Ok, I can’t do any other work on the bus, so I choose to #trymathslive with this problem from @Cshearer41. https://twitter.com/cshearer41/status/1299281163228516352

First I’ll say what I’m noticing. There’s a circle and a triangle around it, whose edges seem to just touch the circle. It doesn’t appear to touch the circle in the centre of each edge.

The three edges are not all that different, so this triangle is not far from being equilateral. I don’t know if that’s important, but it may provide a useful approach later if I get stuck to think about the equilateral triangle’s answer as a ballpark figure.

One thing I know about tangents is that they meet a radius at right angles, so I think I might draw the three radiuses in.

I need to figure out that radius in order to find the area of the circle. Not sure how I’ll do that yet.

I feel like I should draw in the lines from the centre of the circle to the corners of the triangle. Then there will be some right-angled triangles to work with.

Ok now what? I’ll look to see what I notice...
These two triangles are interesting. They have a shared edge between them, and they both have a radius as a second edge. And they’re both right angled. Oh! So they’re congruent! Well that’s cool.

Oh but I KNEW that! When you draw the two tangents to a circle from an external point, the distances from the point to the circle along the two tangents are the same.

So now I’ve got these three pairs of equal edges, three pairs of congruent triangles.

If I call the lengths of those pieces v, n, w to match the marks I made, then I can set up some equations.
v+n=13
n+w=15
w+v=14
That’s three equations in three unknowns so it ought to be enough to find them all. Maybe.

Adding all three equations together I get
2v+2n+2w=13+14+15
2v+2n+2w=42
v+n+w=21

Subtracting each equation from this one gives
w=8
v=6
n=7

Let me put this on the diagram.
I have to say that was pretty satisfying and even if I don’t find the radius I will feel a sense of accomplishment about that.

I’m staring at it and not seeing any way to come up with the radius. But I can see that the area of the big triangle can be calculated using the radius.
If the radius is r, then the three triangles sitting on each edge have areas 1/2*13r, 1/2*14r, 1/2*15r.

So the total area of the triangle is
1/2(13+14+15)r = 21r.
Not sure how this is helpful yet, but it’s pretty cool.

Got to stop for a bit now. May or may not get back to it today.

I guess there is a way to find the area of a triangle from its three edges (Heron’s formula?) but it seems like a bit of overkill.

Oh what the hell.
Heron’s formula goes sqrt[s(s-a)(s-b)(s-c)] if I remember it right, where s is half the perimeter.
I figured out the half-perimeter already. It’s 21.
And I already took the three sides off it too! I got 6, 7, 8.

So the area is sqrt[21*6*7*8].

So 21r=sqrt(21*6*7*8)
21^2 r^2 = 21*6*7*8
r^2 = 6*7*8/21
= 2*8
= 16
Ooh nice.

So the area of the circle is 16π.

As compared to the area of the circle, which I reckon is a whole number too.
Area =sqrt[21*6*7*8]
= sqrt[3*7*2*3*7*8]
= sqrt[3^2*7^2*16]
= 3*7*4
= 84

So the fraction of the triangle the circle takes up is 16π/84=0.598. So about 60% of the triangle.

I wonder if there is a way to do it without Heron. All the numbers in Heron’s formula I had already calculated anyway, so maybe I could do it without.

Or maybe the proof of Heron’s formula uses the ideas here.

But I think all that is for another day. Right now I need to rest because I’m sick.
Thanks for the fun @Cshearer41.

Though a little PS: I went to look at other people’s solutions and lots of people showed how you can cut the triangle in two whole-number right-angled triangles, which is pretty cool.

Coming back to this this morning I am wondering what the height is in the other directions...

Using 15 as the base, the area is 84 = 1/2 * 15 * h, so
h = 2*84/15
= 2*12*7/15
= 2*4*7/5
= 56/5
= 11.2

Using 13 as the base, the area is 84 = 1/2 * 13 * h, so
h = 2*84/13
= 168/13
~= 12.923

How disappointing.