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If $d_{1}$ , $d_{2}, d_{3}$ are the diameters of the three inscribed circles of a triangle ABC, then $d_{1} d_{2}$ + $d_{2} d_{3}$ + $d_{3} d_{1}$ is equal to _________.

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ab+bc+ca

\frac{a}{b} + \frac{b}{c} + \frac{c}{a}

{(a + b + c)}^{2}

a^{2} + b^{2} + c^{2}

answer is C.

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Detailed Solution

According to the problem, we are given that

d_{1}

d_{2}, d_{3}

are the diameters of the three inscribed circles of a triangle ABC. We need to find the value of

d_{1} d_{2}

d_{2} d_{3}

d_{3} d_{1}

.
Let us draw a figure representing the given information.
Question Image

We know that the ex-radii of the triangle is defined as r1=

\frac{∆}{s - a}

, r2=

\frac{∆}{s - b}

and r3=

\frac{∆}{s - c}

.
Where a, b, c are the sides of the triangle,
s = semi perimeter of the triangle =

\frac{a + b + c}{2}

---(1).
Δ = Area of the triangle =

\sqrt{s (s - a) (s - b) (s - c)}

---(2).
We know that the diameter of the circle is twice the radius of the circle.
So, we get

d_{1}

r_{1}

\frac{2 ∆}{s - a}

d_{2}

r_{2}

\frac{2 ∆}{s - b}

and

d_{3}

r_{3}

\frac{2 ∆}{s - c}

. Let us substitute these values in

d_{1} d_{2}

d_{2} d_{3}

d_{3} d_{1}

.
⇒

d_{1} d_{2}

d_{2} d_{3}

d_{3} d_{1}

=((

\frac{2 ∆}{s - a}) (\frac{2 ∆}{s - b}))

+((

\frac{2 ∆}{s - b}

)(

\frac{2 ∆}{s - c}

))+((

\frac{2 ∆}{s - c}

)(

\frac{2 ∆}{s - a}

))
⇒

d_{1} d_{2}

d_{2} d_{3}

d_{3} d_{1}

\frac{4 ∆^{2}}{(s - a) (s - b)}) + (\frac{4 ∆^{2}}{(s - b) (s - c)}) + (\frac{4 ∆^{2}}{(s - c) (s - a)})

---(3).
Let us substitute equation (2) in equation (3).
⇒

d_{1} d_{2}

d_{2} d_{3}

d_{3} d_{1}

(\frac{4 {\sqrt{s (s - a) (s - b) (s - c)}}^{2}}{(s - a) (s - b)})

(\frac{4 {\sqrt{s (s - a) (s - b) (s - c)}}^{2}}{(s - b) (s - c)}) + (\frac{4 {\sqrt{s (s - a) (s - b) (s - c)}}^{2}}{(s - c) (s - a)})

⇒

d_{1} d_{2}

d_{2} d_{3}

d_{3} d_{1}

\frac{4 s (s - a) (s - b) (s - c)}{(s - a) (s - b)}) + (\frac{4 s (s - a) (s - b) (s - c)}{(s - b) (s - c)}) + (\frac{4 s (s - a) (s - b) (s - c)}{(s - c) (s - a)})

⇒

d_{1} d_{2}

d_{2} d_{3}

d_{3} d_{1}

=4s(s−c)+4s(s−a)+4s(s−b)
⇒

d_{1} d_{2}

d_{2} d_{3}

d_{3} d_{1}

=4s(s−c+s−b+s−a)
⇒

d_{1} d_{2}

d_{2} d_{3}

d_{3} d_{1}

=4s(3s−a−b−c).
⇒

d_{1} d_{2}

d_{2} d_{3}

d_{3} d_{1}

=4s(3s−(a+b+c)).
⇒

d_{1} d_{2}

d_{2} d_{3}

d_{3} d_{1}

=4s(3s−2(

\frac{a + b + c}{2}

)) ---(4).
Let us substitute equation (1) in equation (4).
⇒

d_{1} d_{2}

d_{2} d_{3}

d_{3} d_{1}

=4s(3s−2s).
⇒

d_{1} d_{2}

d_{2} d_{3}

d_{3} d_{1}

=4s(s).
⇒

d_{1} d_{2}

d_{2} d_{3}

d_{3} d_{1}

s^{2}

⇒

d_{1} d_{2}

d_{2} d_{3}

d_{3} d_{1}

{(\frac{a + b + c}{2})}^{2}

⇒

d_{1} d_{2}

d_{2} d_{3}

d_{3} d_{1}

=4×

\frac{{(a + b + c)}^{2}}{4}

⇒

d_{1} d_{2}

d_{2} d_{3}

d_{3} d_{1}

{(a + b + c)}^{2}

So, we have found the value of

d_{1} d_{2}

d_{2} d_{3}

d_{3} d_{1}

{(a + b + c)}^{2}

So, the correct answer is “Option 3”.

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