The figure below shows a regular $ABCDE$ pentagon inscribed in an equilateral triangle $MNP$ . Determine the measure of the angle $CMD$.
[img]http://4.bp.blogspot.com/-LLT7hB7QwiA/Xp9fXOsihLI/AAAAAAAAL14/5lPsjXeKfYwIr5DyRAKRy0TbrX_zx1xHQCK4BGAYYCw/s200/2012%2Bobm%2Bl2.png[/img]
1. **Identify the given elements and their properties:**
   - We have a regular pentagon \(ABCDE\) inscribed in an equilateral triangle \(MNP\).
   - We need to determine the measure of the angle \(\angle CMD\).

2. **Establish the coordinates and angles:**
   - Let \(CD = z\), and the midpoint \(O(0,0)\) of \(CD\), so \(D\left(\frac{z}{2}, 0\right)\).
   - The internal angles of a regular pentagon are \(108^\circ\), so \(\angle PDE = 72^\circ\).
   - Since \(\angle EPD = 60^\circ\) (as it is part of the equilateral triangle), we have \(\angle DEP = 48^\circ\).

3. **Determine the coordinates of point \(E\):**
   - Point \(E\) can be expressed as \(E\left(\frac{z}{2} + z \cos 72^\circ, z \sin 72^\circ\right)\).

4. **Equation of the line \(PE\):**
   - The slope of \(PE\) is \(\tan 60^\circ = \sqrt{3}\).
   - The equation of the line \(PE\) is:
     \[
     y - z \sin 72^\circ = -\sqrt{3} \left(x - \left(\frac{z}{2} + z \cos 72^\circ\right)\right)
     \]

5. **Determine the coordinates of point \(M\):**
   - Point \(M\) lies on the x-axis, so its y-coordinate is 0.
   - Using the line equation, we find \(M\left(z \sin 72^\circ + \sqrt{3} \left(\frac{z}{2} + z \cos 72^\circ\right), 0\right)\).

6. **Calculate \(\tan \angle ODM\):**
   - Using the coordinates of \(M\) and \(D\), we have:
     \[
     \tan \angle ODM = \frac{z \sin 72^\circ + \sqrt{3} \left(\frac{z}{2} + z \cos 72^\circ\right)}{\frac{z}{2}}
     \]
   - Simplifying, we get:
     \[
     \tan \angle ODM = 2 \left[\sin 72^\circ + \sqrt{3} \left(\frac{1}{2} + \cos 72^\circ\right)\right]
     \]
   - Further simplification:
     \[
     \tan \angle ODM = 2 \sin 72^\circ + \frac{\sin 60^\circ}{\cos 60^\circ} (1 + 2 \cos 72^\circ)
     \]
     \[
     \tan \angle ODM = 2 \left[2 \sin 72^\circ \cos 60^\circ + \sin 60^\circ + 2 \sin 60^\circ \cos 72^\circ\right]
     \]
     \[
     \tan \angle ODM = 2 \left[\sin 60^\circ + 2 \sin 132^\circ\right]
     \]
     \[
     \tan \angle ODM = 2 \sin 60^\circ + 4 \sin 48^\circ
     \]

7. **Simplify using trigonometric identities:**
   - Using the identity \(\sin x = \cos (90^\circ - x)\):
     \[
     \tan \angle ODM = \frac{(2 \sin 60^\circ + 4 \sin 48^\circ) \sin 36^\circ}{\sin 36^\circ}
     \]
     \[
     \tan \angle ODM = \frac{\cos 24^\circ - \cos 96^\circ + 2 \cos 12^\circ - 2 \cos 84^\circ}{\sin 36^\circ}
     \]
     \[
     \tan \angle ODM = \frac{\cos 24^\circ + 2 \cos 12^\circ - \cos 84^\circ}{\sin 36^\circ}
     \]
     \[
     \tan \angle ODM = \frac{2 \sin 54^\circ \sin 30^\circ + 2 \cos 12^\circ}{\sin 36^\circ}
     \]
     \[
     \tan \angle ODM = \frac{\cos 36^\circ + 2 \cos 12^\circ}{\sin 36^\circ}
     \]
     \[
     \tan \angle ODM = \frac{4 \cos^3 12^\circ - 3 \cos 12^\circ + 2 \cos 12^\circ}{3 \sin 12^\circ - 4 \sin^3 12^\circ}
     \]
     \[
     \tan \angle ODM = \frac{4 \cos^3 12^\circ - \cos 12^\circ}{3 \sin 12^\circ - 4 \sin^3 12^\circ}
     \]
     \[
     \tan \angle ODM = \frac{\cos 12^\circ (4 \cos^2 12^\circ - 1)}{\sin 12^\circ (3 - 4 \sin^2 12^\circ)}
     \]
     \[
     \tan \angle ODM = \frac{\cos 12^\circ}{\sin 12^\circ} = \cot 12^\circ = \tan 78^\circ
     \]

8. **Determine the angle \(\angle OMD\):**
   - Since \(\tan \angle ODM = \tan 78^\circ\), we have \(\angle ODM = 78^\circ\).
   - Therefore, \(\angle OMD = 12^\circ\).

9. **Calculate \(\angle CMD\):**
   - Since \(\angle CMD = 2 \times \angle OMD\), we have:
     \[
     \angle CMD = 2 \times 12^\circ = 24^\circ
     \]

The final answer is \(\boxed{24^\circ}\).