## sexta-feira, 11 de setembro de 2020

(Two squares have been inscribed, one in each of two congruent isosceles right triangles. Which square is larger?)

PCFilho

#### 2 comentários:

1. A variable definition to start:
s = leg length of each large triangle

For the diagram on the left, the square is inscribed so that one of its angles is the right angle of the large triangle and the opposite vertex is the midpoint of that triangle’s hypotenuse. So, the length of its sides is s/2 and its area is (s/2)² = s²/4.

For the diagram on the right, the square is inscribed so that one of its sides lies on the triangle’s hypotenuse and the two other vertices (the ones that aren’t the endpoints of the aforementioned side) lie on the triangle’s legs. This creates three smaller isosceles right triangles. The two isosceles right triangles that use the large triangle’s hypotenuse have legs that are the same length as the square’s sides. That means the square’s sides are one-third as long as the large triangle’s hypotenuse; that is, their length is (√2/3)s. This means that the area of this square is [(√2/3)s]² = 2s²/9.

Since 1/4 is larger than 2/9, the square on the left is larger.

My answer: The square on the left

1. Well done, Jake!

There is also a visual proof of this. It is not hard to draw lines that divide the big triangles into smaller equal triangles.

The big triangle on the left would be divided into 4 smaller equal triangles, 2 of which form the square – therefore, the square occupies 2/4 of the big triangle.

The big triangle on the right would be divided into 9 smaller equal triangles, 4 of which form the square – therefore, the square occupies 4/9 of the big triangle.

As 2/4 > 4/9, we agree: the square on the left is larger. :)

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