Question

A right triangle has legs $a$ and $b$ and hypotenuse $c$. Two segments from the right angle to the hypotenuse are drawn, dividing it into three equal parts of length $x=\frac{c}{3}$. If the segments have length $p$ and $q$, then $p^2+q^2=k{x}^2$. Find the value of $k$.

Answer 1

Consider the two points $A,B$ where the line meets at $D,E$

$CD$ is the median of $△CBE$ to side $BE$

So $CD=\frac{1}{2}\sqrt{{2a}^2+{2q}^2+{\left(\frac{2c}{3}\right)}^2}$

$4{CD}^2={2a}^2+{2q}^2+{\frac{4c}{9}}^2$

$4p^2={2a}^2+{2q}^2+{\frac{4c}{9}}^2...\left(1\right)$

$CE$ is the median of $△CDA$ opposite to side $AD$

So $CE=\frac{1}{2}\sqrt{{2p}^2+{2b}^2-{\left(\frac{2c}{3}\right)}^2}$

$4{CE}^2={2p}^2+{2b}^2-{\frac{4c}{9}}^2$

$4q^2={2p}^2+{2b}^2+{\frac{4c}{9}}^2...\left(2\right)$

Adding the two equations we get

$4p^2+{4q}^2=2a^2+2q^2+2p^2+2b^2-{\frac{4c}{9}}^2-{\frac{4c}{9}}^2$

$2p^2+{2q}^2=2a^2+2b^2-{\frac{8c}{9}}^2$

$2p^2+{2q}^2=2(a^2+b^2)-{\frac{8c}{9}}^2$

$2p^2+{2q}^2=2\left(c^2\right)-{\frac{8c}{9}}^2$

$2p^2+{2q}^2={\frac{10c}{9}}^2$

$p^2+q^2={\frac{5c}{9}}^2$

$p^2+q^2=5{\frac{c}{9}}^2$

Hence, $k=5$.