Discussion
As shown in the figure, circle \(C_{1}\) with center \(O_{1}\) and radius \(r_{1}\) touches the square \(VWXY\) at points \(P\) and \(Q\) while circle \(C_{2}\) with center \(O_{2}\) and radius \(r_{2}\) touches the square \(VWXY\) at points \(R\) and \(S\). The two circles touch each other at \(T\).
Given \(r_{1} = 1\) cm and \(\overline{VY} = \overline{VW} = 4\) cm, \(r_{2}\) = _____ cm.
Answer: C
\(r_{2} = 7 - 4\sqrt{2}\) cm.
Each circle is tangent to the two sides at its corner, so its centre lies on the diagonal WY, at a distance \(r\sqrt{2}\) from the corner. The diagonal is \(4\sqrt{2}\), and along it lie the corner distance of \(C_{1}\), both radii, and the corner distance of \(C_{2}\):
\[ \begin{aligned} r_{1}\sqrt{2} + r_{1} + r_{2} + r_{2}\sqrt{2} &= 4\sqrt{2} \\ (r_{1} + r_{2})(\sqrt{2} + 1) &= 4\sqrt{2} \\ r_{1} + r_{2} &= 4\sqrt{2}(\sqrt{2} - 1) = 8 - 4\sqrt{2} \\ r_{2} &= 7 - 4\sqrt{2} \approx 1.34 \end{aligned} \]
\(4 - 3\sqrt{2}\) is negative and \(1 + 2\sqrt{2} \approx 3.83\) and \(5 + 3\sqrt{2}\) are too big to fit in a 4 cm square beside a 1 cm circle.
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