Find all triangles of which perimeter and area are numerically equal

Question

Find all triangles of which perimeter and area are numerically equal. I have got solution for right angle triangles but not of others

Answer 1

Area = $rs$, where $r=\text{inradius}$ and $s=\text{perimeter}/2$

You can see that $rs=p \implies r=2$

There are infinite triangles with inradius as $2$

Answer 2

The area is $A = \sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is the semiperimeter. Thus we get $$(a+b+c)^2 = \frac{a+b+c}{2}(\frac{a+b+c}{2} - a)(\frac{a+b+c}{2} - b)(\frac{a+b+c}{2} - c).$$ We can further simplify this to $$16(a+b+c) = (-a+b+c)(a-b+c)(a+b-c).$$ Let $u = -a+b+c$, $v = a-b+c$, $w = a+b-c$. Then $$16(u+v+w) = u v w.$$ In particular any $u,v$ such that $uv > 16$ give a solution for $w$: $$w = \frac{16(u+v)}{uv-16}.$$ Now for such $u,v,w$ we have that $a = \frac{v+w}{2}$, $b = \frac{w+u}{2}$ and $c = \frac{u+v}{2}$ are the sides of a triangle whose area is equal to its perimeter.

Answer 3

There is some information at Wikipedia. As another answer notes, these are precisely the triangles with inradius 2. But more information is given, for example, that there are exactly five such triangles with integer sides.

Answer 4

The problem makes not much sense in euclidean plane geometry, for the following reason:

Let $E$ be an euclidean plane and suppose that a unit of length has been chosen by marking two points $A$ and $B$. This automatically defines a unit of area, namely the area of a square with sidelength $|AB|$; in the same way as the length unit "mile" automatically carries with it the area unit "square mile".

In $E$ we have the operation of stretching available. Linear stretching of any figure by a factor $\lambda>0$ multiplies all lengths in this figure by $\lambda$, and all areas by $\lambda^2$.

Now take any (nondegenerate) triangle $T\subset E$. In terms of the chosen length unit $T$ has a certain perimeter $2s>0$ and a certain area $A>0$. Then stretch $T$ by the factor $\lambda:={2s\over A}>0$. You obtain a triangle $T'$ with perimeter $\lambda\cdot 2s={4s^2\over A}$ and area $\lambda^2\cdot A={4s^2\over A}$. It follows that for $T'$ the perimeter and the area are numerically equal.

This implies that there are no distinguished shapes of triangles for which the perimeter and the area are numerically equal.

Answer 5

It should be pointed out that regarding integer sided triangles whose area is numerically equal to its perimeter that no equilateral triangle exists, since that would imply that [sqrt(3)/4]*s^2 = 3s, obviously impossible. For isosceles triangles, using the equation derived in (1), w = 16(u + v)/(uv -16) , suppose we let u = v, which is the same as triangle side a =b. Then we have: w = 32u/(u^2 - 16). This Diophantine equation has the unique solution u = 12, w = 3, and of course v =12. Then p = u + v + w =27, and 2a = p - u =15, so there can be no integer solution for an isosceles triangle. Ed Gray There are 5 triangles with A = p:((5,12,13),(6,8,10),(29,25,6),(20,15,7),(17,10,9). These results are obtained by solving : 16(u+v+w) = uvw