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Calabi triangle
The Calabi triangle is a special triangle found by Eugenio Calabi.
It is the unique triangle that has 3 different placements for the largest square that it contains, and is not the equilateral triangle. It is an isosceles triangle which is obtuse with an irrational but algebraic ratio between the lengths of its sides and its base.
Consider the largest square that can be placed in an arbitrary triangle. It may be that such a square could be positioned in the triangle in more than one way. In the equilateral triangle, the largest such square can be positioned in three different ways. Calabi found that there is exactly one other case, and so it is named the Calabi triangle.
The triangle △ABC is isosceles which has the same length of sides as AB = AC. If the ratio of the base to either leg is x, we can set that AB = AC = 1, BC = x. Then we can consider the following three cases:
Consider the case of AB = AC = 1, BC = x. Then
Let a base angle be θ and a square be □DEFG on base BC with its side length as a. Let H be the foot of the perpendicular drawn from the apex A to the base. Then
Then HB = x/2 and HE = a/2, so EB = x - a/2.
From △DEB ∽ △AHB,
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Calabi triangle
The Calabi triangle is a special triangle found by Eugenio Calabi.
It is the unique triangle that has 3 different placements for the largest square that it contains, and is not the equilateral triangle. It is an isosceles triangle which is obtuse with an irrational but algebraic ratio between the lengths of its sides and its base.
Consider the largest square that can be placed in an arbitrary triangle. It may be that such a square could be positioned in the triangle in more than one way. In the equilateral triangle, the largest such square can be positioned in three different ways. Calabi found that there is exactly one other case, and so it is named the Calabi triangle.
The triangle △ABC is isosceles which has the same length of sides as AB = AC. If the ratio of the base to either leg is x, we can set that AB = AC = 1, BC = x. Then we can consider the following three cases:
Consider the case of AB = AC = 1, BC = x. Then
Let a base angle be θ and a square be □DEFG on base BC with its side length as a. Let H be the foot of the perpendicular drawn from the apex A to the base. Then
Then HB = x/2 and HE = a/2, so EB = x - a/2.
From △DEB ∽ △AHB,