# Math Help - Circle Geometry: Circles in A Equilateral Triangle

1. ## Circle Geometry: Circles in A Equilateral Triangle

Ten equal circular discs can just be placed tightly within a frame in the form of a equilateral triangle. Nine of the them touch the frame; the tenth, in the center, touches six of the other discs. If the internal sides of the frame are each of length measured R, express the radius of each circular disc in terms of R, and show that the outside perimeter of the figure which remains when the frame is removed is:

11/12 (3-sqrt3) (pi) R

Attempted it to no avail. Tried obtaining expression for r (radius of circle) using simultaneous equations. Discovered r = R/2, which is obviously wrong. Please show me how to do this question. Thanks very much.

2. Did you draw another Equilateral Traingle by extending the lines of the three sets of three external circles?

3. Hello, Lukybear!

Ten equal circular discs can just be placed tightly
within a frame in the form of a equilateral triangle.
Nine of the them touch the frame.
The tenth, in the center, touches six of the other discs.

If the internal sides of the triangular frame are each of length $R,$

(a) express the radius of each circular disc $r$ in terms of $R.$

$\begin{array}{c}
/\backslash \\ [-2mm]
/\!\!\bigcirc\!\! \backslash \\ [-2mm]
/ \!\!\bigcirc\!\! \bigcirc\! \backslash \\ [-2mm]
/ \!\!\bigcirc \!\!\bigcirc \!\!\bigcirc \!\!\backslash \\ [-2mm]
/ \!\!\bigcirc \!\!\bigcirc \!\!\bigcirc \!\!\bigcirc \!\!\backslash \\ [-3mm]
-\! -\! -\! -\!-
\end{array}$

The bottom row looks like this:

Code:
                           /     \
* * *   /       \   * * *
*          /*       *\          *
*           /   *   *   \           *
*           /     * *     \           *
/               \
*          /        *        \          *
*         o - - - - * - - - - o - - - - *
*      *  |   r     *     r   |         *
*     |                   |
*        |r       * *      |r         *
*   *       |       *   *       |       *
*        *     |     *       *     |     *
A o---------------*-*-*---------------*-*-*--------------
: - - - r√3 - - - : - - -  2r - - - - :

$A$ is a vertex of the triangular frame.
There are four circles across the bottom row.

We can see that the bottom of the frame is:

. . $R \;=\;\sqrt{3}r + 2r + 2r + 2r + \sqrt{3}r \;=\;2(3+\sqrt{3})r$

Hence: . $r \;=\;\dfrac{R}{2(3 + \sqrt{3})}$

. . which rationalizes to: . $r \;=\;\dfrac{(3-\sqrt{3})}{12}R$

(b) Show that the outside perimeter of the figure which remains
when the frame is removed is: . $P \:=\:\frac{11}{12}(3-\sqrt{3})\pi R$

The perimeter of each circle is: . $2\pi r$

There are three circles at the vertices of the triangle.
. . Each has an exposed perimeter of: . $\frac{5}{6} \times 2\pi r \:=\:\frac{5}{3}\pi r$
The three "corner circles" have a combined perimeter of $5\pi$

There are six circles at the sides of the triangle.
. . Each has an exposed perimeter of: . $\frac{1}{2} \times 2\pi r \:=\:\pi r$
The six "side circles" have a combined perimeter of $6\pi r$

The total perimeter of the figure is: . $P \;=\;5\pi r + 6\pi r \;=\;11\pi r$

Since $r \:=\:\frac{3-\sqrt{3}}{12}R$

. . $P \;=\;11\pi\left(\frac{3-\sqrt{3}}{12}R\right) \;=\;\frac{11}{12}(3-\sqrt{3})\pi R$

4. Wow thanks. Thats fantastic, really appreciated. Can i just ask, what reasoning did you give to state that the corner circumferences were 5/6 x 2(pi)R

5. Hello again, Lukybear!

By what reasoning did you say that the corner circumferences were . $\frac{5}{6} \times 2\pi r$ ?

Consider the three circles in the lower-left corner.
Their centers form an equilateral triangle.

Code:
                        * * *
*             *
*                 *
*                   *

*                     *
*           *         *
*          / \        *
/   \
/     \
* * *   /       \   * * *
*          /*       *\          *
*           /   *   *   \           *
*           /     * *     \           *
/               \
*          / 60     *        \          *
*         * - - - - * - - - - *         *
*              r    *                   *

*                 * *                 *
*               *   *               *
*           *       *           *
* * *               * * *

The exposed circumference of the lower-left circle is $2\pi r$
. . minus the arc subtended by that $60^o$ angle.

And that is why the corner-circumferences are: $\frac{5}{6} \times 2\pi r.$