areageometrytriangles
This geometry problem comes from a recent math test.
The question is the following:
We have a regular hexagon with sides equal to $1$ and six circular arcs with radius equal to $1$ from each vertices of the hexagon in it creating a flower-shaped-like object. Find the area of the flower.
I tried creating an equilateral triangle by connecting the point in the center to the vertices of the boundary but was unable to proceed.
Would be thankful if you could help me out.
Observe a spiral similarity with center at $D$ which takes $A$ to $K$. Then it takes $B$ to $M$ and $C$ to $L$. So this map takes triangle $ABC$ to triangle $KML$ which means that they are similary with dilatation coefficient $k={\sqrt{2}\over 2}$ So the ratio of the areas is $k^2 =1/2$.
OK, I think I was misremembering in my comment, I think the perimeter is easier to work with than the area. So you start with a circle of circumference $\pi$ (meaning a radius of $\frac{1}{2}$). Find the length of the side of the square (it will be $\frac{1}{\sqrt{2}}$), so the initial guess is $4\cdot\frac{1}{\sqrt{2}} = 2\sqrt{2} \approx 2.828427$:
Here is a concrete example where we can use the previous known chord (in this case $\frac{1}{\sqrt{2}}$) to find the next:
This gives $\pi \approx 3.0614674$. Now, here is the general case, where you know the previous chord, $s_n$, and then find the next (knowing that each time, you're bisecting the previous chord so the number of sides doubles). I think this formula is correct, the formula for $s_{n+1}$ given $s_n$ is definitely correct because I tested it, but not entirely certain about the perimeter formula:
Using the above, we can write:
$$ s_{n+1}^2 = \frac{1 - \sqrt{1 - s_n^2}}{2} $$
So we can find:
\begin{align*} P_1 \approx &\ 2.8284271247461903\\ P_2 \approx &\ 3.061467458920718\\ P_3 \approx &\ 3.121445152258053\\ P_4 \approx &\ 3.1365484905459406\\ P_5 \approx &\ 3.140331156954739\\ P_6 \approx &\ 3.141277250932757\\ P_7 \approx &\ 3.1415138011441455\\ P_8 \approx &\ 3.1415729403678827\\ P_9 \approx &\ 3.141587725279961\\ P_{11} \approx &\ 3.141591421504635\\ P_{12} \approx &\ 3.141592345611077\\ P_{13} \approx &\ 3.1415925765450043\\ P_{14} \approx &\ 3.1415926334632482\\ P_{15} \approx &\ 3.141592654807589\\ P_{16} \approx &\ 3.1415926453212153\\ P_{17} \approx &\ 3.1415926073757197\\ P_{18} \approx &\ 3.1415929109396727\\ P_{19} \approx &\ 3.141594125195191\\ P_{20} \approx &\ 3.1415965537048196\\ P_{21} \approx &\ 3.1415965537048196 \end{align*}
This gives five digits of precision: $\pi \approx 3.14159$.
This example was taught to me in my Scientific Computing class way back to display roundoff error in floating point calculations. You'll notice on the last two, we get the same result. That's because the floating point calculations of the computer have essentially hit their limit. The reason for this is because $s_n^2$ has become so small, that $1 - s_n^2$ doesn't generate a "new" number (it just keeps giving the same number that will generate $s_n^2$ again when subtracted from $1$. There is a trick to make this calculation better:
\begin{align*} s_{n+1}^2 =&\ \frac{1 - \sqrt{1 - s_n^2}}{2}\cdot\frac{1 + \sqrt{1 - s_n^2}}{1 + \sqrt{1 - s_n^2}} \\ =&\ \frac{1}{2}\cdot\frac{1 - \left(1 - s_n^2\right)}{1 + \sqrt{1 - s_n^2}}\\ =&\ \frac{1}{2}\cdot\frac{s_n^2}{1 + \sqrt{1 - s_n^2}} \end{align*}
This really does give better results:
\begin{align*} P_1 \approx&\ 2.8284271247461903 \\ P_2 \approx&\ 3.0614674589207183\\ P_3 \approx&\ 3.1214451522580524\\ P_4 \approx&\ 3.1365484905459393\\ P_5 \approx&\ 3.140331156954753\\ P_6 \approx&\ 3.141277250932773\\ P_7 \approx&\ 3.1415138011443013\\ P_8 \approx&\ 3.1415729403670913\\ P_9 \approx&\ 3.1415877252771596\\ P_{10} \approx&\ 3.1415914215111997\\ P_{11} \approx&\ 3.1415923455701176\\ P_{12} \approx&\ 3.1415925765848725\\ P_{13} \approx&\ 3.1415926343385627\\ P_{14} \approx&\ 3.1415926487769856\\ P_{15} \approx&\ 3.141592652386591\\ P_{16} \approx&\ 3.141592653288993\\ P_{17} \approx&\ 3.141592653514593\\ P_{18} \approx&\ 3.141592653570993\\ P_{19} \approx&\ 3.1415926535850933\\ P_{20} \approx&\ 3.141592653588618\\ P_{21} \approx&\ 3.1415926535894996\\ P_{22} \approx&\ 3.1415926535897203\\ P_{23} \approx&\ 3.1415926535897754\\ P_{24} \approx&\ 3.141592653589789\\ P_{25} \approx&\ 3.1415926535897927\\ P_{26} \approx&\ 3.1415926535897936\\ P_{27} \approx&\ 3.1415926535897936\\ \end{align*}
By simply changing the computation, not the algorithm!, we now get twelve digits of precision! $\pi \approx 3.141592653589$--all because of floating point roundoff error.
Best Answer
Assuming those are circular arcs and not "ellipses", you can indeed find the area exactly.
The "flower" has six "petals" Each of those petals has axial symmetry and can be divided into two halves. Each of those halves is the segment subtending a central angle of $\frac{\pi}{3}$ (radian measure) of a circle of radius $1$.
The area of one such segment is $\frac 12 r^2(\theta - \sin\theta) = \frac 12(\frac{\pi}{3} - \frac{\sqrt{3}}{2})$.
There are $12$ such segments, yielding the total area of the "flower" as $2\pi - 3\sqrt 3$.