Coordinate Geometry

We can choose a coordinate frame where the parallel planes are parallel to the \(y-z\) axis. Then we can compute the surface area as an integral of the surface of revolution for \(x^2 + y^2 = r^2\). Using \(y = r \sin t, x = r \cos t\) we have: \begin{align*} S &= 2\pi\int_{\cos^{-1}a}^{\cos^{-1}b}y \sqrt{\left ( \frac{\d x}{\d t} \right)^2+\left ( \frac{\d y}{\d t} \right)^2} \d t \\ &=2\pi\int_{\cos^{-1}a}^{\cos^{-1}b} r^2 \sin t \d t \\ &= 2\pi \cdot r^2 \cdot (a - b) \\ &= 2 \pi \cdot r \cdot (ra-rb) \\ &= 2\pi\times(\mbox{radius of sphere})\times(\mbox{perpendicular distance between the planes}). \end{align*} We can view this surface as a sphere missing a cap of height \(XQ\) and adding a cone of slant height \(OP\) and radius \(PX\) The centre of the circle is at \((c,0)\) and \(OP^2 + a^2 = c^2 \Rightarrow OP = \sqrt{c^2-a^2}\) Since \(OPC \sim OXP\) we must have that \(\frac{OX}{OP} = \frac{OP}{OC} \Rightarrow OX = \frac{c^2-a^2}{c}\) and \(\frac{PX}{OP} = \frac{CP}{OC} \Rightarrow PX = \frac{a}{c}\sqrt{c^2-a^2}\) \(QX = OX - OQ = \frac{c^2-a^2}{c}-(c-a) = \frac{ac-a^2}{c}\) Therefore the surface area is: \begin{align*} S &= 4 \pi a^2 - 2\pi \cdot a \cdot QX+ \pi PX \cdot OP \\ &= 4 \pi a^2 - 2\pi a \cdot \frac{ac-a^2}{c}+\pi \frac{a}{c}\sqrt{c^2-a^2}\cdot \sqrt{c^2-a^2} \\ &= 4\pi a^2 -2\pi \frac{a^2c-a^3}{c}+\pi \frac{ac^2-a^3}{c} \\ &= \pi a \frac{(a+c)^2}{c} \end{align*}

First note that \((\sqrt{6} - \sqrt{2})^2 = 8 - 2\sqrt{12} = 8 - 4\sqrt{3}\) and since \(49 > 16 \times 3\) \(\sqrt{6}-\sqrt{2} > 1\). Therefore we can assume without loss of generality that \(P\) and \(Q\) both do not lie on the same side as each other, a side containing \(O\), otherwise one of those lengths would be \(1 \text{ cm} < (\sqrt{6}-\sqrt{2}) \text{ cm} \). Let \(O = (0,0)\), \(P = (1,x)\), \(Q = (y,1)\), then our lengths squared are: \(1 + x^2, 1 + y^2, (1-x)^2+(1-y)^2\). To maximise the length of the smallest side, each side should be equal in length (otherwise we could increase the length of the smallest side by moving the point between the shortest side and the longest side (without affecting the other side). Therefore \(x = y\) and \(1+x^2 = 2(1-x)^2 \Rightarrow x^2-4x+1 = 0 \Rightarrow x = 2 - \sqrt{3} \). Therefore the distances are all \(\sqrt{1+7-4\sqrt{3}} = \sqrt{8-4\sqrt{3}} = (\sqrt{6}-\sqrt{2}) \text{ cm}\)

The shown isoceles triangle has base \(2r\), and the two side lengths are \(R\). The angle at the center of the circle is \(\frac{2\pi}{n}\). The height of the triangle (by Pythagoras) is \(\sqrt{R^2-r^2}\) and so the area enclosed in the triangle is \(\frac12 2r \sqrt{R^2-r^2}\). The area in the three sectors are: \(\frac{\pi}{n}(R-r)^2\), two sets of \(\frac12(\frac12 \l \pi - \frac{2\pi}{n}\r)r^2 = \l\frac{1}{2} - \frac{1}{n} \r \frac12 \pi r^2\). Therefore the remaining area \(Y/n\) is \(r \sqrt{R^2-r^2} -\frac{\pi}{n}(R-r)^2 - \l\frac{1}{2} - \frac{1}{n} \r \pi r^2\). Multiplying this by \(n\) we get the desired result. For \(X\) we can look at the larger sector, which we obtain using the extension of this triangle. This has area \(\frac12 \frac{2 \pi}{n} (R+r)^2\). The areas to be removed are the area of the triangle: \(r \sqrt{R^2-r^2}\) and the areas of the two sectors, which have radii \(r\) and angles \(\pi - \l \frac12 -\frac1{n} \r\pi = \l \frac{1}{2} + \frac{1}{n} \r \pi\). Therefore the area for \(X/n\) is \(\frac{\pi}{n} (R+r)^2 - r \sqrt{R^2-r^2} - \l \frac{1}{2} + \frac{1}{n} \r \pi r^2\) and so \(X\) has area: \[ X = \pi (R+r)^2 - nr \sqrt{R^2-r^2} - n\pi r^2\l \frac{1}{2} +\frac{1}{n} \r \] \(Z = n\pi r^2\) \begin{align*} \frac{(X+Y)}{Z} &= \frac{\pi (R+r)^2 - nr \sqrt{R^2-r^2} - n\pi r^2\l \frac{1}{2} +\frac{1}{n} \r}{n \pi r^2 }\\ &= \qquad \frac{nr\sqrt{R^{2}-r^{2}}-\pi(R-r)^{2}-n\pi r^{2}\left(\frac{1}{2}-\frac{1}{n}\right)}{n \pi r^2} \\ &= \frac{4\pi R r - n \pi r^2}{n\pi r^2} \\ &= \frac{4R-nr}{n r} \\ &= \frac{4R}{nr} - 1 \end{align*} Since \(\frac{r}{R} = \sin \frac{\pi}{n}\) we have: \begin{align*} \frac{(X+Y)}{Z} &= \frac{4R}{nr} - 1 \\ &= \frac{4}{n \sin \frac{\pi}n} - 1 \\ & \to \frac{4}{\pi} - 1 \end{align*}