Three squares two triangles one circle

I’ve been trying lately to post my solutions to Catriona Shearer’s geometry puzzles in a tweet with no graphics; a severe constraint. For instance, this from yesterday: This puzzle looked daunting at first but turned out to be easier than it looked. The tweet solution may be a bit too terse to be followed easily, though.

Here is a modified diagram: I’ve labeled five points, and the sizes of the three squares: The largest square, touching the circle at point $A$, has size $a$; the medium square, touching at $C$, has size $b$, and the smallest square, touching at $D$, has size $c$. I added a copy of the smallest square, touching the circle at $E$, and drew in the chord $DE$.

The blue triangle is right, with legs $a$ and $c$, so its area is $(ac)/2$. The blue triangle is also right, with legs $a\sqrt{2}$ and $b\sqrt{2}$; its area is $(a\sqrt{2})(b\sqrt{2})/2 = ab$.

By the intersecting chords theorem, the product of lengths $AB$ and $BC$ is equal to the product of $BD$ and $BE$. But $AB = BE = a+c$, so $BC = BD = c$; that means $b = 2c$.

So the red triangle area is $2ac$ which is four times the area of the blue triangle. That area is 5, so the area of the red triangle is 20.

Sometimes, though, you just need to have a picture:

It’s obvious the red triangles’ height is $1/2$ but the blue triangles were less obvious to me. I noted that if you drop a perpendicular from the top vertex of the bottom blue triangle to the bottom line, the resulting right triangle has legs in a ratio of $1:2$ which sum to $1$, so the height is $1/3$. Another approach another poster mentioned is to note that a diagonal of the square is divided into three equal parts by the slanting lines and so the vertical projection of one of those parts has length $1/3$.

Either way, the red and blue triangles both have base = $1/2$ so a red triangle has area $1/8$, a blue triangle has area $1/12$, and four of each add up to $5/6$. Then the octagon’s area is the square’s ( $1$) minus the triangles’ ( $5/6$) which equals $1/6$.