Question

Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0,75).$ Let $O$ and $P$ be two points on the plane with $OP=200.$ Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measure of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR\le 100$ is equal to

• A. $\frac{16}{23}$
• B. $\frac{18}{25}$
• C. $\frac{16}{25}$
• D. $\frac{6}{25}$

Answer 1

The correct option is C $\frac{16}{25}$

$\angle OQP=\angle ORP={90}^{\circ }$
We can draw a semi-circle with diameter $OP$ and points $Q$ and $R$ on the semi-circle.
Radius of semicircle $=100$
Let $QR=x\le 100.$


From figure,
$OQ=200\cos a,PQ=200\sin a$
$OR=200\cos b,PR=200\sin b$
Now, we observe that quadrilateral $OQRP$ is a cyclic quadrilateral.
So, by Ptolemy's theorem,
$200x+\left(200\cos a\right)\left(200\sin b\right)=\left(200\sin a\right)\left(200\cos b\right)$
$\Rightarrow x=|200\sin (a-b)|\le 100$
$\Rightarrow |\sin (a-b)|\le \frac{1}{2}$
$\Rightarrow |a-b|\le 30$
$\Rightarrow -30\le (a-b)\le 30$
Given, $0<a,b<75$

By geometry, probability
$=\frac{\text{Favourable area}}{\text{Total area}}$
$=\frac{75\times 75-45\times 45}{75\times 75}$
$=1-\frac{9}{25}$
$\therefore$ Required probability $=\frac{16}{25}$