Length of Right Triangle

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mathdad

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Suppose that m and n are positive integers with m > n. If a = m^2 - n^2, b = 2mn, and c = m^2 + n^2, show that a, b, and c are the lengths of the sides of a right triangle. Note: Provide the steps leading to the prove.
 

Dr.Peterson

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Assuming c is intended to be the hypotenuse, these lengths will form a right triangle if, and only if, a^2 + b^2 = c^2 (the Pythagorean Theorem). So replace a, b, and c by their expressions, and show that both sides are equal by expanding (distributing).

If this doesn't work (in fact, it will), you would try again taking a different side as the hypotenuse.
 

Harry_the_cat

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Show that a^2 + b^2 = c^2 (ie Pythagoras' Rule which holds for a right triangle).

Sub in the expressions for a and b on the LHS and simplify.
Sub in the expression for c on the RHS and simplify.
Show that they are equal.
 

Harry_the_cat

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Snap, Dr P.
 

mathdad

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a = m^2 - n^2
b = 2mn
c = m^2 + n^2

a^2 - c^2 = (m^2 - n^2)^2 - (m^2 + n^2)^2
a^2 - c^2 = (m^4 - 2m^2n^2 + n^4) - (m^4 + 2m^2n^2 + n^4)
a^2 - c^2 = m^4 - 2m^2n^2 + n^4 - m^4 - 2m^2n^2 - n^4
a^2 - c^2 = - 4m^2n^2
a^2 - c^2 = - (2mn)^2
a^2 - c^2 = - b^2
a^2 + b^2 = c^2

I can recognize the Pythagorean's theorem where c is the hypotenuse.
 

Dr.Peterson

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That worked nicely. I started instead with a^2 + b^2 = (m^2 - n^2)^2 + (2mn)^2, and also simplified c^2 = (m^2 + n^2)^2. Both work out to be the same.
 

mathdad

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That worked nicely. I started instead with a^2 + b^2 = (m^2 - n^2)^2 + (2mn)^2, and also simplified c^2 = (m^2 + n^2)^2. Both work out to be the same.
Good news for a change.
 
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