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\begin{document}

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\section*{circles in rhombii sequel solution}

[[!redirects circles in rhombii sequal solution]]

\hypertarget{solution_to_the_circles_in_rhombii_sequel_puzzle}{}\section*{{Solution to the Circles in Rhombii Sequel Puzzle}}\label{solution_to_the_circles_in_rhombii_sequel_puzzle}

[[CirclesinRhombiiSequal.png:pic]]

\begin{quote}%
A follow up question: hopefully by now you can convince yourself that the red circle covers a greater proportion of the rhombus than the two blue circles. So can you construct a different rhombus where these proportions would be equal?


\end{quote}
\hypertarget{solution_by_trigonometry}{}\subsection*{{Solution by [[Trigonometry]]}}\label{solution_by_trigonometry}

[[CirclesinRhombiiSequalLabelled.png:pic]]

Consider a [[rhombus]] as in the diagram above with $\alpha$ as angle $O \hat{A} C$. For simplicities' sake, set the length $O A$ as $1$ unit. Then $O A$ is the hypotenuse of the [[right-angled triangle]] $O A B$ so the length of $O B$ is $\sin(\alpha)$.

The point labelled $Q$ is the centre of the upper small circle. The point labelled $D$ is where the circle touches the line $E C$, so $Q D$ and $O Q$ are radii of this circle and so of equal length. Therefore, triangles $Q O E$ and $Q D E$ are [[congruent]], so angles $Q \hat{E} O$ and $D \hat{E} Q$ are equal and therefore both equal to $\frac{\alpha}{2}$. Since $O E$ has length $1$, this means that $O Q$ has length $\tan\left(\frac{\alpha}{2}\right)$.

The goal is for the large circle to have the same length as the two smaller circles combined, so the area of the large circle is double that of one of the smaller circles, and so the radius of the larger circle is $\sqrt{2}$ times that of the smaller. This implies:

\begin{displaymath}
\sin(\alpha) = \sqrt{2} \tan \left(\frac{\alpha}{2}\right)
\end{displaymath}
The [[half-angle formulae]] show that with $t = tan\left(\frac{\alpha}{2}\right)$ then $\sin(\alpha) = \frac{2 t}{1 + t^2}$. Substituting in to the above gives:

\begin{displaymath}
\frac{2 t}{1 + t^2} = \sqrt{2} t
\end{displaymath}
which rearranges to $1 + t^2 = \sqrt{2}$ and so $t = \sqrt{\sqrt{2} - 1}$ (taking the positive square root as $\alpha$ is acute).

This gives a value of $\alpha$ as $65.53$ to $4$ significant figures.



\end{document}