3D Pythagoras Theorem, right-angled tetrahedron (2001 UNSW J4 Australia)

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3D Pythagoras Theorem, right-angled tetrahedron (2001 UNSW J4 Australia)

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#1 h Jan 6, 2021, 7:58 AM

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Suppose that a triangle $XYZ$ has side lengths $x, y, z$ with $s = (x + y + z)/2$ . If the area of the triangle is denoted by $|XYZ|$ , then Heron’s Formula states that $|XY Z|^2 = s(s - x)(s - y)(s - z).$ Let $OABC$ be a right-angled tetrahedron, with all angles at the vertex $O$ being right angles as shown. Using Heron’s Formula or otherwise, prove the three dimensional Pythagoras’ Theorem: $|OAB|^2 + |OBC|^2 + |OCA|^2 = |ABC|^2.$

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This post has been edited 3 times. Last edited by parmenides51, Jan 6, 2021, 8:00 AM

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#2 h Jan 6, 2021, 7:58 AM

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posted for the image link

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#3 h Jan 6, 2021, 9:09 AM

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Solution:
We have $|OAB{{|}^{2}}+|OBC{{|}^{2}}+|OCA{{|}^{2}}=|ABC{{|}^{2}}\Leftrightarrow \frac{{{x}^{2}}{{y}^{2}}+{{y}^{2}}{{z}^{2}}+{{z}^{2}}{{x}^{2}}}{4}=|ABC{{|}^{2}}$ (*). Now let $OM\bot BC\Rightarrow OM=\frac{xy}{a},AM=\sqrt{{{z}^{2}}+\frac{{{x}^{2}}{{y}^{2}}}{{{a}^{2}}}}\Rightarrow {{\left| ABC \right|}^{2}}=\frac{B{{C}^{2}}A{{M}^{2}}}{4}=\frac{{{a}^{2}}{{z}^{2}}+{{x}^{2}}{{y}^{2}}}{4}=\frac{({{x}^{2}}+{{y}^{2}}){{z}^{2}}+{{x}^{2}}{{y}^{2}}}{4}$ . Done!

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#4 h Jan 6, 2021, 2:11 PM

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If $\vec{x},\vec{y},\vec{z}$ are the vectors representing the orthogonal sides of the tetrahedron, then $|OAB|^2 + |OBC|^2 + |OCA|^2=\frac{1}{4}\sum_{cyc} |\vec{x}\times\vec{y}|^2$

The area vector of $\Delta ABC$ is $\frac{1}{2}(\vec{z}-\vec{x})\times(\vec{y}-\vec{x})$
$=\frac{1}{2}(\vec{z}\times\vec{y}-\vec{x}\times \vec{y}-\vec{z}\times\vec{x}+\vec{x}\times\vec{x})$
Note that $\vec{x}\times\vec{x}=0$ and that $\vec{z}\times\vec{y},\vec{x}\times \vec{y},\vec{x}\times\vec{x}$ are all orthogonal to eachother (because the pairwise dot products are $0$ )

The square of the area of $ABC$ will be the square of the magnitude of this area vector. Since we have already proven that these vectors are all orthogonal, the magnitude can be found with pythagorean theorem.
$|ABC|^2=\frac{1}{4}\sum_{cyc} |\vec{z}\times \vec{y}|^2$ , so both sides of the equation are equivalent

This post has been edited 2 times. Last edited by natmath, Jan 6, 2021, 2:15 PM

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#5 h Jan 6, 2021, 2:37 PM

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This is straight-up De Gua's Theorem

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#6 h Jan 6, 2021, 2:40 PM

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It's sort of trivial if you do it directly. Set straightforward side lengths, write areas and just verify.

This post has been edited 1 time. Last edited by HamstPan38825, Jan 6, 2021, 2:40 PM

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#7 h Jan 6, 2021, 2:46 PM

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HamstPan38825 wrote:

It's sort of trivial if you do it directly. Set straightforward side lengths, write areas and just verify.

Yes, exactly this is my proof!

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