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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?(a) $ab+bc+ca$(b) $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}$(c) ${{\left( a+b+c \right)}^{2}}$(d) ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}$

Last updated date: 02nd Aug 2024
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Answer
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Hint: We start solving the problem by drawing the figure representing the given figure. We recall the definition of ex-radii of the triangle as ${{r}_{1}}=\dfrac{\Delta }{s-a}$, ${{r}_{2}}=\dfrac{\Delta }{s-b}$ and ${{r}_{3}}=\dfrac{\Delta }{s-c}$. We use the fact that the diameter is equal to twice the radius of the circle and substitute in ${{d}_{1}}{{d}_{2}}+{{d}_{2}}{{d}_{3}}+{{d}_{3}}{{d}_{1}}$. We then use $s=\dfrac{a+b+c}{2}$, $\Delta =\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$ and make necessary calculations to get the required answer.

Complete step by step answer:
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.

We know that the ex-radii of the triangle is defined as ${{r}_{1}}=\dfrac{\Delta }{s-a}$, ${{r}_{2}}=\dfrac{\Delta }{s-b}$ and ${{r}_{3}}=\dfrac{\Delta }{s-c}$.
Where a, b, c are the sides of the triangle,
s = semi perimeter of the triangle = $\dfrac{a+b+c}{2}$ ---(1).
$\Delta$ = Area of the triangle = $\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$ ---(2).
We know that diameter of the circle is twice the radius of the circle.
So, we get ${{d}_{1}}=2{{r}_{1}}=\dfrac{2\Delta }{s-a}$, ${{d}_{2}}=2{{r}_{2}}=\dfrac{2\Delta }{s-b}$ and ${{d}_{3}}=2{{r}_{3}}=\dfrac{2\Delta }{s-c}$. Let us substitute these values in ${{d}_{1}}{{d}_{2}}+{{d}_{2}}{{d}_{3}}+{{d}_{3}}{{d}_{1}}$.
$\Rightarrow {{d}_{1}}{{d}_{2}}+{{d}_{2}}{{d}_{3}}+{{d}_{3}}{{d}_{1}}=\left( \left( \dfrac{2\Delta }{s-a} \right)\left( \dfrac{2\Delta }{s-b} \right) \right)+\left( \left( \dfrac{2\Delta }{s-b} \right)\left( \dfrac{2\Delta }{s-c} \right) \right)+\left( \left( \dfrac{2\Delta }{s-c} \right)\left( \dfrac{2\Delta }{s-a} \right) \right)$.
$\Rightarrow {{d}_{1}}{{d}_{2}}+{{d}_{2}}{{d}_{3}}+{{d}_{3}}{{d}_{1}}=\left( \dfrac{4{{\Delta }^{2}}}{\left( s-a \right)\left( s-b \right)} \right)+\left( \dfrac{4{{\Delta }^{2}}}{\left( s-b \right)\left( s-c \right)} \right)+\left( \dfrac{4{{\Delta }^{2}}}{\left( s-c \right)\left( s-a \right)} \right)$ ---(3).
Let us substitute equation (2) in equation (3).
$\Rightarrow {{d}_{1}}{{d}_{2}}+{{d}_{2}}{{d}_{3}}+{{d}_{3}}{{d}_{1}}=\left( \dfrac{4{{\left( \sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)} \right)}^{2}}}{\left( s-a \right)\left( s-b \right)} \right)+\left( \dfrac{4{{\left( \sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)} \right)}^{2}}}{\left( s-b \right)\left( s-c \right)} \right)+\left( \dfrac{4{{\left( \sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)} \right)}^{2}}}{\left( s-c \right)\left( s-a \right)} \right)$.
$\Rightarrow {{d}_{1}}{{d}_{2}}+{{d}_{2}}{{d}_{3}}+{{d}_{3}}{{d}_{1}}=\left( \dfrac{4s\left( s-a \right)\left( s-b \right)\left( s-c \right)}{\left( s-a \right)\left( s-b \right)} \right)+\left( \dfrac{4s\left( s-a \right)\left( s-b \right)\left( s-c \right)}{\left( s-b \right)\left( s-c \right)} \right)+\left( \dfrac{4s\left( s-a \right)\left( s-b \right)\left( s-c \right)}{\left( s-c \right)\left( s-a \right)} \right)$.
$\Rightarrow {{d}_{1}}{{d}_{2}}+{{d}_{2}}{{d}_{3}}+{{d}_{3}}{{d}_{1}}=4s\left( s-c \right)+4s\left( s-a \right)+4s\left( s-b \right)$.
$\Rightarrow {{d}_{1}}{{d}_{2}}+{{d}_{2}}{{d}_{3}}+{{d}_{3}}{{d}_{1}}=4s\left( s-c+s-b+s-a \right)$.
$\Rightarrow {{d}_{1}}{{d}_{2}}+{{d}_{2}}{{d}_{3}}+{{d}_{3}}{{d}_{1}}=4s\left( 3s-a-b-c \right)$.
$\Rightarrow {{d}_{1}}{{d}_{2}}+{{d}_{2}}{{d}_{3}}+{{d}_{3}}{{d}_{1}}=4s\left( 3s-\left( a+b+c \right) \right)$.
$\Rightarrow {{d}_{1}}{{d}_{2}}+{{d}_{2}}{{d}_{3}}+{{d}_{3}}{{d}_{1}}=4s\left( 3s-2\left( \dfrac{a+b+c}{2} \right) \right)$ ---(4).
Let us substitute equation (1) in equation (4).
$\Rightarrow {{d}_{1}}{{d}_{2}}+{{d}_{2}}{{d}_{3}}+{{d}_{3}}{{d}_{1}}=4s\left( 3s-2s \right)$.
$\Rightarrow {{d}_{1}}{{d}_{2}}+{{d}_{2}}{{d}_{3}}+{{d}_{3}}{{d}_{1}}=4s\left( s \right)$.
$\Rightarrow {{d}_{1}}{{d}_{2}}+{{d}_{2}}{{d}_{3}}+{{d}_{3}}{{d}_{1}}=4{{s}^{2}}$.
$\Rightarrow {{d}_{1}}{{d}_{2}}+{{d}_{2}}{{d}_{3}}+{{d}_{3}}{{d}_{1}}=4{{\left( \dfrac{a+b+c}{2} \right)}^{2}}$.
$\Rightarrow {{d}_{1}}{{d}_{2}}+{{d}_{2}}{{d}_{3}}+{{d}_{3}}{{d}_{1}}=4\times \dfrac{{{\left( a+b+c \right)}^{2}}}{4}$.
$\Rightarrow {{d}_{1}}{{d}_{2}}+{{d}_{2}}{{d}_{3}}+{{d}_{3}}{{d}_{1}}={{\left( a+b+c \right)}^{2}}$.
So, we have found the value of ${{d}_{1}}{{d}_{2}}+{{d}_{2}}{{d}_{3}}+{{d}_{3}}{{d}_{1}}$ as ${{\left( a+b+c \right)}^{2}}$.

So, the correct answer is “Option c”.

Note: We can see that the given problem has a huge amount of calculation, so we need to perform each step carefully. We should not confuse s with the perimeter of the triangle instead of the semi perimeter of the triangle. Whenever we get this type of problem, we should draw the figure first to get the better view of the given information. Similarly, we can expect problems to find the distance between the orthocentre and circumcentre of the triangle.