# Solution to the Five Semi-Circles Puzzle +-- {.image} [[FiveSemiCircles.png:pic]] > A design made from $5$ semicircles. What’s the total shaded area? =-- ## Solution by [[Pythagoras' Theorem]] +-- {.image} [[FiveSemiCirclesLabelled.png:pic]] =-- With the lengths labelled as in the above diagram, [[Pythagoras' theorem]] gives the following identities: $$ \begin{aligned} a^2 &= h^2 + (4 + c)^2 \\ b^2 &= h^2 + (8 - c)^2 \\ d^2 &= h^2 + c^2 \\ e^2 &= h^2 + (4 - c)^2 \\ 4^2 &= e^2 + d^2 \end{aligned} $$ Putting some of these together leads to: $$ \begin{aligned} a^2 + b^2 &= 2 h^2 + (4 + c)^2 + (8 - c)^2 \\ &= 2 h^2 + 4^2 + 8 c + c^2 + 8^2 - 16 c + c^2 \\ &= 2 h^2 + 4^2 + 8^2 - 8 c + 2 c^2 \\ 4^2 &= 2 h^2 + c^2 + (4 - c)^2 \\ &= 2 h^2 + c^2 + 4^2 - 8 c + c^2 \\ &= 2 h^2 + 4^2 - 8 c + 2 c^2 \end{aligned} $$ Combining these gets: $$ a^2 + b^2 = 8^2 + 4^2 = 80 $$ The area of a semi-circle of diameter $D$ is: $$ \frac{\pi (D/2)^2 }{2} = \frac{\pi D^2}{8} $$ And so the areas of the semi-circles add to $10 \pi$. ## Solution by the [[Intersecting Chords Theorem]] (With a smidgeon of [[Pythagoras' Theorem]]) +-- {.image} [[FiveSemiCirclesChords.png:pic]] =-- In the diagram above, the line segment $I H$ is a diameter of the outer circle. As $E$ is on the middle circle, length $g = 8$ and $f = 4$. The [[intersecting chords theorem]] then says that $a d = f g = 32$. Applying [[Pythagoras' theorem]] to the right-angled triangle $E J B$ yields the identity $b^2 = e^2 + d^2$, and applying it to $A J B$ produces: $$ 12^2 = e^2 + (d + a)^2 = e^2 + d^2 + 2 a d + a^2 = b^2 + 64 + a^2 $$ And so $a^2 + b^2 = 144 - 64 = 80$ as before. (*Note: This method was inspired by the [[cosine rule]] and the [[invariance principle]] since if $a^2 + b^2$ is constant then when applying the cosine rule to triangle $A E B$ the term $2 a b cos(C)$ has to be constant. In the above diagram, this is $a d$.*) ## Solution by the [[Parallelogram Law]] +-- {.image} [[FiveSemiCirclesParallelogram.png:pic]] =-- In the above diagram, $K$ is the point on the inner middle circle diametrically opposite to $E$, so that $A K$ is [[congruent]] to $E B$ and $K B$ to $A E$. So $A E B K$ is a parallelogram. According to the [[parallelogram law]], $$ A B^2 + E K^2 = 2 A E^2 + 2 E B^2 $$ Since $A B = 12$ and $E K = 4$, this means that (with $a$ and $b$ as in above sections): $$ a^2 + b^2 = \frac{12^2 + 4^2}{2} = 80 $$ and the solution follows as before. ## Solution by the [[Invariance principles]] The point $E$ can move around on the circumference of the middle circle, and this leads to two special configurations. +-- {.image} [[FiveSemiCirclesInvariantI.png:pic]] =-- The left-hand semi-circle has radius $4$ and the right-hand one has radius $2$. So their areas are: $$ \frac{\pi 4^2}{2} + \frac{\pi 2^2}{2} = 8\pi + 2 \pi = 10 \pi $$ +-- {.image} [[FiveSemiCirclesInvariantII.png:pic]] =-- In this configuration, the diameter of each semi-circle is the hypotenuse of a triangle with other sides $6$ and $2$, so its square is $6^2 + 2^2 = 40$. The sum of the squares of the diameters is therefore $80$ which leads to the area of the semi-circles being $10\pi$ as before.