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

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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.