Both Sue Liu, Madras College, St Andrews and Vassil Vassilev, Lawnswood High School, Leeds solved this one, well done!

Triangle $ABC$ has altitudes $h_{1}$ , $h_{2}$ and $h_{3}$ . The radius of the inscribed circle is $r$ , while the radii of the escribed circles are $r_{1}$ , $r_{2}$ and $r_{3}$ . We prove that

$\frac{1}{r} = \frac{1}{h_{1}} + \frac{1}{h_{2}} + \frac{1}{h_{3}} = \frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} .$

Let $Δ$ be the area, of the triangle $ABC$ and let $X$ be the centre of the inscribed circle.

Clearly,

$Δ = \frac{1}{2} r . AB + \frac{1}{2} r . BC + \frac{1}{2} r . CA,$

so that

$\frac{1}{r} = \frac{AB + BC + CA}{2 Δ} .$

Also,

$Δ = \frac{1}{2} h_{1} . BC = \frac{1}{2} h_{2}, CA = \frac{1}{2} h_{3} . AB,$

thus

$\frac{1}{h_{1}} + \frac{1}{h_{2}} + \frac{1}{h_{3}} = \frac{BC + CA + AB}{2 Δ} = \frac{1}{r} .$

Now let $A'$ , $B'$ and $C'$ be the centres of the escribed circles; see the diagram below. Also, for any triangle with vertices $U$ , $V$ and $W,$ let $Δ (U, V, W)$ denote its area.

Considering the area of the kite $ABA' C$ by splitting it into two triangles in two different ways we get

$Δ (A, B, C) + Δ (B, A', C) = area (ABA' C) = Δ (A, B, A') + Δ (A, C, A') .$

This gives (with $Δ = Δ (A, B, C))$

$Δ + \frac{1}{2} r_{1} . BC = \frac{1}{2} r_{1} . AB + \frac{1}{2} r_{1} . CA,$

and hence

$\frac{2 Δ}{r_{1}} = AB + CA - BC .$

Similarly,

$\frac{2 Δ}{r_{2}} = AB + BC - CA, \frac{2 Δ}{r_{3}} = CA + BC - AB,$

and adding these we get

$\begin{matrix} 2 Δ (\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{C}}) & = (AB + CA - BC) + (AB + BC - CA) + (CA + BC - AB) \\ = (AB + BC + CA) \\ = \frac{2 Δ}{r} \end{matrix}$

as required.