Key Concepts and Formulas

• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
• Homogenization: To find the equation of a pair of straight lines passing through the origin and intersecting a given line, we can homogenize the equation of a conic (or any other curve) using the equation of the line.
• Perpendicular Lines: The condition for the lines $ax^2 + 2hxy + by^2 = 0$ to be perpendicular is $a + b = 0$.
• Pythagorean Theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Step-by-Step Solution

Step 1: Represent the lines OP and OQ

Let the equation of the line $3x + 4y = 12$ be $L = 0$. We want to find the equation of two lines passing through the origin and intersecting $L = 0$ at points P and Q. Since OP and OQ are perpendicular, we can represent them using a homogeneous equation of degree 2.

Step 2: Homogenize the equation

The equation of the line is $3x + 4y = 12$, so we can write $1 = \frac{3x + 4y}{12}$. Let the equation of the pair of lines OP and OQ be $y = m_1x$ and $y = m_2x$, where $m_1m_2 = -1$ since OP and OQ are perpendicular. We can represent the pair of straight lines as $ax^2 + 2hxy + by^2 = 0$. We want to homogenize this using the equation of the line $3x + 4y = 12$. Since we don't have a specific curve to homogenize, we can consider a general second-degree equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$. However, the problem statement implies there's an easier route. We want a pair of lines through the origin intersecting $3x+4y=12$. The key is that $\triangle OPQ$ is right-angled isosceles at $O$, so $OP = OQ$ and $OP \perp OQ$.

Let's consider the pair of lines $y = m x$ and $y = -\frac{1}{m} x$. These lines are perpendicular. Let the combined equation be $(y - mx)(y + \frac{1}{m} x) = 0$. Then $y^2 - mx y + \frac{1}{m} xy - x^2 = 0$, or $my^2 - m^2xy + xy - mx^2 = 0$, so $m x^2 + (m^2 - 1) xy - my^2 = 0$.

Instead, let the combined equation of OP and OQ be $ax^2 + 2hxy + by^2 = 0$. Since OP and OQ are perpendicular, $a + b = 0$, so $b = -a$. The equation becomes $ax^2 + 2hxy - ay^2 = 0$, or $x^2 + \frac{2h}{a}xy - y^2 = 0$. Let $\frac{2h}{a} = k$, so $x^2 + kxy - y^2 = 0$. Now, homogenize $x^2 + kxy - y^2 = 0$ using $3x + 4y = 12$, or $\frac{3x + 4y}{12} = 1$. This doesn't seem like the right approach.

Let $P = (x_1, y_1)$ and $Q = (x_2, y_2)$. Since $P$ and $Q$ lie on the line $3x + 4y = 12$, we have $3x_1 + 4y_1 = 12$ and $3x_2 + 4y_2 = 12$. Also, since $\angle POQ = 90^\circ$, the slopes of OP and OQ multiply to -1. Thus, $\frac{y_1}{x_1} \cdot \frac{y_2}{x_2} = -1$, so $y_1y_2 = -x_1x_2$. Since $OP = OQ$, $x_1^2 + y_1^2 = x_2^2 + y_2^2$.

Step 3: Calculate OP², OQ², and PQ²

We have $OP^2 = x_1^2 + y_1^2$ and $OQ^2 = x_2^2 + y_2^2$. Since $OP = OQ$, $OP^2 = OQ^2$. Also, $PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 = x_1^2 + x_2^2 - 2x_1x_2 + y_1^2 + y_2^2 - 2y_1y_2 = (x_1^2 + y_1^2) + (x_2^2 + y_2^2) - 2(x_1x_2 + y_1y_2) = 2OP^2 - 2(x_1x_2 + y_1y_2)$. Since $y_1y_2 = -x_1x_2$, $PQ^2 = 2OP^2 - 2(x_1x_2 - x_1x_2) = 2OP^2$.

Step 4: Find l

We are given $l = OP^2 + PQ^2 + OQ^2 = OP^2 + 2OP^2 + OP^2 = 4OP^2$. We want to minimize $OP^2$. Let $P = (x_1, y_1)$. Then $OP^2 = x_1^2 + y_1^2$. We want to minimize $x_1^2 + y_1^2$ subject to $3x_1 + 4y_1 = 12$. Using Cauchy-Schwarz inequality, $(3^2 + 4^2)(x_1^2 + y_1^2) \ge (3x_1 + 4y_1)^2$, so $25(x_1^2 + y_1^2) \ge 12^2 = 144$, so $x_1^2 + y_1^2 \ge \frac{144}{25}$. Thus, $OP^2 \ge \frac{144}{25}$. Therefore, $l = 4OP^2 \ge 4 \cdot \frac{144}{25} = \frac{576}{25} = 23.04$. This doesn't match the given answer.

Since $OP = OQ$ and $\angle POQ = 90^\circ$, we have $PQ^2 = OP^2 + OQ^2 = 2OP^2$. Thus, $l = OP^2 + PQ^2 + OQ^2 = OP^2 + 2OP^2 + OP^2 = 4OP^2$. Let $P = (x_1, y_1)$ and $Q = (x_2, y_2)$. Then $OP^2 = x_1^2 + y_1^2$ and $OQ^2 = x_2^2 + y_2^2$. Since $OP = OQ$, we have $x_1^2 + y_1^2 = x_2^2 + y_2^2$. Also, since $OP \perp OQ$, we have $x_1x_2 + y_1y_2 = 0$.

Let $P = (r \cos \theta, r \sin \theta)$ and $Q = (r \cos (\theta + 90^\circ), r \sin (\theta + 90^\circ)) = (-r \sin \theta, r \cos \theta)$. Since $P$ and $Q$ lie on the line $3x + 4y = 12$, we have $3r \cos \theta + 4r \sin \theta = 12$ and $-3r \sin \theta + 4r \cos \theta = 12$. Squaring and adding these equations, we get $9r^2 \cos^2 \theta + 24r^2 \cos \theta \sin \theta + 16r^2 \sin^2 \theta + 9r^2 \sin^2 \theta - 24r^2 \cos \theta \sin \theta + 16r^2 \cos^2 \theta = 144 + 144$, so $25r^2 = 288$. Thus, $r^2 = \frac{288}{25}$. Then $OP^2 = OQ^2 = r^2 = \frac{288}{25}$. $l = 4OP^2 = 4 \cdot \frac{288}{25} = \frac{1152}{25} = 46.08$. The greatest integer less than or equal to $l$ is 46.

Common Mistakes & Tips

• Remember to use the condition for perpendicular lines correctly.
• Applying Cauchy-Schwarz may require clever manipulation to relate to the given conditions.
• Using a parametric representation of the points P and Q based on the given angle can simplify the problem.

Summary

We used the properties of an isosceles right-angled triangle and the equation of a straight line to find the value of $OP^2$. Then we calculated $l = OP^2 + PQ^2 + OQ^2 = 4OP^2$. Using a parametric representation of the points P and Q, we found $OP^2 = \frac{288}{25}$, which gives $l = \frac{1152}{25} = 46.08$. The greatest integer less than or equal to $l$ is 46.

Final Answer

The final answer is \boxed{46}, which corresponds to option (B).