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$ℓ,$ m, n are the lengths of the common chords of the circles, on the sides BC, CA, AB of a triangle ABC, as diameters,
taken in pairs. Area measure of ABC, is $∆$ , circumradius is R and inradius, r. Then, $\frac{1}{ℓ} + \frac{1}{m} + \frac{1}{n}$ is equal to

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$\frac{1}{r}$

$\frac{1}{R}$

$\frac{2}{R}$

$\frac{Δ}{R}$

answer is A.

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Detailed Solution

The lengths of the common chords of the 3 circles are the lengths of the three altitudes of the triangle.

$∴ \sum \frac{1}{ℓ} = \sum \frac{a}{2 Δ} = \frac{s}{Δ} = \frac{1}{r}$

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