Let A, B be points on the two half-lines , at a distance of from their point of intersection P. The line segment AB meets the angle bisector of the given half-lines at the point Q. If and R is the radius of the circumcircle of PAB, then is equal to __

Question

Let A, B be points on the two half-lines $x - \sqrt{3} | y | = α$ , $α > 0$ at a distance of $α$ from their point of intersection P. The line segment AB meets the angle bisector of the given half-lines at the point Q. If $P Q = \frac{9}{2}$ and R is the radius of the circumvircle of $△$ PAB, then $\frac{α^{2}}{R}$ is equal to ________. [2026]

Smart Achievers · Accepted Answer

(9)

Two half-line are $x - \sqrt{3} | y | = α$ , $α > 0$ , meeting at point P.

Points A and B are on these half-lines and each is at distance $α$ from P.

$∴$ Lines are $x - \sqrt{3} | y | = α$ , $α > 0$ , and $x + \sqrt{3} y = α (y < 0)$

Their slopes are $\frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}$

So, angle between them is 60°.

Since PA = PB = $α$ ,

$∴$ $△$ PAB is isosceles triangle with vertex angle 60°.

Hence, PAB is an equilateral triangle

FIGURE

$P Q = \frac{\sqrt{3} α}{2} = \frac{9}{2}$ [ $∵$ Given]

$∴ α = \frac{9}{\sqrt{3}} \Rightarrow α = 3 \sqrt{3}$

Now, $\frac{α}{2 R} = \cos 30 ° \Rightarrow α \frac{\sqrt{3}}{2} \cdot 2 R$

$\Rightarrow α = \sqrt{3} R \Rightarrow \sqrt{3} R = 3 \sqrt{3}$ [ $∵ α = 3 \sqrt{3}$ ]

$\Rightarrow R = 3$

So, $\frac{α^{2}}{R} = \frac{{(3 \sqrt{3})}^{2}}{3} = \frac{27}{3} = 9$

Solution

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