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 $\displaystyle R,$

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

$\displaystyle \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 - - - - :

$\displaystyle A$ 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:

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

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

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

(b) Show that the outside perimeter of the figure which remains

when the frame is removed is: .$\displaystyle P \:=\:\frac{11}{12}(3-\sqrt{3})\pi R$

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

There are three circles at the vertices of the triangle.

. . Each has an exposed perimeter of: .$\displaystyle \frac{5}{6} \times 2\pi r \:=\:\frac{5}{3}\pi r$

The three "corner circles" have a combined perimeter of $\displaystyle 5\pi$

There are six circles at the sides of the triangle.

. . Each has an exposed perimeter of: .$\displaystyle \frac{1}{2} \times 2\pi r \:=\:\pi r$

The six "side circles" have a combined perimeter of $\displaystyle 6\pi r$

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

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

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