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Pentagon with equal-area Triangles

Many of you have come across a certain type of pentagon:

Let us label a pentagon \(ABCDE\). It satisfies that \([ABC]=[BCD]=[CDE]=[DEA]=[EAB]\) where \([XYZ]\) means the area of \(\triangle XYZ\).

The question: What is the least amount of obtuse angles the pentagon must have? Prove your claim.

Note by Daniel Liu
2 years, 10 months ago

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Pentagon

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. Michael Mendrin · 2 years, 10 months ago

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@Michael Mendrin If I may ask, how do you post pictures here? Ariel Gershon · 2 years, 10 months ago

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@Ariel Gershon I picked this up while working on Brilliant. To post a picture, enter the following Latex coding

!\([...title... ]\) \((...http...address...)\) Michael Mendrin · 2 years, 10 months ago

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@Michael Mendrin

Picture

Picture

OK I got it! Thanks!! Ariel Gershon · 2 years, 10 months ago

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@Michael Mendrin Sorry, typoed! I wanted to least amount of obtuse angles the pentagon could have. Daniel Liu · 2 years, 10 months ago

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@Daniel Liu Well, fix it, Daniel. It was a fun problem, once I decided what the problem was. See my motto. Maybe later I'll put up a graphic of such a pentagon with just 3 obtuse angles. Michael Mendrin · 2 years, 10 months ago

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@Michael Mendrin Huh, I fixed it when I commented. Must have been a glitch or something. Fixed it again. Daniel Liu · 2 years, 10 months ago

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@Daniel Liu Here's the case where such a pentagon has 2 right angle vertices

Right Angles Pentagon

Right Angles Pentagon

In working this one out, once one vertex was made a right angle, it works out that there is only one degree of freedom to adjust the angle of the other vertex to make it a right angle, so that making any more vertices a right angle was not possible. As per the proof given above.

The figure given here has a lot of interesting proportions. Michael Mendrin · 2 years, 10 months ago

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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. Ariel Gershon · 2 years, 10 months ago

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@Ariel Gershon That would work, given that the technical definition of an "obtuse angle" is between 90 and 180 degrees, and that the angle of a vertex in any closed polygon, convex or not, is measured on the inside. Michael Mendrin · 2 years, 10 months ago

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@Ariel Gershon Ah, I assumed that the pentagon was convex. Nice job finding a concave pentagon with no obtuse angles though. Daniel Liu · 2 years, 10 months ago

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@Daniel Liu Well, the question didn't specify :P Thanks! Ariel Gershon · 2 years, 10 months ago

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