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Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is 6\sqrt{3}r\cdot

 

 

 

 
 
 
 
 

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Let ABC be an isosceles triangle with AB = AC and a circle with cenre 0 and radius r, touching sides AB,BC,CA at D,F,E resp.
In \Delta ABC let  AD= AE= x.BD= BF= y and CF= CE= Y\left ( \because Tangent\, dream\, from\, an\, external\, point\, are\, equal \right )
Now, ar\left ( \Delta ABC \right )= ar\left ( \Delta AOB \right )+ar\left ( \Delta AOC \right )B+ar\left ( \Delta BOC \right )

= \frac{1}{2}\times 2y\left ( r+\sqrt{r^{2}+x^{2}} \right )=\frac{1}{2}\left ( 2yr+\left ( x+y \right )r+\left ( x+y \right )r \right )
\Rightarrow y\left ( r+\sqrt{r^{2}+x^{2}} \right )= \frac{1}{2}\left ( 2yr+2\left ( x+y \right )r \right )
\Rightarrow y\left ( r+\sqrt{r^{2}+x^{2}} \right )= r\left ( y+x+y \right )
\Rightarrow y\left ( \sqrt{r^{2}+x^{2}} \right )+yr= 2yr+rx
\Rightarrow \left ( \sqrt{r^{2}+x^{2}} \right )y= rx+yr
Squaring both sides, we get
y^{2}\left ( r^{2}+x^{2} \right )= r^{2}x^{2}+y^{2}r^{2}+2r^{2}xy
\Rightarrow y^{2}r^{2}+y^{2}x^{2}= r^{2}x^{2}+y^{2}r^{2}+2r^{2}xy
\Rightarrow y^{2}x= r^{2}x+2r^{2}y
\Rightarrow x= \frac{2r^{2}y}{y^{2}-r^{2}}
Now, P (perimeter of \Delta ABC)= 2x+4y
\Rightarrow p= \frac{4r^{2}y}{y^{2}-r^{2}}+4y
Differentiate above equation w.r.t y, we get
\frac{dp}{dy}= \frac{4r^{2}\left ( y^{2}-r^{2} \right )-4r^{2}y\left ( 2y \right )}{\left ( y^{2}-r^{2} \right )^{2}}+4
\Rightarrow \frac{dp}{dy}= \frac{+4r^{2}\left [ y^{2}-r^{2}-2y^{2} \right ]}{\left ( y^{2}-r^{2} \right )^{2}}+4
\Rightarrow \frac{dp}{dy}= \frac{-4r^{2}\left ( r^{2} +y^{2}\right )}{\left ( y^{2}-r^{2} \right )^2}+4
for maxima and minima of p, put
\Rightarrow \frac{dp}{dy}=0
\Rightarrow \frac{-4r^{2}\left ( r^{2} +y^{2}\right )}{\left ( y^{2}-r^{2} \right )^2}+4= 0
\Rightarrow r^{2}\left ( r^{2}+y^{2} \right )= \left ( y^{2}-r^{2} \right )^{2}
\Rightarrow r^{4}+ r^{2}y^{2}= y^{4}+r^{4}-2y^{2}r^{2}
\Rightarrow 3y^{2}r^{2}= y^{4}
\Rightarrow y^{2}= 3r^{2}\Rightarrow y= \sqrt{3}r
Now, again differentiating w.r.t to y we get \frac{d^{2}p}{dy^{2}}
= \frac{-4r^{2}\left ( 2y \right )\left ( y^{2} -r^{2}\right )+4r^{2}\left ( r^{2}+y^{2} \right )2\left ( y^{2} -r^{2} \right )2y}{\left ( y^{2} -r^{2} \right )^4}

\Rightarrow \frac{d^{2}p}{dy^{2}}= \frac{4r^{2}y\left [ 2y^{2}+6r^{2} \right ]}{\left ( y^{2} -r^{2} \right )^{3}}
\frac{d^{2}p}{dy^{2}}|_{y= \sqrt{3r}}=\frac{6\sqrt{3}}{r}= > 0
Hence, perimeter p of \Delta ABC is least for y= \sqrt{3}r and least perimeter is  p= 4y+\frac{4r^{2}y}{ y^{2} -r^{2}}\Rightarrow 4\sqrt{3}r+\frac{4r^{2}\sqrt{3}r}{2r^{2}}
= 6\sqrt{3}r Hence proved

Posted by

Ravindra Pindel

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