Consider the following brain teaser:
In figure 1, above, find the radius of the quarter circle.
This was featured in Mind Your Decisions (2018-08-06) “A Puzzle Richard Feynman Missed?”. Spoiler alert! The answer is \(b\), but the puzzle is formulated in a way to trick you into going down the route of finding the much more complicated answer:
\[ r = \frac{(a + c) + \sqrt{2b^{2} + 2ac - a^{2} - c^{2}}}{2} \label{a}\tag{Eq.1} \]
This does, however, raise another interesting question, how do we show that this expression in \(\ref{a}\) is equal to \(b\)?
Firstly, find an expression relating \(a\), \(b\) and \(c\) using Pythagoras and rearrange in terms of \(c\).
\[\begin{align} b^{2} & = (b - a)^{2} + (b - c)^{2} \\ b^{2} & = b^{2} + a^{2} - 2ab + b^{2} + c^{2} - 2bc \\ 0 & = c^{2} + c(-2b) + a^{2} + b^{2} - 2ab \\ c & = \frac{2b \pm \sqrt{4b^{2} - 4 \times 1 \times (a^{2} + b^{2} - 2ab)}}{2} \\ c & = b \pm \sqrt{b^{2} - a^{2} - b^{2} + 2ab} \\ c & = b \pm \sqrt{2ab - a^{2}} \end{align}\]
and since \(c < b\)
\[ c = b - \sqrt{2ab - a^{2}} \label{b}\tag{Eq.2} \]
Substitute this expression for \(c\) into \(\ref{a}\).
\[\begin{align} r & = \frac{(a + (b - \sqrt{2ab - a^{2}})) + \sqrt{2b^{2} + 2a(b - \sqrt{2ab - a^{2}}) - a^{2} - (b - \sqrt{2ab - a^{2}})^{2}}}{2} \\ r & = \frac{a + b - \sqrt{2ab - a^{2}} + \sqrt{2(b - a)\sqrt{2ab - a^{2}} + b^{2}}}{2} \end{align}\]
Now attempt to rearrange this expression into the form \(r = \frac{-B \pm \sqrt {B^{2} - 4AC}}{2A}\)
Define \(A\) such that:
\[ A = 1 \]
Define \(B\) such that:
\[\begin{align} -B & = a + b - \sqrt{2ab - a^{2}} \\ B^{2} & = a^{2} + b^{2} + 2ab - a^{2} + 2ab - 2a\sqrt{2ab - a^{2}} - 2b\sqrt{2ab - a^{2}} \label{c}\tag{Eq.3} \end{align}\]
Define \(C\) such that:
\[ B^{2} - 4AC = 2b\sqrt{2ab - a^{2}} - 2a\sqrt{2ab - a^{2}} + b^{2} \label{d}\tag{Eq.4} \]
Subtract \(\ref{d}\) from \(\ref{c}\).
\[\begin{align} 4C & = -2b \sqrt{2ab - a^{2}} + a^{2} + 2ab - a^{2} + 2ab - 2b \sqrt{2ab - a^{2}} \\ & = 4ab - 4b \sqrt {2ab - a^{2}} \\ C & = ab - b \sqrt{2ab - a^{2}} \end{align}\]
Then
\[ r = \frac{-B + \sqrt{B^{2} - 4AC}}{2A} \]
which means (if we ignore the negative solution as the sign in the middle is \(+\), rather than \(\pm\)):
\[ Ar^{2} + Br + C = 0 \]
Substituting back in our expressions for \(A\), \(B\) and \(C\) and factorising:
\[\begin{align} r^{2} + \Big[ \sqrt{2ab - a^{2}} - a - b \Big]r + ab - b \sqrt{2ab - a^{2}} & = 0 \\ (r - b)(r - a + \sqrt{2ab - a^{2}}) & = 0 \\ \end{align}\]
which gives as the positive solution:
\[ r = b \]