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I understand that due to the Pythagorean Theorem, $a^2+b^2=c^2$, given that $a$ and $b$ are legs of a right triangle and $c$ is the hypotenuse of the same right triangle. However, most of the time, $a+b\neq c$. What I have been wondering is, is there any set of values for $a$, $b$, and $c$ that make the statement $a+b=c$ true?

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    $\begingroup$ Try to draw it, $a+b=c$ makes a very flat triangle ... $\endgroup$ – Martin R Apr 9 '16 at 20:42
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    $\begingroup$ Since nobody apparently brought it up: look up the triangle inequality. $\endgroup$ – J. M. is a poor mathematician Apr 10 '16 at 2:23

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Your question has nothing to do with right triangles. If $c=a+b$ in any triangle, then the vertex $C$ must lie on the segment $AB$, and this is called a "degenerate" triangle.

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    $\begingroup$ As in a triangle that looks exactly like a plain line segment? Makes sense. $\endgroup$ – OldBunny2800 Apr 9 '16 at 20:45
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    $\begingroup$ @OldBunny2800: Exactly. It depends on you whether you still want to call such a thing "a triangle". $\endgroup$ – Alex M. Apr 9 '16 at 20:46
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    $\begingroup$ In computer science this is called a corner case. You modeled a triangle and tried to capture every possible case and ended up capturing some cases you never imagined. It happens. It's why we have the number zero in the first place. It's why 0! = 1. A surprising number of things are degenerative cases of other things. $\endgroup$ – candied_orange Apr 9 '16 at 21:45
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    $\begingroup$ Does have relevance the question with right triangles: it corresponds to a leg equal to zero. In contrast, when simply $a+b=c$ the triangle is degenerated and $a$ and $b$ can be nonzero. $\endgroup$ – Piquito Apr 10 '16 at 1:38
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enter image description here

(Addendum: As noted as well by others, this should highlight that your question has little to do with right triangles, as this is in fact a property of triangles in general)

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  • $\begingroup$ this is a beautiful explanation. I'm wondering if it is possible to do such an illustrative drawing to prove that $a^2+b^2=c^2$ if and only if the triangle is right. $\endgroup$ – Surb Apr 15 '17 at 22:15
  • $\begingroup$ @Surb I suspect there is; you should ask this as a new question. $\endgroup$ – MichaelChirico Apr 16 '17 at 22:53
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From the pythagorean theorem, $a^2+b^2=c^2$.

Suppose $a+b=c$, then

$$ a^2+b^2=(c)^2\iff a^2+b^2=(a+b)^2 $$

Can you take it from here?

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    $\begingroup$ a = b = 0. But that's not a triangle any more... $\endgroup$ – Lii Apr 9 '16 at 21:11
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    $\begingroup$ @Lii Well, the question was "is there any set that makes this true"? The answer is yes. Take any of the legs to have $0$ length (not necessarily both, though). If you want to call that a triangle or not depends on you. $\endgroup$ – YoTengoUnLCD Apr 9 '16 at 21:14
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    $\begingroup$ @Lii either a or b, but not necessarily both. $\endgroup$ – njzk2 Apr 9 '16 at 22:59
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    $\begingroup$ @Lii and if it isn't a legitimate triangle what does that tell you about whether a right triangle can have a+b=c? (Hint: if you ask a yes/no question the answer can be no.) $\endgroup$ – fleablood Apr 11 '16 at 1:55
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Substituting $c = a + b$ into $a^2+b^2=c^2$ gives us $a^2+b^2=(a+b)^2$. Multiplying that out gives us $a^2+b^2=a^2+2ab+b^2$ which means $2ab=0$

So to satisfy both $c = a + b$ and $a^2+b^2=c^2$ either $a=0$ or $b=0$.

The question then becomes a bit more philosophical. We have a result that satisfies the Pythagorean equation but can we really consider it a right angle triangle? Is a "side" of length zero really a side at all?

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  • $\begingroup$ that's funny, so a line of length 5cm can be either an unspectacular triangle with side lengths of, say, 5cm, 2cm and 3cm or it could be rectangular, but only if a=0cm, b=c=5cm. And this is the only case I can think of where the hypotenuse is one of the two equally long sides in an isosceles triangle. So which is the hypotenuse then, b or c? Or are they both hypotenuses and there is only one (zero-length) cathetus? Degeneracies allow for many strange results :-) $\endgroup$ – riddleculous Apr 10 '16 at 12:23
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There is another way to formally answer this question, namely using the cosine law.

For any triangle, it is true that $a^2+b^2-2ab\cos\gamma = c^2$ where $\gamma$ is the angle between the $a$ and $b$ sides. On the other hand, if $c=a+b$ then $c^2=(a+b)^2=a^2+2ab+b^2$, thus

$$a^2+2ab+b^2 = a^2+b^2-2ab\cos\gamma,$$

which is possible only if $\cos\gamma=-1$ or $\gamma=\pi$, that is, 180 degrees, or if either $a$ or $b$ is zero. Either way, this is a degenerate triangle, as mentioned in Alex M's answer.

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Like Alex M. said, this doesn't have much to do with right triangles, but it's worth knowing the more general fact, and there's an easy visual!

Imagine you have two sides of a triangle, like an adjustable protractor. As you increase the angle between them, the other side is getting longer. But it won't equal their sum until you have your protractor fully opened up to 180°—and then it's not much of a triangle anymore. You also might wonder if there are any triangles where $a-b=c$ If you decrease the angle between your two sides, the other side is getting shorter. But it won't equal their difference until your protractor is totally closed to 0°—and again, then it's not much of a triangle anymore!

So if we have a triangle and know the length of two of the sides, we know that the length of the other side is less than their sum, and greater than their difference.

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One of the principles that this question illustrates is that in Euclidean geometry any path from point A to point B is not shorter than a straight line from A to B, i.e. a straight line segment A->B is the shortest path from A to B. Any deviation made through a point C that is not on the segment A->B makes a path longer and the only way that (A->C)+(C->B) can be equal to (A->B) is that C is on that segment.

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No, it can't. Although in case of simple triangle, make any one angle 180 degree and then the area of triangle will be zero. It will look like straight line.

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Let $x$ be the angle opposite to altitude. Now,we want $a+b=c$ or Dividing both sides by c we get--- $a/c+b/c=1$ By definition $a/c=sinx$ and $b/c=cosx$ So equation becomes---- $sinx+cosx=1$ or $sinx+(1-(sinx)^2)=1$ By solving we get---- $2(sinx)^2-2sinx=0$ We get 2 solutions that is $sinx=1$ and $sinx=0$ Or as principle solution we get angle $x=0$ and $x=π/2$. Which means $a/c=0$ or $a=0$,$b=c$ and $a/c=1$ or $a=1$,$b=0$. We conclude that either base or the altitude should tend to $0$.

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It has nothing to do with right triangles. There are two line segments and a third one straddling them. Sometimes it is called a $ diangle,$ a needle like shape enclosing no area between them.

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Maybe, the equations $a^2+b^2=c^2$ and $a+b=c$ can be hold simultaneously in a ring or finite field. Say, over $F_{2^m}$, if we set $c=a+b$, then $c^2=a^2+b^2$ always holds. In fact, we know that $(a+b)^p\equiv a^p+b^p\pmod p$ for any prime $p$.

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  • $\begingroup$ If we set c=a+b, wouldn't $c^2=a^2+ab+b^2$? $\endgroup$ – OldBunny2800 May 4 '18 at 14:01

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