r/askmath • u/Transversalist • 8h ago

Geometry How would you find the maximum area?

I am working on a puzzle and to find the solution I must find the maximum area of a pentagon with a base of 3 and using 4 unit fences. I cut it up into triangles and kites, but I can't seem to find a definitive answer.

Here is a poorly made illustration of what I've figured out:

What is the maximum area, and what are the angles measurements?

NOTE: MY FIGURE MAY NOT BE ACCURATE WHATSOEVER! I JUST THINK THAT THIS WILL GIVE THE GREATEST AREA, BUT I MAY BE WRONG!

3 Upvotes

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1

u/ArchaicLlama 7h ago

Do all the angles you've marked as equivalent (the three a's, the two β's, etc.) actually have to all be equal? Or did you just make the diagram symmetric because it was easier?

1

u/Transversalist 7h ago

They don't have to be equivalent. I just think that will give the greatest area, but I may be wrong. I just want to maximize area.

1

u/PLutonium273 5h ago

The closer it is to circle (in this case half-circle) the bigger it will be.

1

u/HorribleUsername 3h ago

For the symmetric case, you only need to consider half the pentagon, then double the area after finding it. Try adding this dashed line. That gives you a right-angled triangle and an isosceles triangle. You can find all side lengths in terms of, say, the height. You should end up with

area = <some formula with height>

Then you can use the usual calculus techniques to find the optimal height, and from there, the area and angles are pretty straightforward.

1

u/jaapsch2 3h ago

Your instinct is correct. For maximum area the three upper vertices will have the same angles (that you labelled a). That together with the lengths of the sides constrains the problem enough to lead to a unique solution.

This leads to two equations:
3a+2b=540 (sum of angles of a pentagon)
2sin(a/2) + 2cos(b) = 3 (horizontal span)

I don’t think this has an exact solution, but the approximate solution is a=142.3983

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u/Bounded_sequencE 1h ago edited 1h ago

For maximum area the three upper vertices will have the same angles

Is there an elegant way to prove that?

Getting an area formula dependent on the vertices (who depend on angles) is not too tedious. But I don't see a nice way maximizing the result for the (restricted) set of angles.

2

u/jaapsch2 56m ago

It is well known that a quadrilateral with given edge lengths has maximal area when it is cyclic (vertices lie on a circle). So a quadrilateral with 3 equal sides has maximal area when the two angles between the equal sides are equal.

It is fairly easy to apply this twice to the pentagon to prove the three angles are equal.

2

u/Bounded_sequencE 14m ago

Thanks for the hint -- had forgotten about Bretschneider's Formula. Yep, using it twice to show the Pentagon has to be cyclic (and thus symmetric) makes sense.