Question

Let the line \(x + y = 1\) meet the axes of x and y at A and B, respectively. A right-angled triangle AMN is inscribed in the triangle OAB, where O is the origin and the points M and N lie on the lines OB and AB, respectively. If the area of the triangle AMN is \(\frac{4}{9}\) of the area of the triangle OAB and \(AN : NB = \lambda : 1\), then the sum of all possible values of \(\lambda\) is:

• A. \(\frac{1}{2}\)
• B. \(\frac{13}{6}\)
• C. 5
• D. 2

Hint:

For geometry problems with ratios and area conditions, breaking the triangle into smaller triangles and applying known area formulas helps simplify calculations.

Answer 1

The Correct Option is B

Step 1: Finding area relations.
Area of triangle \( \Delta OAB = \frac{1}{2} \)
Area of triangle \( \Delta AMN = \frac{4}{9} \times \frac{1}{2} = \frac{2}{9} \)

Step 2: Determining coordinates. Equation of AB is \(x + y = 1\) \[ OA = 1, \quad AM = \sec(45^\circ - \theta) \] \[ AN = \sec(45^\circ - \theta) \cos\theta, \quad MN = \sec(45^\circ - \theta) \sin\theta \] From area conditions: \[ \text{Ar}(AMN) = \frac{1}{2} \sec^2(45^\circ - \theta) \sin\theta \cos\theta = \frac{2}{9} \] \[ \tan\theta = 2 \quad \text{or} \quad \frac{1}{2} \] Since \(\tan\theta = 2\) is rejected, \(\tan\theta = \frac{1}{2}\) From similarity condition, \[ \frac{AN}{NB} = \lambda \quad \Rightarrow \quad \lambda = \frac{13}{6} \]