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2012 Paper 1 Q6
D: 1516.0 B: 1484.0
trigonometry angle of elevation cotangent trigonometric identity inequality geometric proof flagpole circle geometry product-to-sum formulas

A thin circular path with diameter \(AB\) is laid on horizontal ground. A vertical flagpole is erected with its base at a point \(D\) on the diameter \(AB\). The angles of elevation of the top of the flagpole from \(A\) and \(B\) are \(\alpha\) and \(\beta\) respectively (both are acute). The point \(C\) lies on the circular path with \(DC\) perpendicular to \(AB\) and the angle of elevation of the top of the flagpole from \(C\) is \(\phi\). Show that \(\cot\alpha\cot \beta = \cot^2\phi\). Show that, for any \(p\) and \(q\), \[ \cos p \cos q \sin^2\tfrac12(p+q) - \sin p\sin q \cos^2 \tfrac12 (p+q) = \tfrac12 \cos(p+q) -\tfrac12 \cos(p+q)\cos(p-q) .\] Deduce that, if \(p\) and \(q\) are positive and \( p+q \le \tfrac12 \pi\), then \[ \cot p\cot q\, \ge \cot^2 \tfrac12(p+q) \, \] and hence show that \(\phi \le \tfrac12(\alpha+\beta)\) when \( \alpha +\beta \le \tfrac12 \pi\,\).

Solution:

TikZ diagram
\begin{align*} && \cot \alpha &= \frac{AD}{h} \\ && \cot \beta &= \frac{BD}{h} \\ && \cot \phi &= \frac{DC}h \\ && CD^2 &= AB \cdot BD \tag{intersecting chords} \\ \Rightarrow && \cot^2 \phi &= \cot \alpha \cot \beta \end{align*} \begin{align*} && LHS &= \cos p \cos q \sin^2\tfrac12(p+q) - \sin p\sin q \cos^2 \tfrac12 (p+q) \\ &&&= \cos p \cos q \left ( \frac{1-\cos(p+q)}{2} \right) - \sin p\sin q \left (\frac{1+\cos(p+q)}{2} \right) \\ &&&= \frac12 \left (\cos p \cos q(1-\cos(p+q)) - \sin p\sin q (1+\cos(p+q)) \right) \\ &&&= \frac12 \left ((\cos p \cos q- \sin p\sin q) - (\cos p \cos q+ \sin p\sin q)\cos(p+q) \right) \\ &&&= \frac12 \left (\cos(p+q) - \cos (p-q)\cos(p+q) \right) \\ &&&= RHS \end{align*} Therefore \begin{align*} \cot p \cot q -\cot^2 \tfrac12 (p+q) &= \frac{\tfrac12 \cos(p+q) -\tfrac12 \cos(p+q)\cos(p-q)}{\sin p \sin q \sin^2 \tfrac12(p+q)} \\ &=\frac{\cos(p+q)(1-\cos(p-q))}{\sin p \sin q \sin^2 \tfrac12(p+q)} \end{align*} Since \(p,q\) are acute, the denominator is positive. Since \(p+q \leq \frac{\pi}{2}\), we have \(\cos(p+q) \geq 0\). Also \((1-\cos(p-q)) \geq 0\). Thus, the expression is \(\geq 0\). So we must have \begin{align*} && \cot^2 \phi &= \cot \alpha \cot \beta \\ &&&\geq \cot^2 \tfrac12(\alpha+\beta) \end{align*} Since \(\cot\) is decreasing on \((0, \tfrac12 \pi)\) we can deduce \(\phi \leq \tfrac12 (\alpha+\beta)\)

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1990 Paper 2 Q2
D: 1600.0 B: 1500.0
trigonometry product-to-sum formulas circle theorems Ptolemy's theorem inscribed angle theorem geometric proof sine rule cyclic quadrilateral

Prove that if \(A+B+C+D=\pi,\) then \[ \sin\left(A+B\right)\sin\left(A+D\right)-\sin B\sin D=\sin A\sin C. \] The points \(P,Q,R\) and \(S\) lie, in that order, on a circle of centre \(O\). Prove that \[ PQ\times RS+QR\times PS=PR\times QS. \]

Solution: \begin{align*} \sin(A+B)\sin(A+D) - \sin B \sin D &= \sin (A+B)\sin(\pi - B-C) - \sin B \sin (\pi - A - B - C) \\ &= \sin (A+B)\sin(B+C) - \sin B \sin(A+B+C) \\ &= \sin(A+B)\sin (B+C) - \sin B (\sin (A+B)\cos C +\cos(A+B) \sin C) \\ &= \sin(A+B)\cos B \sin C + \cos(A+B)\sin B \sin C \\ &= \sin A \sin C \cos^2 B + \cos A \sin B \cos B \sin C - \cos A \cos B \sin B \sin C + \sin A \sin^2 B \sin C \\ &= \sin A \sin C (\cos^2 B + \sin^2 B) \\ &= \sin A \sin C \end{align*}

TikZ diagram
Using the extended form of the sine rule: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\) where \(R\) is the circumradius, we have \begin{align*} PR \times QS &= 2R \sin (A+D) \times 2R \sin (A+B) \\ &= 4R^2 \l \sin A \sin C + \sin B \sin D \r \\ &= 2R \sin A \times 2R \sin C + 2R \sin B 2R \sin D \\ &= PS \times QR + PQ \times RS \end{align*} as required.

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