Multiple choice

In $\triangle ABC$, If $x,y,z$ are perpendicular distance from the circumcentre to the sides of the $\displaystyle \triangle ABC$ respectively. then $\displaystyle \frac{a}{x}+\frac{b}{y}+\frac{c}{z}$

  1. $\displaystyle \frac{abc}{4xyz}$
  2. $\displaystyle \frac{4abc}{xyz}$
  3. $\displaystyle \frac{xyz}{4abc}$

  4. none of these
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A Correct answer
AI explanation

In a triangle with circumradius $R$, the perpendicular distance from the circumcentre to side $a$ is given by $x = R \cos A$, meaning $\frac{a}{x} = \frac{2R \sin A}{R \cos A} = 2 \tan A$. Similarly, we have $\frac{b}{y} = 2 \tan B$ and $\frac{c}{z} = 2 \tan C$. Using the standard trigonometric identity $\tan A + \tan B + \tan C = \tan A \tan B \tan C$, the sum simplifies to $2(\tan A + \tan B + \tan C) = 2(\frac{a}{2R} \cdot \frac{b}{2R} \cdot \frac{c}{2R}) / (\frac{x}{R} \cdot \frac{y}{R} \cdot \frac{z}{R})$, which reduces to $\frac{abc}{4xyz}$.

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