# Solution to the Circle and Two Semi-Circles in Two Hexagons Puzzle +-- {.image} [[ACircleandTwoSemiCirclesinTwoHexagons.png:pic]] > A circle and two semicircles are trapped by two overlapping regular hexagons. What fraction of the total area is shaded? =-- ## Solution by Angles in an [[Equilateral Triangle]] and [[hexagon|Regular Hexagon]] +-- {.image} [[ACircleandTwoSemiCirclesinTwoHexagonsLabelled.png:pic]] =-- In the above diagram, the edges $A G$ and $B H$ are extended to meet at $C$. The point $O$ is the centre of the larger circle and $E$ of the upper semi-circle. Angles $O \hat{A} G$ and $H \hat{B} O$ are both $60^\circ$ so triangle $A B C$ is an [[equilateral triangle]]. Angle $C \hat{D} O$ is a [[right-angle]] as it is the [[angle between a radius and tangent]] and angle $O \hat{C} D$ is half of $60^\circ$, so triangle $O D C$ is half an [[equilateral triangle]], so the length of $O D$ is half that of $O C$. Hence the length of $E C$ is the same as that of $O D$. A similar argument applies to the triangle $E F C$ so the length of $I C$ is the same as that of $E F$. Putting that together shows that the length of $I C$ is a third of that of $O I$. So the length of $H G$ is a third of the length of $B H$, which is a side of one of the hexagons. This means that dividing each side of the hexagon in three and drawing lines parallel to the sides is a subdivision of each of the regions. +-- {.image} [[ACircleandTwoSemiCirclesinTwoHexagonsDivided.png:pic]] =-- There are $24$ triangles in each of the yellow regions, and $30$ in the white region. So there are $78$ in total and $48$ that are yellow, so the yellow region is $\frac{48}{78} = \frac{24}{39}$ths of the total.