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

The Correct Option is C

 

The line \( 3x + 4y = 12 \) intersects points \( P \) and \( Q \) at coordinates:

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

Substituting these into the line equation:

\[ 3(r \cos \theta) + 4(r \sin \theta) = 12 \implies r(3 \cos \theta + 4 \sin \theta) = 12 \tag{1} \]

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

To find \( r \), we square both sides and add equations (1) and (2):

\[ \left( \frac{12}{r} \right)^2 + \left( \frac{12}{r} \right)^2 = (3 \cos \theta + 4 \sin \theta)^2 + (-3 \sin \theta + 4 \cos \theta)^2 \]

\[ 2 \left( \frac{12}{r} \right)^2 = 25 \implies r^2 = \frac{288}{25} \implies r = \sqrt{\frac{288}{25}} = \frac{12 \sqrt{2}}{5} \]

Now, we find \( \ell = OP^2 + PQ^2 + OQ^2 \):

\[ OP^2 = r^2, \quad OQ^2 = r^2 \]

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

\[ PQ^2 = (r \cos \theta + r \sin \theta)^2 + (r \sin \theta - r \cos \theta)^2 \]

\[ = r^2(\cos \theta + \sin \theta)^2 + r^2(\sin \theta - \cos \theta)^2 = r^2(1 + 1) = 2r^2 \]

Therefore:

\[ \ell = OP^2 + PQ^2 + OQ^2 = r^2 + 2r^2 + r^2 = 4r^2 \]

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

\[ \ell = 4 \cdot \frac{288}{25} = \frac{1152}{25} = 46.08 \]

The greatest integer less than or equal to \( \ell \) is: \[ \lfloor 46.08 \rfloor = 46 \]

Answer 2

Step 1: Set up the geometry
Two rays from the origin \(O\) meet the line \(3x+4y=12\) at points \(P\) and \(Q\). Let the slopes of the rays \(OP\) and \(OQ\) be \(m\) and \(m_2\), respectively. Then the rays have equations \(y=mx\) and \(y=m_2x\). Since \(\angle POQ=90^\circ\), the rays are perpendicular, so \[ m_2=-\frac{1}{m}\quad\text{(with }m\neq 0\text{).} \] Because \(\triangle OPQ\) is isosceles with the right angle at \(O\), the equal sides must be \(OP=OQ\).

Step 2: Find the intersection points and lengths \(OP,\,OQ\)
A point on the line \(3x+4y=12\) and the ray \(y=mx\) satisfies \[ 3x+4(mx)=12\;\;\Longrightarrow\;\;x=\frac{12}{3+4m},\quad y=m\,x=\frac{12m}{3+4m}. \] Hence \[ OP^2=x^2+y^2=\frac{144(1+m^2)}{(3+4m)^2}. \] Similarly, for the perpendicular ray \(y=m_2x\) with \(m_2=-1/m\), \[ OQ^2=\frac{144\bigl(1+m_2^2\bigr)}{(3+4m_2)^2}. \] Impose the isosceles condition \(OP=OQ\) (equivalently \(OP^2=OQ^2\)): \[ \frac{1+m^2}{(3+4m)^2}=\frac{1+m_2^2}{(3+4m_2)^2},\quad m_2=-\frac{1}{m}. \] Compute the right side: \[ 1+m_2^2=1+\frac{1}{m^2}=\frac{1+m^2}{m^2},\qquad 3+4m_2=3-\frac{4}{m}=\frac{3m-4}{m}. \] Thus \[ \frac{1+m_2^2}{(3+4m_2)^2}=\frac{\tfrac{1+m^2}{m^2}}{\left(\tfrac{3m-4}{m}\right)^2} =\frac{1+m^2}{(3m-4)^2}. \] Therefore \[ \frac{1+m^2}{(3+4m)^2}=\frac{1+m^2}{(3m-4)^2} \;\;\Longrightarrow\;\; (3+4m)^2=(3m-4)^2. \] Taking square roots gives two linear cases: \[ \text{(i) }\,3+4m=3m-4\;\Rightarrow\;m=-7,\qquad \text{(ii) }\,3+4m=-(3m-4)=4-3m\;\Rightarrow\;7m=1\;\Rightarrow\;m=\frac{1}{7}. \] These are the expected perpendicular pair, since \(m_2=-1/m\) swaps them.

Step 3: Compute a concrete configuration (use \(m=\tfrac{1}{7}\))
For \(m=\tfrac{1}{7}\), \[ OP^2=\frac{144\bigl(1+\tfrac{1}{49}\bigr)}{\bigl(3+\tfrac{4}{7}\bigr)^2} =\frac{144\cdot \tfrac{50}{49}}{\left(\tfrac{25}{7}\right)^2} =144\cdot\frac{50}{49}\cdot\frac{49}{625} =144\cdot\frac{2}{25} =\frac{288}{25}. \] By symmetry \(OQ^2=\frac{288}{25}\). As a quick coordinate check: with \(m=\tfrac{1}{7}\), \[ P=\left(\frac{12}{3+\tfrac{4}{7}},\,\frac{12(\tfrac{1}{7})}{3+\tfrac{4}{7}}\right) =\left(\frac{84}{25},\,\frac{12}{25}\right),\quad Q\text{ (for }m_2=-7\text{)}=\left(-\frac{12}{25},\,\frac{84}{25}\right), \] so \[ OP^2=\left(\frac{84}{25}\right)^2+\left(\frac{12}{25}\right)^2 =\frac{7056+144}{625} =\frac{7200}{625} =\frac{288}{25}. \]

Step 4: Use the right angle at \(O\) to find \(PQ^2\)
Since \(\angle POQ=90^\circ\), the side opposite this angle is \(PQ\), and by the Pythagorean relation in \(\triangle OPQ\), \[ PQ^2=OP^2+OQ^2=2\cdot\frac{288}{25}=\frac{576}{25}. \] Therefore \[ l=OP^2+PQ^2+QO^2=\frac{288}{25}+\frac{576}{25}+\frac{288}{25} =\frac{1152}{25}=46.08. \]

Step 5: Greatest integer less than or equal to \(l\)
\[ \lfloor l\rfloor=\lfloor 46.08\rfloor=46. \]

Final answer

46