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.
Best Answer
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 $4$.
$m=1\implies n=5\implies (10, 24, 26)$
$m=2\implies n=4 \implies (16, 12,20)$
$m=4 \implies n=5 \implies (40, 9, 41)$