Question 71

A straight line through point P of a triangle PQR intersects the side QR at the point S and the circumcircle of the triangle PQR at the point T. lf S is not the centre of the circumcircle, then which of the following is true?

Solution

Using properties of secant, $$PS \times ST = QS \times SR$$ -------------Eqn(I)

Also, for two numbers, $$PS$$ and $$ST$$, we know that harmonic mean is less than geometric mean.

=> $$\frac{2}{\frac{1}{PS} + \frac{1}{ST}} < \sqrt{PS \times ST}$$

=> $$\frac{1}{PS} + \frac{1}{ST} > \frac{2}{\sqrt{PS \times ST}}$$

Using Eqn(I)

=> $$\frac{1}{PS} + \frac{1}{ST} > \frac{2}{\sqrt{QS \times SR}}$$ ------Eqn(II)

Also, for two numbers, $$QS$$ and $$SR$$, geometric mean is less than arithmetic mean.

=> $$\sqrt{QS \times SR} < \frac{QS + SR}{2}$$

=> $$\frac{1}{\sqrt{QS \times SR}} > \frac{2}{QR}$$  $$(\because QS + SR = QR)$$

Multiplying both sides by $$2$$

=> $$\frac{2}{\sqrt{QS \times SR}} > \frac{4}{QR}$$ ----------Eqn(III)

From eqn(II) and (III)

$$\therefore \frac{1}{PS} + \frac{1}{ST} > \frac{4}{QR}$$