MHB Is Triangle ABC a Right-Angled Triangle in Tetrahedron OABC?

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In tetrahedron OABC, where lines OA, OB, and OC are mutually perpendicular, triangle ABC is proven not to be a right-angled triangle. By using an orthogonal coordinate system with O as the origin, the coordinates of points A, B, and C are defined. The dot product of vectors AB and AC is calculated, resulting in a non-zero value, indicating that angle BAC is not a right angle. This reasoning extends to the other angles in triangle ABC as well. Therefore, triangle ABC cannot be a right-angled triangle.
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Happy New Year! Here is this year's first University-level POTW:

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Tetrahedron $OABC$ is such that lines $OA, OB,$ and $OC$ are mutually perpendicular. Prove that triangle $ABC$ is not a right-angled triangle.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to Opalg and castor28 for their correct solutions to this week's POTW, which was Problem 63 in the MAA Challenges. castor28's solution follows:

[sp]We can take an orthogonal coordinate system with $O$ as origin and one axis through each of $A, B,$ and $C$. The coordinates of the vertices are $A = (a, 0, 0), \; B=(0,b,0),$ and $C=(0,0,c)$.

We compute the dot product of the vectors $AB$ and $AC$:
$$ AB\cdot AC = (-a,b,0)\cdot(-a,0,c) = a^2\ne 0.$$

This shows that $\angle BAC$ is not a right angle; the same argument applies to the other two angles of the triangle $ABC$.
[/sp]
 

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