1990 Paper 2 Q2

Year: 1990
Paper: 2
Question Number: 2

Course: LFM Pure
Section: Trigonometry 2

Difficulty: 1600.0 Banger: 1500.0

trigonometry product-to-sum formulas circle theorems Ptolemy's theorem inscribed angle theorem geometric proof sine rule cyclic quadrilateral

Problem

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*}

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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Difficulty Rating: 1600.0

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Banger Rating: 1500.0

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Problem source

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 source

\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*}

\begin{center}
    \begin{tikzpicture}
        \draw (0,0) circle (2);

        \coordinate (P) at ({2*cos(30)},{2*sin(30)});
        \coordinate (Q) at ({2*cos(100)},{2*sin(100)});
        \coordinate (R) at ({2*cos(220)},{2*sin(220)});
        \coordinate (S) at ({2*cos(290)},{2*sin(290)});

        \filldraw (0,0) circle (1pt) node[left] {$O$};
        \filldraw (P) circle (1pt) node[right] {$P$};
        \filldraw (Q) circle (1pt) node[above] {$Q$};
        \filldraw (R) circle (1pt) node[left] {$R$};
        \filldraw (S) circle (1pt) node[below] {$S$};

        \draw (P) -- (Q) -- (R) -- (S) -- cycle;
        \draw (P) -- (R);
        \draw (Q) -- (S);

        \pic [draw, angle radius=0.8cm, "$B$"] {angle = P--R--Q};
        \pic [draw, angle radius=0.8cm, "$B$"] {angle = P--S--Q};
        \pic [draw, angle radius=0.8cm, "$A$"] {angle = S--R--P};
        \pic [draw, angle radius=0.8cm, "$A$"] {angle = S--Q--P};
        \pic [draw, angle radius=0.8cm, "$D$"] {angle = R--Q--S};
        \pic [draw, angle radius=0.8cm, "$C$"] {angle = Q--S--R};
        % \pic [draw, angle radius=0.8cm, "$D$"] {angle = R--Q--S};

        % \filldraw (0,0) circle (1pt) node[right] {$Q$};
        % \filldraw (0,0) circle (1pt) node[right] {$R$};
        % \filldraw (0,0) circle (1pt) node[right] {$S$};
    \end{tikzpicture} 
\end{center}

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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