Question

If $H$ is the orthocentre of $\Delta ABC,R$ is circumradius and $P=AH+BH+CH$, then:

• A. $P=2(R+r)$
• B. $P=2(R-r)$
• C. Maximum value of $P$ is $3R$
• D. Minimum value of $P$ is $3R$

Answer 1

Answer: (A), (C)

Maximum value of $P$ is $3R$

Given: $H$ is the orthocentre of $\Delta ABC,R$ is circumradius
We know, $AH=2R\cos A,BH=2R\cos B,CH=2R\cos C$
$\therefore P=2R(\cos A+\cos B+\cos C)$
Also, we know $(\cos A+\cos B+\cos C)=(1+\frac{r}{R})$
Hence $P=2R(1+\frac{r}{R})=2(R+r)$
We know, in a triangle $r\le \frac{R}{2}$
Hence $\frac{r}{R}\le \frac{1}{2}$
So, maximum value of $P=2R(1+\frac{1}{2})=3R$