Area i'n it

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 $\Delta$ be the area of the triangle $ABC$ and let $X$ be the centre of the inscribed circle.

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Clearly, $$\mathrm{\Delta }=\frac{1}{2}r.AB+\frac{1}{2}r.BC+\frac{1}{2}r.CA,$$ so that $$\frac{1}{r}=\frac{AB+BC+CA}{2\mathrm{\Delta }}.$$ Also, $$\mathrm{\Delta }=\frac{1}{2}{h}_1.BC=\frac{1}{2}{h}_2,CA=\frac{1}{2}{h}_3.AB,$$ thus $$\frac{1}{h_1}+\frac{1}{h_2}+\frac{1}{h_3}=\frac{BC+CA+AB}{2\mathrm{\Delta }}=\frac{1}{r}.$$ 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 $\Delta(U,V,W)$ denote its area.

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Considering the area of the kite $ABA'C$ by splitting it into two triangles in two different ways we get $$\mathrm{\Delta }(A,B,C)+\mathrm{\Delta }(B,A^{\prime },C)=\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}(AB{A}^{\prime }C)=\mathrm{\Delta }(A,B,A^{\prime })+\mathrm{\Delta }(A,C,A^{\prime }).$$ This gives (with $\Delta = \Delta(A,B,C))$ $$\mathrm{\Delta }+\frac{1}{2}{r}_1.BC=\frac{1}{2}{r}_1.AB+\frac{1}{2}{r}_1.CA,$$ and hence $$\frac{2\mathrm{\Delta }}{r_1}=AB+CA-BC.$$ Similarly, $$\frac{2\mathrm{\Delta }}{r_2}=AB+BC-CA,\frac{2\mathrm{\Delta }}{r_3}=CA+BC-AB,$$ and adding these we get $$\begin{array}{rl}2\mathrm{\Delta }(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_C})& =(AB+CA-BC)+(AB+BC-CA)+(CA+BC-AB)\\ & =(AB+BC+CA)\\ & =\frac{2\mathrm{\Delta }}{r}\end{array}$$ as required.