Estonian Mathematical Olympiad

a) Let triangle $ABC$ be equilateral (Fig. 11). As the medians of this triangle all have equal lengths, the points $D$, $E$, and $F$ are located at equal distances from $M$, i.e., on a circle with centre $M$. It suffices to show that the area of this circle is less than the area of triangle $ABC$. Because the medians of an equilateral triangle are simultaneously altitudes of the triangle and perpendicular to the sides of the triangle, the sides $BC$, $CA$, and $AB$ are tangent to the circle at points $D$, $E$, and $F$, respectively. Therefore the circle is the incircle of the triangle $ABC$. The area of the incircle of the triangle is indeed less than the area of the triangle.

Fig. 11

b) Let $AB=AC$ and $BC=1$ and $AD=6$ (Fig. 12). Because the median drawn from the vertex angle of an isosceles triangle is simultaneously its altitude, the area of this triangle is $\frac{1}{2}\cdot 1\cdot 6=3$. On the other hand, $BE=CF>\frac{1}{2}AD$, because $EF$ is a midsegment of the triangle parallel to $BC$ and $\frac{1}{2}AD$ is the distance between midsegment $EF$ and side $BC$ (shown on Fig. 12 with a dotted line). Therefore $ME=MF>\frac{1}{6}AD=1$. The circles with radii $ME$ and $MF$ have area greater than $\pi$, which in turn is greater than the area of triangle $ABC$, which is 3. The circle with radius $MD$ has area greater than the area of triangle $ABC$ as well, because $MD=\frac{1}{3}AD=2>1$.

Fig. 12

c) If circles with radii $MD$, $ME$, and $MF$ would all have areas equal to triangle $ABC$, their radii should also be equal, therefore $MD=ME=MF$. This would mean that the lengths of the parts of medians that lie on the other side of $M$ are equal as well, in other words, $MA=MB=MC$. We show that then $ABC$ must be equilateral. Indeed, from $ME=MF$ and $MB=MC$ we get that triangles $BMF$ and $CME$ must be equal (Fig. 13), therefore $BF=CE$ and also $AB=AC$. Analogously $AB=BC$. But in part a) we showed that in an equilateral triangle circles with radii $MD$, $ME$, and $MF$ do not have the same area as the triangle.

Fig. 13