Question

If in a triangle $ABC,a,b,c$ are in A.P. and $p_1,p_2,p_3$ are the altitudes from the vertices $A,B,C$ respectively then

• A. $p_1,p_2,p_3$ are in A.P.
• B. $p_1,p_2,p_3$ are in H.P.
• C. $p_1+p_2+p_3\le \frac{3R}{\Delta }$
• D. $\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}\le \frac{3R}{\Delta }$

Answer 1

Answer: (B), (D)

The correct options are
B $p_1,p_2,p_3$ are in H.P.
D $\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}\le \frac{3R}{\Delta }$

$\Delta =\frac{1}{2}a{p}_1=\frac{1}{2}b{p}_2=\frac{1}{2}c{p}_3\Rightarrow p_1=\frac{2\Delta }{a},p_2=\frac{2\Delta }{b},p_3=\frac{2\Delta }{c}$
$\Rightarrow p_1,p_2,p_3$ are in H.P
$\text{Now}{p}_1=\frac{2\Delta }{2R\sin A},p_2=\frac{2\Delta }{2R\sin B},p_3=\frac{2\Delta }{2R\sin C}\Rightarrow \frac{1}{p_1}=\frac{RsinA}{\Delta },\frac{1}{p_2}=\frac{RsinB}{\Delta },\frac{1}{p_3}=\frac{RsinC}{\Delta }$
Since $\sin A,\sin B,\sin C$ are in A.P.
$\sin A+\sin C=2\sin B\therefore \frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}=\frac{R}{\Delta }(\sin A+\sin B+\sin C)=\frac{3R}{\Delta }\sin B\le \frac{3R}{\Delta }$