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

Triangle ABC has altitudes h₁, h₂ and h₃. The radius of the inscribed circle is r, while the radii of the escribed circles are r₁, r₂ and r₃. We prove that

1
r
= 1
h₁
+ 1
h₂
+ 1
h₃
= 1
r₁
+ 1
r₂
+ 1
r₃
.

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

Clearly,

∆ = ¹/₂r.AB+ 1
2
r.BC+ 1
2
r.CA,

so that

1
r
= AB + BC + CA
2∆
.

Also,

∆ = ¹/₂h₁.BC = ¹/₂h₂, CA = ¹/₂h₃.AB,

thus

1
h₁
+ 1
h₂
+ 1
h₃
= BC + CA + AB
2∆
= 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 ∆(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

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

This gives (with ∆ = ∆(A,B,C))

∆+¹/₂r₁.BC = ¹/₂r₁.AB + ¹/₂r₁.CA,

and hence

2∆
r₁
= AB + CA − BC.

Similarly,

2∆
r₂
= AB + BC − CA, 2∆
r₃
= CA + BC − AB,

and adding these we get

2∆ 
 1
r₁
+ 1
r₂
+ 1
r_C



= (AB + CA − BC) + (AB + BC − CA) + (CA + BC − AB)

= (AB + BC + CA)

= 2∆
r

as required.