## Iso-Metrics

In the latest Dell Math Puzzles and Logic Problem (which I recommend that you subscribe to), we have a problem called "Iso-Metrics." We are shown a square and are asked to dissect it into the fewest number of isosceles triangles, none of which contains a right angle, and no two of which are identical. You might want to try it; it's fun. And keep trying to improve your solutions.

On the right is my first attempt, which is not the solution (as it can be improved). I chose a point on the top edge, but not the center. I then drew two lines from that point to the two opposite vertices, forming two right triangles. I drew lines from the two right angles to their hypotenuses, forming four isosceles triangles. Then I drew a perpendicular from one of these hypotenuses to the opposite vertex of the square, forming two more right trangiels. Then I made four isosceles triangles from these two right triangles, just as I had done with the other two right triangles. The result is eight isosceles triangles, which is not the solution.

Solution = 7?: On the left is the solution in the magazine, with seven isosceles triangles. They bisected the top edge, forming two identical right triangles, one of which can be made into two isosceles triangles, as I did above. We can't do the same with the second right triangle, as that would produce identical isosceles triangles, so we dissect that right triangle into two right triangles, by drawing the perpendicular shown. And then we produce four isosceles triangles from those two right triangles, for a total of seven isoscles triangles. Are the two largest obtuse angle triangles really different? Of course they are different; you might want to prove that; it's fairly easy.

Above right is another of my attempts, which ties with their solution. In the big triangle, the two bottom sides are of equal length.

Solution = 6: But their solution is not optimal, and so is not the real solution. On the left is my next attempt, which totals six isosceles triangles. It is the same as theirs, except for the upper right corner, which can be dissected into three isosceles triangles instead of four, as shown. I sent email to Dell Puzzles, but they never responded.

Even better solution = 5?: On the right seems to be an even better solution, with five triangles. Do you think it is a solution?

Of course it is not a solution, as the large triangle cannot be isosceles. Almost. It would work on non-square rectangles.

A different puzzle: If we loosen the restrictions, and allow right isosceles triangles, but keep the restriction that no two are identical, what is the optimal solution then? I thought the answer is still six, my solution above. Then we wouldn't have had to outlaw right triangles. But here is my solution with five isosceles triangles, two of which are right triangles. Is there a better solution?

Better solution: A reader sent me this one with four isosceles triangles. The easiest way to show that the triangles are isosceles is to calculate the angles: (starting with the largest triangle) 45+45+90, 45+67.5+67.5, 45+45+90, and 22.5+22.5+135 (the obtuse angle triangle is the smallest).