Question

Consider an acute angle triangle $\Delta ABC$ with area $S$. Let the area of its pedal triangle is '$p$', satisfies by the relation $\cos B=\frac{2p}{S}$ and $\sin B=\frac{2\sqrt{3}p}{S}$. Then

• A. $\Delta ABC$ is a right angled triangle.
  
• B. $\cos A\cdot \cos B\cdot \cos C=\frac{1}{8}$
• C. $\cos (A-C)=1$
• D. ${\cos }^{2}A\cos B+{\cos }^{2}B\cos C+{\cos }^{2}C\cos A=\frac{3}{8}$

Answer 1

Answer: (B), (C), (D)

${\cos }^{2}A\cos B+{\cos }^{2}B\cos C+{\cos }^{2}C\cos A=\frac{3}{8}$


In figure, $△DEF$ is the pedal triangle of $△ABC$.
$\cos B=\frac{2p}{S},\sin B=\frac{2\sqrt{3}p}{S}...\left(1\right)$

We know that,
${\sin }^{2}B+{\cos }^{2}B=1$
$\Rightarrow {\left(\frac{2p}{S}\right)}^2+{\left(\frac{2\sqrt{3}p}{S}\right)}^2=1$
$\Rightarrow S=4p...\left(2\right)$

$S=\frac{1}{2}ab\sin C$
$=2R^2\sin A\cdot \sin B\cdot \sin C$
$(\because a=2R\sin A,b=2R\sin B)$

Circumradius of a pedal triangle is half the circumradius of the original triangle.
Also, angles of the pedal triangle will be ${180}^o-2A,{180}^o-2b,{180}^o-2C$
Now, area of the pedal triangle will be
$\therefore p=2{\left(\frac{R}{2}\right)}^2\sin 2A\cdot \sin 2B\cdot \sin 2C$
$=\frac{R^2}{2}\sin 2A\cdot \sin 2B\cdot \sin 2C$

From eqn(2)
$S=4p$
$\Rightarrow 2R^2\sin A\cdot \sin B\cdot \sin C=4\times \frac{R^2}{2}\sin 2A\cdot \sin 2B\cdot \sin 2C$
$\Rightarrow \cos A\cdot \cos B\cdot \cos C=\frac{1}{8}...\left(3\right)$

From eqn(1) and (2),
$\cos B=\frac{2p}{S}=\frac{2p}{4p}=\frac{1}{2}$
$\Rightarrow \angle B=\frac{\pi }{3}...\left(4\right)$

From eqn(3),
$\cos A\cdot \cos C=\frac{1}{4}\Rightarrow 2\cos A\cdot \cos C=\frac{1}{2}\Rightarrow \cos (A+C)+\cos (A-C)=\frac{1}{2}\Rightarrow \cos \frac{2\pi }{3}+\cos (A-C)=\frac{1}{2}\Rightarrow \cos (A-C)=1\Rightarrow A=C$

$\therefore \Delta ABC$ is an equilateral triangle.

${\cos }^{2}A\cos B+{\cos }^{2}B\cos C+{\cos }^{2}C\cos A=\frac{1}{4}\times \frac{1}{2}+\frac{1}{4}\times \frac{1}{2}+\frac{1}{4}\times \frac{1}{2}$
$=\frac{3}{8}$