Largest pentagon with side lengths 5, 5, 5, 6, 8

[jemidiah]

I ran across a very interesting question a few days ago. Given a pentagon that has side lengths 5, 5, 5, 6, and 8, what is the largest area the pentagon can enclose?

The answer is 3*(25*sqrt(3)/4) + 12 + 12 = 24 + 75*sqrt(3)/4, or approximately 56.476.

I'll post the reasoning if anyone is interested. Even just finishing off the last bit (as I did just now) was interesting. Whoever came up with this question chose it very carefully; what a wonderful problem. As a hint, I used the fact that for a given perimeter, the largest area you can enclose is enclosed by a circle. [Formally, this is known as the isoperimetric inequality.]

[Milk]

I looked at this when you first posted it. I started by drawing it out where it quickly became apparent that the solution contained some nice rounded numbers with all the points sitting nicely on a circle with a simple integer radius. What comes to me easily in terms of drawing lines, circles and intersections completely eludes me when trying to convert my thinking into math. Can you show us your working? Trying to write out the problem as a formula I kept ending up in either a mess of squares and roots or several arcsines() which I could not resolve.

How did you determine the radius from the 5 chords?

[jemidiah]

Heh, now I wish I had saved my scratch work. Here's my solution:


For any polygon with fixed side lengths, that polygon has the most area when it can be inscribed in a circle. A nice proof can be found here. It essentially relies on the fact that, for a perimeter of a given length, arranging the perimeter into a circle maximizes the area enclosed. This is formally called the isoperimetric inequality (well, the original version; there are many generalizations). So, set the pentagon on a circle. As you suggest, finding the radius of the circle on which the pentagon is inscribed is the key.

Draw segments from the center of the circle to the vertexes of the pentagon. This divides the pentagon into 5 isosceles triangles, one for each side. Label the triangles T1 through T5, where T1 contains the first edge of length 5 and T5 contains the edge of length 8. Label the "peak" angles of T1 through T5 (the ones touching the circle's center) a1 through a5. We can find the ai's in terms of the radius R and the edge the triangle contains using some trig. For an isosceles triangle with two equal legs of length R and a third leg of length W with angle a opposite W, it's pretty easy to show (by dropping an altitude) that sin(a/2) = (W/2)/R = W/(2R), so a/2 = arcsin(W/(2R)), and a = 2arcsin(W/(2R)).

That is,
a1 = 2arcsin(5/(2R))
a2 = 2arcsin(5/(2R))
a3 = 2arcsin(5/(2R))
a4 = 2arcsin(6/(2R))
a5 = 2arcsin(8/(2R))

The angles a1 through a5 go around the full circle once, so we have in radians
a1 + a2 + a3 + a4 + a5 = 2*pi

This gives a horrifically non-linear condition
2asin(5/(2R)) + 2asin(5/(2R)) + 2asin(5/(2R)) + 2asin(6/(2R)) + 2asin(8/(2R)) = 2*pi

You can differentiate the left side to find the derivative is
(2 (-15/Sqrt[4 - 25/R^2] - 4/Sqrt[1 - 16/R^2] - 3/Sqrt[1 - 9/R^2]))/R^2

That is, R must be greater than 4, and when this occurs, the derivative is negative. The left side is therefore strictly monotonic, so there is at most one solution. You can use a numeric solver to find that R=5 is this unique solution. I feel like there should be a less... messy bit of reasoning that gives R=5, but I haven't found one. Wanting to make the 6 and 8 edges into a 6-8-10 triangle gives R=5, if that's any better than using a numerical solver. You can take the sine of both sides of the equation and convert the expression into a long ~polynomial equation, but that doesn't seem helpful.

This gives a1 = a2 = a3 = pi/3. Since the sum of the ai's is 2pi, we have a4+a5 = pi. Now make a triangle out of the edges of length 6 and 8. Since a4+a5 = pi, graphically the hypotenuse of this triangle is just 2R = 10, since this triangle is made by combining T4 and T5. From the law of cosines, this triangle is a right triangle (since 6^2 + 8^2 = 10^2]. Its area, which is the area of T4 and T5 together, is then just 8*6/2 = 24. It's not hard to calculate the area of the triangle T1 either, since now we know it's an equilateral triangle. Using the formula from here, the area of T1 is R^2*sqrt(3)/4 = 25*sqrt(3)/4. In all, the area of the pentagon is then

3*(25*sqrt(3)/4) + 24
= 24 + 75sqrt(3)/4

[Milk]

I'm glad you replied, my reasoning was not so bad.

I could take it as far as...
3asin(5/&#216 + asin(6/&#216 + asin(8/&#216 = pi
...but did not know what to do then.

I thought there might be a way to solve it with bisecting circles ((x - cx)² + (y - cy)² = c&#178 where cx and cy sit on the perimeter of a circle of unknown radius >= 4 and c is the chord, but it quickly got out of hand for me.

[dbasnett]

@jemidah - did you mean a1 + a2 + a3 + a4 + a5 = tau?

[jemidiah]

Haha, yes, I should have used tau instead of pi. I will in this post (so tau = 2*pi).


I asked around a bit and got the following solution. It exactly solves the problem without the need for numeric approximation, though it just barely happens to work in this case (which is cool). I'll refer to the figures I've attached for variable names.

Look at the first figure. From the law of cosines, 8^2 + 6^2 - 2*8*6cos(3t) = x^2. Now consider the triangle bounded by the red and green lines. From the law of sines, sin(u)/x = sin(v)/5, so x = 5sin(u)/sin(v). It turns out that v=t and u=tau/2 - 3t:

Look at the second figure. The three triangles created by adding the two blue lines each have equal peak angles, from the inscribed angle theorem. The key is that the arc subtended in each case is of the same length, since the three sides of the pentagon are each of length 5. Since the three peak angles add to 3t, each angle must be t. Another application of the inscribed angle theorem immediately gives v=t. Two more applications give w=2t. Now u+w+v = tau/2, so u = tau/2 - w - v = tau/2 - 3t. Returning to our previous equation, sin(u) = sin(tau/2 - u) = sin(3t). That is,

x^2 = 100 - 96cos(3t)
x = 5sin(3t)/sin(t)

Squaring the equation and setting them equal gives
100 - 96cos(3t) = 25sin^2(3t)/sin^2(t)

Substitute in sin^2(3t) = 1-cos^2(3t) and sin^2(t) = 1-cos^2(t). Multiply both sides by 1-cos^2(t). Finally, apply the triple angle formula,
cos(3t) = 4cos^3(t) - 3cos(t)

If you do all of this, the equation will be entirely a polynomial in terms of cos(t). Let a=cos(t). Subtracting the right side from the left, the previous algebra then gives the equation

400a^6 + 384a^5 - 600a^4 - 672a^3 + 125a^2 + 288a + 75 = 0

You can verify that the left side factors as
(a-1)(a+1)(4a^2 - 3)(100a^2 + 96a + 25)

[Note that (a-1)(a+1) = a^2-1 = sin^2(t). We could have canceled the sin^2's from both sides and gotten rid of these factors much earlier. That would have given us a quartic equation, which can always be analytically solved. In this particular case, the numbers work out so that it's not utterly horrific to factor. If "6" and "8" were replaced, though, we would need the quartic equation.]

We have to throw out the a-1=0 and a+1=0 solutions, since that gives sin^2(t) = 0, a divide by 0. The quadratic 4a^2 - 3 = 0 implies a = +/- sqrt(3)/2. The other quadratic implies a is imaginary, so it can be ignored.


Now, when a = sqrt(3)/2, arccos(a) = t = tau/12. Thus 3t = tau/4, and 3t is a right angle. Another application of the inscribed angle theorem gives the fact that if an inscribed angle is tau/4, the chord it cuts off is a diameter. That is, the green line is a diameter. From the Pythagorean Theorem, we have x^2 = 6^2 + 8^2 = 100, so x=10, and R=5.