# Solution to the Two Semi-Circles in a Circle Puzzle +-- {.image} [[TwoSemiCirclesinaCircle.png:pic]] > The bases of the semicircles are parallel. What fraction of the circle is shaded? =-- ## Solution by [[Congruent]] Triangles and [[Pythagoras' Theorem]] +-- {.image} [[TwoSemiCirclesinaCircleCongruent.png:pic]] =-- In the above diagram, the points $E$ and $F$ are midpoints of those [[chords]] and the line through $E$ and $F$ is the [[perpendicular bisector]] through those chords so is a [[diameter]] of the circle. The length of $E F$ is the sum of the lengths of $B E$ and $A F$. Let $O$ be the point so that $E O$ has the same length as $A F$, and so $F O$ has the same length as $B E$. Then $O E B$ and $A F O$ are [[right-angled triangles]] with the same side lengths adjacent to the right-angles, so are [[congruent]]. This means that the lengths of $O B$ and $O A$ are the same. As $A$ and $B$ are on the circumference of the circle and $O$ is on a diameter, it is therefore the centre of the circle. This means that $O A$ is a radius and $F O$ has the same length as $B E$, so with $a$ as the length of $A F$, $b$ as the length of $B E$, and $r$ as the length of $F O$, [[Pythagoras' theorem]] applied to the triangle $A F O$ shows that: $$ r^2 = a^2 + b^2 $$ The shaded areas are $\frac{1}{2} \pi a^2 + \frac{1}{2} \pi b^2 = \frac{1}{2} \pi r^2$ and the whole circle has area $\pi r^2$, so the shaded area is $\frac{1}{2}$ of the total circle. ## Solution by [[chord|Midpoints of Chords]] and [[Pythagoras' Theorem]] +-- {.image} [[TwoSemiCirclesinaCircleMidpoints.png:pic]] =-- In the above diagram, $M$ is the midpoint of the [[chord]] $A B$, $E$ of $B C$, and $F$ of $A D$. The line through $E$ and $F$ is the [[perpendicular bisector]] of each of the chords $B C$ and $A D$ and so is a [[diameter]] of the circle. The point $O$, which lies on this diameter, is the centre of the circle. The points $G$, $H$, and $I$ are so that $G M$ is perpendicular to $E F$, $B H$ is perpendicular to $A F$, and $I M$ is also perpendicular to $A F$. As $M$ is the midpoint of $A B$, $G$ is the midpoint of $E F$, and so the length $G F$ is half of the sum of the radii of the two semi-circles. Similarly, $I$ is the midpoint of $A H$ and so the length of $I F$ is the average of the radii of the two semi-circles. Hence $M G$ and $M I$ have the same length. Since angles $G \hat{M} I$ and $O \hat{M} A$ are both right-angles, angles $G \hat{M} O$ and $I \hat{M} A$ are equal. Therefore, triangles $M G O$ and $M I A$ are [[congruent]]. The length of $O G$ is therefore the same as that of $A I$, which is half the difference of the radii of the smaller circles. Since the length of $F G$ is half the sum of these radii, $F O$ is the same length as the smaller radius, namely the length of $B E$. Applying [[Pythagoras' theorem]] to triangle $A F O$ then shows that, with $a$ and $b$ the radii of the semi-circles and $r$ of the outer circle, $$ r^2 = a^2 + b^2 $$ As above, then, the shaded areas are $\frac{1}{2} \pi a^2 + \frac{1}{2} \pi b^2 = \frac{1}{2} \pi r^2$ and the whole circle has area $\pi r^2$, so the shaded area is $\frac{1}{2}$ of the total circle. ## Solution by [[Invariance Principle]] There are two special configurations of this puzzle. In the first, one of the circles is as large as can be and the second is non-existent. In the second, the two circles are of equal size. +-- {.image} [[TwoSemiCirclesinaCircleSpecialA.png:pic]] =-- In this case, it is clear that the shaded area is half the area of the full circle. +-- {.image} [[TwoSemiCirclesinaCircleSpecialB.png:pic]] =-- In this case, the diameters of the smaller circles form two sides of a square and the diagonal of this square is then a diameter of the larger circle. Since the length of the diagonal is $\sqrt{2}$ times the length of a side, the area of the large circle is $2$ times the area of the two semi-circles made into a circle.