*Spoiler for New Scientist Enigma 1751: “Pentagon of squares” *(Follow the link to see the puzzle.)

I’m not happy about my solution to this puzzle. I think I have the right answer, but I had to resort to a computer program to get it. I’m probably thinking about it stupidly.

Here’s a picture of what’s going on:

The red line segments are the pentagon; as required, it doesn’t contain the center of the circle it’s inscribed in . The interior angles (angles between the red line segments) are marked *a*^{2}, *b*^{2}, *c*^{2}, *d*^{2}, and *x*. Since the non-square angle *x* is the smallest, I’ve made it one of the angles on the long side. We know all of the interior angles are less than 180°, and all the ones that are squares must be larger than *x* and therefore larger than 1, so *a*, *b*, *c*, and *d* must be in the range [2..13]. Also marked are the radii through each vertex, and the radial angles (between the radii) α, β, ɣ, and δ.

We know the interior angles of a pentagon must sum to 540°, but that’s not enough of a constraint; we also have to make sure the pentagon can be inscribed in a circle. For that we use a theorem that says interior angles subtending the same chord are equal to one another, and to half the central angle subtending that chord. So for instance interior angle *a*^{2 }and central angle α + β + ɣ subtend the same chord, so *a*^{2 }= (α + β + ɣ)/2. Likewise *x* = (β + ɣ + δ)/2, and similarly (pun intended) *b*^{2 }= 180 – (ɣ + δ)/2, *c*^{2 }= 180 – (β + ɣ)/2, and *d*^{2 }= 180 – (α + β)/2. From these relations we can get expressions for the radial angles:

α = 2(*a*^{2}+*c*^{2})–360

β = 720–2(*a*^{2}+*c*^{2}+*d*^{2})

ɣ = 2(*a*^{2}+*d*^{2})–360

δ = 720–2(*a*^{2}+*b*^{2}+*d*^{2})

So given choices for *a*, *b*, *c*, and *d*, we can calculate α, β, ɣ, δ, and *x*. We require all of these to be positive; *x* < *a*^{2}, *b*^{2}, *c*^{2}, and *d*^{2}; and α+β+ɣ+δ < 180 to make the center lie outside the pentagon.

There may be a clever way from here to find the one and only solution, but I don’t see it. Instead I wrote a Perl script that came up with:

*a*^{2 }= 64, *b*^{2} = 169, *c*^{2} = 144, and *d*^{2 }= 121; then α = 56, β = 62, ɣ = 10, and δ = 12 (summing to 140), and *x *= 42. That in fact is exactly the pentagon I’ve illustrated above. In case you’re suspicious, the interior angles do indeed sum to 540.The answer they’re looking for is: 42, 64, and 121.