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.


6 Answers 6


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 $$\theta = \phi \qquad\text{or}\qquad \theta+\phi=\frac{\pi}{2} \tag{0}$$ (Either makes the triangles similar, which in turn makes them congruent.)

Equating perimeters and areas gives a system we can write as

$$\begin{align} p(1+\sin\theta+\cos\theta) &= q(1+\sin\phi+\cos\phi) \\ p^2 \sin\theta \cos\theta &= q^2 \sin\phi \cos\phi \end{align} \tag{1}$$

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

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

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): $$\begin{align} \tan\frac{\theta}{2}=\tan\frac{\phi}{2} &\quad\to\quad \theta=\phi \\[4pt] \frac{\tan(\theta/2)+\tan(\phi/2)}{1-\tan(\theta/2)\tan(\phi/2)} = 1 &\quad\to\quad \tan\left(\frac{\theta}{2}+\frac{\phi}{2}\right)=\tan\frac{\pi}{4} \quad\to\quad \theta+\phi=\frac{\pi}{2}\end{align} \tag{4}$$ 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.

  • 3
    $\begingroup$ Really nice proof. I have a niggle with your last statement though. In general, $2$ equations with $2$ unknowns may well have multiple solutions if they are not linear. For example, the OP mentions isosceles triangles, where the triangles $(8,8,12)$ and $(6,11,11)$ have the same perimeter and area even though isosceles triangles are also parametrised by two variables. Of course, in your case you know the sum and product of $a$ and $b$, which is a particularly nice pair of equations well known to have a symmetric pair of solutions. $\endgroup$ Apr 10, 2019 at 13:14
  • $\begingroup$ @Jaap Scherphuis -- Thank you for your comment and clarification. $\endgroup$
    – dnqxt
    Apr 10, 2019 at 13:18
  • $\begingroup$ @Dawood ibn Kareem -- Thanks for the comment. I added in the answer the clarification made by Jaap Scherphuis. $\endgroup$
    – dnqxt
    Apr 10, 2019 at 20:21

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.


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 $(233,233,210)$.

  • $\begingroup$ How to scale the 3,4,5 triangle so its area is 1/2? [i.e.$ab=1$ in notation you use.] By "scale" do you mean multiply all sides by some factor $k$ ? $\endgroup$
    – coffeemath
    Apr 11, 2019 at 11:33
  • 1
    $\begingroup$ Yes. Since in this case $ab=12$ you'd divide both sides by $\sqrt{12}$, so that corresponds to $a=3/\sqrt{12}=\sqrt{3}/2$ and $b=4/\sqrt{12}=2\sqrt{3}/3$. $\endgroup$ Apr 11, 2019 at 11:38

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 $z=\left(a^2+b^2\right)$, $x=\left(a^2-b^2\right)$ and $y=2ab$

To expand to all the non primitive Pythagorean triangles we have



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

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: $$\left(a_1^2-b_1^2\right)a_1b_1c_1^2=\left(a_2^2-b_2^2\right)a_2b_2c_2^2$$ $$\frac{a_2+b_2}{a_1+b_1}=\frac{a_1c_1}{a_2c_2}\left(\frac{b_1c_1}{b_2c_2}\frac{ (a_1-b_1) }{ (a_2-b_2) } \right)$$

Therefore combining both gives



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 $$\frac{ (a_2+b_2) a_2 }{ (a_2-b_2) b_2 }=\frac{ (a_1+b_1) a_1 }{ (a_1-b_1) b_1 }$$

  • $\begingroup$ In the version you use for primitive triples I think one needs $a,b$ odd and coprime, $a>b$ to give triples of positives. [That seems it would be known in that version…] I used the other version in my attempt, $p^2-q^2,2pq,p^2+q^2$ with $p,q$ coprime opposite parity and $p>q.$ Still didn't go to a finish in my attempts though. $\endgroup$
    – coffeemath
    Apr 10, 2019 at 12:37
  • 1
    $\begingroup$ updated to your version $\endgroup$ Apr 10, 2019 at 13:10

This a Pythagorean triple treatment of area to perimeter ratios.

We can find Pythagorean triples, if they exist, for any ratio $R$ of area/perimeter by finding the $m,n$(s) that represent them using the following formula which includes a difined finite search for values of $m$. Whenever the $m$ $R$ combination yields a positive integer for $n$, we have the $m,n$ for a triple. We begin by solving the area/perimeter equation for $n$ where area=$D$ so as not to confuse it with $A,B,C$.

$$D=\frac{AB}{2}=\frac{(m^2-n^2)*2mn}{2}=mn(m^2-n^2)\quad P=(m^2-n^2)+2mn+(m^2+n^2)=2m^2+2mn$$

$$\frac{D}{P}=\frac{mn(m^2-n^2)}{2m^2+2mn}=\frac{mn(m-n)(m+n)}{2m(m+n)}=\frac{n(m-n)}{2}=R\qquad n^2-mn+2R=0$$


$$n=\frac{m\pm \sqrt{m^2-8R}}{2}\text{ where }\lceil\sqrt{8R}\space \rceil \le m \le 2R+1$$

Example: For $R=0.5\quad \sqrt{8*0.5}=2\le m \le 2*0.5+1=2$

$$n=\frac{2\pm \sqrt{2^2-8*0.5}}{2}=1\qquad f(2,1)=(3,4,5)$$ But you are asking about $R=1$ and there only two such Pythagorean triples to be found.

$$R=1\rightarrow 3\le m \le 3\quad f(3,2)=(5,12,13)\quad f(3,1)=(8,6,10)$$ For other ratios: $$R=1.5\rightarrow 4\le m \le 4\quad f(4,3)=(7,24,25)\quad f(4,1)=(15,8,17)$$

$$R=2\rightarrow 4\le m \le 5\quad f(4,2)=(12,16,20)\quad f(5,4)=(9,40,41)\quad f(5,1)=(24,10,26)$$

$$R=2.5\rightarrow 4\le m \le 6\quad f(6,5)=(11,60,61)\quad f(6,1)=(35,12,37)$$

$$R=3\rightarrow 4\le m \le 7\quad f(5,3)=(16,30,34)\quad f(5,2)=(21,20,29)\quad f(7,6)=(13,84,85)\quad f(7,1)=(48,14,50)$$

$$R=18\rightarrow 12\le m \le 37\quad f(12,6)=(108,144,180)\quad f(13,9)=(88,234,250)\quad f(13,4)=(153,104,185)\quad f(15,12)=(81,360,369)\quad f(15,3)=(216,90,234)\quad f(20,18)=(76,720,724)\quad f(20,2)=(396,80,404)\quad f(37,36)=(73,2664,2665)\quad f(37,1)=(1368,74,1370)$$

Aside: you can always find one primitive triple for a given $R$ if you let $(m,n)=(2R+1,2R)$.


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