Pentagon with equal-area Triangles

Pentagon

For all those triangles to have equal areas, then all the chords must be parallel to the opposite side. From this, all the angles ${1, 2, 3, 4, 5}$ must be as shown in the figure above. Now, given a vertex, say, $A$, which is the sum of angles ${4, 1, 3}$, for this to be obtuse, then angle $a$ must be $<90$. If this is so, then angles $b$ and $e$ cannot be $<90$. If one of the remaining angles $b, c$ is $<90$, then the other cannot be. Thus, at most, only $2$ vertices can be acute, which means at least $3$ of the triangles must be obtuse.

I'm not sure if this is what is to be proven, but certainly no triangle can have more than one obtuse angle.

I have found a way to construct such a pentagon which has NO obtuse angles: http://i60.tinypic.com/2yocn4w.png

This pentagon has 3 acute angles and 2 reflex angles, none of which are obtuse (an obtuse angle is an angle $\theta$ such that $90 < \theta < 180$).

In the diagram, let $\angle BAE = \angle ABC = x = 30$, $\angle CDE = 2y = arcsin \left(\frac{\sqrt{3}}{4}\right)$, $z = 90+x+y$, $AB = 1$, $AE = BC = m = \frac{sin(x)sin(2y)}{cos^2 (x+y)}$, and $CD = DE = n = \frac{sin(x)}{cos(x+y)}$.

By symmetry we see that $[ABC] = [EAB]$ and $[DEA] = [BCD]$.

Now, we see that $[EAB] = \frac{m*1*sin(x)}{2}$ and, by definition of $z$ and $n$, we have:

$[DEA] = \frac{m*n*sin(z)}{2} = \frac{m*n*sin(90+x+y)}{2} = \frac{m*(n*cos(x+y))}{2} = \frac{m*sin(x)}{2} = [EAB]$

Thus $[DEA] = [EAB]$.

Now by definition of $m$, $n$ and $z$, notice that $m*sin(z) = m*cos(x+y) = \frac{sin(x)sin(2y)}{cos(x+y)} = n*sin(2y)$. Therefore,

$[DEA] = \frac{n*(m*sin(z))}{2} = \frac{n*(n*sin(2y))}{2} = [CDE]$

Therefore, $[ABC] = [BCD] = [CDE] = [DEA] = [EAB]$. Finally we need to show that this pentagon is "anatomically correct" - i.e. that the defined side lengths and angles create a correct pentagon. To do this, we simply need to show that $m*cos(x) + n*sin(y) = \frac{1}{2}$ (so that the horizontal components match up correctly).

$m*cos(x) + n*sin(y) = \frac{sin(x)cos(x)sin(2y) + sin(x)sin(y)cos(x+y)}{cos^2(x+y)}$

$= \frac{\frac{1}{2} * \frac{\sqrt{3}}{2} * \frac{\sqrt{3}}{4} + \frac{1}{2}sin(y)cos(30+y)}{cos^2(30+y)}$

$= \frac{3 + 8sin(y) \left[\frac{\sqrt 3}{2} cos(y) - \frac{1}{2} sin(y) \right]}{16 \left[\frac{\sqrt{3}}{2} cos(y) - \frac{1}{2} sin(y) \right]^2}$

$= \frac{3 + 2\sqrt 3 sin(2y) - 4sin^2(y)}{12cos^2(y) - 8 \sqrt 3 cos(y)sin(y) + 4sin^2(y)}$

$= \frac{3 + \frac{3}{2} - 4sin^2(y)}{12 - 4 \sqrt 3 sin(2y) - 8sin^2(y)}$

$= \frac{\frac{9}{2} - 4sin^2(y)}{9 - 8sin^2(y)} = \frac{1}{2}$

Therefore, this pentagon satisfies the necessary conditions and has no obtuse angles. Therefore the least possible amount of obtuse angles is zero.