# Math Help - Homologic (perspective) pedal triangle

1. ## Homologic (perspective) pedal triangle

Hello everyone!

I have some serious problems with the next trigonometry problem.

I have to find point P such that its pedal triangle is perspective to given triangle ABC (see picture). If I understood correcty, triangles are prespective if extensions of their sides collide in three collinear points. I think that this point P should be in incenter of triangle ABC. Now I have some trouble proving this statement.

I tried to start with two obvious facts:

$\angle C'DE + \angle BDC' + \angle BDF = 180^{\circ}$
$\angle BAP = \angle BC'P = \angle AP'B = 90^{\circ}$

Using these relations, I tried to prove that
$\angle A'C'B = \angle BA'C$
or $\angle AB'C' = \angle AC'B$
or $\angle A'B'C = \angle B'A'C$.

I'm trying to do it for a few last days but I just can't find the solution. I keep turning around the same thing again and again. I'm not even sure if my initial assumptions are enough to prove this or maybe I'm going in completely wrong direction.

I managed to prove this in opposite direction (if I assume that one of these three equations I wrote above are true) but it doesn't actually help a lot.

Please give me some clues, because I'm completely lost. Thanks in advance!

2. Hello Dr. Jekyll
Originally Posted by Dr. Jekyll
Hello everyone!

I have some serious problems with the next trigonometry problem.

I have to find point P such that its pedal triangle is perspective to given triangle ABC (see picture). If I understood correcty, triangles are prespective if extensions of their sides collide in three collinear points. I think that this point P should be in incenter of triangle ABC. Now I have some trouble proving this statement.
...
I think the point you want is the orthocentre, not the incentre. If the orthocentre is denoted by $P$, then $AA', BB', CC'$ all intersect at $P$. Hence the pedal triangle $A'B'C'$ is in perspective with $\triangle ABC$, $P$ being the perspective centre.

You are right about the points of intersection of corresponding sides being collinear, when two triangles are in perspective. This is known as Desargue's Theorem.

3. Originally Posted by Grandad
I think the point you want is the orthocentre, not the incentre.
You're right about the orthocentre. It does give a perspective pedal triangle, but I'm still pretty sure that the incenter is also an option. I know it's not the proper mathematical way, but if I draw it on computer, it perfectly fits (incentre as well as the orthocentre).

Let's assume that point P is located at the incentre of the triangle ABC. Then we know that $\angle A'C'B = \angle BA'C'$ and $\angle AC'B' = \angle AB'C'$ and $\angle B'A'C = \angle A'B'C$ (see the picture). Then we have these relations:

$\angle BDF + \angle BFD = 2 \angle A'C'B \Rightarrow \angle BDF = 2 \angle A'C'B - \angle BFD$

$\angle BDC' + 2 \angle A'C'B + \angle AC'B' = 180^{\circ} \Rightarrow \angle BDC' = 180^{\circ} - 2 \angle A'C'B - \angle AC'B'$

$\angle C'DE + (180^{\circ} - \angle A'C'B -\angle AC'B') + \angle C'ED = 180^{\circ}$
$\Rightarrow \angle C'DE = \angle A'C'B + \angle A'C'B - \angle C'ED$

If we sum these equations we get:

$\angle BDF +\angle BDC' + \angle C'DE =$
$= 2 \angle A'C'B - \angle BFD + 180^{\circ} - 2 \angle A'C'B - \angle AC'B' + \angle A'C'B + \angle A'C'B - \angle C'ED$
$= 180^{\circ} + \angle A'C'B - (\angle BFD + \angle C'ED)$

Since $\angle BFD + \angle C'ED =\angle A'C'B$ we get:

$\angle BDF +\angle BDC' + \angle C'DE = 180^{\circ}$.

This means that points of intersection E, D and F are collinear.

I'm not so sure that all of this is correct but it seems logical to me. But, going the other way around seems a bit tricky...

4. OK, I have another idea how to solve this problem. I've been experimenting with Sketchpad a bit and I believe that there are more possible configurations where pedal triangle of point P is perspective to given triangle ABC than just orthocentre and incentre. (I attached a picture, of course.)

My idea is to somehow express the angles of triangle DEF using some other angles in triangle ABC. Then, if two angles of triangle DEF are 0, we get lines AA', BB' and CC' to collide in one point (perspector). Using these conditions I'm hoping to find all possible locations of point P.

I just want to know if this would be a right method. Everything I tried before had no results. As always, just a few tips, please!

5. Nobody has any idea???

My last post is not so good idea after all... I've just discovered Ceva's theorem and I think it could be helpful, but I still get some confusing equations which don't help me much. They involve trigonometric functions of angles that I am looking for: angles of triangle ABC and angles between sides of triangle and point P.

6. OK, even though nobody's answering, I'll still keep you posted about my progress. Using the Ceva's theorem and numerous trigonometric transformations I got this equation:

$\cos{\alpha}\sin{(2\alpha_{1}-\alpha)}+\cos{\beta}\sin{(2\beta_{1}-\beta)}+\cos{(\alpha+\beta)}\sin{(2\alpha_{1}-\alpha+2\beta_{1}-\beta)}=0$

Where $\alpha$ and $\beta$ are anges of triangle ABC, $\alpha_{1}=\angle C'AD$ and $\beta_{1}=\angle A'BD.$ (See attached pcture!)

Trivial solution of that equation is:
$\alpha_{1}=\frac{1}{2}\alpha, \beta_{1}=\frac{1}{2}\beta.$
This is the case when P is incentre.

Now, I'm trying to find more general solution, e.g. to express angle $\beta_{1}$ in terms of other angles in equation.

7. Hi!

I'm new here, but I'm really interested in your problem, (it's my problem too), so please keep on writing if you find something

8. Have you seen Darboux Cubic -- from Wolfram MathWorld ? It might be useful.

9. OK, I managed to solve this problem using analytical geometry, but there's still a problem. I found equation for a curve that is the locus of point P and when I draw it in Sketchpad everything seems alright. I believe it's the Darboux cubic.

Now, I heard of Neuberg's cubic, which could also be an option. Now, I'm a bit confused about this. Definition says that pedal triangle of point P is in perspective with triangle ABC if and only if it is located on Neuberg's cubic, for which I can't find equation.