# What is the greatest possible perimeter of a right-angled triangle with integer side lengths if one of the sides has length 12?

What is the greatest possible perimeter of a right-angled triangle with integer side lengths if one of the sides has length 12 ?

I know this question can be done in a variety of ways, the answer comes out to be $$84$$.

However, my friend and I were trying this question using QM$$\ge$$AM, and we didn't get $$84$$ through that, so I decided to ask here.

Let $$x,y,z$$ be the sides, and $$x=12$$.

We have:

$$\sqrt{\frac{x^2+y^2+z^2}{3}} \ge \frac{x+y+z}{3}$$

And $$x,y,z$$ are +ve because length can't be negative.

Since we want the maximum value the perimeter, the equality of the expression holds at $$x=y=z$$

plugging in the values, I get the maximum perimeter as $$36$$. Why am I getting the wrong answer solving this way?

I would be grateful if someone helped. Thanks.

• $x=y=z$ gives you an equilateral triangle. Your question is about right triangle, that doesn't get addressed in your approach. Commented Jul 19, 2021 at 2:45
• Yes, you're right. Thanks, however using this inequality shouldn't we be getting the maximum perimeter in all possible cases (including all triangles)? Commented Jul 19, 2021 at 2:47
• Your approach can't be right because if you scale up a $3$-$4$-$5$ right triangle by a factor of $4$, you get a perimeter of $48$. Commented Jul 19, 2021 at 2:48
• @Mathematica The issue is that the inequality you are using is about three positive numbers $x,y,z$, whereas in your problem you want $x,y,z$ to be the sides of a triangle, so they are not just any numbers, they should satisfy constraints like triangle inequality etc. and that is why your approach won't yield the results. Commented Jul 19, 2021 at 2:54
• yeahh got it thankss Commented Jul 19, 2021 at 2:59

Alternative approach.

Either $$(12)$$ is the hypotenuse or it is one of the legs. The analysis below, which assumes that $$(12)$$ is one of the legs, will clearly show that setting $$(12)$$ as the hypotenuse will not maximize the perimeter.

Therefore, you have that $$144 = (12)^2 = (x^2 - y^2) = (x + y)(x - y)$$.

Here, it is desired to maximize $$(x + y)$$, so the possibility of $$(x + y) = 144$$ must be examined. This can be seen as impossible, because it requires that $$x,y$$ have the same odd/even parity, which makes it impossible for $$(x - y)$$ to equal $$1$$.

Therefore, the maximum value for $$(x + y)$$ is $$72$$, which requires that $$(x - y) = 2.$$ This is achieved via $$(x,y) = (37,35).$$

• Just a quick note: $12$ cannot be the length of the hypotenuse because $12$ cannot be written as sum of squares of two integers. So that case can be ruled out easily. Commented Jul 19, 2021 at 4:15
• @AnuragA true, but requires some analysis to prove. My way, no such corresponding analysis is required, because it becomes irrelevant whether $(12)$ can be a hypotenuse. Commented Jul 19, 2021 at 4:18

Using Euclid algorithm. $$a=2mn=12$$, max possible with $$m=6$$ and $$n=1$$ gives $$b=m^2-n^2=35$$ and $$c=m^2+n^2=37$$ so perimeter $$=84$$.

• Trying $b=m^2-n^2=12$ leads to $m=4$ and $n=2$ as the only possibility, making $a=2mn=16$ and $c=m^2+n^2=20$ for a total of $48 \lt 84$. Commented Jul 19, 2021 at 17:32
• The Euclid algorithm does not generate all Pythagorean triangles. It produces all the primitive ones, but it will not produce $9,12,15$, for example. In this case it finds the one we want. Commented Oct 12, 2021 at 4:44