Reuleaux Rosie

Relevant wiki: Length and Area - Composite Figures

A Reuleaux pentagon represents the combination of a regular pentagon and $5$ minor segments. For this problem, we generalize the solution, given the following parameters:

• $d$ is the diagonal length of a pentagon. Since a polygon is regular, $d = \dfrac{1 + \sqrt{5}}{2}s$ where $s$ is the side length of a regular pentagon. This is easily proven by law of cosines .

• $c$ is the circumradius of a pentagon. By law of cosines , we find that $c = \dfrac{1}{10}\sqrt{50 + 10\sqrt{5}}$

• $A$ denotes the area of a Reuleaux pentagon. The area of a pentagon is $\dfrac{1}{4}\sqrt{25 + 10\sqrt{5}}s^2$ , and the area of a minor segment is $\dfrac{\pi}{10}d^2 - \dfrac{1}{2}d^2\sin\left(36^{\circ}\right)$ . Combining those expressions in terms of $d$ , the area of a Reuleaux pentagon is $A = \dfrac{1}{2}d^2\left(\pi - \sqrt{25 - 10\sqrt{5}}\right)$

Recurrence Relation Construction

Consider the following diagram

where $F,G,H,I$ and $J$ are midpoints of circular segments, and $O$ is the centroid of two pentagons $ABCDE$ and $FGHIJ$ . Since the circumradius of the new pentagon decreases after each $n$ th iteration by $d - c$ (indicated by $|GO| = |FO| = |FC| - |CO| = |GD| - |DO|$ ), the recurrence relation is $c_{n + 1} = d_n - c_n$ for each positive integer $n \geq 1$ , where $d_1 = 1$ and $c_1 = \dfrac{\sqrt{50 + 10\sqrt{5}}}{5 + 5\sqrt{5}}$ . Because the lengths of both $ABCD$ and $FGHIJ$ are directly proportional, the relationship is $\dfrac{d_n}{c_n} = \dfrac{d_{n + 1}}{c_{n + 1}} = \dfrac{d_{n + 2}}{c_{n + 2}} = \cdots$ Then, since we want to determine the area of all red petals, the total area is $\begin{array}{rl} A_{\text{red}} &= \sum\limits_{n=1}^{\infty} \dfrac{1}{2}\left(d_n\right)^2\left(\pi - \sqrt{25 - 10\sqrt{5}}\right)(-1)^{n - 1}\\ &= \sum\limits_{n=0}^{\infty} \dfrac{1}{2}\left(c_{n + 2} + c_{n + 1}\right)^2\left(\pi - \sqrt{25 - 10\sqrt{5}}\right)(-1)^{n} \end{array}$

Computation

The fun part of this problem is that the computation involves two variables - $d_n$ and $c_n$ . Instead of time-consuming substitution with the values of $c$ 's and $d$ 's, we can approach this the elegant way. Given that $d_1 = 1$ and $c_1 = \dfrac{\sqrt{50 + 10\sqrt{5}}}{5 + 5\sqrt{5}}$ , then $\begin{array}{rl} \dfrac{d_n}{c_n} &= \dfrac{1}{\dfrac{\sqrt{50 + 10\sqrt{5}}}{5 + 5\sqrt{5}}}\\ d_n &= \dfrac{5 + 5\sqrt{5}}{\sqrt{50 + 10\sqrt{5}}}c_n \end{array}$ which shows that since $c_{n + 1} = c_n\left(\dfrac{5 + 5\sqrt{5}}{\sqrt{50 + 10\sqrt{5}}} - 1\right)$ then the recurrence relation in terms of $c$ only is the following geometric series $c_n = \dfrac{\sqrt{50 + 10\sqrt{5}}}{5 + 5\sqrt{5}}\left(\dfrac{5 + 5\sqrt{5}}{\sqrt{50 + 10\sqrt{5}}} - 1\right)^{n-1}$ Thus, the total area of the red petals is approximately $0.418$ squared feet. The answer to the problem is $\left\lfloor A \cdot 10^3 \right\rfloor = \boxed{418}$ .