Brilliant – a lovely puzzle app and a source of many little puzzlers if you follow their Facebook page. The other day, I came across this:

It looked like it might be interesting so I screen shot it and thought, “I’ll have a go at that later, when I’ve got a pen. It’s bound to be nice using a bit of trigonometry and angle reasoning.”

But as I thought about it I realised I didn’t need paper. The hypotenuse of the large triangle is easy enough to find (6rt2) using Pythagoras’s Theorem. You can deduce the size of the green square is then 2rt2 as the big triangle is isosceles meaning the angles are 90, 45 and 45, as the square is only right angles then the little blue triangles in the 45 degree corners must also be isosceles. Thus the two blue and the green segments of the hypotenuse are equal.

The area of the square is then way to find (8) by squaring 2rt2. A nice easy puzzle.

My first thought had been that it would take a bit of working out, but it didn’t, it was a very straightforward question once I got going. That got me thinking, problem solving is something that I would love my students to get better at and I’m hoping to launch a puzzle of the month in January. This sort of puzzle is ideal. It will require then to build their perseverance skills as well as their problem solving skills and will give them a mental workout. I’m going to use this as a starter this week to warm them up.

*This post was cross posted to the blog Cavmaths here.*

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I got to 8 a different way: the two big triangles are half the square each, and the little triangle has to be half of half the square, so the square makes up 1/ (2.25) of the total area: 4/9 of 1/2 of 6*6 is 8.

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Nice, I love that these things can be tackled in different ways.

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Now generalise the problem: Any right triangle, corresponding picture. First question: “Does my method for the simple problem shed any light on the general problem?”.

The it really become math.

More so if you try to construct such a square.

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Aye, certainly want to ask them to generalise next, or maybe give a different sized similar triangle first then generalise.

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A similar-yet-different approach:

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Love it.

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A bit of care is needed to show that the figure formed by rotating copies of the triangle 90 degrees is truly a square partitioned into equally sized squares, but I think the picture gives an indication of one way in which someone can actually

formulatesuch a problem! (I find some value in thinking about how such problems arise in the first place.)Since Howard mentioned generalizing, one can note (as above) that the original problem tiled a square with a 3×3 checkerboard, then cut along the diagonals, and, finally, removed a single triangle to ask for a color’s covered area. Note that the figure has area 72 and the number of tiles in the full square is 9.

An analogous problem can created by, say, tiling the full square as a 5×5 checkerboard and slicing it along the diagonals. In this case, I labelled the two legs of the isosceles right triangle as each having length 10; I will leave the reasoning behind such a choice without explanation. A last image, for which one might again ask about the area of the portion shaded green:

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N.B.It seems that one cannot edit comments on betterQs. Suffice to say that the latter challenge is not quite analogous. Rather, a picture of the following nature would be closer to the most natural generalization:All of these problems are tractable (with the proper phrasing).

I’ll leave it to others to pin down the most exciting paths to chase down!

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Nice comments thanks. And I love both suggested questions!

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