By Jacques Ozanam

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The base of each triangle becomes the base of the corresponding rectangle. If we perform this procedure for both of the given triangles, then we obtain two rectangles with base 2s and height r=2. We re-cut the rectangle from the first triangle to match the cuts for the second triangle, and then form the three subtriangles for the second triangle. This solves the problem. Bonus: A Family of Triangle Pairs Let’s find an infinite family of non-similar pairs of triangles with integer side lengths such that the two triangles in each pair have the same perimeter and the same area.

Denote a rotation with center O with angle !. ˛r Â ˛/rÂ D rÂ rÂ D r2Â : (b) Let’s say that the rotation at P has angle 2Â and the rotation at P 0 has angle 2Â 0 . From (a), we know that the rotation at P is the product of two reflections, ˛ and ˇ, intersecting at an angle Â, as in the figure on the right. ) Similarly, the rotation at P 0 is the product of two reflections, ˇ and , intersecting at an angle Â 0 . Â C Â 0 /, since the angle between ˛ and ˇ is Â C Â 0 . Bonus: Tiling With Triangles Which Euclidean triangles have the property that reflections in their sides produce tilings of the Euclidean plane?

I leave it to you to prove the fact that a box of given volume and minimum surface area is a cube. You can do it with two applications of the isoperimetric inequality that the rectangle of given perimeter having the greatest area is a square (see p. 36). Composting—A Hot Topic? Suppose that a compost bin is to be made so that the structure looks from above like the letter E, with three parallel equal-length pieces of fence in one direction and two equallength pieces of fence in a line in the perpendicular direction.