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Prove that the sum of distances of the orthocenter from the sides of an acute triangle is less than or equal to $$3r$$, where $$r$$ is the inradius of the same triangle.

Note by Aaron Jerry Ninan
1 year, 5 months ago

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If $$D, E, F$$ are the feet of the altitudes from $$A, B, C$$, respectively, to the opposite sides, and $$H$$ the orthocentre of $$\triangle ABC$$, $2\Delta = a\cdot HD + b\cdot HE + c\cdot HF = r(a + b + c)$

$$\Delta$$ is the area of $$\triangle ABC$$, and $$r$$ is its inradius.

$r = \frac{a\cdot HD + b\cdot HE + c\cdot HF}{a + b + c}$

Note that $$(a, b, c)$$ and $$(HD, HE, HF)$$ are similarily sorted. (As $$HD = 2R \cos B \cos C$$ and so on.)
Then, we can apply Chebyshev's inequality of the means, (basically the Rearrangement Inequality added many times)

$\frac{a\cdot HD + b\cdot HE + c\cdot HF}{3} \geq \frac{a + b + c}{3} \cdot \frac{HD + HE + HF}{3}$ This gives, $r = \frac{a\cdot HD + b\cdot HE + c\cdot HF}{a + b + c} \geq \frac{HD + HE + HF}{3}$ as required.

Equality iff $$a = b = c \iff HD = HE = HF$$ as is apparent from the Rearrangement Inequality condition.

- 1 year, 5 months ago