Balti Tee 2011 · Valikvooru ülesanne

Call the triangle $ABC$ and let $AP$ be the angle bisector, with $P$ on $BC$, $BP = 3$ and $CP = 1$. By the Angle Bisector Theorem, we get $\frac{AB}{AC} = 3$. Fixing the points $B$ and $C$, the locus of all points $A$ satisfying this is an Apollonius circle, whose centre lies on the line $BC$. This circle passes through $P$ itself and a point on $BC$, that lies $2$ units beyond $C$. Consequently that circle has radius $\frac{3}{2}$.

It is clear that the maximal height of the triangle, as measured from the base line $BC$ of length $4$, is $\frac{3}{2}$, but that there is no minimal height. The area of the triangle may therefore take any positive value that is at most $3$.