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Going in circles

- The radius of the big circle is 10.
- You mean the circle which has the four yellow circle and five green circles within it?
- Is that green? When did you have your eyes checked the last time?
- What is the radius of circle 1?
- Let me find out.
- What is the radius of circle 2?
- Don’t talk. I am busy.

The diagram is taken from Math Magic Packing Archive.

This entry was posted on Thursday, January 26th, 2012 at 12:04 am and is filed under Problem. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

One Response to “Going in circles”

  1. Richard Sabey Says:
    January 26th, 2012 at 6:56 pm

    Set up Cartesian coordinates such that circle 1′s centre O is (0, 0) and the rightmost point of circle 2 is (10, 0).

    Let r be the radius of a yellow circle.

    Let A be the centre of the upper right yellow circle, and P the foot of the perpendicular from A to the x-axis. Then AP=r. A is on y=x, so A=(r, r). Let OA produced meet the large circle at B. Then
    \left. \begin{array}{crcll} {} & OA & = & r\surd 2 \\ {} & AB & = & r \\ {} & OB & = & 10 \\ \Rightarrow & r(\surd 2+1) & = & 10 \\ \Rightarrow & r & = & \frac{10}{\surd 2+1} \end{array} \right.

    Let r_1, r_2 be the radii of circles 1, 2. Let that yellow circle cut OA at C. Then
    \left. \begin{array}{crcll} {} & AC & = & r \\ \Rightarrow & r_1 & = & OA-CA \\ {} & = & r\surd 2-r \\ {} & = & r(\surd 2-1) \\ {} & = & \frac{10(\surd 2-1)}{\surd 2+1} \\ {} & = & \frac{10(\surd 2-1)^2}{(\surd 2+1)(\surd 2-1)} \\ {} & = & 10(2-2\surd 2+1)/1 \\ {} & = & 10(3-2\surd 2) \end{array} \right.

    I couldn’t see how to find r_2 by classical geometry, so I applied Descartes’ theorem to the 3 circles that touch circle 2, namely 2 yellow ones and the enclosing circle.
    \left. \begin{array}{crcll} {} & \frac 1{r_2} & = & \frac 1 r + \frac 1 r - \frac 1{10} \\ {} & = & \frac2r-\frac1{10} \\ {} & = & \frac{2(\surd 2+1)}{10}-\frac1{10} \\ {} & = & \frac{2\surd 2+2-1}{10} \\ {} & = & \frac{2\surd 2+1}{10} \\ \Rightarrow & r_2 & = & \frac{10}{2\surd2+1} \end{array} \right.

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