# How many right triangles are there with these conditions?

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.

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.

• 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. Aug 16 '20 at 22:02
• Ah! I'm so, so sorry. Aug 17 '20 at 12:21

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.

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?

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.