Firstly, the equal total area thing is just a cute way to give the relative side length - let the small squares be area 1 with side length 1, then the large has area 3 and therefore side length sqrt 3.
Now look at the triangle formed by the top of the large square and the small edges touching its corners. Those small edges are both 1, so the triangle is isosceles. Bisect it to get a right triangle with hypotenuse 1, horizontal long edge half sqrt 3, and therefore vertical short edge 1/2 by Pythagoras.
Continuing the vertical line down, we can traverse the edge of the second small square easily by similar triangles, which moves us half sqrt 3 down and 1/2 right. Combined, we have moved 1/2 + half sqrt 3 in both directions, so that corner (indicated by the angle) is actually on the diagonal of the large square.
In any case, draw a horizontal line through that corner. The resulting right triangle involves the angle, where its opposite is half sqrt 3 + 1/2 and adjacent is sqrt 3 minus that, which is half sqrt 3 - 1/2. The angle is the arctan of that ratio, or arctan(2 + sqrt 3), which is 75 degrees.
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fugg I have no clue. :D
I get 75 degrees. Neat little puzzle!
Firstly, the equal total area thing is just a cute way to give the relative side length - let the small squares be area 1 with side length 1, then the large has area 3 and therefore side length sqrt 3.
Now look at the triangle formed by the top of the large square and the small edges touching its corners. Those small edges are both 1, so the triangle is isosceles. Bisect it to get a right triangle with hypotenuse 1, horizontal long edge half sqrt 3, and therefore vertical short edge 1/2 by Pythagoras.
Continuing the vertical line down, we can traverse the edge of the second small square easily by similar triangles, which moves us half sqrt 3 down and 1/2 right. Combined, we have moved 1/2 + half sqrt 3 in both directions, so that corner (indicated by the angle) is actually on the diagonal of the large square.
In any case, draw a horizontal line through that corner. The resulting right triangle involves the angle, where its opposite is half sqrt 3 + 1/2 and adjacent is sqrt 3 minus that, which is half sqrt 3 - 1/2. The angle is the arctan of that ratio, or arctan(2 + sqrt 3), which is 75 degrees.
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