# math contest geometry problem

Consider a triangle $ABC$ with circumcircle $\omega$. Let $O$ be the center of $\omega$ and let $D, E, F$ be the midpoints of minor arcs $BC, CA, AB$ respectively. Let $DO$ intersect $\omega$ again at a point $A'$. Define $B'$ and $C'$ similarly. Prove that

$[ABC] \leq [A'B'C']$

$[X]$ denotes area of $X$.

• Note that $[DEF]=[A'B'C']$ – vrugtehagel Jan 26 '16 at 15:30
• If $A',B',C'$ are just the opposing points of $D,E,F$, then isn't $A'B'C'$ congruent to $DEF$? – Thomas Andrews Jan 26 '16 at 15:31
• @ThomasAndrews, yes. Since the perpendicular bisector of $BC$ goes through $O$ and $D$, $\triangle A'B'C'$ is just a 180 degree rotation of $\triangle DEF$ around center $O$ – vrugtehagel Jan 26 '16 at 15:34
• It has nothing to do with $BC$, @vrugtehagel You take any triangle $DEF$ on $\omega$ and pick the anti-podal points on the circle, you get a congruent triangle to $DEF$. – Thomas Andrews Jan 26 '16 at 15:42

Let $\angle AOB=2\gamma$, $\angle BOC=2\alpha$, $\angle COA=2\beta$, so that: $$[ABC]={1\over2}r^2(\sin2\alpha+\sin2\beta+\sin2\gamma),$$ where $r$ is the radius of $\omega$. On the other hand it follows from the definition of $DEF$ that $$[DEF]={1\over2}r^2(\sin(\alpha+\beta)+\sin(\beta+\gamma)+\sin(\gamma+\alpha)).$$ As $[A'B'C']=[DEF]$, we must prove $[ABC]\le[DEF]$, that is $$\sin2\alpha+\sin2\beta+\sin2\gamma\le \sin(\alpha+\beta)+\sin(\beta+\gamma)+\sin(\gamma+\alpha).$$ As $\alpha+\beta+\gamma=\pi$ we can eliminate $\gamma$ from the above inequality, which then becomes: $$\sin2\alpha+\sin2\beta-\sin2(\alpha+\beta)\le \sin(\alpha+\beta)+\sin\alpha+\sin\beta.$$ This can also be put in the form $F(\alpha,\beta)\ge0$, where: $$F(\alpha,\beta)=\sin\alpha+\sin\beta +\sin(\alpha+\beta)(1-4\sin\alpha\sin\beta)$$ This inequality must be proved for $0\le\alpha,\beta\le\pi/3$, because we can take as $\gamma$ the largest among the angles. The proof can be then carried out with the standard techniques of calculus, to show that the minimum of $F$ is zero.