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 which of the following is/are correct?

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

Answer 1

Answer: (A), (D)

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

$\Delta =\frac{1}{2}a{p}_1=\frac{1}{2}b{p}_2=\frac{1}{2}c{p}_3$

$\Rightarrow p_1=\frac{2\Delta }{a},p_2=\frac{2\Delta }{b},p_3=\frac{2\Delta }{c}$

$\Rightarrow p_1,p_2,p_3$ are in H.P. (since a,b,c are in A.P)

Now
$p_1=\frac{2\Delta }{2RsinA},p_2=\frac{2\Delta }{2RsinB},p_3=\frac{2\Delta }{2RsinC}$
$\Rightarrow \frac{1}{p_1}=\frac{RsinA}{\Delta },\frac{1}{p_2}=\frac{RsinB}{\Delta },\frac{1}{p_3}=\frac{RsinC}{\Delta }$

$\Rightarrow$ $\sin A,\sin B,\sin C$ are in $A.P.$

$\Rightarrow sinA+sinC=2sinB$

$\therefore \frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}=\frac{R}{\Delta }(sinA+sinB+sinC)$

$=\frac{3R}{\Delta }sinB\le \frac{3R}{\Delta }$