Given perimeter of triangle and one side, find other two sides

In triangle ABC, all three sides have integer lengths. If AB = 21, the perimeter is 54, and the area is a positive integer, what are the lengths of BC and AC?

I tried using Heron's Formula, but I couldn't really get anywhere. Any suggestions? This question was in my Math Challenge II Number Theory packet.

• Heron's formula will go wild. Draw a height, it will be much nicer. – chubakueno May 8 '14 at 1:22

If the area is an integer, the height to $AB$ has to be an even integer, and because the other sides are an integer, it has to be a member of two different pythagorean triangles (since we can't divide $21$ in a half). $8,12$ are possible candidates then.$12$ works:

$$12^2+5^2=13^2,12^2+16^2=20^2, 13+20+21=54$$

Just to show that Heron's formula isn't so bad ...

In $$\text{area}^2 = \frac{1}{16}(a+b+c)(-a+b+c)(a-b+c)(a+b-c)$$ you know $c=21$ and $a+b+c=54$, so that $b=33-a$. So, the above becomes \begin{align} \text{area}^2 &= \frac{1}{16}(54)(54-2a)(2a-12)(12) \\[4pt] &= 2\cdot 3^4 \cdot (27-a)(a-6) \end{align} Thus, $a$ is somewhere between $6$ and $27$. Since the right-hand side needs to be a perfect square, the factors $(27-a)$ and $(a-6)$ need to contribute an odd number of $2$s, and an even number of any other prime, to the factorization. You can tick through cases pretty quickly, using the latter fact to instantly weed-out cases with obvious singleton prime factors before expending any real mental energy counting other primes:

$$\begin{array}{rccll} a & (27-a) & ( a - 6 ) & & \\ 7 & 20 & 1 & \text{single } 5\\ 8 & 19 & 2 & \text{single } 19 \\ 9 & 18 & 3 & - & 3^\text{odd} \to \text{nope!}\\ 10 & 17 & 4 & \text{single } 17\\ 11 & 16 & 5 & \text{single } 5\\ 12 & 15 & 6 & \text{single } 5 \\ \color{red}{13} & \color{red}{14} & \color{red}{7} & \color{red}{-} & \color{red}{2^\text{odd} \cdot 7^\text{even} \to \text{yes!}} \\ 14 & 13 & 8 & \text{single } 13\\ 15 & 12 & 9 & - & 3^\text{odd}, 2^\text{even} \to \text{nope!}\\ 16 & 11 & 10 & \text{single } 11 \\ 17 \dots 26 & - & - & \text{don't need to check} & \text{(why?)} \\ \end{array}$$

Therefore, the other sides have length $13$ and $33-13=20$. $\square$

protected by user642796Apr 27 '17 at 17:21

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?