Problems

A cyclic quadrilateral \(ABCD\) has sides \(AB\), \(BC\), \(CD\) and \(DA\) of lengths \(a\), \(b\), \(c\) and \(d\), respectively. The area of the quadrilateral is \(Q\), and angle \(DAB\) is \(\theta\). Find an expression for \(\cos\theta\) in terms of \(a\), \(b\), \(c\) and \(d\), and an expression for \(\sin\theta\) in terms of \(a\), \(b\), \(c\), \(d\) and \(Q\). Hence show that \[ 16Q^2 = 4(ad+bc)^2 - (a^2+d^2-b^2-c^2)^2 \,, \] and deduce that \[ Q^2 = (s-a)(s-b)(s-c)(s-d)\,, \] where \(s= \frac12(a+b+c+d)\). Deduce a formula for the area of a triangle with sides of length \(a\), \(b\) and \(c\).

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Solution:

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\begin{align*} && BD^2 &= a^2+d^2 - 2ad \cos \theta \\ && BD^2 &= b^2+c^2-2bc \cos (\pi - \theta) \\ \Rightarrow && a^2+d^2 - 2ad \cos \theta &= b^2+c^2+2bc \cos \theta \\ \Rightarrow && 2(ad+bc)\cos \theta &= a^2+d^2-b^2-c^2 \\ \Rightarrow && \cos \theta &= \frac{a^2+d^2-b^2-c^2}{2(ad+bc)} \\ \\ && Q &= \frac12 ad \sin \theta + \frac12 bc \sin (\pi - \theta) \\ &&&= \frac12 (ad+bc) \sin \theta \\ \Rightarrow && \sin \theta &= \frac{2Q}{ad+bc} \\ \\ && 1 &= \sin^2 \theta + \cos^2 \theta \\ &&&= \frac{4Q^2}{(ad+bc)^2} + \frac{(a^2+d^2-b^2-c^2)^2}{4(ad+bc)^2} \\ \Rightarrow && 4(ad+bc)^2 &= 16Q^2 + (a^2+d^2-b^2-c^2)^2 \\ \Rightarrow && 16Q^2 &= 4(ad+bc)^2- (a^2+d^2-b^2-c^2)^2 \\ \Rightarrow && 16Q^2 &= (2ad+2bc - a^2-d^2+b^2+c^2)(2ad+2bc+a^2+d^2-b^2-c^2) \\ &&&= ((b+c)^2-(a-d)^2)((a+d)^2-(b-c)^2) \\ &&&= (b+c-a+d)(b+c+a-d)(a+d+b-c)(a+d-b+c) \\ \Rightarrow && Q^2 &= (s-a)(s-b)(s-c)(s-d) \end{align*} Since all triangles are cyclic, we can place \(D\) at the same point as \(A\) to obtain Heron's formula \(A = \sqrt{s(s-a)(s-b)(s-c)}\) where \(s = \frac12(a+b+c)\)

1. If \(x+y+z=\alpha,\) \(xy+yz+zx=\beta\) and \(xyz=\gamma,\) find numbers \(A,B\) and \(C\) such that \[ x^{3}+y^{3}+z^{3}=A\alpha^{3}+B\alpha\beta+C\gamma. \] Solve the equations \begin{alignat*}{1} x+y+z & =1\\ x^{2}+y^{2}+z^{2} & =3\\ x^{3}+y^{3}+z^{3} & =4. \end{alignat*}
2. The area of a triangle whose sides are \(a,b\) and \(c\) is given by the formula \[ \mathrm{area}=\sqrt{s(s-a)(s-b)(s-c)} \] where \(s\) is the semi-perimeter \(\frac{1}{2}(a+b+c).\) If \(a,b\) and \(c\) are the roots of the equation \[ x^{3}-16x^{2}+81x-128=0, \] find the area of the triangle.

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 \(R\).