Question

Carpenter Rajesh has a circular piece of plywood of diameter 30 feet. He has cut out two disks of diameter 20 feet and 10 feet. What is the diameter of the largest disk that can be cut out from the remaining portion of the plywood piece?

• A. >8.00$ feet and $\le 8.20$ feet
• B. >8.21$ feet and $\le 8.40$ feet
• C. >8.41$ feet and $\le 8.60$ feet
• D. >8.61$ feet and $\le 8.80$ feet
• E. >8.81$ feet and $\le 9.00$ feet

Hint:

When several circles are packed inside a larger circle with all tangencies, the maximal configuration occurs when they are mutually tangent.
Descartes' Circle Theorem (kissing circles) quickly relates the radii via curvatures: \((\sum b)^2=2\sum b^2\) with the containing circle taken negative.

Answer 1

The Correct Option is C

Step 1: Let the outer circle (plywood) have radius \(R=15\) ft. The two removed disks have radii \(r_1=10\) ft and \(r_2=5\) ft. To maximize the radius \(r\) of another disk that can fit in the remaining portion, place all three inner disks mutually tangent to each other and tangent to the outer boundary.
Step 2: Use Descartes' Circle Theorem for four mutually tangent circles. If \(b_i=\frac{1}{r_i}\) is the curvature (take the outer circle with negative curvature \(b_0=-\frac{1}{R}\)), then \[ (b_0+b_1+b_2+b_3)^2=2\,(b_0^2+b_1^2+b_2^2+b_3^2). \] Here \(b_0=-\frac{1}{15},\ b_1=\frac{1}{10},\ b_2=\frac{1}{5},\ b_3=\frac{1}{r}\). Solving, \[ \left(-\frac{1}{15}+\frac{1}{10}+\frac{1}{5}+\frac{1}{r}\right)^2 =2\left(\frac{1}{15^2}+\frac{1}{10^2}+\frac{1}{5^2}+\frac{1}{r^2}\right). \] This gives \(r=\dfrac{30}{7}\approx 4.2857\ \text{ft}\). Step 3: The largest possible disk therefore has diameter \[ 2r=\frac{60}{7}\approx 8.57\ \text{ft}, \] which lies in the interval \(>8.41\) ft and \(\le 8.60\) ft. \(\boxed{\text{Option (C)}}\).