TITLE: Find an angle in the figure defined by a equilateral triangle and a regular pentagon
QUESTION [6 upvotes]: Given the figure below, were the triangle MNP is equilateral and the pentagon ABCDE is regular, find the angle $\angle$CMD.


Background: I am a 9th grader that has some experience in math contests. This is question 4 (level 2) from the third stage of the 2012 Brazilian Math Olympic (OBM). The answer was not given.
My attempt:
(1)The triangle MNP is equilateral, so:
$$\angle MNR=\angle NRM=\angle RMN=60°$$
$$\overline {MN}=\overline {NR}=\overline {RM}$$
(2)The pentagon ABCDE is regular, so:
$$\angle A=\angle B=\angle C=\angle D=\angle E=a$$
$$a=\frac {180°(n-2)}{n}$$
$$n=5$$
$$a=\frac{180°(3)}{5}=108°$$
(3)The figure is symmetric, with the axis of symmetry defined by M and the midpoint of $\overline {NP}$, so:
$$\triangle BCN=\triangle EDR$$
$$\overline {MB}=\overline {MD}$$
$$\overline {MC}=\overline {MD}$$
Using (1), (2), (3) and the knowledge that the sum of the internal angles of a triangle is 180°, I deduced that:
$$\angle NBC=\angle DER=48°$$
$$\angle BCN=\angle EDR=72°$$
$$\angle MBA=\angle MEA=24°$$
I tried to prove that the $\overline {CD}$ is congruent to $\overline {ND}$ and $\overline {DR}$, but discovered it couldn't be true because $\triangle BNC$ and $\triangle EDR$ are scalene (the angles are different), and the side $\overline {BN}$ of $\triangle BNC$ and the side $\overline {ER}$ of $\triangle EDR$ are already congruent to $\overline{CD}$, and because the triangles are scalene, the other sides cannot be equal to $\overline {CD}$.
Then I got stuck. Any help is appreciated.

REPLY [5 votes]: Note that $BM=BE=BD$ so $B$ is the circumcenter of $\triangle DEM$. It follows that
$$\angle EDM = \frac 12 \angle EBM = \frac 12 \cdot 60^\circ = 30^\circ.$$
Now,
$$\angle MDC = \angle EDC - \angle EDM = 108^\circ - 30^\circ = 78^\circ.$$
By symmetry $\angle DCM = 78^\circ$. Therefore
$$\angle CMD = 180^\circ - \angle MDC - \angle DCM = 24^\circ.$$