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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 :)


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closed as off-topic by Joel Reyes Noche, Mario Carneiro, Claude Leibovici, Bookend, Arctic Char Mar 17 '15 at 8:10

  • This question does not appear to be about math within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

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
I'm voting to close this question as off-topic because it was asked while it was a question from an on-going contest. – Joel Reyes Noche Mar 17 '15 at 4:14
up vote 5 down vote accepted

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. enter image description here

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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

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