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Given a line segment $s$, there are exactly two right triangles which have $s$ as a hypotenuse.

Is there a name for this theorem?

Assuming this line segment lies on a Cartesian plane, how can we compute the points at which the legs of these two triangles intersect?

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    $\begingroup$ There are more than $2$ right triangles with hypotenuse $s$. $\endgroup$ – André Nicolas Mar 15 '15 at 17:34
  • $\begingroup$ That's not obvious to me. Can you please elaborate? $\endgroup$ – Isaac Kleinman Mar 15 '15 at 17:36
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    $\begingroup$ Let $A$ and $B$ be the endpoints of $s$, and let $O$ be the midpoint of $s$. Draw the circle with centre $O$ and radius $OA$. Let $X$ be any point on this circle other than $A$ or B$. Then $\triangle $AXB$ is a right triangle with hypotenuse $s$. If you replace "right triangle" by isosceles right triangle, and stick to a plane, there will only be two. $\endgroup$ – André Nicolas Mar 15 '15 at 17:39
  • $\begingroup$ And does the latter have a name? $\endgroup$ – Isaac Kleinman Mar 15 '15 at 17:48
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    $\begingroup$ By latter do you mean the remark about isosceles right triangles? No name as far as I know, too obvious to have a name. The one I referred to earlier (about $\triangle AXB$) is sometimes called the theorem of Thales. $\endgroup$ – André Nicolas Mar 15 '15 at 17:52
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If your line segment is length 65, then there are right triangles with this hypotenuse with legs $a,b$ with $(a,b)=(39,52), (25,60), (33,56)$. This actually yields 12 different right triangles with the same hypotenuse and integer leg lengths.

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It's not a theorem because it's false. Given the segment, erect a perpendicular of any length to it at one end. Then connect the two open ends to make a triangle. It's a right triangle because of the perpendicularity, and one of the sides was arbitrarily chosen.

EDIT: The above construction is faulty because the wrong side is the hypotenuse as pointed out in comments. Correction follows.

Correction: Given the segment, draw a ray from one end of the segment at any acute angle to the segment. Choose the unique line perpendicular to the ray that passes through the other end of the original segment. The triangle bounded by the original segment, the ray, and the line has the original segment as hypotenuse and is a right triangle because of the perpendicularity.

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  • $\begingroup$ You don't draw the perpendicular to the hypotenuse. $\endgroup$ – Fred Kline Mar 15 '15 at 17:42
  • $\begingroup$ Yes, while you are correct that the theorem is wrong, you've disproved some other theorem. $\endgroup$ – Thomas Andrews Mar 15 '15 at 17:48
  • $\begingroup$ You're right, of course, @FredKline. Corrected. $\endgroup$ – MPW Mar 15 '15 at 17:53
  • $\begingroup$ Why leave the wrong answer in such a way that people reading your answer wade through it only to find out at the end that it is wrong? Delete the wrong part, or leave it at the end with a preamble indicating it is wrong. $\endgroup$ – Thomas Andrews Mar 15 '15 at 17:57
  • $\begingroup$ @ThomasAndrews: Because removing the error completely renders any comments about it meaningless and mystifying to later readers. I will add strikethrough for the faulty portion. $\endgroup$ – MPW Mar 15 '15 at 19:40

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