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\begin{document}

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\section*{two overlapping squares iii solution}

\hypertarget{solution_to_the_two_overlapping_squares_iii_puzzle}{}\section*{{Solution to the [[Two Overlapping Squares III]] Puzzle}}\label{solution_to_the_two_overlapping_squares_iii_puzzle}

[[TwoOverlappingSquaresIII.png:pic]]

\begin{quote}%
Is more of this design green or blue (and by how much)?


\end{quote}
\hypertarget{solution_by_properties_of_area_and_transformations}{}\subsection*{{Solution by Properties of [[Area]] and [[Transformations]]}}\label{solution_by_properties_of_area_and_transformations}

[[TwoOverlappingSquaresIIIAnnotated.png:pic]]

Label the points as above. To compare the blue and green areas, we add in to both the quadrilateral $F J D E$. In this, the point $I$ is such that $F I$ is [[perpendicular]] to $D H$ and $E$ is on the continuation of $B D$ so that $F E$ is perpendicular to $B E$.

Rotating $90^\circ$ anticlockwise about $A$ takes $A B$ to $A H$ and $A C$ to $A G$, therefore triangles $A B C$ and $A H G$ are [[congruent]] and, in particular, have the same area.

This also means that angle $A \hat{H} G$ is $90^\circ$ and so $H G$ is a continuation of $D H$, so $D H G$ is a straight line.

Translating $A C$ along $A G$ to $G$ takes $C$ to $F$ and $B$ onto the line $G D$. Since angle $C \hat{B} A$ is $90^\circ$, and angle $F \hat{I} G$ is also $90^\circ$, this translation takes $B$ to $I$ and so triangles $A B C$ and $G I F$ are also congruent. Further translating it so that $G$ translates to $H$ takes $G I F$ onto $H D E$, establishing those as congruent.

A similar argument shows that triangle $C E F$ is congruent to $A H G$ via the translation that takes $A$ to $C$ and $G$ to $F$. Since triangles $A B C$ and $A H G$ are congruent, this means that triangles $C E F$ and $H D E$ are also congruent.

This means that by including the quadrilateral $F J D E$ and considering it a part of both the blue and green regions means that they can be matched except for the triangle $H E F$ which is the excess blue.

This triangle is half of the [[parallelogram]] $G H E F$ and so its area is the same as the pink region, which is $5$.

Hence the blue area is larger than the green by $5$.

\hypertarget{solution_by_invariance_principle}{}\subsection*{{Solution by [[Invariance Principle]]}}\label{solution_by_invariance_principle}

The angle at which the squares are to each other can be varied, with the restriction that $C$ stay on the edge $B D$. Putting it at $B$ doesn't work since then the squares overlap exactly and all the regions are empty (even the one set at $5$). That leaves putting it at $D$ which results in the diagram below.

[[TwoOverlappingSquaresIIIInvariance.png:pic]]

In this version of the diagram, the triangles are all congruent so the blue region is one triangle more than the green, and that triangle is the same as the pink so has area $5$.



\end{document}