# Two Congruent Rectangles in a Square +-- {.image} [[TwoCongruentRectanglesinaSquare.png:pic]] > Two congruent rectangles are packed into a square. What's the angle? =-- ## Solution by [[Similar Triangles]] and a lot of algebra +-- {.image} [[TwoCongruentRectanglesinaSquareLabelled.png:pic]] =-- With the points labelled as above, work in units so that the outer square has side length $1$. With this convention, let $a$ be the length of $A B$, $b$ of $C E$, and $c$ of $E G$. Triangles $C E G$ and $F B C$ are [[similar]], with scale factor $a$, so $B C$ has length $a c$. Also, triangle $G H I$ is [[congruent]] to $F B C$ so $H G$ has length $a b$. Therefore, $E H$ has length $c + a b$ and $A E$ has length $a + a c + b$. These are both equal to $1$, so: $$ \begin{aligned} c + a b &= 1 \\ a + a c + b &= 1 \end{aligned} $$ Added these together gives $a + (a + 1)(c + b) = 2$, so $c + b = \frac{2 - a}{1 + a}$. Likewise, subtracting gives $c - b = \frac{a}{1 - a}$. So: $$ \begin{aligned} 2c &= \frac{2 - a}{1 + a} + \frac{a}{1 - a} \\ c &= \frac{1 - a + a^2}{1 - a^2} \\ 2b &= \frac{2 - a}{1 + a} - \frac{a}{1 - a} \\ b &= \frac{1 - 2 a}{1 - a^2} \end{aligned} $$ Triangle $C E G$ is [[right-angled triangle|right-angled]], and $C G$ has length $1$, so applying [[Pythagoras' theorem]] shows that $b^2 + c^2 = 1$ which leads to: $$ \begin{aligned} \left(\frac{1 - a + a^2}{1 - a^2}\right)^2 + \left(\frac{1 - 2 a}{1 - a^2}\right)^2 &= 1 \\ (1 - a + a^2)^2 + (1 - 2 a)^2 &= (1 - a^2)^2 \\ 1 + a^2 + a^4 - 2 a + 2 a^2 - 2 a^3 + 1 - 4 a + 4 a^2 &= 1 - 2 a^2 + a^4 \\ 1 - 6 a + 9 a^2 - 2 a^3 &= 0 \end{aligned} $$ Either by inspection or by noting that there is a solution where the two rectangles each occupy half the square, this equation can be seen to have a root at $a = \frac{1}{2}$. Dividing out by $1 - 2 a$ results in the quadratic: $$ 1 - 4 a + a^2 = 0 $$ which has roots $2 \pm \sqrt{3}$. As $2 + \sqrt{3}$ is greater than $1$, the solution must be $2 - \sqrt{3}$. Substituting in for $b$ gives $b = \frac{1}{2}$. This means that $C E$ is half the length of $C G$, and so triangle $C E G$ is half an [[equilateral triangle]] so angle $E \hat{C} G$ is $60^\circ$. ## Solution by [[rectangle|Lengths in Rectangles]] and [[Equilateral Triangles]] With the points labelled as in the diagram above, $C I$ is a diagonal of the blue rectangle so is the same length as $A K$. The [[quadrilateral]] $A C I K$ therefore has two sides the same length ($A K$ and $C I$) and the other two sides [[parallel]]. It is therefore a [[parallelogram]] and in particular $C I$ is parallel to $A K$. This means that reflecting rectangle $F C G I$ in $C I$ produces the rectangle $C D I J$ with $C D$ along $A E$ and $I J$ along $H L$. So this is a horizontal translate of rectangle $A B K L$. In particular, $C D$ has the same length as $A B$ and $D$ is directly below $I$. Triangles $F B C$ and $G H I$ are [[congruent]], so $B C$ and $I H$ have the same length, and so $B C$ and $D E$ have the same length. This means that $A C$ and $C E$ have the same length, so $C E$ is half the length of $A E$ and thus half the length of $C G$. As above, this means that $G C E$ is half an [[equilateral triangle]] so angle $E \hat{C} G$ is $60^\circ$.