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\begin{document}

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\section*{three semi-circles in a rectangle solution}

\hypertarget{solution_to_the_three_semicircles_in_a_rectangle_puzzle}{}\section*{{Solution to the [[Three Semi-Circles in a Rectangle]] Puzzle}}\label{solution_to_the_three_semicircles_in_a_rectangle_puzzle}

[[ThreeSemiCirclesinaRectangle.jpeg:pic]]

\begin{quote}%
Is more of this rectangle red or blue?


\end{quote}
\hypertarget{solution_by_pythagoras_theorem_and_the_area_of_a_circle}{}\subsection*{{Solution by [[Pythagoras' Theorem]] and the [[Area of a Circle]]}}\label{solution_by_pythagoras_theorem_and_the_area_of_a_circle}

[[ThreeSemiCirclesinaRectangleLabelled.jpeg:pic]]

In the diagram above, $A$ is the midpoint of the upper edge so is the centre of the upper semi-circle, $E$ is the midpoint of the lower edge, and $D$ is the centre of the lower semi-circle.

Let $a$ be half of the width of the rectangle and $b$ half of its height. Then the radii of the various circles are:

\begin{enumerate}%
\item Upper semi-circle: $a$
\item Right quarter circle: $2 b$
\item Left semi-circle: $b$
\item Lower semi-circle: $\frac{2a - 2b}{2} = a - b$

\end{enumerate}
By including the white petal regions into both the blue and red, the difference in the areas of the red and blue regions is the same as the differences in the corresponding partial circles.

The upper and lower semi-circles have area:

\begin{displaymath}
\frac{1}{2} \pi a^2 + \frac{1}{2} \pi (a - b)^2 = \frac{1}{2} \pi (2 a^2 - 2 a b + b^2)
\end{displaymath}
The left and right partial circles have area:

\begin{displaymath}
\frac{1}{2} \pi b^2 + \frac{1}{4} \pi (2 b)^2 = \frac{3}{2} \pi b^2
\end{displaymath}
Triangle $A D E$ is a [[right-angled triangle]] with sides of lengths $2 b$, $a - (a - b) = b$, and $a + (a - b) = 2a - b$. Applying [[Pythagoras' theorem]] yields:

\begin{displaymath}
\begin{aligned}
(2a - b)^2 &= (2 b)^2 + b^2 \\
4 a^2 - 4 a b + b^2 &= 5 b^2 \\
a^2 - a b &= b^2
\end{aligned}
\end{displaymath}
Substituting in to the expression for the area of the upper and lower semi-circles, this shows that that area is:

\begin{displaymath}
\frac{1}{2} \pi (2 a^2 - 2 a b + b^2) = \frac{1}{2} \pi (2 b^2 + b^2) = \frac{3}{2} \pi b^2
\end{displaymath}
which is exactly the area of the side circle segments. Hence the red and blue areas are the same.



\end{document}