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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?

Isaac Kleinman
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    There are more than $2$ right triangles with hypotenuse $s$. – André Nicolas Mar 15 '15 at 17:34
  • That's not obvious to me. Can you please elaborate? – Isaac Kleinman Mar 15 '15 at 17:36
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    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. – André Nicolas Mar 15 '15 at 17:39
  • And does the latter have a name? – Isaac Kleinman Mar 15 '15 at 17:48
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    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. – André Nicolas Mar 15 '15 at 17:52

2 Answers2

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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.

MPW
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  • You don't draw the perpendicular to the hypotenuse. – Fred Daniel Kline Mar 15 '15 at 17:42
  • Yes, while you are correct that the theorem is wrong, you've disproved some other theorem. – Thomas Andrews Mar 15 '15 at 17:48
  • You're right, of course, @FredKline. Corrected. – MPW Mar 15 '15 at 17:53
  • 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. – Thomas Andrews Mar 15 '15 at 17:57
  • @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. – MPW Mar 15 '15 at 19:40
  • That's the nature of comments, @MPW The goal of the site the site is to give good answers. Comments are ephemeral. Bad answers left at the start, without any warning, waste people's time. – Thomas Andrews Mar 15 '15 at 19:42
  • @ThomasAndrews: Please see the edit--better? Personally, I find it instructive to see an error in reasoning that has been corrected. – MPW Mar 15 '15 at 19:44
  • It's better, yes. Sometimes it is useful to keep wrong answers around, but a simple misreading of the problem isn't really that interesting. At least here, you are no longer forcing readers to read before they find out that it is wrong. But it is still a lot of noise to the answer, making it read inelegantly for a very very simple problem. I like to keep supplementary information at the end - there is no reason to read an answer as a narrative of answers, with early wrong answers first. Good writing puts the primary information up front. – Thomas Andrews Mar 15 '15 at 19:47
  • If the wrong answer is valuable, then reading that crossed-out text is actually kind of a pain. – Thomas Andrews Mar 15 '15 at 19:47
  • @ThomasAndrews: I appreciate the advice. Will leave this post as is, considering it's only three short sentences, but will avoid doing this in the future. Cheers. – MPW Mar 15 '15 at 23:39
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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.

Thomas Andrews
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