UPSC CDS I 2024 Math Paper

Concept:Use the relation between radii of two circles inscribed in a right‑angled triangle and the angle at the vertex where they are tangent to the hypotenuse.Explanation:Let the radius of the circle with center $O_{1}$ be $r_{1}$.In $\triangle ABC$, right‑angled at $B$, the radius of the circle with center $O$ is given as $2$ cm.Given $MA : MC = 2 : 3$, let $MA = 2x$ and $MC = 3x$.Using Pythagoras theorem in $\triangle ABC$: $(AB)^2 + (BC)^2 = (AC)^2$ Here $AB = AM + MB = 2x + 2$, $BC = CM + MB = 3x + 2$, $AC = 5x$.So $(2x+2)^2 + (3x+2)^2 = (5x)^2$.Expanding: $4x^2 + 8x + 4 + 9x^2 + 12x + 4 = 25x^2$ $\Rightarrow 13x^2 + 20x + 8 = 25x^2$ $\Rightarrow 12x^2 - 20x - 8 = 0$ Divide by $4$: $3x^2 - 5x - 2 = 0$ Factor: $(3x+1)(x-2)=0$ Thus $x=2$ (positive). Then $AM = 4$ cm.In $\triangle AOM$, $AO = \sqrt{2^2 + 4^2} = \sqrt{20} = 2\sqrt{5}$.Let $\angle OAM = \theta$, then $\sin\theta = \frac{OM}{AO} = \frac{2}{2\sqrt{5}} = \frac{1}{\sqrt{5}}$.For two circles inscribed in a right triangle tangent to the same leg and to the hypotenuse, the radii satisfy$\frac{r_1}{r} = \frac{1 - \sin\theta}{1 + \sin\theta}$.Substitute $r=2$, $\sin\theta = \frac{1}{\sqrt{5}}$:$\frac{r_1}{2} = \frac{1 - \frac{1}{\sqrt{5}}}{1 + \frac{1}{\sqrt{5}}} = \frac{\sqrt{5} - 1}{\sqrt{5} + 1}$.Rationalise: $\frac{r_1}{2} = \frac{(\sqrt{5} - 1)^2}{(\sqrt{5})^2 - 1^2} = \frac{5 + 1 - 2\sqrt{5}}{5 - 1} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}$.Hence $r_1 = 2 \times \frac{3 - \sqrt{5}}{2} = 3 - \sqrt{5}$ cm.Answer:$3 - \sqrt{5}$ (option D).