BEHc$,$ADHc$ and $BAP$ are pairwaise similar and $BP/AP=AH/BH$.Now,let $F$ be such that $AOF$ is similar to $ABH$($F$ and $H$ are from distinct sides of $AO$,now we have $<FOB=<APB$(easy angle chase) and $AH/BH=AO/OF=BO/OF=BP/AP$ so we get $BOF$ and $BAP$ are similar.Now,spiral similarity with center $A$ sends $AOH$ to $AFB$ and spiral similarity with center $B$ sends $BAF$ to $BOP$ so we get triangles $AOH$ and $BOP$ are similar and from this we get $OP*OH=OA*OB=R^2$,so done.

Let $AD\cap OH=P'$ and $A'$ be the reflection of $A$ wrt $H_BH_C.$ Now $A,H,O$ are $H_A$-excenter, incenter, Bevan point of $\triangle H_AH_BH_C$ so $HOA'H_A$ is cyclic (well-known). Thus $\angle OHA=\angle OA'H_A=\angle PA'O \Rightarrow \triangle OAH \sim \triangle OP'A \Rightarrow OH\cdot OP=OA^2=R^2.$ Similarly, let $BE\cap OH=P''$ then $OH\cdot OP''=R^2$ hence we conclude $P''$ coincides with $P',$ implying $P'$ coincides with $P$ and we get $OP\cdot OH=R^2$.