Question

If in a triangle R and r are the circumradius and inradius respectively, then the harmonic mean of the exradii of the triangle is
(a)3r
(b)2R
(c)R+r
(d)None of these

Answer 1

Hint: Use the formula that the exradii of the circle are given by $\dfrac{\Delta }{s-a},\text{ }\dfrac{\Delta }{s-b}\text{ and }\dfrac{\Delta }{s-c}$ . Also, the harmonic mean of the tree number a, b and c is $\dfrac{3}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}$ . You should also know that the inradius of a triangle is equal to $\dfrac{\Delta }{s}$ .

Complete step-by-step answer:
Before starting with the solution, let us draw a diagram for better visualisation.

Now, starting with the solution of the question given in the figure. We know that a circle is touching the side BC of the triangle ABC and touching AB and Ac when produced is called the excircle of the triangle. There are three excircles to a triangle and there radii is given by $\dfrac{\Delta }{s-a},\text{ }\dfrac{\Delta }{s-b}\text{ and }\dfrac{\Delta }{s-c}$ . So, the harmonic mean of these three exradii is is:
$\dfrac{3}{\dfrac{1}{\dfrac{\Delta }{s-a}}+\dfrac{1}{\dfrac{\Delta }{s-b}}+\dfrac{1}{\dfrac{\Delta }{s-c}}}=\dfrac{3\Delta }{s-a+s-b+s-c}$
Now, we know that s is called the semi-perimeter of the triangle and is equal to $\dfrac{a+b+c}{2}$ . So, we can say that a+b+c is equal to 2s.
$\dfrac{3\Delta }{3s-a-b-c}=\dfrac{3\Delta }{3s-2s}=\dfrac{3\Delta }{s}$
And we know that the inradius of a triangle is equal to $\dfrac{\Delta }{s}$ , making the above expression to be equal to 3 times the inradius.
$\therefore \dfrac{3\Delta }{s}=3r$
Therefore, the answer to the above question is option (a).

Note: Remember that the centre of the excircle is the point of intersection of the interior angle bisector of the opposite angle and the exterior angle bisectors of the other two angles of the triangle. While the centre of the largest circle that fits inside a triangle is called the incentre and is defined as the meeting point of all three angle bisectors of the triangle.