Find an angle in the figure defined by a equilateral triangle and a regular pentagon

Question

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.

Answer 1

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 $$\angle CMD = 180^\circ - \angle MDC - \angle DCM = 24^\circ.$$

Answer 2

Hint: without loss of generality, we can set $MN=1$, then the altitude $MK$ of $\triangle{NMP}=\frac{\sqrt{3}}{2}$. This altitude will also be the altitude of the $\triangle{CMD}$ and the bisector of the angle in question. Now using law of sines and $\triangle{NBC}$, we can obtain that $$CD=\frac{1}{1+4\frac{sin \: 48°}{\sqrt{3}}}$$ Now that we know base and altitude in the $\triangle{CMD}$, it's not difficult to find the required angle.