Multiple choice

If $O$ is the circumcentre of the $\Delta ABC$ and $R_1,R_2$ and $R_3$ are the radii of the circumcircles $OBC,\,\,OCA$ and $OAB$ repectively then $\dfrac{a}{{{R_1}}} + \dfrac{b}{{{R_2}}} + \dfrac{c}{{{R_3}}}$ has the value equal to:

  1. $\dfrac{{abc}}{{2{R^3}}}$
  2. $\dfrac{{{abc}}}{{R^3}}$
  3. $\dfrac{{4\Delta }}{{{R^2}}}$
  4. $\dfrac{\Delta }{{4{R^2}}}$
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B Correct answer
Explanation

In any triangle, the ratio a/R1 = 2*sin(A), b/R2 = 2*sin(B), c/R3 = 2*sin(C). Using the properties of circumcenters and circumradii, the sum simplifies to abc / R^3.

AI explanation

Using the extended sine rule, the side lengths of triangle ABC are $a = 2R \sin A$, $b = 2R \sin B$, and $c = 2R \sin C$. For triangle OBC, the angle at the center $O$ is $2A$, so its circumradius $R_1$ is found by $BC / (2 \sin(2A)) = a / (4 \sin A \cos A)$. Substituting $a = 2R \sin A$, we get $R_1 = R / (2 \cos A)$, so $a / R_1 = 4 \sin A \cos A = 2 \sin(2A)$. Applying this cyclically and summing gives $2(\sin(2A) + \sin(2B) + \sin(2C))$, which simplifies to $4 \sin A \sin B \sin C$. Substituting $\sin A = a/(2R)$, this becomes $a b c / R^3$.

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