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Let $D$ be the footpoint of the altitude from $B$ in the triangle $A B C$, where $A B=1$. The incentre of triangle $B C D$ coincides with the centroid of triangle $A B C$. Find the lengths of $A C$ and $B C$.
Answer: $A C=B C=\sqrt{\frac{5}{2}}$.

The centroid of $A B C$ lies on the median $C C^{\prime}$. It will also, by the assumption, lie on the angle bisector through $C$. Since the median and the angle bisector coincide, $A B C$ is isosceles with $A C=$ $B C=a$.

Furthermore, the centroid lies on the median $B B^{\prime}$ and the bisector of $\angle D B C$, again by hypothesis. By the Angle Bisector Theorem,

$$
\frac{B^{\prime} D}{B D}=\frac{B^{\prime} C}{B C}=\frac{a / 2}{a}=\frac{1}{2}
$$

The triangles $A B D \sim A C C^{\prime}$ since they have equal angles, whence

$$
\frac{1}{a}=\frac{A B}{A C}=\frac{A D}{A C^{\prime}}=\frac{a / 2-B^{\prime} D}{1 / 2}=a-B D
$$

Using the fact that the length of the altitude $C C^{\prime}$ is $\sqrt{a^{2}-\frac{1}{4}}$, this leads to

$$
a^{2}-1=a B D=2|A B C|=\sqrt{a^{2}-\frac{1}{4}},
$$

or, equivalently,

$$
a^{2}-1=\sqrt{a^{2}-\frac{1}{4}}
$$

Clearly, $a>1$, and the only solution is $a=\sqrt{\frac{5}{2}}$.

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