Complex Numbers

Let $|a|=a_1+a_2i$ , $|b|=b_1+b_2i$ , and $|c|=c_1+c_2i$ , so that $a_1^2+a_2^2+b_1^2+b_2^2+c_1^2+c_2^2=250$ . The sides of the triangle have length $|b-a|$ , $|c-b|$ , and $|c-a|$ . WLOG let $h=|c-a|$ represent the hypotenuse. By the Pythagorean Theorem, $|b-a|^2+|c-b|^2=|c-a|^2$ .

Therefore, $|c-a|^2=\frac{|c-a|^2+|c-a|^2}2=\frac{|b-a|^2+|c-b|^2+|c-a|^2}2$ , which becomes $\frac{(b_1-a_1)^2+(b_2-a_2)^2+(c_1-b_1)^2+(c_2-b_2)^2+(c_1-a_1)^2+(c_2-a_2)^2}2$ . Multiplying this out, we get $\frac{2(a_1^2+a_2^2+b_1^2+b_2^2+c_1^2+c_2^2)-2(a_1b_1+a_2b_2+a_1c_1+a_2c_2+b_1c_1+b_2c_2)}2$ . This equals $(a_1^2+a_2^2+b_1^2+b_2^2+c_1^2+c_2^2)-(a_1b_1+a_2b_2+a_1c_1+a_2c_2+b_1c_1+b_2c_2)$ , or $250-(a_1b_1+a_2b_2+a_1c_1+a_2c_2+b_1c_1+b_2c_2)$ .

Now, by Vieta, the sum of the solutions is $0$ , so $(a_1+b_1+c_1)+(a_2+b_2+c_2)i=0$ . Equating coefficients, $a_1+b_1+c_1=0$ and $a_2+b_2+c_2=0$ . Squaring both and adding, we get $a_1^2+a_2^2+b_1^2+b_2^2+c_1^2+c_2^2+2(a_1b_1+a_2b_2+a_1c_1+a_2c_2+b_1c_1+b_2c_2)=0$ , so $250+2(a_1b_1+a_2b_2+a_1c_1+a_2c_2+b_1c_1+b_2c_2)=0$ . Thus $a_1b_1+a_2b_2+a_1c_1+a_2c_2+b_1c_1+b_2c_2=-125$ .

Therefore, we have $h^2=|c-a|^2=250-(a_1b_1+a_2b_2+a_1c_1+a_2c_2+b_1c_1+b_2c_2)=250-(-125)$ , or $\boxed{375}$ .

By Vieta's formula , the sum of the roots is equal to $0,$ or $a+b+c=0.$ Therefore, $\dfrac{(a+b+c)}{3}=0$ . Because the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin. Suppose one leg of the right triangle is $x$ and the other leg is $y$ .

WLOG, suppose $\overline{ac}$ is the hypotenuse. The magnitudes of $a, b,$ and $c$ are just $\dfrac{2}{3}$ of the medians because the origin, or the centroid in this case, cuts the median in a ratio of $2:1.$ So, $|a|^2=\dfrac{4}{9}\cdot \left\{\left(\dfrac{x}{2}\right)^2+y^2 \right\}=\dfrac{x^2}{9}+\dfrac{4y^2}{9}$ because $|a|$ is two thirds of the median from $a$ . Similarly, $|c|^2=\dfrac{4}{9}\cdot(x^2+(\dfrac{y}{2})^2)=\dfrac{4x^2}{9}+\dfrac{y^2}{9}$ . The median from $b$ is just half the hypotenuse because the median of any right triangle is just half the hypotenuse. So, $|b|^2=\frac{4}{9}\cdot\frac{x^2+y^2}{4}=\dfrac{x^2}{9}+\dfrac{y^2}{9}$ . Hence, $|a|^2+|b|^2+|c|^2=\dfrac{6x^2+6y^2}{9}=\dfrac{2x^2+2y^2}{3}=250.$ Therefore, $h^2=x^2+y^2=\dfrac{3}{2}\cdot250=\boxed{375}$ .