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STEP II 1990 question 2 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1990 question 2 solution

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Using the product to sum and compound angle formulae:

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This is because cos(x) is an even function.

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For the second part, draw the cyclic quadrilateral with vertices P, Q, R, S, and also the diagonals PR and QS.

Let angle(PRQ) = angle(PSQ) = A, angle(QSR) = angle(QPR) = B, angle(RQS) = angle(RPS) = C, angle(SRP) = angle(SQP) = D. Each pair of angles is equal as they are angles subtended by the same arc.

It is clear that A + B + C + D = \pi, as the angles make up the angles of a triangle.

\frac{PQ}{\sin A} = \frac{QR}{\sin B} = \frac{RS}{\sin C} = \frac{SR}{\sin D} = \frac{PR}{\sin (A + B)} = \frac{QS}{\sin (A + D)} = 2R

This is a result of the (full) Sine Rule. R is the radius of the circle.

Using the result of the first part, you get:

\sin A\sin C + \sin B\sin D = \sin (A + B)\sin (A + D)

\Rightarrow \frac{PQ . RS}{4R^{2}} + \frac{QR . SP}{4R^{2}} = \frac{PR . QS}{4R^{2}}

\Rightarrow PQ . RS + QR . SP = PR . QS

Solution by Dystopia.

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