Ndtyhij

I found there are two incongruent isosceles triangles with integer sides and areas, where both have same perimeter, same area.

I looked around Dickson's History of Number Theory but couldn't find where the right triangle version is treated. [I thought if a nonexistence proof was simple it would pop up in my search, but found none.]

It may be simple to show none exist, but I had no luck, only filled few notebook pages with formulas going nowhere. Reference/example/proof appreciated. Thanks.

Consider two right triangles, with hypotenuses $p$ and $q$ and respective acute angles $theta$ and $phi$. To see that having equal perimeter and area makes them congruent, it suffices to show that
$$theta = phi qquadtext{or}qquad theta+phi=frac{pi}{2} tag{0}$$
(Either makes the triangles similar, which in turn makes them congruent.)

Equating perimeters and areas gives a system we can write as

$$begin{align}
p(1+sintheta+costheta) &= q(1+sinphi+cosphi) \
p^2 sintheta costheta &= q^2 sinphi cosphi
end{align} tag{1}$$

Defining $u:=tan(theta/2)$, we "know" that
$$sintheta = frac{2u}{1+u^2} qquad costheta=frac{1-u^2}{1+u^2} quadtoquad 1+costheta+sintheta= frac{2 (1 + u)}{1 + u^2}$$
and likewise for $v:=tan(phi/2)$. Thus, $(1)$ can be rewritten as
$$begin{align}
pfrac{(1+u)}{1+u^2} &= qfrac{(1+v)}{1+v^2} \[4pt]
p^2frac{u(1+u)(1-u)}{(1+u^2)^2} &= q^2frac{v(1+v)(1-v)}{(1+v^2)^2}
end{align}tag{2}$$
Dividing the second equation by the square of the first ...

$$frac{u(1-u)}{1+u} = frac{v(1-v)}{1+v} quadtoquad (u-v)(uv+u+v-1)=0 quadtoquad u=v, text{ or } frac{u+v}{1-uv}=1 tag{3}$$

Therefore, we have one of the following situations (bearing in mind that $theta/2$ and $phi/2$ are each at most $pi/4$, so that we may draw appropriate conclusions from these tangent inequalities):
$$begin{align}
tanfrac{theta}{2}=tanfrac{phi}{2} &quadtoquad theta=phi \[4pt]
frac{tan(theta/2)+tan(phi/2)}{1-tan(theta/2)tan(phi/2)} = 1 &quadtoquad
tanleft(frac{theta}{2}+frac{phi}{2}right)=tanfrac{pi}{4} quadtoquad theta+phi=frac{pi}{2}end{align} tag{4}$$
which match the sufficient conditions in $(0)$. $square$

Let $S$ be the circumference and $A$ twice the area of a triangle.

Then,
$$a_i+b_i+sqrt{a_i^2+b_i^2}=S
text{ and } a_ib_i=A. tag{1}$$
After squaring, $S^2+2A-2S(a_i+b_i)=0$ and from here $a_i+b_i=S/2+A/S$. Thus, $c_i=S-(a_i+b_i)=S/2-A/S=c$, that is, triangles have the same hypotenuse.

Then, $a_i+b_i=S-c=T$ and $a_ib_i=A$, which results in a solutions for $a_i$ and $b_i$ expressed in terms of constants $A$ and $T$. Although one of the resulting equations is quadratic, there is a symmetric pair of solutions (thank you for comments below). Hence, all sides must be the same.

Consider a right-angled triangle with sides $a$ and $b$.

The hypotenuse has length $c=sqrt{a^2+b^2}$.
Its area is $ab/2$, and perimeter is $a+b+c$.

I will allow $a,b,c$ to be any positive real numbers, not restrict them to positive integers.

Suppose we scale the triangles such that $ab=1$, (i.e. an area of $1/2$). Is it possible to have two of these triangles that are distinct but with the same perimeter?

We may assume that $a$ is the longer side, i.e. $a>b$, so we must have $a>1$.

The perimeter is
$$P(a) = a+b+c\ = a+b+sqrt{a^2+b^2}\ = a+frac{1}{a}+sqrt{a^2+frac{1}{a^2}}$$

This is an increasing function on the interval $[1,infty)$ because its derivative w.r.t. $a$ is positive for $a>1$. This is tedious to check by hand, so I used Wolfram alpha. You can however understand why this is the case by noticing that if you increase $a$, then the rate at which $a$ increases is larger than the rate at which $1/a$ decreases, and the same holds for $a^2$ versus $1/a^2$.

This means that there are no two values of $a$, both with $a>1$ for which you get the same perimeter.

Bringing it back to the original problem, it means that there are no two right-angled triangles with the same perimeter and area, unless they have the same sides. Basically, given an area and a perimeter, their two equations uniquely determine the two triangle sides because the lines those equations represent are not curved enough to intersect multiple times.

P.S. By the way, the OP mentioned that there are pairs of isosceles triangles with matching areas and perimeters, even when all sides and the area are integers. Two examples are:
$(29,29,40)$ and $(37,37,24)$,

and also
$(218,218,240)$ and $(233,233,210)$.

The area and the perimeter uniquely define the radius of the inscribed circle because $A=frac12Pr$, and the hypotenuse because $r=frac{P}{2}-c$. That fixes both $a+b=P-c$ and $ab=2A$ so $a$ and $b$ are also unique up to permutation, QED.

Long Comment:

You can at least give the simple formulae for the perimeter $P$ and area $A$ of a right angle triangle.

If $z^2=x^2+y^2$ is a primitive Pythagorean triangle with $x$ being the base of the triangle and $y$ being the height (due to the right angle), then $P=x+y+z$ and $A=frac{1}{2}xy$

Then you can quote the formulae for primitive Pythagorean Triples, where $z=left(a^2+b^2right)$, $x=left(a^2-b^2right)$ and $y=2ab$

To expand to all the non primitive Pythagorean triangles we have

$$(cz)^2=(cx)^2+(cy)^2$$

Therefore

$$P=a(2a+2b)c$$
$$A=frac{1}{2} left(a^2-b^2right)(2ab)c^2=left(a^2-b^2right)abc^2$$

For two incongruent Pythagorean triangles 1 and 2 the condition is

$$P_1=P_2 ;;text{and} ;; A_1=A_2$$

For Perimeter:
$$a_1(2a_1+2b_1)c_1=a_2(2a_2+2b_2)c_2$$
$$frac{a_2+b_2}{a_1+b_1}=frac{a_1c_1}{a_2c_2}tag{1}$$
For Area:
$$left(a_1^2-b_1^2right)a_1b_1c_1^2=left(a_2^2-b_2^2right)a_2b_2c_2^2$$
$$frac{a_2+b_2}{a_1+b_1}=frac{a_1c_1}{a_2c_2}left(frac{b_1c_1}{b_2c_2}frac{ (a_1-b_1) }{ (a_2-b_2) } right)$$

Therefore combining both gives

$$b_1c_1(a_1-b_1)=b_2c_2(a_2-b_2)tag{2}$$

Update:

Using (1) and (2) we can eliminate the variables $c_1$ and $c_2$ eventually giving
$$frac{ (a_2+b_2) a_2 }{ (a_1+b_1) a_1 }=frac{ (a_2-b_2) b_2 }{ (a_1-b_1) b_1 }$$
or
$$frac{ (a_2+b_2) a_2 }{ (a_2-b_2) b_2 }=frac{ (a_1+b_1) a_1 }{ (a_1-b_1) b_1 }$$