Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.7 AIME Prove it Belarus

A triangle $ABC$ is inscribed in the parabola $y = x^2$ . Let $a$ , $b$ , $c$ be the abscissae of the midpoints of its sides.
Find the radius of the circumcircle of $\triangle ABC$ .

Solution

Answers: $R = 0.5\sqrt{(1 + 4a^2)(1 + 4b^2)(1 + 4c^2)}$ .

Let $A(l; l^2)$ , $B(m; m^2)$ , $C(n; n^2)$ be coordinates of the vertices of the triangle $ABC$ . Then
$AB = \sqrt{(m-l)^2 + (m^2-l^2)^2} = |m-l|\sqrt{1+(m+l)^2} = |m-l|\sqrt{1+4c^2},$
since $c = 0.5(m + l)$ .
Similarly, $BC = |m - n|\sqrt{1 + 4a^2}$ and $AC = |n - l|\sqrt{1 + 4b^2}$ .

Further, it is easy to calculate that $S(ABC) = 0.5|(m - l)(l - n)(n - m)|$ .

Now we find the circumradius using the formula

R = $\text{AB BC}$ CA}{4S(ABC)} = $\frac{1}{2} \sqrt{(1}$ + 4a^2)(1 + 4b^2)(1 + 4c^2)}.

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