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.
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Place the square in a coordinate system with W at the origin. Both circle centers then lie on the diagonal WY, at perpendicular distances r1 and r2 from the two nearby sides. Set the center to center distance equal to r1 plus r2.
Step 1: Set up coordinates for the square.
Place corner \(W\) at the origin \((0,0)\), with side \(WX\) along the positive x-axis and side \(WV\) along the positive y-axis. Since \(\overline{VW} = \overline{VY} = 4\) cm, the square has side length 4, so the four corners are \(W(0,0)\), \(X(4,0)\), \(Y(4,4)\), \(V(0,4)\).
Step 2: Locate the center of the smaller circle \(C_1\).
Circle \(C_1\) touches the square at \(P\), on side \(WV\), and \(Q\), on side \(WX\), so it sits in the corner near \(W\), tangent to both the left side and the bottom side. A circle tangent to both these sides has its center at a distance \(r_1\) from each, so \(O_1 = (r_1, r_1) = (1,1)\).
Step 3: Locate the center of the larger circle \(C_2\).
Circle \(C_2\) touches the square at \(S\), on side \(VY\), the top, and \(R\), on side \(YX\), the right side, so it sits in the corner near \(Y\), tangent to the top and right sides. Its center is a distance \(r_2\) in from each of these sides, so \(O_2 = (4 - r_2,\ 4 - r_2)\).
Step 4: Use the tangency condition between the two circles.
The two circles touch each other externally at \(T\), so the distance between their centers must equal the sum of their radii:
\[ O_1O_2 = r_1 + r_2 \]
Using the coordinates from Steps 2 and 3, and \(r_1=1\):
\[ O_1O_2 = \sqrt{(3-r_2)^2 + (3-r_2)^2} = \sqrt{2}\,(3-r_2) \]
Step 5: Solve for \(r_2\).
Setting the two expressions for \(O_1O_2\) equal:
\[ \sqrt{2}\,(3-r_2) = 1 + r_2 \]
\[ 3\sqrt{2} - \sqrt{2}\,r_2 = 1 + r_2 \]
\[ 3\sqrt{2} - 1 = r_2(1+\sqrt{2}) \]
\[ r_2 = \frac{3\sqrt{2}-1}{1+\sqrt{2}} \]
Rationalize by multiplying top and bottom by \((\sqrt{2}-1)\):
\[ r_2 = \frac{(3\sqrt{2}-1)(\sqrt{2}-1)}{(1+\sqrt{2})(\sqrt{2}-1)} = \frac{6 - 3\sqrt{2} - \sqrt{2} + 1}{1} = 7 - 4\sqrt{2} \]
Step 6: Sanity check the value.
Numerically, \(\sqrt{2} \approx 1.4142\), so \(r_2 \approx 7 - 5.657 = 1.343\) cm, a positive number smaller than the square's side, which fits a circle sitting snugly in the opposite corner.
