# Solution to the [[Three Congruent Rectangles]] Puzzle +-- {.image} [[ThreeCongruentRectangles.png:pic]] > Three congruent rectangles overlap. How long is the diagonal? =-- ## Solution by [[Dissection]] and [[Congruent]] Triangles +-- {.image} [[ThreeCongruentRectanglesLabelled.png:pic]] =-- In the above diagram, $G$, $I$, $B$, and $D$ are where the the yellow rectangle meets the edges of the figure. Line segments $L B$ and $G K$ are horizontal, while $I L$ and $D K$ are vertical. Triangles $G M I$ and $G H I$ are [[congruent]], as are triangle $B L I$ and $B A I$, triangles $B N D$ and $B C D$, and triangles $G K D$ and $G F D$. Therefore the sum of the areas of triangles $G H I$, $B L I$, $B N D$, and $G K D$ is the same as that of triangles $G M I$, $B A I$, $B C D$, and $G F D$. The latter comprises the region of the figure that is outside the yellow rectangle, but since the yellow rectangle is [[congruent]] to the blue and red, its area is half of the full diagram. Therefore triangles $G H I$, $B L I$, $B N D$, and $G K D$ cover the same area as the yellow rectangle. But those triangles cover the yellow rectangle with an overlap, namely the rectangle $L M K N$. Therefore, that rectangle must have zero area and so points $B$ and $G$ are at the same height. The distance between $B$ and $G$ is therefore the same as the bottom side of the diagram, which is $F$. This is also the diagonal of the yellow rectangle, so is also the length of line segment $I D$. ## Solution by [[Similar Triangles]] and [[Pythagoras' Theorem]] With the points labelled as above, let line segment $H G$ have length $a$, $A C$ have length $b$, and $H I$ have length $c$. Then $I A$ has length $4 - c$. Triangles $G H I$ and $I A B$ are [[similar]] since they are both [[right-angled triangles|right-angled]] and angles $G \hat{I} H$ and $A \hat{I} B$ add up to $90^\circ$. Therefore: $$ \frac{a}{c} = \frac{4 - c}{b} $$ which rearranges to $c(4 - c) = a b$. Since the rectangles are [[congruent]], $G I$ has length $2$ and $I B$ has length $a + b$. Therefore, applying [[Pythagoras' theorem]] to triangles $G H I$ and $I A B$ produces: $$ \begin{aligned} a^2 + c^2 &= 4 \\ \text{So } c^2 &= 4 - a^2 \\ (4 - c)^2 + b^2 &= (a + b)^2 = a^2 + 2 a b + b^2 \\ \text{So } (4 - c)^2 &= a^2 + 2 a b \end{aligned} $$ Putting those together: $$ \begin{aligned} 4^2 &= (c + 4 - c)^2 = c^2 + 2 c(4 - c) + (4 - c)^2 \\ &= 4 - a^2 + 2 a b + a^2 + 2 a b \\ 12 & = 4 a b \\ 3 &= a b \end{aligned} $$ Therefore $c(4 - c) = 3$ which rearranges to $c^2 - 4 c + 3 = 0$ and this has solutions $c = 1$ and $c = 3$. Note that $4 - c$ is the other solution, so without loss of generality take $c = 1$. Then $a^2 = 4 - c^2 = 3$, so $a = \sqrt{3}$ and likewise $b = \sqrt{3}$. This means that $a + b = 2 \sqrt{3}$, so the rectangles have side lengths $2$ and $2 \sqrt{3}$. Their diagonal is then $\sqrt{2^2 + (2 \sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4$.