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Everyone knows the Pythagorean theorem: In a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

The Pythagorean Theorem: $$a^2+b^2=c^2$$

Recently, professor Halfbrain has proved a theorem that he calls the Non-Pythagorean theorem. This theorem considers a right-angled triangle, in which the two sides adjacent to the right angle have lengths $a$ and $b$ (just as in the classical Pythagorean theorem), together with a mysterious quantity that the professor calls $x$.

The Non-Pythagorean Theorem: $$a^{-2}+b^{-2}=x^{-2}$$

This puzzle asks: What's the geometric role of this mysterious quantity $x$ in the right-angled triangle?

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up vote 53 down vote accepted

$x$ is the altitude $h$ from side $c$.


$a^{-2} + b^{-2} = \frac{1}{a^2} + \frac{1}{b^2} = \frac{a^2+b^2}{a^2 b^2} = \frac{c^2}{a^2 b^2},$ hence $x = \frac{ab}{c}$. The area of the triangle is $\frac{ab}{2}=\frac{ch}{2}$, so $x=h$.

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Vey nicely done! – Xander Feb 9 at 11:55

If $a^2+b^2=c^2$, then $1/a^2+1/b^2=(c/ab)^2$. Using the area formula, we can find out that $ab/c=h_c$, so $(c/ab)^2=(1/h_c)^2=1/x^2$, which means $x=h_c$.

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