Area of a circumcenter triangle equals area of medial triangle

Let $X$, $Y$, $Z$ be the midpoints of sides $BC$, $AC$, $AB$ respectively in triangle $ABC$. Let $O_{A}$, $O_{B}$, and $O_{C}$ be the circumcenters of triangles $AZX$, $BXY$, and $CYZ$ respectively. Prove that the area of triangle $O_{A} O_{B} O_{C}$ equals the area of triangle $XYZ$.

Basically, I've done a horrific coordinate bash, and the statement is definitely true. But I'd be very interested in seeing a purely Euclidean solution.

• What is the source of this problem? – Ahaan S. Rungta Dec 20 '14 at 5:51
• This is something I conjectured was true, then confirmed via coordinates, then could not solve with pure geometry. – Sameer Kailasa Dec 20 '14 at 15:39
• Then why does this have the contest-math tag? Is this from an ongoing contest? I'm afraid nobody will want to help you without further details. – Ahaan S. Rungta Dec 20 '14 at 15:58
• It's not from an ongoing contest. The contest-math tag has description "problems from or inspired by mathematics competitions." This is certainly inspired by olympiad geometry problems. – Sameer Kailasa Dec 20 '14 at 15:59
• Would you like to show us your work in “co-ordinate bash” supporting your claim is definitely true? – Mick Dec 21 '14 at 2:57

We assign coordinates $A(0,0)$, $B(b,0)$, $C(c,d)$ to the vertices of triangle $ABC$. This gives $X = ((b+c)/2, d/2)$, $Y = (c/2, d/2)$, $Z = (b/2, 0)$. $O_A$ has coordinates corresponding to the intersection of the perpendicular bisector of $AZ$ and $AX$. These bisectors have equations $x = b/4$ and $$y = -\frac{b+c}{d} \left(x-\frac{b+c}{2} \right)+ \frac{d}{4}$$ Solving the system gives $$O_{A} = \left(\frac{b}{4}, \frac{d}{4} - \frac{bc+c^2}{4d} \right)$$ Similarly, we compute $O_{B}$ and $O_{C}$: $$O_{B} = \left(\frac{2c+b}{4}, \frac{3bc+d^2-c^2-2b^2}{4d} \right)$$ $$O_{C} = \left(\frac{b^2 + 2(c^2 + d^2)}{4b}, -\frac{b^2 c+ (2c-3b)(c^2 + d^2)}{4bd} \right)$$ Plugging these in to the shoelace formula and simplifying with a CAS yields $$[O_{A} O_{B} O_{C}] = \frac{1}{4} \left(\frac{1}{2} bd\right) = \frac{1}{4} [ABC] = [XYZ]$$ as desired.