Third side
What are the possible lengths for the third side of this right-angled triangle?
Age
14 to 16
Challenge level
Problem
The lengths of the sides of a right-angled triangle are all integers.
The sum of the lengths of the hypotenuse and one of the other sides is 49.
What are the three possible lengths for the third side?
This problem is adapted from the World Mathematics Championships
Student Solutions
Answers: $7, 21, 35$
Image
$a^2+b^2=c^2$
$c+a=49$
Find $b$
The difference of two squares
$\begin{split}b^2&=c^2-a^2\\
&=(c+a)(c-a)\\
&=\ \ 49\ \ (c-a)\\
&=7^2(c-a)\end{split}$
$\therefore b=7\sqrt{c-a}$ and so $(c-a)$ is a square number
Also $c+a=49\ \ \therefore c$ and $a$ are one even, one odd $\therefore \ c-a$ is odd
And $b\lt c\Rightarrow b\lt49=7\times7$ so $\sqrt{c-a} \lt 7$
So $c-a = 1^2, 3^2$ or $5^2$
$c-a=1^2\Rightarrow b=7\times1=7$
$c-a=3^2\Rightarrow b=7\times3=21$
$c-a=5^2\Rightarrow b=7\times5=35$
Expanding and factorising
$\begin{align} c&=49-a\\
\therefore a^2+b^2 &= (49-a)^2\\
\Rightarrow a^2+b^2 &= 49^2-98a+a^2\\
\Rightarrow b^2 &=49^2-98a\end{align}$
Need $49^2-98a$ to be a square number
$=49(49-2a)\\
=7^2(49-2a)$
$\therefore 49-2a$ is also a square number, and it is odd$-$even$=$odd
Odd squares below $49:$
$25\ \ \Rightarrow b^2=7^2\times25\Rightarrow b=35\\
9\ \ \Rightarrow b^2=7^2\times9\Rightarrow b=21\\
1\ \ \Rightarrow b^2=7^2\times1\Rightarrow b=7$
Expanding and factorising
$\begin{align} c&=49-a\\
\therefore a^2+b^2 &= (49-a)^2\\
\Rightarrow a^2+b^2 &= 49^2-98a+a^2\\
\Rightarrow b^2 &=49^2-98a\end{align}$
Need $49^2-98a$ to be a square number
$=49(49-2a)\\
=7^2(49-2a)$
$\therefore 49-2a$ is also a square number, and it is odd$-$even$=$odd
Odd squares below $49:$
$25\ \ \Rightarrow b^2=7^2\times25\Rightarrow b=35\\
9\ \ \Rightarrow b^2=7^2\times9\Rightarrow b=21\\
1\ \ \Rightarrow b^2=7^2\times1\Rightarrow b=7$