Problems

A right circular cone has base radius \(r\), height \(h\) and slant length \(\ell\). Its volume \(V\), and the area \(A\) of its curved surface, are given by \[ V= \tfrac13 \pi r^2 h \,, \ \ \ \ \ \ \ A = \pi r\ell\,. \]

1. Given that \(A\) is fixed and \(r\) is chosen so that \(V\) is at its stationary value, show that \(A^2 = 3\pi^2r^4\) and that \(\ell =\sqrt3\,r\).
2. Given, instead, that \(V\) is fixed and \(r\) is chosen so that \(A\) is at its stationary value, find \(h\) in terms of \(r\).

{\it Note that the volume of a tetrahedron is equal to \(\frac1 3\) \(\times\) the area of the base \(\times\) the height.} The points \(O\), \(A\), \(B\) and \(C\) have coordinates \((0,0,0)\), \((a,0,0)\), \((0,b,0)\) and \((0,0,c)\), respectively, where \(a\), \(b\) and \(c\) are positive.

1. Find, in terms of \(a\), \(b\) and \(c\), the volume of the tetrahedron \(OABC\).
2. Let angle \(ACB = \theta\). Show that \[ \cos\theta = \frac {c^2} { { \sqrt{\vphantom{ \dot b} (a^2+c^2)(b^2+c^2)} } ^{\vphantom A} \ } \] and find, in terms of \(a\), \(b\) and \(c\), the area of triangle \(ABC\). % is %\(\displaystyle \tfrac12 \sqrt{ \vphantom{\dot A } a^2b^2 +b^2c^2 + c^2 a^2 \;} \;\).

A pyramid stands on horizontal ground. Its base is an equilateral triangle with sides of length~\(a\), the other three sides of the pyramid are of length \(b\) and its volume is \(V\). Given that the formula for the volume of any pyramid is $ \textstyle \frac13 \times \mbox{area of base} \times \mbox {height} \,, $ show that \[ V= \frac1{12} {a^2(3b^2-a^2)}^{\frac12}\;. \] The pyramid is then placed so that a non-equilateral face lies on the ground. Show that the new height, \(h\), of the pyramid is given by \[ h^2 = \frac{a^2(3b^2-a^2)}{4b^2-a^2}\;. \] Find, in terms of \(a\) and \(b\,\), the angle between the equilateral triangle and the horizontal.

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Solution: First let's consider the area of the base. It is an equilateral triangle with side length \(a\), so \(\frac12 a^2 \sin 60^\circ = \frac{\sqrt{3}}4a^2\).

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Let's consider the height. The distance to the centre \(\frac23 \frac{\sqrt{3}}2 a = \frac{a}{\sqrt{3}}\) so \(h = \sqrt{b^2 - \frac{a^2}{3}}\) and therefore the volume is: \begin{align*} V &= \frac13 \times \mbox{area of base} \times \mbox {height} \\ &= \frac13 \frac{\sqrt{3}}{4}a^2 \sqrt{\frac{3b^2-a^2}{3}} \\ &= \frac1{12}a^2 (3b^2-a^2)^{\frac12} \end{align*} The area of an isoceles triangle with sides \(a,b,b\) can be found by considering the perpendicular:

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ie \(\frac{a}{4} \sqrt{b^2-\frac{a^2}{4}} = \frac{a\sqrt{4b^2-a^2}}{8}\). Therefore by considering the volume, we must have \begin{align*} && V &= \frac13 \times \mbox{area of base} \times \mbox {height} \\ \Rightarrow && \frac1{12}a^2 (3b^2-a^2)^{\frac12} &= \frac13 \frac{a\sqrt{4b^2-a^2}}{8} h \\ \Rightarrow && h &= \frac{2a(3b^2-a^2)}{(4b^2-a^2)^{\frac12}} \\ \Rightarrow && h^2 &= \frac{4a^2(3b^2-a^2)}{4b^2-a^2} \end{align*}

The cuboid \(ABCDEFGH\) is such \(AE\), \(BF\), \(CG\), \(DH\) are perpendicular to the opposite faces \(ABCD\) and \(EFGH\), and \(AB =2, BC=1, AE={\lambda}\). Show that if \(\alpha\) is the acute angle between the diagonals \(AG\) and \(BH\) then $$\cos {\alpha} = |\frac {3-{\lambda}^2} {5+{\lambda}^2} |$$ Let \(R\) be the ratio of the volume of the cuboid to its surface area. Show that \(R<\frac{1}{3}\) for all possible values of \(\lambda\). Prove that, if \(R\ge \frac{1}{4}\), then \(\alpha \le \arccos \frac{1}{9}\).

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Solution:

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Set \(A\) to be the origin, then \(B = \langle 2, 0, 0 \rangle, G = \langle 2, 1, \lambda \rangle, H = \langle 0, 1, \lambda \rangle\), in particular \begin{align*} && AG&= \langle 2, 1, \lambda \rangle \\ && BH &= \langle -2, 1, \lambda \rangle \\ \Rightarrow && \cos \alpha &= |\frac{-4+1+\lambda^2}{\sqrt{2^2+1^2+\lambda^2}\sqrt{(-2)^2+1^2+\lambda^2}}| \\ &&&= |\frac{-3+\lambda^2}{5+\lambda^2}| \end{align*} \begin{align*} && \text{Volume} &= 2\lambda \\ && \text{Surface area} &= 2\cdot2\lambda + 2\cdot\lambda + 2\cdot2 \\ \Rightarrow && R&= \frac{\lambda}{3\lambda + 2} < \frac{1}{3} \\ && \frac14 &\leq R \\ \Rightarrow && 3\lambda +2 &\leq 4\lambda \\ \Rightarrow &&2 & \leq \lambda \end{align*} Then \(\frac{\lambda^2-3}{5+\lambda^2}\) is increasing as \(\lambda\) increases, in particularly the smallest value is \(\frac{1}{9}\).

1. Show that four vertices of a cube, no two of which are adjacent, form the vertices of a regular tetrahedron. Hence, or otherwise, find the volume of a regular tetrahedron whose edges are of unit length.
2. Find the volume of a regular octahedron whose edges are of unit length.
3. Show that the centres of the faces of a cube form the vertices of a regular octahedron. Show that its volume is half that of the tetrahedron whose vertices are the vertices of the cube.

A cylindrical biscuit tin has volume \(V\) and surface area \(S\) (including the ends). Show that the minimum possible surface area for a given value of \(V\) is \(S=3(2\pi V^{2})^{1/3}.\) For this value of \(S\) show that the volume of the largest sphere which can fit inside the tin is \(\frac{2}{3}V\), and find the volume of the smallest sphere into which the tin fits.

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Solution: Suppose we have height \(h\) and radius \(r\), then: \(V = \pi r^2 h\) and \(S = 2\pi r^2 + 2\pi r h\). \(h = \frac{V}{\pi r^2}\), so \begin{align*} S &= 2 \pi r^2 + 2 \pi r\frac{V}{\pi r^2} \\ &= 2\pi r^2 +V \frac1{r}+V \frac1{r} \\ &\underbrace{ \geq }_{\text{AM-GM}} 3 \sqrt[3]{2\pi r^2 \frac{V^2}{r^2} } = 3 (2 \pi V^2)^{1/3} \end{align*} Equality holds when \(r = \sqrt[3]{\frac{V}{2 \pi}}, h = \frac{V}{\pi (V/2\pi)^{2/3}} = \sqrt[3]{\frac{4V}{\pi}}\) Since \(h > r\) the sphere has a maximum radius of \(r\) and so it's largest volume is \(\frac43 \pi r^3 = \frac43 \pi \frac{V}{2 \pi} = \frac23 V\).

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The radius of the sphere is \(\sqrt{\left (\frac{r}{2} \right)^2 + \left (\frac{h}{2} \right)^2 } = \frac12 \sqrt{r^2+h^2}\) \begin{align*} V_{sphere} &= \frac43 \pi (r^2+h^2)^{3/2} \\ &= \frac43 \pi \left (\left( \frac{V}{2 \pi} \right)^{2/3}+\left( \frac{4V}{ \pi} \right)^{2/3} \right)^{3/2} \\ &= \frac43 \pi \frac{V}{ \pi} \left ( 2^{-2/3}+4^{2/3}\right)^{3/2} \\ &= \frac 43 V \left ( \frac{1+4}{2^{2/3}} \right)^{3/2} \\ &= \frac43 \frac{5^{3/2}}{2} V \\ &= \frac{2 \cdot \sqrt{125}}{3} V \end{align*}

My house has an attic consisting of a horizontal rectangular base of length \(2q\) and breadth \(2p\) (where \(p < q\)) and four plane roof sections each at angle \(\theta\) to the horizontal. Show that the length of the roof ridge is independent of \(\theta\) and find the volume of the attic and the surface area of the roof.

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Solution:

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The distance to the top of the house (viewed from above) from the long side will be \(p\). The distance from the short side will also be the same, since the roof sections are climbing at the same angle, so they will take just as far to reach the top. Therefore the length of the roof ridge will be \(2q - 2p\) which is independent of \(\theta\). \vspace{1em} The height of the roof will be \(h = p \tan \theta\). The attic can be split into a prism (along the roof ridge) and a pyramid (along the sloping sides). The pyramid will have volume \(\frac13 p \tan\theta (2p)^2 = \frac83 \tan\theta p^3\). The prism will have volume \(2(q-p)p^2 \tan\theta\). Therefore the total volume will be \(\l \frac{2}{3}p + 2q \r p^2\tan\theta \) The distance (along the plane) to the roof of the house will be \(\frac{p}{\cos \theta}\) and therefore the two end roof-sections will be triangles of area \(\frac{p^2}{\cos \theta}\). The two side roof-sections will be trapiziums will area \(\frac{1}{2} \l 2q + 2(q-p) \r \frac{p}{\cos \theta}\) Therefore the total area will be \(\frac{1}{\cos \theta} \l 2p^2 + 4pq - 2p^2 \r = \frac{4pq}{\cos \theta}\)

The diagram shows a coffee filter consisting of an inverted hollow right circular cone of height \(H\) cm and base radius \(a\) cm. \noindent

Frosty the snowman is made from two uniform spherical snowballs, of initial radii \(2R\) and \(3R.\) The smaller (which is his head) stands on top of the larger. As each snowball melts, its volume decreases at a rate which is directly proportional to its surface area, the constant of proportionality being the same for both snowballs. During melting each snowball remains spherical and uniform. When Frosty is half his initial height, find the ratio of his volume to his initial volume. If \(V\) and \(S\) denote his total volume and surface area respectively, find the maximum value of \(\dfrac{\mathrm{d}V}{\mathrm{d}S}\) up to the moment when his head disappears.