Question

Let two straight lines drawn from the origin $O$ intersect the line $3x + 4y = 12$ at the points $P$ and $Q$ such that $\triangle OPQ$ is an isosceles triangle and $\angle POQ = 90^\circ$. If $l = OP^2 + PQ^2 + QO^2$, then the greatest integer less than or equal to $l$ is:

• A. 44
• B. 48
• C. 46
• D. 42

Answer 1

Answer: (C)

46

Answer 2

The line equation:

$$3x + 4y = 12$$

intersects points $P$ and $Q$ at coordinates:

$$P(r \cos \theta, r \sin \theta), \quad Q(r \cos (90^\circ + \theta), r \sin (90^\circ + \theta)) = (-r \sin \theta, r \cos \theta)$$

Substituting into the line equation:

$$3(r \cos \theta) + 4(r \sin \theta) = 12$$ $$\implies r(3 \cos \theta + 4 \sin \theta) = 12 \tag{1}$$ $$3(r \sin \theta) + 4(r \cos \theta) = 12$$ $$\implies r(-3 \sin \theta + 4 \cos \theta) = 12 \tag{2}$$

Finding $r$:

Squaring and adding equations (1) and (2):

$$\left( \frac{12}{r} \right)^2 + \left( \frac{12}{r} \right)^2 = (3 \cos \theta + 4 \sin \theta)^2 + (-3 \sin \theta + 4 \cos \theta)^2$$ $$2 \left( \frac{12}{r} \right)^2 = 25$$ $$\implies r^2 = \frac{288}{25}$$ $$\implies r = \sqrt{\frac{288}{25}} = \frac{12 \sqrt{2}}{5}$$

Finding $\ell = OP^2 + PQ^2 + OQ^2$:

$$OP^2 = r^2, \quad OQ^2 = r^2$$

To find $PQ^2$, we use the distance formula between $P(r \cos \theta, r \sin \theta)$ and $Q(-r \sin \theta, r \cos \theta)$:

$$PQ^2 = (r \cos \theta + r \sin \theta)^2 + (r \sin \theta - r \cos \theta)^2$$ $$= r^2(\cos \theta + \sin \theta)^2 + r^2(\sin \theta - \cos \theta)^2$$ $$= r^2(1 + 1) = 2r^2$$

Thus:

$$\ell = OP^2 + PQ^2 + OQ^2 = r^2 + 2r^2 + r^2 = 4r^2$$

Substituting $r^2 = \frac{288}{25}$:

$$\ell = 4 \times \frac{288}{25} = \frac{1152}{25} = 46.08$$

Final Answer:

The greatest integer less than or equal to $\ell$ is:

$$\lfloor 46.08 \rfloor = 46$$