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

Question

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

Answer 1

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 $\angle BPA=\angle BAP = 70^\circ$.

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

Answer 2

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