Find the $PMC$ angle in the construction below

Question

In an isosceles triangle $ABC$, where $AB=AC$, angle $A$ measures $40$ degrees, $BP$ is plotted with $P$ in $A$C, and the angle $ABP$ measures $20$ degrees. $A$ point $M$ is taken in $BP$ so that $AP=PM$, let $"x"$ be the $PMC$ angle. The sum of digits of $"x"$ is?(Answer:8)

I found a trigoometric solution. Would there be a geometric solution?

$x = AP, l=AB=A, \alpha = \angle PMC$

$\dfrac{\sin(120^o )}{l}=\dfrac{\sin(20^ o)}{x} \Longrightarrow x=\dfrac{2\sqrt{3}l}{3}\sin(20^o)$

$\triangle CPM:\dfrac{\sin(\alpha)}{l-x}=\dfrac{\sin(120^o - \alpha)}{x} \Longrightarrow \sin(120 ^o - \alpha)-\dfrac{2\sqrt{3}}{3}\sin(120 ^o - \alpha) \sin(20^o)=\dfrac{2\sqrt{3}}{3} \sin(20 ^o) \sin(\alpha)$

$\sin(120^o - \alpha)=\sin(60 ^o + \alpha)$

$\sin(60 ^o+\alpha)(\sin(60^o)-\sin(20^o))=\sin(20^o)\sin(\alpha)$

$\sin(60^o)-\sin(20^o)=2\sin(20^o)\cos(40^o)$

$\sin(60^o+\alpha)\cos(40^o)=\dfrac{1}{2}(\sin(20^o+\alpha)+\sin(100^o +\alpha))$

$\sin(100^o+\alpha)=\cos(10^o+\alpha)$

$\therefore \cos(10^o +\alpha)=\sin(\alpha)-\sin(20^o +\alpha) $

$\sin(\alpha)-\sin(20^o+\alpha)=2\sin(-10^o)\cos(10^o+\alpha)$

$\therefore-2\sin(10^o)\cos(10^o+\alpha)=\cos(10^o+\alpha)$

$If \cos(10^o+\alpha) \neq 0 \implies \sin(10^o)=-\dfrac{1}{2} \nexists$

$\therefore \cos(10^o+\alpha)=0 \Longrightarrow 10^o+\alpha=90^o \Longrightarrow \boxed{\alpha=80^o}$

Answer 1

Consider $O$, the circumcentre of $\triangle AMB$. $\triangle OAB$ is equilateral.

Now, $\angle AOC=40^{\circ}= \angle AOM$($AO=AC$) and so, $M$ lies on $OC$. From here, $\angle PMC=80^{\circ}$

Answer 2

I like @Limestone's solution. But I got curious about finding another and came up with this:

Since we have an isosceles triangle, we can use its symmetries to create the general construct shown below. Note the presence of several isosceles triangles, especially the 80-80-20s which will come in handy, and the formation of a 60-degree angle in each base.

Embed the problem information: we have a cevian marking off 20 degrees from a base angle hitting the side at P with angles of 60 and 120. Within the construct we see an equilateral has been formed. Rather than knowing that AP=MP and proving the angle, I will assume the angle of 80 and prove that AP=MP. I just realized I didn't mark the relevant 80 in the figure, but fill in corners and it is 80. You don't need that particular angle for the proof.

The angle at P is ripe for bisection. Do so and we have an equilateral triangle with AP=AZ and we also see that AD and BP are parallel, and that the parallelogram DXPZ implies that DZ=XP. If we can remove AP from DZ, we can use the parallels to see what remains of XP, hoping that XM remains (so that as the problem states, removal of AP from BP leaves BM).

We need a way to take another AP off of DZ. The parallelogram shows AP=ZP=DX. We can create another DX: Use the symmetry of the construction to create another cevian from the upper triangle, establishing DK=AP. So DZ-AP=DZ-DK=KZ. If KZ=XM, we have our proof.

By the symmetry of the construction the central triangles are isosceles, and the angles of KX and ZM are both 60 with the parallels DA and BP. Hence KZMX is a parallelogram and KZ=XM.

Thus, the cevian BP cuts off MP=AP.