# Solution to the Rectangles in Concentric Rings Puzzle +-- {.image} [[RectanglesinConcentricRings.png:pic]] > Each rectangle is tangent to two of the concentric circles. If the inner blue right has area $4$, what's the purple area? =-- ## Solution by [[Angle Between a Radius and Tangent]] and [[Pythagoras' Theorem]] +-- {.image} [[RectanglesinConcentricRingsLabelled.png:pic]] =-- Let $a$, $b$, $c$, $d$ be the radii of the circles in increasing order. That the blue ring has area $4$ means that $\pi b^2 - \pi a^2 = 4$. Let $e$ and $f$ be the half heights of the two rectangles, with $e$ the length of $A D$ and $f$ the length of $E F$. Using the fact that the [[angle between a radius and tangent]] is $90^\circ$, there are many [[right-angled triangles]] in the diagram. Applying [[Pythagoras' theorem]] to these triangles leads to the identities: $$ \begin{aligned} b^2 &= a^2 + e^2 && \triangle O A D \\ c^2 &= b^2 + e^2 && \triangle O B C \\ c^2 &= a^2 + f^2 && \triangle O E F \\ d^2 &= c^2 + f^2 && \triangle O H G \end{aligned} $$ The purple ring has area $ \pi d^2 - \pi c^2$, and using the above identities gives: $$ \begin{aligned} d^2 - c^2 &= f^2 \\ &= c^2 - a^2 \\ &= b^2 + e^2 - a^2 \\ &= b^2 + b^2 - a^2 - a^2 \\ &= 2(b^2 - a^2) \end{aligned} $$ Putting this together shows that the purple ring has twice the area of the blue, so has area $8$.