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:
Step-by-Step Solution
Key Concept: Let $P=(r\cos\theta,r\sin\theta)$ and $Q=(-r\sin\theta,r\cos\theta)$ (perpendicular, equal distance from origin since isosceles with $OP=OQ$). Both lie on $3x+4y=12$: $r(3\cos\theta+4\sin\theta)=12$ and $r(-3\sin\theta+4\cos\theta)=12$.
Step 1: Understand the geometric properties of $\triangle OPQ$.
The problem states that $\triangle OPQ$ is an isosceles triangle and $\angle POQ=90^\circ$. This implies that the sides $OP$ and $OQ$ are equal in length. Let $OP=OQ=r$. Since $\angle POQ$ is $90^\circ$, $\triangle OPQ$ is an isosceles right-angled triangle.
Step 2: Express the quantity $l$ in terms of $r$.
The given quantity is $l=OP^2+PQ^2+QO^2$.
From Step 1, we know $OP=OQ=r$.
In an isosceles right-angled triangle, by the Pythagorean theorem, the square of the hypotenuse $PQ$ is the sum of the squares of the other two sides:
$$PQ^2 = OP^2 + OQ^2 = r^2 + r^2 = 2r^2$$
Now, substitute these values into the expression for $l$:
$$l = r^2 + (2r^2) + r^2 = 4r^2$$
Step 3: Calculate the perpendicular distance from the origin to the line $3x+4y=12$.
The line $PQ$ is given by the equation $3x+4y=12$. Let $h$ be the perpendicular distance from the origin $O(0,0)$ to this line. The formula for the perpendicular distance from a point $(x_0, y_0)$ to a line $Ax+By+C=0$ is $\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$.
Here, $(x_0, y_0) = (0,0)$, $A=3$, $B=4$, $C=-12$.
$$h = \frac{|3(0)+4(0)-12|}{\sqrt{3^2+4^2}} = \frac{|-12|}{\sqrt{9+16}} = \frac{12}{\sqrt{25}} = \frac{12}{5}$$
Step 4: Relate $r$ to the perpendicular distance $h$.
In an isosceles right-angled triangle $OPQ$ with $OP=OQ=r$ and $\angle POQ=90^\circ$, the altitude from the vertex $O$ to the hypotenuse $PQ$ (which is $h$) is half the length of the hypotenuse $PQ$.
We found in Step 2 that $PQ^2 = 2r^2$, so $PQ = r\sqrt{2}$.
Therefore, $h = \frac{PQ}{2} = \frac{r\sqrt{2}}{2} = \frac{r}{\sqrt{2}}$.
From this, we can express $r$ in terms of $h$:
$$r = h\sqrt{2}$$
Step 5: Calculate $r^2$.
Substitute the value of $h$ from Step 3 into the expression for $r$:
$$r = \frac{12}{5}\sqrt{2}$$
Now, square both sides to find $r^2$:
$$r^2 = \left(\frac{12\sqrt{2}}{5}\right)^2 = \frac{12^2 \times (\sqrt{2})^2}{5^2} = \frac{144 \times 2}{25} = \frac{288}{25}$$
Step 6: Calculate $l$.
Using the relationship $l = 4r^2$ from Step 2 and the value of $r^2$ from Step 5:
$$l = 4 \times \frac{288}{25} = \frac{1152}{25}$$
Now, perform the division:
$$l = 46.08$$
Step 7: Find the greatest integer less than or equal to $l$.
We need to find $\lfloor l \rfloor$, which is the floor of $l$.
$$\lfloor l \rfloor = \lfloor 46.08 \rfloor = 46$$
The final answer is $\boxed{46}$.
Correct Answer: 2