Particular Pentagon Point

Matt, We let the side of the pentagon equal 1, then used law of sines to find EM. After we had EM, we used the law of cosines to find DM. It turned out that DM=1. This, by symmetry of the figure, meant that CM was 1, which makes the triangle DCM equilateral. Thanks for sharing this great problem and great website. Jeff (PROMYS 2002-2009)

Drawing a sketch, you might suspect that the desired angle measure is 60°. This happens to be correct, but how to prove it? We can do so by reasoning "backwards"--that is, begin with a point N inside the pentagon such that $\Delta DNC$ is an equilateral triangle, then show that N coincides with our point M .

Let N be such a point. Since the angles of $\Delta DNC$ are all 60°, both $\angle BCN$ and $\angle EDN$ measure $108^{\circ}-60^{\circ}=48^{\circ}$ . Since the segments $BC$ , $CN$ , $ED$ , and $DN$ are all congruent, $\Delta BCN$ and $\Delta EDN$ are both isosceles. This allows us to find that $\angle CBN = \angle DEN = 66^{\circ}$ , and therefore $\angle NBA = \angle NEA = 108^{\circ}-66^{\circ}=42^{\circ}$ .

Since there is only one point inside the pentagon satisfying the given properties (forming the two 42° angles), the points M and N must coincide. (That claim may require more justification--feel free to provide it for me!) Therefore the desired angle is, indeed, 60°.

From the properties of a regulat pentagon: < EAB =108 and from symmetry <MAB=54. In addition <CBM=108-42=66. Consequently from the triangle MAB we get <AMB=84. Since in the triangle MAB:AB/MB =sin(84)/sin(54) and in the triangle CMB: BC/MB=sin(<CMB)/sin(180-66-<CMB). Therefore we get sin(84)/sin(54)=sin(<CMB)/sin(114-<CMB)=1.2293. By rearranging this expression we get that tan(<CMB)=1.2293 sin(114)/(1+1.2293 cos(114)). then <CMB=66. Since 2 <CMB+2 <BMA+<DMC=360. Then <DMC=360-168-132=60.

Let Angle DME=CMB=m Let AngleDMC=n then 2m+n=192. Let ED=a. Then DM=a/2 cosec n/2 Now In triangle DME DM=a/2 cosec n/2 & ED=a Angle DME=m & Angle DEM=66. Apply sin rule

Let side length of pentagon be 2 units for convenience.

R = 1/ Cos 54 d

H = Tan 54 d

r = 2 Sin 42 d/ Sin 84 d

h = R + H - r = Tan 54 d + Sec 54 d - 2 Sin 42 d/ Sin 84 d = Sqrt (3)

Tan (x/ 2) = 1/ h = 1/ Sqrt (3)

x = 2 Atan [1/ Sqrt(3)] = 60 degrees.

In this question, only manipulation of angles could not give the answer wanted. I think 42 degrees for 60 degrees is a made figure since our logic tells that only whole numbers will be there. Not obvious and answer in a minute made me incorrect. Unless we knew this, I think an honest man won't be able to answer in a minute.