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.

  • 2
    $\begingroup$ What is the source of this problem? $\endgroup$ – Ahaan S. Rungta Dec 20 '14 at 5:51
  • $\begingroup$ This is something I conjectured was true, then confirmed via coordinates, then could not solve with pure geometry. $\endgroup$ – Sameer Kailasa Dec 20 '14 at 15:39
  • $\begingroup$ 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. $\endgroup$ – Ahaan S. Rungta Dec 20 '14 at 15:58
  • $\begingroup$ 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. $\endgroup$ – Sameer Kailasa Dec 20 '14 at 15:59
  • $\begingroup$ Would you like to show us your work in “co-ordinate bash” supporting your claim is definitely true? $\endgroup$ – Mick Dec 21 '14 at 2:57

As requested, here is a solution via coordinates. However, as mentioned above, I'd be interested in seeing a solution via pure geometry instead.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.