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![enter image description here

Given a right triangle with sides $ABC$ make two more right triangles using sides $A$ and $C$ (long side) and a new long side $x$ (same for both new triangles). By Pythagoras the implied third sides will have lengths $a$ and $c$ such that $a^2+A^2 = x^2 = c^2+C^2$.

Now using some algebra I can show that if we can form a triangle with sides $aBc$ it must be right, too, viz.: $B^2+c^2 = B^2 + x^2 - C^2 = x^2 - A^2 = a^2$

But that feels just wrong like instrument flying on a bright day.

Can you

  • either rearrange the figure in such a way as to make it ($aBc$ is right) obvious
  • or make a direct geometric argument
  • or a combination of both?

Note on the figure. By unfortunate coincidence (pun intended) the purple circle appears to pass through $\angle AB$. That is not necessarily the case. The circle is the one of radius $c$ around $\angle BC$

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  • $\begingroup$ In case you want to improve the diagram, here are some observations: 1: it's very hard to follow the text and find the corresponding parts in the image, there's too much clutter. Maybe start with a much simpler diagram, and then add the relevant elements to the picture as the text explanation progresses? 2: The points aren't labeled, so it's cumbersome to write about the diagram. 3: The circle with radius c doesn't pass through the single point (at the angle of x and c) that's initially known to be exactly at distance c from the centre. $\endgroup$ – Bass Aug 18 '20 at 12:44
  • $\begingroup$ @Bass thanks, these are valid observations. $\endgroup$ – Paul Panzer Aug 18 '20 at 14:23
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Consider the third dimension.

Suppose we choose a point in the plane through $B$ perpendicular to the plane of the triangle. This creates three new triangles. The triangle upon $A$ is always right. (This is $Axa$.) The triangle upon $C$ is right at $BC$ if and only if the the point is directly above vertex $BC$ (ie the line through the new point and vertex $BC$ is perpendicular to the plane of the original triangle). (This is $Cxc$.) In this case, the triangle upon $B$ is clearly right as well (also at $BC$). (This is $aBc$.)

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    $\begingroup$ Not sure I follow your every detail but the underlying idea is brilliant! Perhaps an even more intuitive way of presenting it would be rot13("Gnxr gur guerr ynetr gevnatyrf qrcvpgrq naq renfr gur fznyy bar. Orpnhfr gur bhgrezbfg fvqrf ner obgu k jr ina tyhr gurz gbtrgure sbezvat n gevnathyne pbar. Natyrf Nn naq ON ner evtug, fb gur arjyl sbezrq onfr vf crecraqvphyne gb N. fvqr p vf crecraqvphyne gb P naq yvrf va n cynar gung vf crecraqvphyne gb N. Pbafrdhragyl vg zhfg or crecraqvphyne gb gur cynar fcnaarq ol N naq P juvpu pbagnvaf O.") $\endgroup$ – Paul Panzer Aug 18 '20 at 16:24

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