## Three rectangles

Catriona Shearer posted this problem to Twitter and it got a lot of discussion, so I thought I’d post my solution here in more detail than Twitter permits.

This design is made of three 2×1 rectangles. What fraction of it is shaded?

Label vertices and draw in a couple line segments:

Considering triangles CFJ and CKJ, angles CFJ and CKJ are right, line segment CJ is common, and CF = CK = 2, so the two triangles are congruent and FJ = JK = 1. But then angles CGE and JGK are equal, angles CEG and GKJ are right, and CE = JK, so those triangles are congruent and CG = GJ.

So JK = 1, and if KG = x, GJ = CG = 2-x, so 1²+x² = (2-x)². The solution is x = 3/4. The area of each shaded triangle is 3/8. The area of the whole pattern is three rectangles minus the shaded area: 6-3/4 = 21/5. The shaded area is 1/7 of the area of the whole pattern.

Notice the shaded triangles are 3:4:5.

Can the shaded area be dissected into pieces, seven of each of which will fill the pattern? Yes.

Here are the relative sizes of the pieces, in case you’re interested:

• Blue: 9:12:15
• Green: 12:16:20
• Red: 9:25:30:38
• Magenta: 10:16:38:40
• Shaded triangles (blue+red and green+magenta): 30:40:50

## 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 $1/6$.