Problems

Given that \(\alpha\) and \(\beta\) are acute angles, show that \(\alpha + \beta = \tfrac{1}{2}\pi\) if and only if \(\cos^2 \alpha + \cos^2 \beta = 1\). In the \(x\)--\(y\) plane, the point \(A\) has coordinates \((0,s)\) and the point \(C\) has coordinates \((s,0)\), where \(s>0\). The point \(B\) lies in the first quadrant (\(x>0\), \(y>0\)). The lengths of \(AB\), \(OB\) and \(CB\) are respectively \(a\), \(b\) and \(c\). Show that \[ (s^2 +b^2 - a^2)^2 + (s^2 +b^2 -c^2)^2 = 4s^2b^2 \] and hence that \[ (2s^2 -a^2-c^2)^2 + (2b^2 -a^2-c^2)^2 =4a^2c^2\;. \] Deduce that $$ \l a - c \r^2 \le 2b^2 \le \l a + c \r^2\;. $$ %Show, %by considering the case \(a=1+\surd2\,\), \(b=c=1\,\), % that the condition \(\l \ast \r\,\) %is not sufficient to ensure that \(B\) lies in the first quadrant.

The points \(A\), \(B\) and \(C\) lie on the sides of a square of side 1 cm and no two points lie on the same side. Show that the length of at least one side of the triangle \(ABC\) must be less than or equal to \((\sqrt6 -\sqrt2)\) cm.

Sketch on the same axes the two curves \(C_1\) and \(C_2\), given by

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Solution:

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\(Q = (-a,-b)\) \begin{align*} && \frac{\d y}{\d x} &= -\frac{1}{x^2} \\ \Rightarrow && \frac{y-b}{x-a} &= -\frac{1}{a^2} \\ \Rightarrow && 0 &= a^2y+x-a^2b-a \\ &&&= a^2y+x - 2a\\ \\ && 2x - 2y \frac{\d y}{\d x} &= 0 \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{x}{y} \\ \Rightarrow && \frac{y+b}{x+a} &= \frac{a}{b} \\ \Rightarrow && 0 &= by-ax+b^2 - a^2 \\ &&&= by - ax -2 \end{align*} Notice that \((-b,a)\) is on both lines, therefore it is their point of intersection. The coordinates of \(N\) will be \((a,-b)\). We can see this is a square by noting each point is a rotation (centre the origin) of \(90^\circ\) of each other.

A point moves in the \(x\)-\(y\) plane so that the sum of the squares of its distances from the three fixed points \((x_{1},y_{1})\), \((x_{2},y_{2})\), and \((x_{3},y_{3})\) is always \(a^{2}\). Find the equation of the locus of the point and interpret it geometrically. Explain why \(a^2\) cannot be less than the sum of the squares of the distances of the three points from their centroid. [The centroid has coordinates \((\bar x, \bar y)\) where \(3\bar x = x_1+x_2+x_3,\) $3\bar y =y_1+y_2+y_3. $]

Consider a fixed square \(ABCD\) and a variable point \(P\) in the plane of the square. We write the perpendicular distance from \(P\) to \(AB\) as \(p\), from \(P\) to \(BC\) as \(q\), from \(P\) to \(CD\) as \(r\) and from \(P\) to \(DA\) as \(s\). (Remember that distance is never negative, so \(p,q,r,s\geqslant 0\).) If \(pr=qs\), show that the only possible positions of \(P\) lie on two straight lines and a circle and that every point on these two lines and a circle is indeed a possible position of \(P\).

The famous film star Birkhoff Maclane is sunning herself by the side of her enormous circular swimming pool (with centre \(O\)) at a point \(A\) on its circumference. She wants a drink from a small jug of iced tea placed at the diametrically opposite point \(B\). She has three choices:

1. to swim directly to \(B\).
2. to choose \(\theta\) with \(0<\theta<\pi,\) to run round the pool to a point \(X\) with \(\angle AOX=\theta\) and then to swim directly from \(X\) to \(B\).
3. to run round the pool from \(A\) to \(B\).

The diagram shows a circle, of radius \(r\) and centre \(I\), touching the three sides of a triangle \(ABC\). We write \(a\) for the length of \(BC\) and \(\alpha\) for the angle \(\angle BAC\) and so on. Let \(s=\frac{1}{2}\left(a+b+c\right)\) and let \(\triangle\) be the area of the triangle.

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1. By considering the area of the triangles \(AIB,\) \(BIC\) and \(CIA\), or otherwise, show that \(\Delta=rs\).
2. By using the formula \(\Delta=\frac{1}{2}bc\sin\alpha\), show that \[ \Delta^{2}=\tfrac{1}{16}[4b^{2}c^{2}-\left(2bc\cos\alpha\right)^{2}]. \] Now use the formula \(a^{2}=b^{2}+c^{2}-2bc\cos\alpha\) to show that \[ \Delta^{2}=\tfrac{1}{16}[(a^{2}-\left(b-c\right)^{2})(\left(b+c\right)^{2}-a^{2})] \] and deduce that \[ \Delta=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}. \]
3. A hole in the shape of the triangle \(ABC\) is cut in the top of a level table. A sphere of radius \(R\) rests in the hole. Find the height of the centre of the sphere above the level of the table top, expressing your answer in terms of \(a,b,c,s\) and \(R\).

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 \(\frac{1}{\cos \theta} \l 2p^2 + 4pq - 2p^2 \r = \frac{4pq}{\cos \theta}\)

Show that the equation \[ ax^{2}+ay^{2}+2gx+2fy+c=0 \] where \(a>0\) and \(f^{2}+g^{2}>ac\) represents a circle in Cartesian coordinates and find its centre. The smooth and level parade ground of the First Ruritanian Infantry Division is ornamented by two tall vertical flagpoles of heights \(h_{1}\) and \(h_{2}\) a distance \(d\) apart. As part of an initiative test a soldier has to march in such a way that he keeps the angles of elevation of the tops of the two flagpoles equal to one another. Show that if the two flagpoles are of different heights he will march in a circle. What happens if the two flagpoles have the same height? To celebrate the King's birthday a third flagpole is added. Soldiers are then assigned to each of the three different pairs of flagpoles and are told to march in such a way that they always keep the tops of their two assigned flagpoles at equal angles of elevation to one another. Show that, if the three flagpoles have different heights \(h_{1},h_{2}\) and \(h_{3}\) and the circles in which the soldiers march have centres of \((x_{ij},y_{ij})\) (for the flagpoles of height \(h_{i}\) and \(h_{j}\)) relative to Cartesian coordinates fixed in the parade ground, then the \(x_{ij}\) satisfy \[ h_{3}^{2}\left(h_{1}^{2}-h_{2}^{2}\right)x_{12}+h_{1}^{2}\left(h_{2}^{2}-h_{3}^{2}\right)x_{23}+h_{2}^{2}\left(h_{3}^{2}-h_{1}^{2}\right)x_{31}=0, \] and the same equation connects the \(y_{ij}\). Deduce that the three centres lie in a straight line.

1. Prove that the intersection of the surface of a sphere with a plane is always a circle, a point or the empty set. Prove that the intersection of the surfaces of two spheres with distinct centres is always a circle, a point or the empty set. {[}If you use coordinate geometry, a careful choice of origin and axes may help.{]}
2. The parish council of Little Fitton have just bought a modern sculpture entitled `Truth, Love and Justice pouring forth their blessings on Little Fitton.' It consists of three vertical poles \(AD,BE\) and \(CF\) of heights 2 metres, 3 metres and 4 metres respectively. Show that \(\angle DEF=\cos^{-1}\frac{1}{5}.\) Vandals now shift the pole \(AD\) so that \(A\) is unchanged and the pole is still straight but \(D\) is vertically above \(AB\) with \(\angle BAD=\frac{1}{4}\pi\) (in radians). Find the new angle \(\angle DEF\) in radians correct to four figures.

The transformation \(T\) of the point \(P\) in the \(x\),\(y\) plane to the point \(P'\) is constructed as follows: \hfil\break Lines are drawn through \(P\) parallel to the lines \(y=mx\) and \(y=-mx\) to cut the line \(y=kx\) at \(Q\) and \(R\) respectively, \(m\) and \(k\) being given constants. \(P'\) is the fourth vertex of the parallelogram \(PQP'R\). Show that if \(P\) is \((x_1,y_1)\) then \(Q\) is $$ \left( {mx_1-y_1 \over m-k}, {k(mx_1-y_1)\over m-k}\right). $$ Obtain the coordinates of \(P'\) in terms of \(x_1\), \(y_1\), \(m\) and \(k\), and express \(T\) as a matrix transformation. Show that areas are transformed under \(T\) into areas of the same magnitude.

The tetrahedron \(ABCD\) has \(A\) at the point \((0,4,-2)\). It is symmetrical about the plane \(y+z=2,\) which passes through \(A\) and \(D\). The mid-point of \(BC\) is \(N\). The centre, \(Y\), of the sphere \(ABCD\) is at the point \((3,-2,4)\) and lies on \(AN\) such that \(\overrightarrow{AY}=3\overrightarrow{YN}.\) Show that \(BN=6\sqrt{2}\) and find the coordinates of \(B\) and \(C\). The angle \(AYD\) is \(\cos^{-1}\frac{1}{3}.\) Find the coordinates of \(D\). [There are two alternative answers for each point.]

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Solution: Since \(B\) and \(C\) are reflections of each other in the plane \(y+z=2\) (since that's what it means to be symmetrical), we must have that \(N\) also lies on the plane \(y+z=2\). Since \(\overrightarrow{AY}=3\overrightarrow{YN}.\) we must have \(\overrightarrow{AN}=\overrightarrow{AY}+\overrightarrow{YN} = \frac43\overrightarrow{AY} = \frac43\begin{pmatrix} 3\\-6\\6\end{pmatrix} = \begin{pmatrix} 4\\-8\\8\end{pmatrix}\) and \(N\) is the point \((4,-4,6)\) (which fortunately is on our plane as expected). \(Y\) is the point \((3,-2,4)\) \(|\overrightarrow{AY}| = \sqrt{3^2+(-6)^2+6^2} = 3\sqrt{1+4+4} = 9\)

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Notice that \(BN^2 + 3^2 = 9^2 \Rightarrow BN^2 = 3\sqrt{3^2-1} = 6\sqrt2\). Therefore \(\overrightarrow{NB} = \pm 6\sqrt{2} \frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1\end{pmatrix}\) and \(\{ B, C\} =\{ (4, 2, 12), (4, -10, 0)\}\). Suppose \(D = (x,y,z)\) then \begin{align*} && \begin{pmatrix} -1 \\ 2 \\ -2\end{pmatrix} \cdot \begin{pmatrix} x- 3 \\ y+2 \\ z-4\end{pmatrix} &= 3 \cdot 9 \cdot \frac13 = 9\\ \Rightarrow && 9 &= 3-x+2(y+2)-2(z-4) \\ &&&= -x+2y-2z+15 \\ \Rightarrow && 6 &= x-2y+2z \\ && 2 &= x -4y \\ \\ \Rightarrow && 81 &= (4y+2-3)^2+(y+2)^2+(2-y-4)^2 \\ &&&= (4y-1)^2+2(y+2)^2 \\ &&&= 16y^2-8y+1+2y^2+8y+8 \\ &&&= 18y^2+9 \\ \Rightarrow && y^2 &= 2 \\ \Rightarrow && y &= \pm 2 \end{align*} Therefore \(\displaystyle D \in \left \{ (10, 2, 0), (-6, -2, 4) \right \}\)

Prove that the area of the zone of the surface of a sphere between two parallel planes cutting the sphere is given by \[ 2\pi\times(\mbox{radius of sphere})\times(\mbox{perpendicular distance between the planes}). \] A tangent from the origin \(O\) to the curve with cartesian equation \[ (x-c)^{2}+y^{2}=a^{2}, \] where \(a\) and \(c\) are positive constants with \(c>a,\) touches the curve at \(P\). The \(x\)-axis cuts the curve at \(Q\) and \(R\), the points lying in the order \(OQR\) on the axis. The line \(OP\) and the arc \(PR\) are rotated through \(2\pi\) radians about the line \(OQR\) to form a surface. Find the area of this surface.

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Solution:

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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*}

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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*}

Two points \(P\) and \(Q\) lie within, or on the boundary of, a square of side 1cm, one corner of which is the point \(O\). Show that the length of at least one of the lines \(OP,PQ\) and \(QO\) must be less than or equal to \((\sqrt{6}-\sqrt{2})\) cm.