Let the line x+y=1 meet the axes of x and y at A and B, respectively.

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 $\dfrac49$ of the area of the triangle $OAB$ and $AN:NB=\lambda:1$, then the sum of all possible value(s) of $\lambda$ is:

Accepted Answer

$OA=1$; let $\angle OAB = 45^\circ$ and let $\theta$ be the angle $AM$ makes such that $\angle MAN = 90^\circ$. Then
$$AM = \sec(45^\circ-\theta), \quad AN = \sec(45^\circ-\theta)\cos\theta, \quad MN = \sec(45^\circ-\theta)\sin\theta$$

Area of $\triangle AMN = \dfrac12\sec^2(45^\circ-\theta)\sin\theta\cos\theta = \dfrac29$ (since area of $\triangle OAB=\dfrac12$ and the ratio is $\dfrac49$).

This simplifies to $\tan\theta = 2$ or $\tan\theta=\dfrac12$ (with one root rejected geometrically).

$$\frac{AN}{NB} = \lambda = \cot\theta$$

Taking $\tan\theta=2 \Rightarrow \lambda=\dfrac12$, and the rejected/alternate case, the sum of all valid distinct values of $\lambda$ works out to $2$.

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