Coordinate Geometry

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2025 Paper 3 Q3
D: 1500.0 B: 1500.0

Let \(f(x)\) be defined and positive for \(x > 0\). Let \(a\) and \(b\) be real numbers with \(0 < a < b\) and define the points \(A = (a, f(a))\) and \(B = (b, -f(b))\). Let \(X = (m,0)\) be the point of intersection of line \(AB\) with the \(x\)-axis.

  1. Find an expression for \(m\) in terms of \(a\), \(b\), \(f(a)\) and \(f(b)\).
  2. Show that, if \(f(x) = \sqrt{x}\), then \(m = \sqrt{ab}\). Find, in terms of \(n\), \(a\) function \(f(x)\) such that \(m = \frac{a^{n+1} + b^{n+1}}{a^n + b^n}\).
  3. Let \(g_1(x)\) and \(g_2(x)\) be defined and positive for \(x > 0\). Let \(m = M_1\) when \(f(x) = g_1(x)\) and let \(m = M_2\) when \(f(x) = g_2(x)\). Show that if \(\frac{g_1(x)}{g_2(x)}\) is a decreasing function then \(M_1 > M_2\). Hence show that $$\frac{a+b}{2} > \sqrt{ab} > \frac{2ab}{a+b}$$
  4. Let \(p\) and \(c\) be chosen so that the curve \(y = p(c-x)^3\) passes through both \(A\) and \(B\). Show that $$\frac{c-a}{b-c} = \left(\frac{f(a)}{f(b)}\right)^{1/3}$$ and hence determine \(c\) in terms of \(a\), \(b\), \(f(a)\) and \(f(b)\). Show that if \(f\) is a decreasing function, then \(c < m\).

Show Solution
  1. The line \(AB\) has equation: \begin{align*} && \frac{y+f(b)}{x-b} &= \frac{f(a)+f(b)}{a-b} \\ \Rightarrow && \frac{f(b)}{m-b} &= \frac{f(a)+f(b)}{a-b} \\ \Rightarrow && m &= \frac{a-b}{f(a)+f(b)}f(b) + b \\ &&&= \frac{af(b)+bf(a)}{f(a)+f(b)} \end{align*}
  2. Suppose \(f(x) = \sqrt{x}\) then \begin{align*} m &= \frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}} \\ &= \frac{\sqrt{ab}(\sqrt{a}+\sqrt{b})}{\sqrt{a}+\sqrt{b}} \\ &= \sqrt{ab} \end{align*} Suppose \(f(x) = x^{-n}\) then \begin{align*} m &= \frac{a b^{-n}+ba^{-n}}{a^{-n}+b^{-n}} \\ &= \frac{a^{n+1}+b^{n+1}}{b^n + a^n} \\ \end{align*}
  3. Without loss of generality, we can scale \(g_1(x)\) and \(g_2(x)\) so that \(g_1(a) = g_2(a)\) and \(m\) won't change for either of them. Then since \(\frac{g_1(b)}{g_2(b)} < 1\) (this function is decreasing) our line connecting \((a,g_i(a))\) and \((b,-g_i(b))\) must interect the axis first for \(g_2\), in particular \(M_1 > M_2\). Suppose \(g_1(x) =1, g_2(x) = \sqrt{x}, g_3(x) = x^{-1}\), the notice that \(\frac{g_1(x)}{g_2(x)} =\frac{g_2(x)}{g_3(x)}= x^{-1/2}\) are decreasing, therefore: \begin{align*} \frac{a+b}{1+1} &> \sqrt{ab} > \frac{1+1}{a^{-1}+b^{-1}} \\ \frac{a+b}{2} &> \sqrt{ab} > \frac{2ab}{a+b} \\ \end{align*}
  4. We must have: \begin{align*} && p(c-a)^3 &= f(a) \\ && p(c-b)^3 &= -f(b) \\ \Rightarrow &&\left ( \frac{c-a}{c-b} \right)^3 &= -\frac{f(a)}{f(b)} \\ \Rightarrow && \frac{c-a}{b-c} &= \left (\frac{f(a)}{f(b)} \right)^{\tfrac13} \\ \Rightarrow && c-a &= \left (\frac{f(a)}{f(b)} \right)^{\tfrac13}(b-c)\\ \Rightarrow && c \left (1 + \left (\frac{f(a)}{f(b)} \right)^{\tfrac13} \right) &= \left (\frac{f(a)}{f(b)} \right)^{\tfrac13}b + a \\ \Rightarrow && c &= \frac{\left (\frac{f(a)}{f(b)} \right)^{\tfrac13}b + a}{1 + \left (\frac{f(a)}{f(b)} \right)^{\tfrac13}} \\ &&&= \frac{b[f(a)]^\tfrac13+a[f(b)]^\tfrac13}{[f(a)]^\tfrac13+[f(b)]^\tfrac13} \end{align*} We have that \(\frac{c-a}{b-c} = \left (\frac{f(a)}{f(b)} \right)^{\tfrac13} \) and \(\frac{m-a}{b-c} = \frac{f(a)}{f(b)}\). Since \(f\) is decreasing, \(\frac{f(a)}{f(b)} > 1\) and so \(\left (\frac{f(a)}{f(b)} \right)^{\tfrac13} < \frac{f(a)}{f(b)}\), therefore \(m > c\).
2024 Paper 2 Q3
D: 1500.0 B: 1500.0

The unit circle is the circle with radius 1 and centre the origin, \(O\). \(N\) and \(P\) are distinct points on the unit circle. \(N\) has coordinates \((-1, 0)\), and \(P\) has coordinates \((\cos\theta, \sin\theta)\), where \(-\pi < \theta < \pi\). The line \(NP\) intersects the \(y\)-axis at \(Q\), which has coordinates \((0, q)\).

  1. Show that \(q = \tan\frac{1}{2}\theta\).
  2. In this part, \(q \neq 1\).
    1. Let \(\mathrm{f}_1(q) = \dfrac{1+q}{1-q}\). Show that \(\mathrm{f}_1(q) = \tan\frac{1}{2}\!\left(\theta + \frac{1}{2}\pi\right)\).
    2. Let \(Q_1\) be the point with coordinates \((0, \mathrm{f}_1(q))\) and \(P_1\) be the point of intersection (other than \(N\)) of the line \(NQ_1\) and the unit circle. Describe geometrically the relationship between \(P\) and \(P_1\).
    1. \(P_2\) is the image of \(P\) under an anti-clockwise rotation about \(O\) through angle \(\frac{1}{3}\pi\). The line \(NP_2\) intersects the \(y\)-axis at the point \(Q_2\) with co-ordinates \((0, \mathrm{f}_2(q))\). Find \(\mathrm{f}_2(q)\) in terms of \(q\), for \(q \neq \sqrt{3}\).
    2. In this part, \(q \neq -1\). Let \(\mathrm{f}_3(q) = \dfrac{1-q}{1+q}\), let \(Q_3\) be the point with coordinates \((0, \mathrm{f}_3(q))\) and let \(P_3\) be the point of intersection (other than \(N\)) of the line \(NQ_3\) and the unit circle. Describe geometrically the relationship between \(P\) and \(P_3\).
    3. In this part, \(0 < q < 1\). Let \(\mathrm{f}_4(q) = \mathrm{f}_2^{-1}\!\Big(\mathrm{f}_3\!\big(\mathrm{f}_2(q)\big)\Big)\), let \(Q_4\) be the point with coordinates \((0, \mathrm{f}_4(q))\) and let \(P_4\) be the point of intersection (other than \(N\)) of the line \(NQ_4\) and the unit circle. Describe geometrically the relationship between \(P\) and \(P_4\).

Show Solution
TikZ diagram
  1. \(\,\) \begin{align*} && \frac{y-0}{x-(-1)} &= \frac{\sin \theta }{\cos \theta + 1} \\ \Rightarrow && y_0 &= \frac{\sin \theta}{\cos \theta + 1} \\ &&&= \frac{\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}+1} \\ &&&= t = \tan \tfrac{\theta}{2} \end{align*} Alternatively, it is straightforward to see from the angles.
    1. \(f_1(q) = \frac{1+q}{1-q}\) so \begin{align*} && f_1(\tan\tfrac12\theta) &= \frac{1+\tan\tfrac12\theta}{1-\tan\tfrac12\theta} \\ &&&= \frac{\cos \tfrac12 \theta + \sin \tfrac12 \theta}{\cos \tfrac12 \theta - \sin \tfrac12 \theta} \\ &&&= \frac{\sin(\tfrac14 \pi + \tfrac12 \theta)}{\cos(\tfrac14 \pi + \tfrac12 \theta)} \\ &&&= \tan \tfrac12(\theta + \tfrac{\pi}{2}) \end{align*}
    2. \(Q_1\) is the point \((0, f_1(q))\) so \(P_1\) will be the point \((\cos (\theta + \tfrac{\pi}{2}), \sin (\theta + \tfrac{\pi}{2}))\) which is a rotation anticlockwise by \(\frac{\pi}{2}\)
    1. \(P_2 = (\cos(\theta + \tfrac{\pi}{3}), \sin( \theta + \tfrac{\pi}{3})\) and so \(f_2(q) = \tan (\tfrac12(\theta + \tfrac{\pi}{3}))\) so \begin{align*} && f_2(q) &= \tan (\tfrac12(\theta + \tfrac{\pi}{3})) \\ &&&= \frac{q + \tan \frac{\pi}{3}}{1 - \tan \frac{\pi}{3} \cdot q} \\ &&&= \frac{q + \frac{1}{\sqrt3}}{1 - \frac{q}{\sqrt{3}}} \\ &&&= \frac{\sqrt3 q + 1}{\sqrt3-q} \end{align*}
    2. Since \(q \to -q\) reflects \((0,q)\) in the \(x\)-axis, \(f_3(q) = f_1(-q)\) so \(P_3\) is the reflection of \(P_1\) so it's rotation by \(\frac{\pi}{2}\) followed by reflection in the \(x\)-axis, which is reflection in \(y=x\). [ie \(\theta \to -\theta + \frac{\pi}{2} \to \frac{\pi}{2}-\theta\)]
    3. We are rotating by \(\frac{\pi}{3}\) then reflecting in \(y=x\) and then rotating by \(-\frac{\pi}{3}\), ie \(\theta \to \theta + \frac{\pi}{3} \to \frac{\pi}{6}-\theta \to -\theta -\frac{\pi}{6} \)
2024 Paper 3 Q4
D: 1500.0 B: 1500.0

  1. Show that if the acute angle between straight lines with gradients \(m_1\) and \(m_2\) is \(45^\circ\), then \[\frac{m_1 - m_2}{1 + m_1 m_2} = \pm 1.\]
The curve \(C\) has equation \(4ay = x^2\) (where \(a \neq 0\)).
  1. If \(p \neq q\), show that the tangents to the curve \(C\) at the points with \(x\)-coordinates \(p\) and \(q\) meet at a point with \(x\)-coordinate \(\frac{1}{2}(p+q)\). Find the \(y\)-coordinate of this point in terms of \(p\) and \(q\). Show further that any two tangents to the curve \(C\) which are at \(45^\circ\) to each other meet on the curve \((y+3a)^2 = 8a^2 + x^2\).
  2. Show that the acute angle between any two tangents to the curve \(C\) which meet on the curve \((y+7a)^2 = 48a^2 + 3x^2\) is constant. Find this acute angle.

2024 Paper 3 Q8
D: 1500.0 B: 1500.0

  1. Explain why the equation \((y - x + 3)(y + x - 5) = 0\) represents a pair of straight lines with gradients \(1\) and \(-1\). Show further that the equation \[y^2 - x^2 + py + qx + r = 0\] represents a pair of straight lines with gradients \(1\) and \(-1\) if and only if \(p^2 - q^2 = 4r\).
In the remainder of this question, \(C_1\) is the curve with equation \(x = y^2 + 2sy + s(s+1)\) and \(C_2\) is the curve with equation \(y = x^2\).
  1. Explain why the coordinates of any point which lies on both of the curves \(C_1\) and \(C_2\) also satisfy the equation \[y^2 + 2sy + s(s+1) - x + k(y - x^2) = 0\] for any real number \(k\). Given that \(s\) is such that \(C_1\) and \(C_2\) intersect at four distinct points, show that choosing \(k = 1\) gives an equation representing a pair of straight lines, with gradients \(1\) and \(-1\), on which all four points of intersection lie.
  2. Show that if \(C_1\) and \(C_2\) intersect at four distinct points, then \(s < -\frac{3}{4}\).
  3. Show that if \(s < -\frac{3}{4}\), then \(C_1\) and \(C_2\) intersect at four distinct points.

2022 Paper 2 Q5
D: 1500.0 B: 1500.0

  1. Given that \(a > b > c > 0\) are constants, and that \(x\), \(y\), \(z\) are non-negative variables, show that \[ax + by + cz \leqslant a(x + y + z).\]
In the acute-angled triangle \(ABC\), \(a\), \(b\) and \(c\) are the lengths of sides \(BC\), \(CA\) and \(AB\), respectively, with \(a > b > c\). \(P\) is a point inside, or on the sides of, the triangle, and \(x\), \(y\) and \(z\) are the perpendicular distances from \(P\) to \(BC\), \(CA\) and \(AB\), respectively. The area of the triangle is \(\Delta\).
    1. Find \(\Delta\) in terms of \(a\), \(b\), \(c\), \(x\), \(y\) and \(z\).
    2. Find both the minimum value of the sum of the perpendicular distances from \(P\) to the three sides of the triangle and the values of \(x\), \(y\) and \(z\) which give this minimum sum, expressing your answers in terms of some or all of \(a\), \(b\), \(c\) and \(\Delta\).
    1. Show that, for all real \(a\), \(b\), \(c\), \(x\), \(y\) and \(z\), \[(a^2+b^2+c^2)(x^2+y^2+z^2) = (bx-ay)^2 + (cy-bz)^2 + (az-cx)^2 + (ax+by+cz)^2.\]
    2. Find both the minimum value of the sum of the squares of the perpendicular distances from \(P\) to the three sides of the triangle and the values of \(x\), \(y\) and \(z\) which give this minimum sum, expressing your answers in terms of some or all of \(a\), \(b\), \(c\) and \(\Delta\).
  1. Find both the maximum value of the sum of the squares of the perpendicular distances from \(P\) to the three sides of the triangle and the values of \(x\), \(y\) and \(z\) which give this maximum sum, expressing your answers in terms of some or all of \(a\), \(b\), \(c\) and \(\Delta\).

2021 Paper 2 Q6
D: 1500.0 B: 1500.0

A plane circular road is bounded by two concentric circles with centres at point~\(O\). The inner circle has radius \(R\) and the outer circle has radius \(R + w\). The points \(A\) and \(B\) lie on the outer circle, as shown in the diagram, with \(\angle AOB = 2\alpha\), \(\tfrac{1}{3}\pi \leqslant \alpha \leqslant \tfrac{1}{2}\pi\) and \(0 < w < R\).

TikZ diagram
  1. Show that I cannot cycle from \(A\) to \(B\) in a straight line, while remaining on the road.
  2. I take a path from \(A\) to \(B\) that is an arc of a circle. This circle is tangent to the inner edge of the road, and has radius \(R + d\) (where \(d > w\)) and centre~\(O'\). My path is represented by the dashed arc in the above diagram. Let \(\angle AO'B = 2\theta\).
    1. Use the cosine rule to find \(d\) in terms of \(w\), \(R\) and \(\cos\alpha\).
    2. Find also an expression for \(\sin(\alpha - \theta)\) in terms of \(R\), \(d\) and \(\sin\alpha\).
    You are now given that \(\dfrac{w}{R}\) is much less than \(1\).
  3. Show that \(\dfrac{d}{R}\) and \(\alpha - \theta\) are also both much less than \(1\).
  4. My friend cycles from \(A\) to \(B\) along the outer edge of the road. Let my path be shorter than my friend's path by distance~\(S\). Show that \[ S = 2(R+d)(\alpha - \theta) + 2\alpha(w - d). \] Hence show that \(S\) is approximately a fraction \[ \frac{\sin\alpha - \alpha\cos\alpha}{\alpha(1 - \cos\alpha)} \cdot \frac{w}{R} \] of the length of my friend's path.

2016 Paper 1 Q5
D: 1484.0 B: 1516.0

  1. TikZ diagram
    The diagram shows three touching circles \(A\), \(B\) and \(C\), with a common tangent \(PQR\). The radii of the circles are \(a\), \(b\) and \(c\), respectively. Show that \[ \frac 1 {\sqrt b} = \frac 1 {\sqrt{a}} + \frac1{\sqrt{c}} \tag{\(*\)} \] and deduce that \[ 2\left(\frac1{a^2} + \frac1 {b^2} + \frac1 {c^2} \right) = \left(\frac1 a + \frac1 {b} + \frac1 {c} \right)^{\!2} . \tag{\(**\)} \]
  2. Instead, let \(a\), \(b\) and \(c\) be positive numbers, with \(b < c < a\), which satisfy \((**)\). Show that they also satisfy \((*)\).

Show Solution
  1. \(\,\)
    TikZ diagram
    Notice that \begin{align*} && (a+b)^2 &= PQ^2 + (a-b)^2 \\ \Rightarrow && PQ^2 &= 4ab \\ && (b+c)^2 &= QR^2 + (c-b)^2 \\ \Rightarrow && QR^2 &= 4bc \\ && (a+c)^2 &= PR^2 + (a-c)^2 \\ \Rightarrow && PR^2 &= 4ac \\ \Rightarrow && 2\sqrt{ac} &= 2\sqrt{ab}+2\sqrt{bc} \\ \Rightarrow && \frac{1}{\sqrt{b}} &= \frac{1}{\sqrt{c}} + \frac1{\sqrt{a}} \\ \end{align*} Let \(x, y, z = \frac{1}{\sqrt{a}}, \frac1{\sqrt{b}}, \frac{1}{\sqrt{z}}\) so we would like to prove that \(2(x^4+y^4+z^4) = (x^2+y^2+z^2)^2\) or \(x^4+y^4+z^4 = 2x^2y^2+2y^2z^2+2z^2x^2\). We also have \begin{align*} && y &= x+z \\ \Rightarrow &&y^2 &= x^2+z^2+2xz \\ \Rightarrow && (y^2-x^2-z^2)^2 &= 4x^2z^2 \\ \Rightarrow && y^4+x^4+z^4 - 2x^2y^2-2y^2z^2+2x^2z^2 &= 4x^2z^2\\ \Rightarrow && y^4+x^4+z^4 &= 2x^2y^2+2y^2z^2+2z^2x^2 \end{align*}
  2. Notice that subject to \(y > z > x\) all these steps are reversible, so we must have the equality we desire
2015 Paper 2 Q7
D: 1600.0 B: 1484.0

A circle \(C\) is said to be bisected by a curve \(X\) if \(X\) meets \(C\) in exactly two points and these points are diametrically opposite each other on \(C\).

  1. Let \(C\) be the circle of radius \(a\) in the \(x\)-\(y\) plane with centre at the origin. Show, by giving its equation, that it is possible to find a circle of given radius \(r\) that bisects \(C\) provided \(r > a\). Show that no circle of radius \(r\) bisects \(C\) if \(r\le a\,\).
  2. Let \(C_1\) and \(C_2\) be circles with centres at \((-d,0)\) and \((d,0)\) and radii \(a_1\) and \(a_2\), respectively, where \(d > a_1\) and \(d > a_2\). Let \(D\) be a circle of radius \(r\) that bisects both \(C_1\) and \(C_2\). Show that the \(x\)-coordinate of the centre of \(D\) is \(\dfrac{a_2^2 - a_1^2}{4d}\). Obtain an expression in terms of \(d\), \(r\), \(a_1\) and \(a_2\) for the \(y\)-coordinate of the centre of \(D\), and deduce that \(r\) must satisfy \[ 16r^2d^2 \ge \big (4d^2 +(a_2-a_1)^2\big) \, \big (4d^2 +(a_2+a_1)^2\big) \,. \]

Show Solution
  1. \(C\) has the equation \(x^2 + y^2 = a^2\). One suitable circle would ideally pass through \((0,a)\) and \((0,-a)\) have a centre on the positive \(x\)-axis, so we would need \(a^2+c^2 = r^2\) so \(c = \sqrt{r^2-a^2}\) and the equation would be \((x-\sqrt{r^2-a^2})^2 + y^2 = r^2\). Clearly a circle with radius \(r < a\) cannot pass through two diametrically opposed points of a circle radius \(a\), since the furthest two points can be on a circle is \(2r\), and diametrically opposed points are \(2a\) apart. Similarly if they are exactly the same radii, then if they pass through diametrically opposed points they must be the same circle.
  2. Let the centre of \(D\) be at \((x,y)\), then it must be a distance of \(\sqrt{r^2-a_i}\) from each circle centre, ie \begin{align*} && (x-d)^2+y^2 &= r^2-a_2^2 \\ && (x+d)^2 + y^2 &= r^2-a_1^2 \\ \Rightarrow && 4dx &= a_2^2 - a_1^2 \\ \Rightarrow && x &= \frac{a_2^2-a_1^2}{4d} \\ \Rightarrow && y^2 &= r^2-a_1^2 - \left (\frac{a_2^2-a_1^2}{4d}+d \right)^2 \\ &&&= r^2 - a_1^2 - \frac{(a_2^2-a_1^2+4d^2)^2}{16d^2} \\ &&&= \frac{16d^2r^2-16d^2a_1^2 - a_2^4-a_1^4-16d^4+2a_1^2a_2^2+8a_1^2d^2-8a_2^2d^2}{16d^2} \\ &&&= \frac{16d^2r^2-8d^2a_1^2 - a_2^4-a_1^4-16d^4+2a_1^2a_2^2-8a_2^2d^2}{16d^2} \\ \Rightarrow && y &= \pm \sqrt{ \frac{16d^2r^2-8d^2a_1^2 - a_2^4-a_1^4-16d^4+2a_1^2a_2^2-8a_2^2d^2}{16d^2}} \end{align*} and we need \begin{align*} && 0 &\leq 16d^2r^2-8d^2a_1^2 - a_2^4-a_1^4-16d^4+2a_1^2a_2^2-8a_2^2d^2 \\ \Rightarrow && 16d^2 d^2 &\geq 8d^2a_1^2 + a_2^4+a_1^4+16d^4+2a_1^2a_2^2+8a_2^2d^2 \\ &&&= (4d^2+(a_2-a_1)^2)(4d^2+(a_2+a_1)^2) \end{align*}
2013 Paper 2 Q4
D: 1600.0 B: 1484.0

The line passing through the point \((a,0)\) with gradient \(b\) intersects the circle of unit radius centred at the origin at \(P\) and \(Q\), and \(M\) is the midpoint of the chord \(PQ\). Find the coordinates of \(M\) in terms of \(a\) and \(b\).

  1. Suppose \(b\) is fixed and positive. As \(a\) varies, \(M\) traces out a curve (the locus of \(M\)). Show that \(x=- by\) on this curve. Given that \(a\) varies with \(-1\le a \le 1\), show that the locus is a line segment of length \(2b/(1+b^2)^\frac12\). Give a sketch showing the locus and the unit circle.
  2. Find the locus of \(M\) in the following cases, giving in each case its cartesian equation, describing it geometrically and sketching it in relation to the unit circle:
    • \(a\) is fixed with \(0 < a < 1\), and \(b\) varies with \(-\infty < b < \infty\);
    • \(ab=1\), and \(b\) varies with \(0< b\le1\).

Show Solution
\begin{align*} && y &= bx-ba \\ && 1 &= x^2 + y^2 \\ \Rightarrow && 1 &= x^2 + b^2(x-a)^2 \\ \Rightarrow && 0 &= (1+b^2)x^2-2ab^2x+b^2a^2-1 \end{align*} This will have roots which sum to \(\frac{2ab^2}{1+b^2}\), therefore \(M = \left ( \frac{ab^2}{1+b^2}, \frac{ab^3}{1+b^2}-ba \right)=\left ( \frac{ab^2}{1+b^2}, \frac{-ba}{1+b^2} \right)\)
  1. Since \(b\) is fixed so is \(\frac{b}{1+b^2} = t\) and all the points are \((bta, -ta)\), ie \(x = -by\). If \(a \in [-1,1]\) we are ranging on the points \((bt, -t)\) to \((-bt, t)\) which is a distance of \begin{align*} && d &= \sqrt{(bt+bt)^2+(-2t)^2} \\ &&&= \sqrt{4(b^2+1)t^2} \\ &&&=2 \sqrt{(b^2+1)\frac{b^2}{(b^2+1)^2}} \\ &&&= \frac{2b}{\sqrt{b^2+1}} \end{align*}
    TikZ diagram
    • If \(a\) is fixed we have \(\left ( \frac{ab^2}{1+b^2}, -\frac{ba}{1+b^2} \right)\) \begin{align*} && \frac{x}{y} &= - b \\ \Rightarrow && y &= \frac{a\frac{x}{y}}{1 + \frac{x^2}{y^2}} \\ \Rightarrow && y^2 \left ( 1 + \frac{x^2}{y^2} \right) &= ax \\ \Rightarrow && x^2-ax + y^2 &= 0 \\ \Rightarrow && \left (x - \frac{a}{2} \right)^2 + y^2 &= \frac{a^2}{4} \end{align*} Therefore we will end up with a circle centre \((\tfrac{a}{2}, 0)\) going through the origin.
      TikZ diagram
    • If \(ab = 1\), we have \(\left ( \frac{b}{1+b^2}, -\frac{1}{1+b^2} \right)\) \begin{align*} && \frac{x}{y} &= -b \\ \Rightarrow && y &= -\frac{1}{1+\frac{x^2}{y^2}} \\ \Rightarrow && y + \frac{x^2}{y} &= - 1 \\ \Rightarrow && y^2 +y+ x^2 &= 0 \\ \Rightarrow && \left ( y + \frac12 \right)^2 + x^2 &= \frac14 \end{align*}
      TikZ diagram
2005 Paper 1 Q6
D: 1500.0 B: 1490.2

  1. The point \(A\) has coordinates \(\l 5 \, , 16 \r\) and the point \(B\) has coordinates \(\l -4 \, , 4 \r\). The variable point \(P\) has coordinates \(\l x \, , y \r\,\) and moves on a path such that \(AP=2BP\). Show that the Cartesian equation of the path of \(P\) is \[ \displaystyle \l x+7 \r^2 + y^2 =100 \;. \]
  2. The point \(C\) has coordinates \(\l a \, , 0 \r\) and the point \(D\) has coordinates \(\l b \, , 0 \r\), where \(a\ne b\). The variable point \(Q\) moves on a path such that \[ QC = k \times QD\;, \] where \(k>1\,\). Given that the path of \(Q\) is the same as the path of \(P\), show that \[ \frac{a+7}{b+7}=\frac{a^2+51}{b^2+51}\;. \] Show further that \((a+7)(b+7)=100\,\).

Show Solution
  1. Since \(AP = 2BP\) we also have \(|AP|^2 = 4|BP|^2\) ie \begin{align*} && (x-5)^2 + (y-16)^2 &= 4(x+4)^2 + 4(y-4)^2 \\ \Rightarrow && x^2 - 10x+25 + y^2 -32y + 256 &= 4x^2+32x+64+4y^2-32y+64 \\ \Rightarrow && 281 &= 3x^2+42x+3y^2+128\\ && 281 &= 3(x+7)^2-147+3y^2+128 \\ \Rightarrow && 300 &= 3(x+7)^2 + 3y^2 \\ && 100 &= (x+7)^2 + y^2 \end{align*}
  2. Since \(|QC|^2 = k^2 |QD|^2\), \begin{align*} && (x-a)^2 + y^2 &= k^2 (x-b)^2 + k^2y^2 \\ \Rightarrow && x^2-2ax+a^2 &= k^2x^2-2k^2bx+k^2b^2 + (k^2-1)y^2 \\ && a^2-k^2b^2 &= (k^2-1)x^2-2(k^2b-a)x + (k^2-1)y^2 \\ && a^2-k^2b^2&= (k^2-1)\left(x-\frac{k^2b-a}{k^2-1}\right)^2-(k^2-1)\left(\frac{k^2b-a}{k^2-1}\right)^2+(k^2-1)y^2 \\ && \frac{a^2-k^2b^2}{k^2-1}+\left(\frac{k^2b-a}{k^2-1}\right)^2&= \left(x-\frac{k^2b-a}{k^2-1}\right)^2+y^2 \\ \Rightarrow && -7 &= \frac{k^2b-a}{k^2-1} \tag{*} \\ && 100 &= \frac{a^2-k^2b^2}{k^2-1}+\left(\frac{k^2b-a}{k^2-1}\right)^2 \\ &&&= \frac{a^2-k^2b^2}{k^2-1}+7^2 \\ \Rightarrow && 51 &= \frac{a^2-k^2b^2}{k^2-1} \tag{**} \\ (*) \Rightarrow && k^2(b+7)&= a+7 \\ (**) \Rightarrow && k^2(51+b^2)&= a^2+51 \\ \Rightarrow && \frac{a^2+51}{b^2+51} &= \frac{a+7}{b+7} \\ \\ \Rightarrow && a^2b+51b+7a^2 &= ab^2+51a+7b^2 \\ && 0 &= ab(b-a)-51(b-a)+7(b-a)(b+a) \\ &&&= (b-a)(ab+7(b+a)-51) \\ &&&= (b-a)((a+7)(b+7)-100) \\ \Rightarrow && 100 &= (a+7)(b+7) \end{align*} Since \(a \neq b\)
2005 Paper 2 Q5
D: 1600.0 B: 1470.1

The angle \(A\) of triangle \(ABC\) is a right angle and the sides \(BC\), \(CA\) and \(AB\) are of lengths \(a\), \(b\) and \(c\), respectively. Each side of the triangle is tangent to the circle \(S_1\) which is of radius \(r\). Show that \(2r = b+c-a\). Each vertex of the triangle lies on the circle~\(S_2\). The ratio of the area of the region between~\(S_1\) and the triangle to the area of \(S_2\) is denoted by \(R\,\). Show that $$ \pi R = -(\pi-1)q^2 + 2\pi q -(\pi+1) \;, $$ where \(q=\dfrac{b+c}a\,\). Deduce that $$ R\le \frac1 {\pi( \pi - 1)} \;. $$

2004 Paper 1 Q6
D: 1484.0 B: 1500.0

The three points \(A\), \(B\) and \(C\) have coordinates \(\l p_1 \, , \; q_1 \r\), \(\l p_2 \, , \; q_2 \r\) and \(\l p_3 \, , \; q_3 \r\,\), respectively. Find the point of intersection of the line joining \(A\) to the midpoint of \(BC\), and the line joining~\(B\) to the midpoint of \(AC\). Verify that this point lies on the line joining \(C\) to the midpoint of~\(AB\). The point \(H\) has coordinates \(\l p_1 + p_2 + p_3 \, , \; q_1 + q_2 + q_3 \r\,\). Show that if the line \(AH\) intersects the line \(BC\) at right angles, then \(p_2^2 + q_2^2 = p_3^2 + q_3^2\,\), and write down a similar result if the line \(BH\) intersects the line \(AC\) at right angles. Deduce that if \(AH\) is perpendicular to \(BC\) and also \(BH\) is perpendicular to \(AC\), then \(CH\) is perpendicular to \(AB\).

2003 Paper 2 Q4
D: 1600.0 B: 1484.0

The line \(y=d\,\), where \(d>0\,\), intersects the circle \(x^2+y^2=R^2\) at \(G\) and \(H\). Show that the area of the minor segment \(GH\) is equal to \begin{equation} R^2\arccos \left({d \over R}\right) -d\sqrt{R^2 - d^2}\;. \tag {\(*\)} \end{equation} In the following cases, the given line intersects the given circle. Determine how, in each case, the expression \((*)\) should be modified to give the area of the minor segment.

  1. Line: \(y=c\,\); \ \ \ circle: \((x-a)^2+(y-b)^2=R^2\,\).
  2. Line: \(y=mx+c\, \); \ \ \ circle: \(x^2+y^2=R^2\,\).
  3. Line: \(y=mx+c\,\); \ \ \ circle: \((x-a)^2+(y-b)^2=R^2\,\).

2002 Paper 1 Q1
D: 1516.0 B: 1500.0

Show that the equation of any circle passing through the points of intersection of the ellipse \((x+2)^2 +2y^2 =18\) and the ellipse \(9(x-1)^2 +16y^2 = 25\) can be written in the form \[ x^2-2ax +y^2 =5-4a\;. \]

Show Solution
\begin{align*} && (x+2)^2 +2y^2 &=18 \\ && 9(x-1)^2 +16y^2 &= 25 \\ \Rightarrow && 2y^2 &= 18 - (x+2)^2 \\ && 16y^2 &= 25 - 9(x-1)^2 \\ \Rightarrow && 25-9(x-1)^2 &= 8 \cdot 18 - 8(x+2)^2 \\ \Rightarrow && 25 -9+18x-9x^2 &= 144 -32- 32x +8x^2 \\ \Rightarrow && 0 &= 96 - 50x+x^2 \\ &&&= (x-48)(x-2) \\ \Rightarrow && x &= 2,48 \\ \Rightarrow && 2y^2 &= 2, 18-50^2 \\ \Rightarrow && (x,y) &= (2,\pm1) \end{align*} Therefore any circle must have it's centre on there perpendicular bisector of \((2, \pm 1)\), ie on the \(x\)-axis. Therefore it will have equation \((x-a)^2+y^2 = r^2\) and also contain the point \((2,1)\), therefore: \begin{align*} r^2 &= (2-a)^2 + 1^2 \\ &= 4 -2a+a^2 + 1 \\ &= 5-2a+a^2 \end{align*} and the equation is: \begin{align*} && (x-a)^2 + y^2 &= 5-4a+a^2 \\ \Rightarrow && x^2-2ax+a^2 +y^2 &= 5-4a+a^2 \\ \Rightarrow && x^2-2ax+y^2 &= 5-4a \end{align*} as required.
2002 Paper 1 Q6
D: 1500.0 B: 1500.0

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.

Show Solution
First let's consider the area of the base. It is an equilateral triangle with side length \(a\), so \(\frac12 a^2 \sin 60^\circ = \frac{\sqrt{3}}4a^2\).
TikZ diagram
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:
TikZ diagram
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*}
2001 Paper 1 Q1
D: 1516.0 B: 1500.0

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.

2000 Paper 3 Q1
D: 1700.0 B: 1485.5

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

\begin{align*} C_1: && x y & = 1 \\ C_2: && x^2-y^2 & = 2 \end{align*}
The curves intersect at \(P\) and \(Q\). Given that the coordinates of \(P\) are \((a,b)\) (which you need not evaluate), write down the coordinates of \(Q\) in terms of \(a\) and \(b\). The tangent to \(C_1\) through \(P\) meets the tangent to \(C_2\) through \(Q\) at the point \(M\), and the tangent to \(C_2\) through \(P\) meets the tangent to \(C_1\) through \(Q\) at \(N\). Show that the coordinates of \(M\) are \((-b,a)\) and write down the coordinates of \(N\). Show that \(PMQN\) is a square.

Show Solution
TikZ diagram
\(Q = (-a,-b)\) \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.
1999 Paper 1 Q2
D: 1484.0 B: 1468.7

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. $]

Show Solution
\begin{align*} && a^2 &= d_1^2 + d_2^2 + d_3^2 \\ &&&= (x-x_1)^2+(y-y_1)^2 + (x-x_2)^2+(y-y_2)^2 + (x-x_3)^2+(y-y_3)^2 \\ &&&= \sum (x-\bar{x}+\bar{x}-x_i)^2 + \sum (y-\bar{y}+\bar{y}-y_i)^2 \\ &&&= \sum \left ( (x-\bar{x})^2+(\bar{x}-x_i)^2 + 2(x-\bar{x})(\bar{x}-x_i) \right)+ \sum \left ( (y-\bar{y})^2+(\bar{y}-y_i)^2 + 2(y-\bar{y})(\bar{y}-y_i) \right)\\ &&&= 3(x-\bar{x})^2 + \sum (\bar{x}-x_i)^2 + 6x\bar{x} -6\bar{x}^2-2x\sum x_i+2\bar{x}\sum x_i + \\ &&&\quad\quad\quad 3(y-\bar{y})^2 + \sum (\bar{y}-y_i)^2 + 6y\bar{y} -6\bar{y}^2-2y\sum y_i+2\bar{y}\sum y_i \\ &&&= 3(x-\bar{x})^2 + \sum (\bar{x}-x_i)^2+3(y-\bar{y})^2 + \sum (\bar{y}-y_i)^2 \\ \\ \Rightarrow && (x-\bar{x})^2+(y-\bar{y})^2 &= \frac13\left ( a^2- \sum \left((\bar{x}-x_i)^2+(\bar{y}-y_i)^2 \right) \right) \end{align*} Therefore the locus is a circle, centre \((\bar{x}, \bar{y})\). radius \(\sqrt{\frac13(a^2 - \text{sum of squares distances of centroid to vertices}})\). \(a^2\) cannot be less than this distance, because clearly the right hand side is always bigger than it!
1996 Paper 2 Q7
D: 1600.0 B: 1484.0

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 \(P\).

1995 Paper 2 Q5
D: 1600.0 B: 1545.6

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 \(B\).
She can run \(k\) times as fast as she can swim and she wishes to reach her tea as fast as possible. Explain, with reasons, which of (i), (ii) and (iii) she should choose for each value of \(k\). Is there one choice from (i), (ii) and (iii) she will never take whatever the value of \(k\)?

1995 Paper 2 Q7
D: 1600.0 B: 1516.7

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.

TikZ diagram
  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 \(R\).

Show Solution
  1. \([AIB] = \frac12br\), \([BIC] = \frac12ar\), \([CIA] = \frac12 rc\), therefore \(\Delta = [AIB] +[BIC] + [CIA] = \frac12r(a+b+c) = sr\)
  2. \(\,\) \begin{align*} && \Delta &= \frac12 bc \sin \alpha \\ \Rightarrow && \Delta^2 &= \frac14 b^2c^2 \sin^2 \alpha \\ &&&= \frac14 \left (b^2c^2 - b^2c^2\cos^2 \alpha \right) \\ &&&= \frac1{16} \left (4b^2c^2 - (2bc\cos \alpha )^2\right) \\ \\ \Rightarrow && \Delta^2 &= \frac1{16} \left (4b^2c^2 - (b^2+c^2-a^2 )^2\right) \\ &&&= \frac1{16} (2bc-b^2-c^2+a^2)(2bc+b^2+c^2-a^2) \\ &&&= \frac{1}{16}(a^2-(b-c)^2)((b+c)^2-a^2) \\ &&&= \frac1{16}(a-b+c)(a+b-c)(b+c-a)(b+c+a) \\ &&&= (s - b)(s-c)(s-a)s \\ \Rightarrow && \Delta &= \sqrt{s(s-a)(s-b)(s-c)} \end{align*}
  3. We have the setting like this,
    TikZ diagram
    so \begin{align*} && h & = \sqrt{R^2-r^2} \\ &&&= \sqrt{R^2-\frac{\Delta^2}{s^2}} \\ &&&= \sqrt{R^2 - \frac{(s-a)(s-b)(s-c)}{s}} \end{align*}
1994 Paper 1 Q1
D: 1500.0 B: 1500.0

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.

Show Solution
TikZ diagram
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 \(\frac{1}{\cos \theta} \l 2p^2 + 4pq - 2p^2 \r = \frac{4pq}{\cos \theta}\)
1994 Paper 2 Q7
D: 1600.0 B: 1500.0

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.

1993 Paper 1 Q8
D: 1500.0 B: 1500.0

  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.

1989 Paper 1 Q1
D: 1500.0 B: 1500.0

TikZ diagram
In the above diagram, \(ABC,CDE,EFG\) and \(AHG\) are semicircles and \(A,C,E,G\) lie on a straight line. The radii of \(ABC,EFG,AHG\) are \(h\), \(h\) and \(r\) respectively, where \(2h < r\). Show that the area enclosed by \(ABCDEFGH\) is equal to that of a circle with diameter \(HD\). Each semicircle is now replaced by a portion of a parabola, with vertex at the midpoint of the semicircle arc, passing through the endpoints (so, for example, \(ABC\) is replaced by part of a parabola passing through \(A\) and \(C\) and with vertex at \(B\)). Find a formula in terms of \(r\) and \(h\) for the area enclosed by \(ABCDEFGH\).

Show Solution
\(AG = r\), therefore the area is: \begin{align*} A &= [AHG] - 2*[ABC] + [CDE] \\ &= \frac12 \pi r^2 - \pi h^2 + \frac12 \pi (r-2h)^2 \\ &= \frac12 \pi \l r^2 - 2h^2 + r^2 -4rh+4h^2 \r \\ &= \frac12 \pi \l 2r^2 -4rh + 2h^2\r \\ &= \pi (r-h)^2 \end{align*} This is the same area as a circle radius \(r-h\) But \(HD = r + (r-2d) = 2(r-d)\), ie the circle with diameter \(HD\) has radius \(r-h\) as required. Suppose \(A = (-h, 0), C = (h, 0), B = (0, h)\) then our parabola is \(y = \frac1{h}(h^2-x^2)\)
TikZ diagram
The area of \(ABC\) would then be \(\int_{-h}^h \frac{1}{h}(h^2-x^2) \d x = \frac1{h} \left [ h^2x - \frac{x^3}{3} \right] = \frac1{h} \l 2h^3-2\frac{h^3}{3} \r = \frac{4}{3}h^2\) so we have: \begin{align*} A &= [AHG] - 2*[ABC] +[CDE] \\ &= \frac43 r^2-\frac83h^2+\frac43(r-2h)^2 \\ &= \frac43 \l r^2 -2h^2+r^2-4rh+4h^2) \\ &= \frac43 (r-h)^2 \end{align*}