Let A,B be points on the two half-lines x-√3|y|= , 0 at a distance of from their point of…

Question

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

Answer: 9

Accepted Answer

The two half-lines are given by

$x-\sqrt{3}|y|=\alpha ,\, \alpha >0$

We will first understand their geometry.

1. Find the two half-lines and their point of intersection
Since $|y|$ is present, we split into two cases:

Case 1: $y\ge 0$
$x-\sqrt{3}y=\alpha $

Case 2: $y\le 0$
$x+\sqrt{3}y=\alpha $

These are two lines meeting at the point $P$.

To find their intersection, put $y=0$ in either equation:

$x=\alpha $

So,

$P=(\alpha ,0)$

Thus the two half-lines start from $P$ and go outward along these two lines.

2. Find points $A$ and $B$
Points $A$ and $B$ lie on these two half-lines and are each at distance $\alpha $ from $P$.

Let $A$ be on the upper line $x-\sqrt{3}y=\alpha $ and $B$ on the lower line $x+\sqrt{3}y=\alpha $.

The slope of the upper line is

$y=\frac{x-\alpha }{\sqrt{3}}\Rightarrow m=\frac{1}{\sqrt{3}}$

So this line makes angle ${30}^{\circ }$ with the positive $x$-axis.

Similarly, the lower line makes angle $-{30}^{\circ }$ with the positive $x$-axis.

Hence the angle between the two half-lines is

${60}^{\circ }$

Since $PA=\alpha $ and $PB=\alpha $, the points are obtained by moving distance $\alpha $ from $P=(\alpha ,0)$ along directions $\pm {30}^{\circ }$.

Therefore,

$A=(\alpha +\alpha \cos {30}^{\circ },\alpha \sin {30}^{\circ })$

$B=(\alpha +\alpha \cos {30}^{\circ },-\alpha \sin {30}^{\circ })$

Using $\cos {30}^{\circ }=\frac{\sqrt{3}}{2}$ and $\sin {30}^{\circ }=\frac{1}{2}$,

$A=(\alpha +\frac{\sqrt{3}\alpha }{2},\frac{\alpha }{2})$

$B=(\alpha +\frac{\sqrt{3}\alpha }{2},-\frac{\alpha }{2})$

3. Find the angle bisector and point $Q$
The figure is symmetric about the $x$-axis, so the angle bisector of the two half-lines is the positive $x$-axis.

Also, $A$ and $B$ have the same $x$-coordinate, so $AB$ is a vertical segment.

Hence $AB$ meets the $x$-axis at

$Q=(\alpha +\frac{\sqrt{3}\alpha }{2},0)$

So,

$PQ=(\alpha +\frac{\sqrt{3}\alpha }{2})-\alpha =\frac{\sqrt{3}\alpha }{2}$

Given that

$PQ=\frac{9}{2}$

therefore,

$\frac{\sqrt{3}\alpha }{2}=\frac{9}{2}$

So,

$\sqrt{3}\alpha =9$

$\alpha =\frac{9}{\sqrt{3}}=3\sqrt{3}$

Hence,

${\alpha }^{2}=(3\sqrt{3}{)}^{2}=27$

4. Find circumradius $R$ of $	rianglePAB$
In $	rianglePAB$,

$PA=PB=\alpha $

and the included angle

$\angleAPB={60}^{\circ }$

So the triangle is actually equilateral, because

$A{B}^{2}={\alpha }^{2}+{\alpha }^{2}-2{\alpha }^{2}\cos {60}^{\circ }=2{\alpha }^{2}-{\alpha }^{2}={\alpha }^{2}$

Thus,

$AB=\alpha $

Hence $	rianglePAB$ is equilateral with side $\alpha $.

For an equilateral triangle of side $a$, circumradius is

$R=\frac{a}{\sqrt{3}}$

So here,

$R=\frac{\alpha }{\sqrt{3}}$

Using $\alpha =3\sqrt{3}$,

$R=\frac{3\sqrt{3}}{\sqrt{3}}=3$

5. Compute $\frac{{\alpha }^{2}}{R}$
$\frac{{\alpha }^{2}}{R}=\frac{27}{3}=9$

Therefore, the required value is

$9$

Solutions · 1