The Perpendiculars

Geometry Level 4

In acute $\triangle ABC,$ $D, E, F$ are the feet of perpendiculars from $A, B, C$ to $BC, CA, AB$ respectively. Let $P, Q, R$ be the feet of perpendiculars from $A, B, C$ to $EF, FD, DE$ respectively. It turns out that lines $AP, BQ, CR$ are concurrent at a point $X$ within $\triangle ABC.$ Then, $X$ is the:

Details and assumptions

This problem is not original; I got this from one of my friends. I don't know its original source.

Gergonne Point of

\triangle ABC.

circumcenter of

\triangle ABC.

incenter of

\triangle ABC.

Nagel point of

\triangle ABC.

3 solutions

Eloy Machado
Apr 14, 2014

Since $\triangle FDE$ is the orthic triangle of $\triangle ABC$ , the orthocenter of $\triangle ABC$ is the incenter of $\triangle FDE$ . Then, $BE$ is perpendicular to $AC$ (we already knew that!) and angle bissector of $\angle FED$ . It gives $\angle AEP = \angle CER$ .

Since $\triangle AEP$ and $\triangle CER$ are right triangles, $\angle PAE = \angle RCE$ and then we get $\triangle AXC$ is isosceles.

Analogously, we get $\triangle AXB$ is isosceles too.

Therefore, $AX = BX = CX$ and $X$ is the circumcenter of $\triangle ABC$ .

Nice! My solution:

Note that $BCEF$ is cyclic with diameter $BC,$ so $\angle AFP = \angle BCA.$ It follows that $\angle PAB = 90^{\circ} - \angle BCA.$ The circumcenter of $\triangle ABC,$ thus, lies on $AP.$ Analogously the circumcenter of $\triangle ABC$ also lies on $BQ$ and $CR,$ which shows that $X$ is the circumcenter of $\triangle ABC.$

Sreejato Bhattacharya - 7 years, 1 month ago

Raphael Tsiamis
May 19, 2014

It is a corollary of Nagel's Theorem. Q.E.D.

Phil Peters
Apr 13, 2014

It's just an angle chase really, but that is a pretty good problem. Maybe we can say some homothecy stuff about triangle PQR? Can we start up on discussion on that?

Also, can I say, I'm not really 23. I'm 15, and I'm in Olympiad training, and would like to improve my maths abilities, and find out really cool stuff about maths, because I love it, but I can't seem to change my account settings, so , Calvin Lin can you help me with this?

Phil Peters - 7 years, 2 months ago

Hint: Angle chase and show that $\angle PAB= 90^{\circ} - \angle C$ (find the other angles similarly). What can you say about the point of concurrency from this? I'll post my full solution if necessary.

Sreejato Bhattacharya - 7 years, 2 months ago

No, don't worry, I had the full solution when I posted the comment, I was just wondering if the triangle PQR is homothetic to some other major triangles related to ABC, so we could say some cool stuff about it.

Phil Peters - 7 years, 2 months ago

Also, my next comment is that as EF is anitparallel to BC with respect to A, and that the Orthocentre is the isogonal conjugate of the Circumcentre, so the result is trivial.

Phil Peters - 7 years, 2 months ago

@Phil Peters – But to find out stuff about homothecies, we would need to know the angles of the triangle PQR. Since triangles ABC and PQR are in perspective, we maybe can say some Desargues stuff.

Phil Peters - 7 years, 2 months ago

The Perpendiculars

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