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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 $${1\over r} = {1\over h_1} + {1\over h_2} + {1\over h_3} = {1\over r_1} + {1\over r_2} + {1\over r_3}.$$ Let $\Delta$ be the area of the triangle $ABC$ and let $X$ be the centre of the inscribed circle.