# $P$ is a point in the interior of a square $ABCD$, such that $\angle DCP = \angle CAP = 25^\circ$. What is $\angle BPA$?

Moderator Note: this is a question from the Federal Mathematics Competition 2013.

So here's another quite complex problem: $P$ is a point in the interior of a square $ABCD$, such that $\angle DCP = \angle CAP = 25^\circ$. What is $\angle ***PBA***$?

Does anybody have any ideas on this problem? I tried to find as much angles as I could, but I just got stuck...

Hope for some good answers :)

Markus

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P.S.: I'm the German guy, so don't wonder about my language :D –  user55214 Jan 4 '13 at 22:01
Using Geogebra I found 70 degrees. Lets try to prove geometrically. –  Sigur Jan 4 '13 at 22:11
yes, it is really important for me to give good evidence for my answer... I need it for my a levels :/ –  user55214 Jan 4 '13 at 22:13

Consider the circumcircle of $CAP$. Since $\angle CAP = \angle PCD$, it follows that $CD$ is tangential of the circumcircle. Hence, the circumcenter lies on $BC$, which is perpendicular to $CD$ at $D$. Also, the circumcenter lies on the perpendicular bisector of $AC$, which is the line $BD$. Thus, $B$ is the circumcenter of $APC$.

This shows that $BA=BP=BC$, so $BAP$ is an isosceles triangle, which gives that $\angle BPA=\angle BAP = 70^\circ$.

It is easy to figure out $\angle PBA$ given the above.

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great answer! thank you :) –  user55214 Jan 4 '13 at 23:00
Ohhhhhooho I did a big mistake!!! We are looking for "angle PBA" !!! but the rest of the task was correct –  user55214 Jan 4 '13 at 23:07
Well, I'm sure you can figure out what $\angle PBA$ is now ... Hint: sum of angles in triangle is ?? –  Calvin Lin Jan 4 '13 at 23:39
It has to be 40° :D It was too late in Germany to think properly :D –  user55214 Jan 5 '13 at 9:56
@robjohn Is there a way for me to delete my answer, since this problem arises from a live competition? I tried deleting it manually, but it refused to let me because it was accepted. –  Calvin Lin Jan 11 '13 at 9:28

A pictorial proof. An image to illustrate the solution given by Calvin Lin.

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+1 for the edit :) You comment that the answer was 70 helped me figure this out quickly. –  Calvin Lin Jan 4 '13 at 23:05
Ohhhhhooho I did a big mistake!!! We are looking for "angle PBA" !!! but the rest of the task was correct –  user55214 Jan 4 '13 at 23:08
@user55214, $\angle PBA$ is $180-70-25-45$. –  Sigur Jan 4 '13 at 23:14
=40° :D I'm a f'' –  user55214 Jan 4 '13 at 23:32
=40° :D I'm a fu**in' idiot :D –  user55214 Jan 4 '13 at 23:32