Trig Ceva kills this problem immediately!

Moreover it can be easily proven that $p_A,p_B,p_C$ are passing through the circumcenter (the case when $ABC$ is right-angled triangle is a special case). The 2nd official solution makes use of the Carnot theorem

By simple angle-chasing we can conclude that $CH_c$ and $p_c$ are symmetric with respect to the angle bisector of $\angle{C}$, and similary for the other ones. Hence, the lines$p_a, p_b, p_c$ meet at the isogonal conjugate of the orthocentre, which is exactly the circumcentre of $ABC$.

yea, trig ceva is insta-solve!! and as stated by stefan its also trivial to prove that each of $p_a,p_b,p_c$ pass through circumcentre of triangle $ABC$

Dear Mathlinkers, by considering the tangent to the circumcircle of ABC at A, and using the Reim's theorem, we have a nice proof... Sincerely Jean-Louis

[asy][asy] size(12cm); pair C=(0,0); pair B=(9,0); pair A=(3,7); pair Ha=foot(A,B,C); pair Hb=foot(B,C,A); pair Hc=foot(C,A,B); pair H=orthocenter(A,B,C); pair O=circumcenter(A,B,C); pair Hap=foot(A,Hb,Hc); D(A--B--C--cycle); D(A--Ha); D(B--Hb); D(C--Hc); D(circumcircle(B,C,Hc)); D(Hb--Hc); D(A--extension(A,foot(A,Hb,Hc),B,C)); D(rightanglemark(Hc,foot(A,Hb,Hc),A)); D(rightanglemark(A,Ha,C)); D(rightanglemark(C,Hb,B)); D(rightanglemark(C,Hc,B)); D(MA(C,A,Ha,.8,blue)); D(MA(O,A,Hc,.8,blue)); pair point = origin; pair[] p={A,B,C,Ha,Hb,Hc,H}; string s = "A,B,C,H_A,H_B,H_C,H"; int size = p.length; real[] d; real[] mult; for(int i = 0; i<size; ++i) { d[i] = 0; mult[i] = 1;} string[] k= split(s,","); for(int i = 0;i<p.length;++i) { dot("$"+k[i]+"$",p[i],dir(O--p[i]),red); } dot("$O$",O,dir(0),red); dot("$H_{A'}$",Hap,dir(240),red); [/asy][/asy] Note that $\angle CH_BC=\angle CH_CB$ so $BCH_BH_C$ is cyclic, hence $\triangle AH_BH_C\sim\triangle ABC$. Let $H_{A'}$ be the foot of $A$ on $H_BH_C$. It follows from the similarity that $\angle H_CAH_{A'}=\angle CAH_A$, so $AH_A$ and $AH_{A'}$ are reflections over the $A$ angle bisector. The same is true for $B,C$. Since $AH_A,BH_B,CH_C$ concur at $H$, their reflections over their respective angle bisectors must concur at the isogonal conjugate of $H$, which happens to be the circumcenter of $\triangle ABC$.

Use that $O,H$ are isogonal and $EF$ is anti-parallel to $BC$

This is a useful result for IMO 2013 Problem 3.

Let $A_1, B_1, C_1$ be the intersection of lines $p_A, p_B, p_C$ with $H_BH_C, H_CH_A, H_AH_B$, respectively. By Carnot's theorem for $\triangle H_AH_BH_C$ and points $A_1, B_1, C_1$, it is enough to prove that ${A_1H_C}^{2}+{B_1H_A}^{2}+{C_1H_B}^{2}={A_1H_B}^{2}+{B_1H_C}^{2}+{C_1H_A}^{2}$. However, the Pythagorean theorem gives us the following relations: ${A_1H_C}^{2}={AH_C}^{2}-{AA_1}^{2}$, ${A_1H_B}^{2}={AH_B}^{2}-{AA_1}^{2}$, ${B_1H_C}^{2}={BH_C}^{2}-{BB_1}^{2}$, ${B_1H_A}^{2}={BH_A}^{2}-{BB_1}^{2}$, ${C_1H_B}^{2}={CH_B}^{2}-{CC_1}^{2}$ and ${C_1H_A}^{2}={CH_A}^{2}-{CC_1}^{2}$. By plugging them in, our claim is equivalent to ${AH_C}^{2}+{BH_A}^{2}+{CH_B}^{2}={AH_B}^{2}+{BH_C}^{2}+{CH_A}^{2}$, which is true by Carnot's theorem for $\triangle ABC$ and points $H_A, H_B, H_C$.

Without loss of generality assume that the altitude from $C$ has unit length, and set the origin at point $H_C$. Then $C(0,1), A(a,0), B(b,0)$, where $a$ and $b$ are real parameters. Now, the equations of $AB, BC$ and $CA$ are $y=0$, $y=\frac{-x}{b}+1$ and $y=\frac{-x}{a}+1$, respectively. Next, we obtain that the equations of $AH_A, BH_B$ and $CH_C$ are $y=bx-ab$, $y=ax-ab$ and $x=0$, respectively. By solving two systems of linear equations, it follows that $H_A(\frac{a{b}^{2}+b}{{b}^{2}+1}, \frac{{b}^{2}-ab}{{b}^{2}+1})$ and $H_B(\frac{{a}^{2}b+a}{{a}^{2}+1}, \frac{{a}^{2}-ab}{{a}^{2}+1})$. Lines $H_AH_B, H_BH_C$ and $H_CH_A$ have slopes $\frac{a+b}{1-ab}$, $\frac{a-b}{ab+1}$ and $\frac{b-a}{ab+1}$. Finally, the equations of lines $p_A, p_B, p_C$ are $y=\frac{ab+1}{b-a}x+\frac{{a}^{2}b+a}{a-b}$, $y=\frac{ab+1}{a-b}x+\frac{a{b}^{2}+b}{b-a}$ and $y=\frac{ab-1}{a+b}x+1$, so it is clear that they pass through a common point $X(\frac{a+b}{2},\frac{ab+1}{2})$.

Why hasn't someone said this yet: Trivial by Carnot's theorem? I know this useless post deserves many down votes, feel free to.

One way to do this is to notice that , if $p_A \cap BC=X; \angle BAH_A =\angle XAC =90^{\circ}-B $ Hence, $p_A \text { and } H_A $ are isogonal lines and hence their intersection, ie, orthocenter is the isogonal conjugate of circumcenter which is indeed where the lines, $ p_A,p_B, p_C $ concur Other solution is: Notice $\Delta H_AH_BH_C $ is orthic triangle, hence, $CH_C,BH_B,AH_A $ are angle bisectors in $\Delta H_AH_BH_C $ Now, angle chase, $$\angle BCH_C =\angle BH_BH_A=90^{\circ}-B= \angle BH_BH_C=\angle BAH_A $$And, $$\angle H_AAC=\angle CH_CH_B=\angle 90^{\circ}-C \implies \angle AH_CH_B=C $$Hence, $\angle BAX =90^{\circ}-C \implies \angle XAC=A+C-90^{\circ}=90^{\circ}-B $ By sine rule, $$\frac {BX}{\sin \angle BAX}=\frac {BX}{\cos C}=\frac {AX}{\sin B} \text { and, } \frac {AX}{\sin C}=\frac {XC}{\sin \angle XAC}=\frac {XC}{\cos B} $$Combining them, $$\frac {CX}{BX}=\frac {\sin 2B}{\sin 2C} $$Similar procedure for, $p_B \cap AC=Y $ and, $p_C \cap AB =Z $, we can either asiign barycentric coordinates for $X,Y,Z $ and get the concurrence point as circumcenter or we can use ceva and prove that $p_A,p_B,p_C $ are concurrent lines

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%201.pdf p. 59... Sincerely Jean-Louis

Fully different Solution from Others.... Stefan4024 wrote: Let $AH_A, BH_B$ and $CH_C$ be altitudes in $\triangle ABC$. Let $p_A,p_B,p_C$ be the perpendicular lines from vertices $A,B,C$ to $H_BH_C, H_CH_A, H_AH_B$ respectively. Prove that $p_A,p_B,p_C$ are concurrent lines. Solution:- Let $AP\perp H_BH_C, BQ\perp H_CH_A$ and $CR\perp H_AH_B$ where $P,Q,R\in H_BH_C,H_CH_A,H_AH_B$ respectively. Now notice that $P$ is the $A-extouch$ point of the $H_A$ excircle of $\triangle H_AH_BH_C$. Similarly we get that $Q,R$ are the $B,C$ extouch point of the $H_B,H_C$ excircles of the the triangle $H_AH_BH_C$ respectively. So, $PH_A,QH_B,RH_C$ concur on the Nagel Point of the orthic triangle of $\triangle ABC$. also $AH_A,BH_BCH_C$ are concurrent. Hence, by Cevian Nest Theorem we get that $p_A,p_B,p_C$ are concurrent. $\blacksquare$. BTW Why is this link not working??? This is #9 of this thread. jlammy wrote: This is a useful result for IMO 2013 Problem 3.

Nice... THANKS

It is well known that $A,B,C$ are excenters of the orthic triangle so the lines just concur at the bevan point of the orthic triangle or the circumcenter of $\Delta ABC$.