  Question

In $$\displaystyle \triangle ABC,$$ the incircle touches the sides $$BC, CA$$ and $$AB$$, respectively, at $$D, E$$ and $$F$$. If the radius of the incircle is $$4$$ units and $$BD, CE$$ and $$AF$$ are consecutive integers, then the value of $$\dfrac{s}{3}$$, where $$s$$ is a semi-perimeter of the triangle, is?

Solution

Given in radius $$r=4$$$$BD,CE,AF$$ are in A.P of common difference $$1$$$$s=\dfrac { \triangle }{ r } =\dfrac { \sqrt { s(s-a)(s-b)(s-c) } }{ r } \\$$$$=2y=x+z\\$$$$z-y=y-x=1\\$$$$a=x+z,b=y+z,c=x+y\\$$$$S=\left( \dfrac { a+b+c }{ 2 } \right) =(3y)\\$$$$\triangle =\sqrt { (3y)\left[ 3y-(x+z) \right] \left[ 3y-(y+z) \right] \left[ 3y-(x+4) \right] } \\$$$$=\sqrt { (3y)(y)+(2y-z)(2y-x) } \\$$$$=\sqrt { (3y)y(y-1)(y+1) } \\$$$$=\dfrac { \sqrt { 3{ y }^{ 2 }({ y }^{ 2 }-1) } }{ r } \\$$$$s=\dfrac { \triangle }{ r } ={ s }^{ 2 }=\dfrac { { \triangle }^{ 2 } }{ { r }^{ 2 } } \\$$$$9{ y }^{ 2 }=\dfrac { 3{ y }^{ 2 }({ y }^{ 2 }-1) }{ 16 } \Rightarrow 3\times 16=({ y }^{ 2 }-1)\\$$$${ y }^{ 2 }=49\Rightarrow y=7\\$$$$\dfrac { s }{ 3 } =\dfrac { 3y }{ 3 } =7$$ Maths

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