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# Good o' Pythagoreas

## Question

Is there any triangle other than a right-angled triangle that possesses the sum of squares property (pythagorean theorem)?

What about other polygons constructed similarly on three sides of a right-angled triangle?  Is the area of the figure constructed on the hypotenuse equal to the sum of the figures constructed on the other two sides or is the result only for squares?

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What immediately comes to mind is triangles with one side length zero. As for the second question, it probably holds for any figure as long as they are similar and the shared side is the same.

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Wouldn't a triangle with one side = 0 collapse into a line segment or 2
co-located line segments ??

I don't have an answer, but I suspect that is NOT the answer that BMAD is
looking for - but it made me chuckle [thanks].

What immediately comes to mind is triangles with one side length zero. As for the second question, it probably holds for any figure as long as they are similar and the shared side is the same.

Edited by dgreening
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Wouldn't a triangle with one side = 0 collapse into a line segment or 2

co-located line segments ??

I don't have an answer, but I suspect that is NOT the answer that BMAD is

looking for - but it made me chuckle [thanks].

What immediately comes to mind is triangles with one side length zero. As for the second question, it probably holds for any figure as long as they are similar and the shared side is the same.

Either a degenerate triangle, as gavinksong suggests, where a2 + 0 = c2,

or a triangle with an infinitesimally long side.

Otherwise, if Pythagoras holds, the largest of the three sides is an hypotenuse.

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it actually does hold for non-90 degree triangles but there is an added x or -x that makes the statement true. I.e. A^2 +b^2 +- x = c^2. Figure out the x.

Wouldn't a triangle with one side = 0 collapse into a line segment or 2

co-located line segments ??

I don't have an answer, but I suspect that is NOT the answer that BMAD is

looking for - but it made me chuckle [thanks].

What immediately comes to mind is triangles with one side length zero. As for the second question, it probably holds for any figure as long as they are similar and the shared side is the same.

Either a degenerate triangle, as gavinksong suggests, where a2 + 0 = c2,

or a triangle with an infinitesimally long side.

Otherwise, if Pythagoras holds, the largest of the three sides is an hypotenuse.

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it actually does hold for non-90 degree triangles but there is an added x or -x that makes the statement true. I.e. A^2 +b^2 +- x = c^2. Figure out the x.

Wouldn't a triangle with one side = 0 collapse into a line segment or 2

co-located line segments ??

I don't have an answer, but I suspect that is NOT the answer that BMAD is

looking for - but it made me chuckle [thanks].

What immediately comes to mind is triangles with one side length zero. As for the second question, it probably holds for any figure as long as they are similar and the shared side is the same.

Either a degenerate triangle, as gavinksong suggests, where a2 + 0 = c2,

or a triangle with an infinitesimally long side.

Otherwise, if Pythagoras holds, the largest of the three sides is an hypotenuse.

Hmm... technically any 3 numbers a, b, and c can be part of this equation a2+b2+x=c2. They don't even have to be sides of a triangle and x=c2-a2-b2. But x=0 only when a, b and c are sides of a right triangle and c is the hypotenuse.

The formula will work for all regular polygons and not just squares. The area of regular polygon is always proportionate to the square of its side with a fixed coefficient.

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it actually does hold for non-90 degree triangles but there is an added x or -x that makes the statement true. I.e. A^2 +b^2 +- x = c^2. Figure out the x.

Well, if you're referring to the law of cosines:

c2 = a2 + b2 - x where x = 2 ab cos(θ)

In this case, x is dependent on a and b, but this is an absolute geometric law and there's no way around it unless cos(θ) = 0 (aka. a right triangle with hypotenuse c).

As for the second question, despite k-man's answer, I stand by my claim that the equality holds for any three similar polygons constructed on the sides of a right triangle.

Edited by gavinksong
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it actually does hold for non-90 degree triangles but there is an added x or -x that makes the statement true. I.e. A^2 +b^2 +- x = c^2. Figure out the x.

Well, if you're referring to the law of cosines:

c2 = a2 + b2 - x where x = 2 ab cos(θ)

In this case, x is dependent on a and b, but this is an absolute geometric law and there's no way around it unless cos(θ) = 0 (aka. a right triangle with hypotenuse c).

As for the second question, despite k-man's answer, I stand by my claim that the equality holds for any three similar polygons constructed on the sides of a right triangle.

Sure, gavinksong. You're right. I had a brain fart with the second part.

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