# Solution to the [[Two Semi-Circles and Two Quarter Circles]] Puzzle +-- {.image} [[TwoSemiCirclesandTwoQuarterCircles.jpeg:pic]] > This design is made from two semicircles and two quarter circles. Is more of it shaded yellow or blue? =-- ## Solution by [[Pythagoras' Theorem]] and the [[Area of a Circle]] +-- {.image} [[TwoSemiCirclesandTwoQuarterCirclesLabelled.jpeg:pic]] =-- In the above diagram, $O$ is the centre of the two semi-circles and $D$ is such that angle $C \hat{D} O$ is a [[right-angle]]. By including the white regions to both the yellow and blue, the comparison is between the two quarter circles and the two semi-circles. Let $R$ be the radius of the larger blue semi-circle and $r$ of the smaller. Let $b$ be the radius of the left quarter circle and $a$ of the right. So the sides of the blue triangle are $a$, $b$, and $2 r$. Let $d$ be the length of the line segment $O D$ and $h$ of $C D$. There are three right-angled triangles, namely $C D B$, $C D A$, and $C D O$. Applying [[Pythagoras' theorem]] to all three yields: $$ \begin{aligned} a^2 &= h^2 + (r + d)^2 \\ b^2 &= h^2 + (r - d)^2 \\ R^2 &= h^2 + d^2 \end{aligned} $$ Adding the first two equations together gives: $$ a^2 + b^2 = 2 h^2 + (r + d)^2 + (r - d)^2 = 2 h^2 + 2 r^2 + 2 d^2 $$ And then substituting in the third shows that $a^2 + b^2 = 2 r^2 + 2 R^2$. Multiplying by $\frac{1}{4} \pi$ gives: $$ \frac{1}{4} \pi a^2 + \frac{1}{4} \pi b^2 = \frac{1}{2} \pi r^2 + \frac{1}{2} \pi R^2 $$ and so the blue and yellow regions have the same area. ## Solution by the [[Cosine Rule]] and [[Parallelogram Identity]] The above can be shortened by using either the [[cosine rule]] or the [[parallelogram identity]]. Using the cosine rule avoids needing point $D$. Applied to triangles $C O B$ and $C O A$ it yields: $$ \begin{aligned} a^2 &= r^2 + R^2 - 2 r R \cos( B \hat{O} C) \\ b^2 &= r^2 + R^2 - 2 r R \cos( C \hat{O} A) \begin{aligned} $$ Then, since $\cos(B \hat{O} C) = - \cos(C \hat{O} A)$, adding the equations gives: $$ a^2 + b^2 = 2r^2 + 2R^2 $$ This is also the result of applying the [[parallelogram identity]] to the parallelogram obtained by duplicating triangle $C A B$ and rotating it about $O$ (so that $A B$ as a diameter). ## Solution by [[Invariance Principle]] The positions of point $C$ on the circumference and $A$ and $B$ on the diameter can be varied, leading to two positions where the result is obvious. In both, $C$ is positioned at the midpoint of the arc. +-- {.image} [[TwoSemiCirclesandTwoQuarterCirclesInvarianceA.jpeg:pic]] =-- Here, $A$ and $B$ are at the ends of the diameter, meaning that the blue and green is a full circle. The triangle is an [[isosceles]] [[right-angled triangle]] so its sides are in the ratio $1 : 1: \sqrt{2}$. So the yellow and green regions comprise half a circle of twice the area of the blue and green. +-- {.image} [[TwoSemiCirclesandTwoQuarterCirclesInvarianceB.jpeg:pic]] =-- Here, $A$ and $B$ are at the centre of the blue semi-circle, and the yellow quarter circles combine to be a semi-circle of exactly the same size.