Mathematics - Properties of Triangle Question with Solution | TestHub

If $\mathrm{a},\mathrm{b},\mathrm{c}$ are the lengths of the sides of the $\Delta$ and $\mathrm{x},\mathrm{y},\mathrm{z}$ are the length of perpendicular from the point on the sides $BC,CA,AB$ respectively we have to minimize $x^2+y^2+z^2=t$ we have $\frac{1}{2}ax+\frac{1}{2}by+\frac{1}{2}cz=\Delta$ $\Rightarrow \mathrm{ax}+\mathrm{by}+\mathrm{cz}=2\Delta$ Where $\Delta$ is the area of the $\Delta \mathrm{ABC}$ we have the identity; $\Rightarrow \left({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2\right)\left({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\right)-(\mathrm{ax}+\mathrm{bx}+\mathrm{cz}{)}^2=(\mathrm{ax}-\mathrm{by}{)}^2+(\mathrm{by}-\mathrm{cz}{)}^2+(\mathrm{cz}+\mathrm{ax}{)}^2$ $\Rightarrow \left({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2\right)\left({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\right)\ge (\mathrm{ax}+\mathrm{by}+\mathrm{cz}{)}^2$ $\Rightarrow \left({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2\right)\left({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\right)\ge 4{\Delta }^2$ $\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2\ge \frac{4{\Delta }^2}{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}$ and equality only when $\frac{\mathrm{x}}{\mathrm{a}}-\frac{\mathrm{y}}{\mathrm{b}}-\frac{\mathrm{z}}{\mathrm{c}}=\frac{\mathrm{ax}+\mathrm{by}+\mathrm{cz}}{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}=\frac{2\Delta }{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}$ The minimum value of $t$ is $\frac{4{\Delta }^2}{a^2+b^2+c^2}$

$$\mathrm{t}\min =\frac{4(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})\mathrm{s}}{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}\text{ Ans. }$$

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