Let $M$ be the middle of the segment $[BC]$ and $P$ the middle of the segment $[BM]$. A point $A$ in the exterior of the segment $[BC]$ such that $\angle BAP=\angle PAM=\angle MAC$. Find the measure of the following angles: $\angle BAC, \angle CBA, \angle BCA$?

Can you find the measure of the angles without using Pythagora's theorem?

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Does such point $A$ with those hypothesis exist? How to prove this? – Sigur Sep 15 '12 at 15:17
Don't know about whether Pythagorean Theorem is necessary. I certainly used it in solving the problem. Also used somewhat fancier tools. – André Nicolas Sep 15 '12 at 15:29
Yes, can avoid using Pythagoreaan Theorem directly. Don't know about indirectly. – André Nicolas Sep 15 '12 at 15:39
please, can you write a solution:) thanks – Iuli Sep 15 '12 at 15:41
@Sigur: just take half of an equilateral triangle. Picture can be found below in my answer. – tomasz Sep 15 '12 at 15:58

Denote $BAC$ by $3\alpha$, $CBA$ by $\beta$, $ACB$ by $\gamma$.

I'm not sure if it qualifies, but you can see that from the fact that the angular bisector divides the third side of a triangle in proportion equal to the proportion of the two other sides (angle bisector theorem), so the triangle $AMB$ must be isosceles, by applying it to the triangle $PAC$ you see that the cosine of $2\alpha$ is $\frac {1}{2}$.

So $2\alpha=\pi/3$, $\beta=\pi/2-\alpha=\pi/3$, $\gamma=\pi-\beta-3\alpha=\pi/6$

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Very nice. how did you draw this ? – Belgi Sep 15 '12 at 16:00
@Belgi: Geogebra. It's a good tool to cheat with to solve high school geometry problems. ;-) – tomasz Sep 15 '12 at 16:01
Why $\cos(2\alpha)=1/2$? We know that it is $AP/AC$. – Sigur Sep 15 '12 at 16:04
@Sigur: by angular bisector theorem. Notice that $MC$ is twice as long as $PM$. – tomasz Sep 15 '12 at 16:07

Let $\alpha=\angle BAP$. At least one of the triangles $BPA$, $PMA$ has an angle $\ge\frac\pi2$ at $P$ (and a nonzero angle at $B$ or $M$, respectively). From the sum of angles in a triangle, we conclude $\alpha<\frac\pi2$. Construct $X$ such that $\angle PBX=\frac\pi2-\alpha$ and $XP\perp BC$. Then $\angle BXP=\alpha$ and $A,X$ are on the same side of $BC$. Therefore by the inverse of the inscribed angle theorem, $A$ is on the circumscribed circle of $BPX$. Reflecting the figure at $BP$, we find $\angle PXM=\alpha$, hence $A$ is also on the circumscribed circle of $PMX$. These two circles intersect exactly in $P$ and $X$. From $A\ne P$ we conclude $A=X$ and from this $|AB|=|AM|$

Translate the figure by $\overrightarrow{BM}$. Then $B\mapsto M$, $M\mapsto C$, $A\mapsto A''$. Since $|A''M|=|A''C|$, let $c$ be the circle araound $A''$ through $M$ and $C$. Because $\angle MA''C=2\alpha$, the circle $c$ is the locus of all points $X$ with $\angle MXC=\alpha$. Hence $A\in c$. We conclude that $|BM|=|AA''|=|A''M|=|AB|$, i.e. $BMA$ is equilateral, hence $\alpha=\frac\pi6$. The rest is easy: $\angle BAC=3\alpha=\frac\pi2$, $\angle CBA=\frac\pi2-\alpha=\frac\pi3$, $\angle BCA=\frac\pi2-2\alpha=\frac\pi6$.

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My friend with I work a lot gave me a very interesting solution. For the beginning I will introduce the draw.

Notations: $$BP=x.$$

Because $BP=x$ it follows that $PM=x$ and $MC=2x.$ $M \in[AM)$ and $[AM)$ is bisector of the angle $\angle PAC$ $\Rightarrow$ $MQ \perp AC$ and $MQ=PM$, so $MQ=x.$ So knowing that the triangle $\triangle MQC$ is a rectangular angle and $MC=2MQ$ $\Rightarrow$ the measure of the angle $\angle C=30^{\circ}$. What it remained to prove is easy to demonstrate it. ($\triangle ABM$ is isosceles because $P$ is a middle of $BM$ and $\angle BAP=\angle PAM$ from here the measure of the angle $\angle APM =90^{\circ}$).

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