# Solution to the Two Half-Squares and a Circle Puzzle +-- {.image} [[TwoHalfSquaresandaCircle.png:pic]] > Two half squares, and a circle of radius $1$. What’s the total shaded area? =-- ## Solution by [[chord|Properties of Chords]] and [[Pythagoras' Theorem]] +-- {.image} [[TwoHalfSquaresandaCircleLabelled.png:pic]] =-- In the above diagram, point $O$ is the middle of the circle, $I$ is the midpoint of $E D$ and $H$ is the midpoint of $A B$. The point $G$ is where the extension of $E F$ meets $A B$, and likewise $K$ is where the extension of $E D$ meets $B C$. As $E D$ is a [[chord]] of the circle, the centre of the circle lies on the [[perpendicular line]] through its midpoint (namely, $I$). The same argument applied to $A B$ shows that the centre also lies on the perpendicular line through $H$. As $E D$ and $A B$ are parallel, these perpendicular lines are also [[parallel]]. Since the centre lies on both, they must therefore be the same line. So the points $I$, $O$, and $H$ lie on a straight line that is perpendicular to both $E D$ and $A B$. This establishes $E D$ as lying symmetrically over $A B$, and so the excess on one side, which is the length of $A G$, is equal to the excess on the other, which is the length of $D K$. Since $D K C$ is an [[isosceles]] [[right-angled triangle]], $C K$ has the same length as $D K$. The line $A J$ is a chord with the property that angle $J \hat{B} A$ is a [[right-angle]], so as the [[angle in a semi-circle]] is a right-angle, $A J$ must be a diameter. By the same result, angle $A \hat{D} J$ is also a right-angle, so since angle $E \hat{D} F$ is $45^\circ$, angle $J \hat{D} K$ is $45^\circ$. This means that triangle $D K J$ is also an [[isosceles]] [[right-angled triangle]] and so $K J$ also has the same length as $D K$. Then the length of $B J$ is that of $B C$ with two lengths of $D K$ removed. This is the same as the length of $E D$. Let $a$ be the length of $E D$ (and thus of $B J$) and $b$ the length of $A B$. The shaded region has area $\frac{1}{2}(a^2 + b^2)$. Triangle $A B J$ is a [[right-angled triangle]] with [[hypotenuse]] $2$ and other side lengths $a$ and $b$, so by [[Pythagoras' theorem]], $a^2 + b^2 = 2^2 = 4$. Hence the shaded area is $2$. ## Solution by [[Invariance Principle]] There are two special cases of this diagram. One in which both triangles are the same size, and one where one triangle has shrunk to zero size. +-- {.image} [[TwoHalfSquaresandaCircleEqual.png:pic]] =-- In this version, the total shape makes a square inside the circle. The diagonal of the square coincides with a diameter, since the area of a square is half the square of its diagonal, this establishes the area as $\frac{1}{2} \cdot 2^2 =2 $. +-- {.image} [[TwoHalfSquaresandaCircleVanish.png:pic]] =-- In this version, the total shape is an [[isosceles]] [[right-angled triangle]] with one side as a diameter of the circle, so the area is $\frac{1}{2} \cdot 2 \cdot 2 = 2$.