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How many right triangles are there with the following conditions:

  • the sides $a$, $b$, and $c$ have an integer length (Pythagorean triplets)

  • the amounts of area and perimeter are the same for each triangle.

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Here's another approach which I mention not because it's necessarily better than hexomino's, but because the technique is useful to know.

Every integer right-angled triangle

has sides $2kmn, k(m^2-n^2), k(m^2+n^2)$ for positive integers $k,m,n$. (And for any positive integers $k,m,n$ these three are the sides of an integer right-angled triangle.) Such a triangle has area $\frac12\cdot2kmn\cdot k(m^2-n^2)=k^2mn(m^2-n^2)$ and perimeter $2kmn+k(m^2+n^2)+k(m^2-n^2)=2km(m+n)$. These are equal iff $kn(m-n)=2$.

So

one of $k,n,m-n$ is 2 and the other two are 1. Taking $k=2,n=1,m-n=1$ gives (8,6,10). Taking $k=1,n=2,m-n=1$ gives (12,5,13). And taking $k=1,n=1,m-n=2$ gives (6,8,10) which of course is just the first solution with its legs the other way around.

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  • $\begingroup$ Aaah I misread your first line and thought that somewhere further down the page there would be a post solving the problem by tiling the triangle with hexominos. After that moment all the beautiful algebraic solutions became mere disappointments. $\endgroup$
    – Vincent
    Aug 16 '20 at 22:02
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    $\begingroup$ Ah! I'm so, so sorry. $\endgroup$
    – Gareth McCaughan
    Aug 17 '20 at 12:21
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I think there are

Two such triangles, up to switching labels

Which are the following

$(a,b,c) = (5,12,13)$
$(a,b,c) = (6,8,10)$

Proof

The conditions are $$ a^2 + b^2 = c^2 \,\,\,\,,\,\,\,\, \frac{1}{2}ab = a+b+c $$ The second condition may be reformulated as $$c = \frac{1}{2}ab - a -b$$ which when substituted into the first equation yields $$a^2 + b^2 = a^2 + b^2 + \frac{1}{4}a^2b^2 - ab(a+b) + 2ab$$ $$\Rightarrow ab(ab+8) = 4ab(a+b)$$ Given that $a$ and $b$ must be positive we can divide across by $ab$ and rearrange to get $$ (a-4)(b-4) = 8$$ Since $a$ and $b$ are positive, it quickly follows that $a-4$ and $b-4$ must be positive and factors of $8$ (since, otherwise, one of them will be $\leq -4$).
Up to switching $a$ and $b$, the only possibilities for $(a-4, b-4)$ are $(1,8)$ and $(2,4)$. This leaves $(a,b)$ as $(5,12)$ or $(6,8)$, both of which form Pythagorean triples.

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I have a solution, as follows:

$$a^2+b^2=c^2\text{ and }\frac12ab=a+b+c$$Multiplying the second equation by $4$ and add it to the first equation yields$$\begin{split}(a+b)^2&=4(a+b+c)+c^2\\(a+b+c)(a+b-c)&=4(a+b+c)\end{split}$$As $a+b+c\ne0$, $a+b=c+4$, or $c=a+b-4$. We substitute this in the second equation above.\begin{split}\frac12ab&=2a+2b-4\\ab-4a-4b+8&=0\\(a-4)(b-4)&=8\end{split}So $(a-4,b-4)=(1,8),(2,4),(4,2),(8,1)$, or $(a,b)=(5,12),(6,8),(8,6),(12,5)$. We notice that these values of $a,b$ are pythagorean triples, so we have $(a,b,c)=(5,12,13),(12,5,13),(6,8,10),(8,6,10)$


Edit: Sorry that this is similar to @hexomino's solution. 2nd edit: How could I forget to spoilorise it?

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Here is an economical solution:

The area of any triangle is half its incircle radius times its circumference, hence the requirement wrt area and circumference can be simply rephrased as incircle radius $R = 2$. Let $a$ be the shortest side. It must (1) touch the right angle and (2) be larger than $2R=4$, (3) once its length is chosen, the triangle is fully determined because the middle side must form a right angle which fully determines the incircle's position which in turn determines the long sides position (it must touch $a$ at the far end and it must touch the incircle.) With these constraints we can enumerate: $a=5 \overset {(3)} \Rightarrow b=12,c=13$ $a=6 \overset {(3)} \Rightarrow b=8,c=10$ $a=7$ does not work ($b=7$ too short, $b=8$ too long) $a=8$ not shortest side. And that's all.

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