Answer to Question #95123 in Calculus for Uday chadar

Question

If x,y,z are length of the perpandicular dropped from a point inside the triangle of given area A,on the three sides of the triangle then the minimum value of x^2+y^2+z^2 is 4A^2/a^2+b^2+c^2

Answer 1

Let ABC and H be given triangle and point inside it.

Let $x=HA_1,y=HB_1,z=HC_1$ be lengths of the perpendicualars to the sides $a=BC,b=AC,c=AB$ respectively. We have $\frac{ax}{2},\frac{by}{2},\frac{cz}{2}$ are areas of the triangles $BCH, ACH, ABH$ respectively, so $ax+by+cz=2A$. We can express $z$: $z=\frac{2A-ax-by}{c}$.

Then $F(x,y)=x^2+y^2+z^2=x^2+y^2+\bigl(\frac{2A-ax-by}{c}\bigr)^2=$

$=\frac{1}{c^2}(x^2c^2+y^2c^2+4A^2+a^2x^2+$

$+b^2y^2-4Aax-4Aby+2abxy)$ .

Since $x\ge 0$, $y\ge 0$ and $z=\frac{2A-ax-by}{c}\ge 0$, $F$ is defined on the set $U=\{(x,y)\in\mathbb R^2: x\ge0, y\ge0,ax+by\le 2A\}$. Since $U$ is bounded ($U\subset\bigl[0,\frac{2A}{a}\bigr]\times\bigl[0,\frac{2A}{b}\bigr]$ ) and closed, $U$ is a compact set by criterion of compactness in $\mathbb R^n$ .

Moreover, $F(x,y)=x^2+y^2+z^2\ge 0$ for all $x$ and $y$ . So $F(U)$ is bounded from below.

Since $U$ is a compact set, $F$ is a continuous function bounded from below on $U$ , we have that there is the minimum of $F$ on $U$.

If we want to minimize $F(x,y)$ , then we need such $x$ and $y$ that $\frac{\partial F}{\partial x}(x,y)=0$ and $\frac{\partial F}{\partial y}(x,y)=0$.

So $0=\frac{\partial F}{\partial x}(x,y)=\frac{2}{c^2}(xc^2+xa^2-2Aa+aby)$ and $0=\frac{\partial F}{\partial y}(x,y)=\frac{2}{c^2}(yc^2+yb^2-2Ab+abx)$ .

We obtain the linear equation system

$\begin{cases} xc^2+xa^2-2Aa+aby=0\\ yc^2+yb^2-2Ab+abx=0 \end{cases}$

The system has the only one solution $x_0=\frac{2aA}{a^2+b^2+c^2}$ and $y_0=\frac{2bA}{a^2+b^2+c^2}$ (so $F(x_0,y_0)$ is minimum of $F(x,y)$) . Value of $z_0$ is $z_0=\frac{2A-ax_0-by_0}{c}=\frac{2cA}{a^2+b^2+c^2}$ . So minimum of $F$ on $U$ is $F(x_0,y_0)=x_0^2+y_0^2+z_0^2=\frac{4A^2}{a^2+b^2+c^2}$