Question

If r is the inradius and R is the circumradius, then $2R\le r$

• A. True
• B. False

Answer 1

The correct option is B

False

$r=4Rsin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}Nowconsider,cos\left(A\right)+cos\left(B\right)+cos\left(C\right)=2cos\left(\frac{(A+B)}{2}\right)cos\left(\frac{(A-B)}{2}\right)-(2co{s}^2\left(\frac{(A+B)}{2}\right)-1)=2\left(cos\left(\frac{(A-B)}{2}\right)\right)-cos\left(\frac{(A+B)}{2}\right)cos\left(\frac{(A+B)}{2}\right)+1=4sin\left(\frac{A}{2}\right)sin\left(\frac{B}{2}\right)sin\left(\frac{C}{2}\right)+1$
It is useful to remember the identity, cos(A) + cos(B) + cos(C) $\le \frac{3}{2}$ though the proof is a little beyond the scope of this chapter
Therefore, $4sin\left(\frac{A}{2}\right)sin\left(\frac{B}{2}\right)sin\left(\frac{C}{2}\right)+1\le \frac{3}{2}Sin\frac{A}{2}Sin\frac{B}{2}Sin\frac{C}{2}\le \frac{1}{2}.\frac{1}{2}.\frac{1}{2}=\frac{1}{8}\Rightarrow r<4R\left(\frac{1}{8}\right)=\frac{R}{2}$
$\therefore$ Given statement is false