P on altitude and concurrent lines

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P on altitude and concurrent lines

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Source: Moldova TST 2017, B3

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Snakes

#1 h Mar 6, 2017, 7:42 PM • 3 Y

Y by microsoft_office_word, Adventure10, Mango247

Let $\omega$ be the circumcircle of the acute nonisosceles triangle $\Delta ABC$ . Point $P$ lies on the altitude from $A$ . Let $E$ and $F$ be the feet of the altitudes from P to $CA$ , $BA$ respectively. Circumcircle of triangle $\Delta AEF$ intersects the circle $\omega$ in $G$ , different from $A$ . Prove that the lines $GP$ , $BE$ and $CF$ are concurrent.

This post has been edited 1 time. Last edited by Snakes, Mar 6, 2017, 7:50 PM

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#2 h Mar 6, 2017, 7:56 PM • 1 Y

Y by Adventure10

Outline of the solution

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#4 h Mar 6, 2017, 10:29 PM • 1 Y

Y by Adventure10

Let $H=AP\cap BC$ , since $PFBH$ and $AEPF$ is cyclic we get $\measuredangle ABC=\measuredangle FPA=\measuredangle FEA$ $\Longrightarrow$ $BCEF$ is cyclic, let $A'$ the antipode of $A$ wrt $\omega$ since $PF\parallel A'B$ and $PE\parallel CA'$ we get the perpendicular bisectors of $BF$ and $CE$ intersect in $PA'$ be that point $O$ $\Longrightarrow$ $O$ is the circumcenter of $\odot (BCEF)$ $\Longrightarrow$ $O,P, A'$ are collinear, it so easy note that $G=\overline {OPA'}\cap \omega$ . On the other hand by radical axis in $\odot (AFE)$ , $\odot (BCEF)$ and $\omega$ we get $AG$ , $FE$ , $BC$ are concurrent in $X$ , let $Y$ $=$ $BE$ $\cap$ $CF$ $\Longrightarrow$ by Brocard's theorem in $\odot (BCEF)$ we get $OY$ $\perp$ $AX$ , but $OG$ $\perp$ $AY$ hence $O$ $,$ $G$ $,$ $Y$ are collinear $\Longrightarrow$ $O$ $,$ $P$ $,$ $G$ $,$ $A'$ $,$ $Y$ are collinear hence $GP$ , $BE$ and $CF$ are concurrent in $Y$ .

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#5 h Apr 3, 2017, 6:05 PM • 1 Y

Y by Adventure10

This is a very common miquel configuration.
$\angle AEP + \angle AFP = 90^{\circ}+90^{\circ} = 180^{\circ}$ impies $AEPF$ cyclic.
Similarly, we get $PDCE$ and $PDBF$ cyclic where $D$ is the foot of the perpendicular from $A$ to $BC$ . From this we get $BCEF$ cyclic and $G$ its miquel point.
Let $BE \cap CF = X$ . Also, let the center of $BCEF$ be $O$ . From the miquel configuration, we have $G, X, O$ colinear. So, we reduce the problem into showing $X, P, O$ colinear.
$\angle FEP = \angle FEB - \angle PEX = \angle FCB - 90^{\circ} + \angle BEC = \angle FCB - \angle BCO = \angle FCO$ . So, $EP$ meets $CO$ on the circle. Let this point be $Y$ . Analogously, $FP$ meets $BO$ at $Z$ which is also on the circle.
Now we apply pascal's theorem to hexagon $BZFCYE$ . We get our conclusion.

This post has been edited 1 time. Last edited by Itticantdomath, Apr 3, 2017, 6:06 PM
Reason: Mistake in latex

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#6 h Aug 19, 2020, 9:42 PM

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Cool problem.
Similar solution as above.
Firstly we have that
$\angle PEA =\angle PFA = 90^\circ$ from where we obtain that $PEFA$ is cyclic. Let $H$ be feet of altitude from $A$ . Now from
$\angle PFB = \angle PHB = 90^\circ$ and $\angle PEC = \angle PHC = 90^\circ$ we obtain that $PFBH$ and $PEHC$ are cyclic too.
Now we are doing some angle chase.
$\angle FBC = 180^\circ - \angle FPH =\angle FPA = \angle FEA = 180^\circ - \angle FEC$ With this we get that $BCFE$ is cyclic. Now let $T = BE \cap CF$ . Let $O$ be circumcenter of $BCEF$ . $G$ is Miquel point of the quadrilateral and the rest is the same as above so there is no point in writing it all over again

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#7 h Oct 2, 2020, 3:55 AM • 1 Y

Y by math31415926535

See my solution on my Youtube channel here. It is very similar to FabrizioFelen's solution, but with a slight difference in the middle (I think his way is actually simpler).

https://www.youtube.com/watch?v=Rl6S_sU7rK4

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#8 h Oct 2, 2020, 8:58 AM

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You can work this up using Hong Kong 2001 Test 1 Q1.
$\definecolor{A}{RGB}{0,90,255}\color{A}\fbox{Solution}$
As I proved there, if you denote by $O$ the circumcenter of quadrilateral $BFEC$ then points $G,P,O$ lie on line perpendicular to $AG$ . Now take $X$ as the intersection of lines $BE,CF$ . By radical axis theorem and Brocard's theorem $OX$ is the line perpendicular to $AG$ . There's only one line perpendicular to $AG$ passing through point $O$ , thus $X\in GP.\blacksquare$
#1802

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#9 h Oct 3, 2022, 12:39 PM

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$BEFC$ is cyclic.So from radical center $AG,EF,BC$ concur at $S$ .
If $BE,CF$ concur at $T$ then $GT,GP$ are perpedicular to $SA$ so $G,T,P$ collinear so we are done

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