question_answer

Let \[{{d}_{1}},\,\,{{d}_{2}}\] and \[{{d}_{3}}\] be the lengths of perpendiculars from circumventer of \[\Delta ABC\] on the sides BC, AC and AB, respectively, if \[\lambda \left( \frac{a}{{{d}_{1}}}+\frac{b}{{{d}_{2}}}+\frac{c}{{{d}_{3}}} \right)=\frac{abc}{{{d}_{1}}{{d}_{2}}{{d}_{3}}}\] then \[\lambda \] equals

A) 1

B) 2

C) 3

D) 4  

Correct Answer: D

Solution :

[d] We have \[\tan A=\frac{a}{2{{d}_{1}}};\]

\[{{d}_{1}}=R\cos A\] etc.

Similarly \[\tan B=\frac{b}{2{{d}_{2}}}\]

and \[\tan C=\frac{C}{2{{d}_{3}}}\]

In \[\Delta ABC,\tan A+\tan B+\tan C\]

\[=\,\,\,\,\tan A\cdot \tan B\cdot \tan C\]

\[\Rightarrow \frac{a}{2{{d}_{1}}}+\frac{b}{2{{d}_{2}}}+\frac{c}{2{{d}_{3}}}=\frac{abc}{8{{d}_{1}}{{d}_{2}}{{d}_{3}}}\]

\[\therefore \,\,\,\,\,\,\,4\left( \frac{a}{{{d}_{1}}}+\frac{b}{{{d}_{2}}}+\frac{c}{{{d}_{3}}} \right)=\frac{abc}{{{d}_{1}}{{d}_{2}}{{d}_{3}}}\]

\[\Rightarrow \lambda =4\]