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.



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