Question

P is a point inside the triangle ABC from which the length of perpendiculars drawn of the sides of lengths a,b,c are respectively p,q and r. Determine the position of p if $\frac{a}{p}+\frac{b}{q}+\frac{c}{r}$ is minimum.

• A. circumcentre of the triangle
• B. incentre of the triangle
• C. centroid of the triangle
• D. orthocentre of the triangle

Answer 1

The correct option is B incentre of the triangle

$\Delta =\Delta PBC+\Delta PCA+\Delta PAB$ $=\frac{1}{2}(ap+bq+cr)=k$ =constant as $\Delta$ is given ($\Delta$ means area of triangle)

Now $z=\frac{a}{p}+\frac{b}{q}+\frac{c}{r}$ will be minimum

if $y=k(\frac{a}{p}+\frac{b}{q}+\frac{c}{r})$

$=\frac{1}{2}(ap+bq+cr)(\frac{a}{p}+\frac{b}{q}+\frac{c}{r})$ is minimum.

$\Rightarrow y=\frac{1}{2}[a^2+b^2+c^2+ab(\frac{p}{q}+\frac{q}{p})+bc(\frac{q}{r}+\frac{r}{q})+ca(\frac{r}{p}+\frac{p}{r})]$

Now we know that

A.M.>G.M.

$\frac{p}{q}+\frac{q}{p}\ge 2{(\frac{p}{q}\times \frac{q}{p})}^{1/2}=2$

Equality when $\frac{p}{q}+\frac{q}{p}=0\Rightarrow p^2=q^2\Rightarrow p=q$

$\therefore y\ge \frac{1}{2}[a^2+b^2+c^2+2ab+2bc+2ca]$

Hence the minimum value of $y$ is $\frac{1}{2}{(a+b+c)}^2$ when $p=q=r$.

In this case point $P$ will be at the incentre of $\Delta ABC.$