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\begin{document}

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\section*{semi-circle in a square solution}

\hypertarget{solution_to_the_semicircle_in_a_square_puzzle}{}\section*{{Solution to the Semi-Circle in a Square Puzzle}}\label{solution_to_the_semicircle_in_a_square_puzzle}

[[SemiCircleinaSquare.png:pic]]

\begin{quote}%
Is more of this square pink or blue? How would you prove it?


\end{quote}
\hypertarget{solution_by_angle_between_a_radius_and_tangent_isosceles_rightangled_triangles_area_of_a_circle_and_area_of_a_square}{}\subsection*{{Solution by [[Angle Between a Radius and Tangent]], [[Isosceles]] [[Right-Angled Triangles]], [[Area of a Circle]], and [[Area of a Square]]}}\label{solution_by_angle_between_a_radius_and_tangent_isosceles_rightangled_triangles_area_of_a_circle_and_area_of_a_square}

[[SemiCircleinaSquareLabelled.png:pic]]

In the above diagram, the point labelled $O$ is the centre of the [[semi-circle]]. The point labelled $H$ is where the semi-circle touches the top edge of the square, so as the [[angle between a radius and tangent]] is $90^\circ$, $H C$ is [[perpendicular]] to $I G$. Similarly, $J E$ is perpendicular to $I A$.

The line segments $O H$ and $O J$ are radii of the circle, so are of the same length. Therefore, $O E$ and $O C$ are of the same length. Triangles $O E F$ and $B C O$ are [[congruent]] since $J E$ is [[parallel]] to $A D$ and so angles $C \hat{B} O$ and $E \hat{O} F$ are [[corresponding angles]], and $O B$ and $O F$ have the same length. So $O E$ and $E F$ have the same length so triangle $O E F$ is [[isosceles]].

Let $r$ be the radius of the semi-circle. Then as $O E F$ is an [[isosceles]] [[right-angled triangle]] with hypotenuse of length $r$, its shorter sides both have length $\frac{r}{\sqrt{2}}$. The side length of the square is therefore $r\left(1 + \frac{1}{\sqrt{2}}\right)$. Its area is thus

\begin{displaymath}
r^2 \left(1 + \frac{1}{\sqrt{2}} \right)^2
\end{displaymath}
The area of the [[semi-circle]] is $\frac{1}{2} \pi r^2$.

To compare these two, consider the decimal representations of the circle area and half the area of the square, since if the blue area is larger than the pink then it will be over half the square and vice-versa:

\begin{displaymath}
\begin{aligned}
\frac{\pi}{2} &\simeq 1.5708 \\
\frac{1}{2}\left(1 + \frac{1}{\sqrt{2}}\right)^2 &\simeq 1.4571
\end{aligned}
\end{displaymath}
Therefore the blue area is larger.



\end{document}