National Olympiad Final Round

Assume w.l.o.g. that $BC$ is the shortest side of the triangle. Then the least angle of the triangle is by vertex $A$. Denote $a=BC$, $b=CA$, $c=AB$, $\alpha =\angle BAC$ and let $m$ be the length of the median drawn from vertex $A$. By assumptions made at the beginning of the solution, $\alpha \le 60^{\circ }$. Let $D,E,F$ be the midpoints of sides $BC,CA,AB$, respectively (Fig. 21).

As $\angle AFD=180^{\circ }-\alpha$ because of $DF\parallel CA$, the law of cosines in triangle $AFD$ implies $$m^2={\left(\frac{b}{2}\right)}^2+{\left(\frac{c}{2}\right)}^2-2\cdot \frac{b}{2}\cdot \frac{c}{2}\cos \angle AFD=\frac{b^2}{4}+\frac{c^2}{4}+\frac{bc}{2}\cos \alpha$$ $$\ge \frac{b^2}{4}+\frac{c^2}{4}+\frac{bc}{2}\cos 60^{\circ }=\frac{b^2+c^2+bc}{4}.$$ As $b^2+c^2\ge 2bc$, we obtain $m^2\ge \frac{3bc}{4}$.

On the other hand, let $S$ be the area of the triangle $ABC$. Then $S=\frac{1}{2}bc\sin \alpha \le \frac{1}{2}bc\sin 60^{\circ }\le \frac{\sqrt{3}bc}{4}$. Before we obtained the inequality $m^2\ge \frac{3bc}{4}=\sqrt{3}\cdot \frac{\sqrt{3}bc}{4}$. Consequently, $m^2\ge \sqrt{3}S$.

Fig. 21

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Alternative solution.

Let the median drawn from vertex $A$ of the triangle $ABC$ be $AD$ and the centroid of the triangle be $M$. W.l.o.g., let $AD$ be the longest median of triangle $ABC$; denote $AD=m$. Draw to both sides of the median $AD$ triangles $ADX$ and $ADY$ that have right angles by vertex $D$ and angles of size $30^{\circ }$ by vertex $A$; then $AXY$ is an equilateral triangle having median $AD$ and centroid $M$ in common with triangle $ABC$ (Fig. 22). We show next that the area of triangle $AXY$ is at least as large as the area of triangle $ABC$. Since a median divides the triangle into two parts of equal area, it suffices to show that the area of triangle $ADX$ is at least as large as the area of triangle $ADB$.

If point $B$ lies inside the triangle $ADX$ or on its side then this claim holds obviously. If point $C$ lies inside the triangle $ADY$ or on its side then this claim holds by symmetry. It remains to handle the case where both $B$ and $C$ lie outside the triangle $AXY$ (Fig. 23). Suppose w.l.o.g. that line segments $DB$ and $AX$ intersect (the other case where line segments $DC$ and $AY$ intersect is symmetric). Let $l$ be the line parallel to $AX$ passing through $Y$. As line $l$ is symmetric w.r.t. point $D$ with line $AX$ and $DC=DB$, line segment $DC$ intersects line $l$. Therefore $MC>MY=MA$, which contradicts the assumption that $AD$ is the longest median of the triangle $ABC$. Hence this case cannot appear.

As $XD=YD=\frac{m}{\sqrt{3}}$, the area of the triangle $AXY$ is $m\cdot \frac{m}{\sqrt{3}}=\frac{m^2}{\sqrt{3}}$. By the argumentation above, the area of the triangle $ABC$ must be at most $\frac{m^2}{\sqrt{3}}$.

Fig. 22 Fig. 23

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Alternative solution.

Let the median of the triangle $ABC$ drawn from point $A$ be $AD$ and the centroid of the triangle be $M$. W.l.o.g., let $AD$ be the longest median of the triangle $ABC$; denote $AD=m$. Then $A$ is the vertex of the triangle with largest distance from point $M$. Thus vertices $B$ and $C$ lie in circle $c$ with centre $M$ and radius $MA$. Let $B^{\prime }$ and $C^{\prime }$ be the second intersection points of rays $AB$ and $AC$, respectively, with circle $c$ (Fig. 24); then the area of the triangle $A{B}^{\prime }{C}^{\prime }$ is at least as large as the area of triangle $ABC$. Let $X$ and $Y$ be points on circle $c$ such that triangle $AXY$ is equilateral (Fig. 25); then the area of $AXY$ is at least as large as the area of triangle $A{B}^{\prime }{C}^{\prime }$ and also at least as large as the area of triangle $ABC$. The triangle $AXY$ can be divided into three equal triangles $MAX$, $MXY$ and $MYA$ whose total area is $\frac{3}{2}\cdot (\frac{2}{3}m{)}^2\sin 120^{\circ }$, which equals $\frac{m^2}{\sqrt{3}}$. Hence the area of the triangle $ABC$ is at most $\frac{m^2}{\sqrt{3}}$.

Fig. 24 Fig. 25

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Alternative solution.

Let $M$ be the centroid of the triangle $ABC$ and $m$ be the length of its longest median. The area of the triangle $ABC$ is $$S=\frac{1}{2}\cdot (MB\cdot MC\cdot \sin \alpha +MC\cdot MA\cdot \sin \beta +MA\cdot MB\cdot \sin \gamma ),$$ where $\alpha =\angle BMC$, $\beta =\angle CMA$ and $\gamma =\angle AMB$. As $MA\le \frac{2}{3}m$, $MB\le \frac{2}{3}m$ and $MC\le \frac{2}{3}m$, we obtain $S\le \frac{1}{2}\cdot (\frac{2}{3}m{)}^2\cdot (\sin \alpha +\sin \beta +\sin \gamma )$. As $\alpha ,\beta$ and $\gamma$ are less than $180^{\circ }$, Jensen's inequality applies and gives $$\frac{\sin \alpha +\sin \beta +\sin \gamma }{3}\le \sin \frac{\alpha +\beta +\gamma }{3}=\sin \frac{360^{\circ }}{3}=\sin 120^{\circ }=\frac{\sqrt{3}}{2}.$$ Consequently, $S\le \frac{1}{2}\cdot \frac{4}{9}{m}^2\cdot 3\cdot \frac{\sqrt{3}}{2}=\frac{m^2}{\sqrt{3}}$, directly implying the claim.