# A right-angled triangle has sides of integer length. Its area (in square metres) is twice its perimeter (in metres). What are the lengths of the sides

A right-angled triangle has sides of integer length. Its area (in square metres) is twice its perimeter (in metres). What are the lengths of the sides?

The equations I have made so far is:

Using Pythagoras' Theroem and Area Formula: $$\frac{ab}{4}=a+b+\sqrt{a^2+b^2}$$

Using Heron's formula:

$$s(s-a)(s-b)(s-c)=4(a+b+c)^2$$ For LHS:$$(\frac{a+b+c}{2})(\frac{a+b+c-2a}{2})(\frac{a+b+c-2b}{2}) (\frac{a+b+c-2c}{2})$$ $$\frac{1}{16}((a+b+c)(b+c-a)(a+c-b)(a+b-c))$$ $$\frac{1}{16}(((b+c)^2-a^2)(a^2-(b-c)^2))$$

Please find the original link for the question here: https://www.maths.uq.edu.au/qamt/papers/Year9-10-2022-Paper.pdf Question 4

I believe that even though this question is simple, it is quite tricky and does need some consideration. Additionally, I needed this question's answer while I was preparing for the Year 9-10 UQ/QAMT Paper, as it still holds relevance as it still only is a 2-year old question!

My question is often compared to A right triangle with integer sides has area equal to twice its perimeter. Find sum of all possible circumradii., but I believe the question is different.

• I wonder if the person who posed this problem realized how many different answers are possible. Apr 25 at 3:01
• This is kind of a general equation with many possible answers. Are you sure no other data has been provided? If you're familiar with Heron's formula, it'll help you get the equation in a clean factorised form
– Gwen
Apr 25 at 3:02
• This question comes from the UQ/QAMT Maths Competition. Find the link here: maths.uq.edu.au/qamt/papers/Year9-10-2022-Paper.pdf Question 4 Apr 25 at 3:05
– Gwen
Apr 25 at 4:12
• Does this answer your question? A right triangle with integer sides has area equal to twice its perimeter. Find sum of all possible circumradii. Apr 26 at 2:15

I don't think you need Heron's Formula. Parameterize the triangle:

$$a=2mn, b=n^2-m^2, c=n^2+m^2$$

$$A=\frac{ab}{2}$$. $$\ \$$ $$P=a+b+c.$$

$$A=mn\cdot (n^2-m^2)$$

$$P=2mn+n^2-m^2+m^2+n^2=2mn+2n^2=2n(m+n)$$

$$A=2P\implies mn(n^2-m^2)=4nm+4n^2$$

$$m(n-m)=4$$

So $$m$$ must be a factor of $$4$$.

$$m=1\implies n=5\implies (10, 24, 26)$$

$$m=2\implies n=4 \implies (16, 12,20)$$

$$m=4 \implies n=5 \implies (40, 9, 41)$$

• Heron's formula would probably be known to students sitting this paper, but I do prefer the solutions that don't use it (just personal preference). Apr 25 at 5:25

$$\frac{ab}{4}=a+b+\sqrt{a^2+b^2}\\ \implies ab=4a+4b+4\sqrt{a^2+b^2}\\ \implies -4\sqrt{a^2+b^2}=4a+4b-ab \\ \implies (-4\sqrt{a^2+b^2})^2=(4a+4b-ab)^2 \\ \implies 16a^2+16b^2=16a^2+16b^2+a^2 b^2 + 32ab-8a^2b-8b^2a \\ \implies 8a^2b+8b^2a=a^2b^2+32ab\\ \implies 8a+8b-ab=32\\ \implies a+b-\frac{ab}{8}=4\\ \implies b=\frac{8(a-4)}{(a-8)}$$ From here, we can get integer $$b$$ when $$(a-8)$$ is either a factor or multiple of 8. Thus $$a=9,10,12,16,24,40$$. Consequently $$b=40,24,16,12,10,9$$.

I don't take further $$a$$ values because in the expression, $$a-4>a-8,\frac{a-4}{a-8}>1,\frac{8(a-4)}{a-8}>8,b>8$$. Thus when I saw that at $$a=40,b=9$$ this meant all other greater $$a$$ values gave non integer $$b,8. ($$b$$ is decreasing as $$a$$ increases)

Magically enough, all these pairs are Pythagorean triplets. $$9^2+40^2=41^2,12^2+16^2=20^2,24^2+10^2=26^2$$

Let $$a$$ and $$b$$ be the legs. Then the area $$ab/2$$ doubles the perimeter $$a+b+\sqrt{a^2+b^2}$$. So

$$ab=4(a+b+\sqrt{a^2+b^2}).$$

Multiply by the conjugate $$a+b-\sqrt{a^2+b^2}$$ and apply the difference of squares factorization to simplify the right side:

$$ab(a+b-\sqrt{a^2+b^2})=8ab$$

$$a+b-\sqrt{a^2+b^2}=8$$

$$a+b-8=\sqrt{a^2+b^2}$$

$$a^2+2ab-16a+b^2-16b+64=a^2+b^2$$

$$2ab-16a-16b+64=2(ab-8a-8b+32)=0$$

So

$$(a-8)(b-8)=ab-8a-8b+64=32,$$

and we identify three distinct triangles:

$$a-8=1,b-8=32\implies a=9,b=40,c=\sqrt{a^2+b^2}=a+b-8=41$$

$$a-8=2,b-8=16\implies a=10,b=24,c=26$$

$$a-8=4,b-8=8\implies a=12,b=16,c=20.$$

First we apply a lemma

Lemma. Let $$p,r,A$$ be the perimeter, inscribed radius and the area of the triangle respectively. Then we have $$pr=2A$$.

The proof is easy, just because radius of inscribed circle is perpendicular to triangle side, so we can divide it into three slices, and $$\cdots$$

Now with this fact, we have $$pr=2A=4p$$, so we get $$r=4$$.

From now on, we let the triangle be $$\Delta ABC$$ with $$\angle A$$ is right angle. set $$I_C,I_B$$ be the projection of the incenter on the line $$AB$$ and $$AC$$ respectively. Note $$AI_B=AI_C=r=4$$. Let $$I_BC=t,I_CB=s$$. By tangent property we have $$BC=s+t$$. By pythagorus theorem, we have $$(4+s)^2+(4+t)^2=(s+t)^2$$ Rearranging we get (Here we know $$c\ne4$$, can you do it yourself?)$$s=\dfrac{4t+16}{t-4}=4+\dfrac{32}{t-4}$$ Thus, we have $$t-4$$ is a factor of $$32$$, i.e. $$t-4=1,2,4,8,16,32\iff (s,t)=(36,5),(20,6),(12,8),(8,12),(6,20),(5,36)$$

Now I give you a task, is to verify how much of them is actually a solution.

• OP has already given meaning to $b$ and $c$, perhaps you should use some other letters if your $b$ and $c$ aren't the same as those of OP. Apr 25 at 4:01
• yah that's a good idea, i will adjust it. Apr 25 at 4:04
• Check your ordered pairs for $s$ and $t$. They should be symmetrical, that is a given $(s,t)$ should be accompanied by $(t,s)$. A correct rendering gives no extraneous solutions but counts the good ones twice due to this symmetry. Apr 26 at 20:12
• arh yes I'm too lazy to check it, thanks for pointing out. Apr 27 at 1:23