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If \triangle ABC is a equalateral triangle of side a then find the minimum distance between two excentres ?

Option: 1

a


Option: 2

\frac{3a}{2}


Option: 3

2a


Option: 4

\frac{5a}{2}


Answers (1)

best_answer

 

 

Excenters of Triangle -

Excenters of Triangle

An excenter is a point at which the line bisecting one interior angle meets the bisectors of the two exterior angles on the opposite side.

The circle opposite to the vertex A is called the escribed circle or the circle escribed to the side BC . If I1 is the point of intersection of the internal bisector of ∠BAC and external bisector of ∠ABC and ∠ACB then,

Coordinates of I1 , I2 and I3 is given by

\\\mathrm{I_1\equiv \left(\frac{-a x_{1}+b x_{2}+c x_{3}}{-a+b+c}, \frac{-a y_{1}+b y_{2}+c y_{3}}{-a+b+c}\right)}\\\\\mathrm{I_2\equiv\left(\frac{a x_{1}-b x_{2}+c x_{3}}{a-b+c}, \frac{a y_{1}-b y_{2}+c y_{3}}{a-b+c}\right)}\\\\\mathrm{I_3\equiv\left(\frac{a x_{1}+b x_{2}-c x_{3}}{a+b-c}, \frac{a y_{1}+b y_{2}-c y_{3}}{a+b-c}\right)}

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 \text{Let } in\ \triangle ABC \text{ A(0,0),}B(\frac{a}{2},\frac{\sqrt{3}a}{2}),C(a,0)\\ \begin{array}{l}{\mathrm{I}_{1} \equiv\left(\frac{-\mathrm{ax}_{1}+\mathrm{bx}_{2}+\mathrm{cx}_{3}}{-\mathrm{a}+\mathrm{b}+\mathrm{c}}, \frac{-\mathrm{ay}_{1}+\mathrm{by}_{2}+\mathrm{cy}_{3}}{-\mathrm{a}+\mathrm{b}+\mathrm{c}}\right)} \\ {\mathrm{I}_{2} \equiv\left(\frac{\mathrm{ax}_{1}-\mathrm{bx}_{2}+\mathrm{cx}_{3}}{\mathrm{a}-\mathrm{b}+\mathrm{c}}, \frac{\mathrm{ay}_{1}-\mathrm{by}_{2}+\mathrm{cy}_{3}}{\mathrm{a}-\mathrm{b}+\mathrm{c}}\right)} \\ {\mathrm{I}_{3} \equiv\left(\frac{\mathrm{ax}_{1}+\mathrm{bx}_{2}-\mathrm{cx}_{3}}{\mathrm{a}+\mathrm{b}-\mathrm{c}}, \frac{\mathrm{ay}_{1}+\mathrm{by}_{2}-\mathrm{cy}_{3}}{\mathrm{a}+\mathrm{b}-\mathrm{c}}\right)}\end{array}\\ I_{1}=\left(\frac{-a \cdot 0 +a \cdot a+a \cdot \frac{a}{2}}{-a+a+a}, \frac{-a \cdot 0 +a \cdot 0+a \cdot \frac{\sqrt{3}a}{2}}{-a+a+a}\right)\\ I_{1}=(\frac{3a}{2}.\frac{\sqrt{3}a}{2})\\

I_{2}=\left(\frac{a \cdot 0 -a \cdot a+a \cdot \frac{a}{2}}{a-a+a}, \frac{a \cdot 0 -a \cdot 0+a \cdot \frac{\sqrt{3}a}{2}}{a-a+a}\right)\\ I_{2}=(\frac{-a}{2}.\frac{\sqrt{3}a}{2})\\ I_{3}=\left(\frac{a \cdot 0 +a \cdot a-a \cdot \frac{a}{2}}{a+a-a}, \frac{a \cdot 0 +a \cdot 0-a \cdot \frac{\sqrt{3}a}{2}}{a+a-a}\right)\\ I_{3}=(\frac{a}{2}.\frac{-\sqrt{3}a}{2})\\

Distance between any two excentres is 2a.

Note:-

The excentres of an equilateral triangle is equidistant from each other and this distance is twise of side of triangle 

 

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avinash.dongre

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