Question

Two points $P_1$ and $P_2$ are at distances $r_1$ and $r_2$ respectively from the origin $O$ and $OP_1$ and $OP_2$ makes angle $\theta_1$ and $\theta_2$ respectively with the x-axis. Let there be a point $P$ on $P_1P_2$ such that $OP$ makes an angle $\frac{\theta_2+\theta_1}{2}$ with the x-axis. Then $OP$ is:

• A. $\frac{2r_1r_2}{r_1+r_2}cos\frac{\theta_2-\theta_1}{2}$
• B. $\frac{2r_1r_2}{r_1+r_2}sin\frac{\theta_2-\theta_1}{2}$
• C. $\frac{r_1r_2}{r_1+r_2}cos\frac{\theta_2+\theta_1}{2}$
• D. $\frac{r_1r_2}{r_1+r_2}sin\frac{\theta_2+\theta_1}{2}$

Answer 1

Answer: (A)

$\frac{2r_1r_2}{r_1+r_2}cos\frac{\theta_2-\theta_1}{2}$

Answer 2

The problem asks for the length of the line segment $OP$, where $O$ is the origin, $P_1$ and $P_2$ are two points, and $P$ is a point on the line segment $P_1P_2$. We are given the distances of $P_1$ and $P_2$ from the origin as $r_1$ and $r_2$ respectively, and their angles with the x-axis as $\theta_1$ and $\theta_2$. The point $P$ is special because $OP$ makes an angle $\frac{\theta_2+\theta_1}{2}$ with the x-axis.

Let $r = OP$.

The coordinates of the points can be written in polar form:

$P_1 = (r_1, \theta_1)$

$P_2 = (r_2, \theta_2)$

$P = (r, \theta)$, where $\theta = \frac{\theta_1 + \theta_2}{2}$.

First, let's analyze the angles.

The angle that $OP_1$ makes with the x-axis is $\theta_1$.

The angle that $OP_2$ makes with the x-axis is $\theta_2$.

The angle that $OP$ makes with the x-axis is $\frac{\theta_1 + \theta_2}{2}$.

Consider the angle $\angle P_1OP$. This angle is the difference between the angle of $OP$ and the angle of $OP_1$:

$\angle P_1OP = \frac{\theta_1 + \theta_2}{2} - \theta_1 = \frac{\theta_2 - \theta_1}{2}$.

Consider the angle $\angle P_2OP$. This angle is the difference between the angle of $OP_2$ and the angle of $OP$:

$\angle P_2OP = \theta_2 - \frac{\theta_1 + \theta_2}{2} = \frac{\theta_2 - \theta_1}{2}$.

Since $\angle P_1OP = \angle P_2OP = \frac{\theta_2 - \theta_1}{2}$, the line segment $OP$ is the angle bisector of $\angle P_1OP_2$.

Now, consider the triangle $\triangle OP_1P_2$. The point $P$ lies on the side $P_1P_2$, and $OP$ is the angle bisector of $\angle P_1OP_2$.

We can use the property that the area of $\triangle OP_1P_2$ is the sum of the areas of $\triangle OP_1P$ and $\triangle OP_2P$.

Area of a triangle with two sides $a, b$ and included angle $\phi$ is $\frac{1}{2}ab\sin\phi$.

1. Area($\triangle OP_1P_2$) = $\frac{1}{2} OP_1 \cdot OP_2 \sin(\angle P_1OP_2)$

   $\angle P_1OP_2 = \theta_2 - \theta_1$ (assuming $\theta_2 > \theta_1$, otherwise it's $|\theta_2 - \theta_1|$ or $2\pi - |\theta_2 - \theta_1|$ depending on the convention, but $\sin(\theta)$ is same as $\sin(-\theta)$ for this context).

   Area($\triangle OP_1P_2$) = $\frac{1}{2} r_1 r_2 \sin(\theta_2 - \theta_1)$.

2. Area($\triangle OP_1P$) = $\frac{1}{2} OP_1 \cdot OP \sin(\angle P_1OP)$

   Area($\triangle OP_1P$) = $\frac{1}{2} r_1 r \sin\left(\frac{\theta_2 - \theta_1}{2}\right)$.

3. Area($\triangle OP_2P$) = $\frac{1}{2} OP_2 \cdot OP \sin(\angle P_2OP)$

   Area($\triangle OP_2P$) = $\frac{1}{2} r_2 r \sin\left(\frac{\theta_2 - \theta_1}{2}\right)$.

Now, set up the area equation:

Area($\triangle OP_1P_2$) = Area($\triangle OP_1P$) + Area($\triangle OP_2P$)

$\frac{1}{2} r_1 r_2 \sin(\theta_2 - \theta_1) = \frac{1}{2} r_1 r \sin\left(\frac{\theta_2 - \theta_1}{2}\right) + \frac{1}{2} r_2 r \sin\left(\frac{\theta_2 - \theta_1}{2}\right)$

Cancel $\frac{1}{2}$ from all terms:

$r_1 r_2 \sin(\theta_2 - \theta_1) = r_1 r \sin\left(\frac{\theta_2 - \theta_1}{2}\right) + r_2 r \sin\left(\frac{\theta_2 - \theta_1}{2}\right)$

Factor out $r \sin\left(\frac{\theta_2 - \theta_1}{2}\right)$ from the right side:

$r_1 r_2 \sin(\theta_2 - \theta_1) = r(r_1 + r_2) \sin\left(\frac{\theta_2 - \theta_1}{2}\right)$

Use the trigonometric identity $\sin(2A) = 2 \sin A \cos A$. Let $A = \frac{\theta_2 - \theta_1}{2}$.

So, $\sin(\theta_2 - \theta_1) = 2 \sin\left(\frac{\theta_2 - \theta_1}{2}\right) \cos\left(\frac{\theta_2 - \theta_1}{2}\right)$.

Substitute this into the equation:

$r_1 r_2 \left[2 \sin\left(\frac{\theta_2 - \theta_1}{2}\right) \cos\left(\frac{\theta_2 - \theta_1}{2}\right)\right] = r(r_1 + r_2) \sin\left(\frac{\theta_2 - \theta_1}{2}\right)$

Assuming $\sin\left(\frac{\theta_2 - \theta_1}{2}\right) \neq 0$ (i.e., $\theta_1 \neq \theta_2$), we can cancel it from both sides:

$2 r_1 r_2 \cos\left(\frac{\theta_2 - \theta_1}{2}\right) = r(r_1 + r_2)$

Solve for $r$:

$r = \frac{2 r_1 r_2 \cos\left(\frac{\theta_2 - \theta_1}{2}\right)}{r_1 + r_2}$

This matches option (A).