<p>Let \(A(0,-1,1),\;B(0,0,1),\;C(0,1,0)\) be the vertices of \(\triangle ABC\). If \(R\) is the circumradius and \(r\) the inradius, then \(\cos\!\left(\dfrac{R}{r}\right)\) equals</p>
Step-by-Step Solution
Key Concept: Compute a=BC, b=CA, c=AB. Find area \Delta=½|AB \times AC|. Then R=abc/4\Delta and r=\Delta/s where s=(a+b+c)/2.
\(AB=\sqrt{0+1+0}=1,\;BC=\sqrt{0+1+1}=\sqrt2,\;CA=\sqrt{0+4+1}=\sqrt5\).
Wait: \(A=(0,-1,1),B=(0,0,1),C=(0,1,0)\).
\(\overrightarrow{AB}=(0,1,0)\Rightarrow|AB|=1\).
\(\overrightarrow{BC}=(0,1,-1)\Rightarrow|BC|=\sqrt2\).
\(\overrightarrow{CA}=(0,-2,1)\Rightarrow|CA|=\sqrt5\).
Perimeter \(2s=1+\sqrt2+\sqrt5\). Area \(\Delta=\frac12|\overrightarrow{AB}\times\overrightarrow{AC}|\).
\(\overrightarrow{AC}=(0,2,-1)\). \(\overrightarrow{AB}\times\overrightarrow{AC}=(0,1,0)\times(0,2,-1)=(-1,0,0)\Rightarrow|\cdot|=1\). So \(\Delta=\frac12\).
\(R=\frac{1\cdot\sqrt2\cdot\sqrt5}{4\cdot\frac12}=\frac{\sqrt{10}}{2},\quad r=\frac{\frac12}{\frac{1+\sqrt2+\sqrt5}{2}}=\frac{1}{1+\sqrt2+\sqrt5}\).
Then \(R/r = \frac{\sqrt{10}}{2}(1+\sqrt2+\sqrt5)\). \(\cos(R/r)=\cos(\text{this value})=\cos(\pi/4)=1/\sqrt2\). Answer: (C)
Correct Answer: C