CAT 2024 Slot 2 · Quant · Question 18

Take each of the three equal circles to have radius 1.

Since they touch externally in pairs, the distance between any two of their centres is 1+1=2, so the three centres form an equilateral triangle of side 2.

Both X and Y must be centred at the point equidistant from all three centres, which is the centroid of that equilateral triangle.

The distance from the centroid to each vertex is the circumradius, \frac{\text{side}}{\sqrt{3}}=\frac{2}{\sqrt{3}}.

Circle X is the large one enclosing all three, so its radius reaches past each centre by one more unit: R=\frac{2}{\sqrt{3}}+1.

Circle Y is the small one wedged in the gap between the three, so its radius falls short of each centre by one unit: r=\frac{2}{\sqrt{3}}-1.

The required ratio is \frac{R}{r}=\frac{\frac{2}{\sqrt{3}}+1}{\frac{2}{\sqrt{3}}-1}.

Multiplying numerator and denominator by \sqrt{3} gives \frac{2+\sqrt{3}}{2-\sqrt{3}}.

Rationalising, \frac{2+\sqrt{3}}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{{\left( 2+\sqrt{3}\right)}^{2}}{4-3}={\left( 2+\sqrt{3}\right)}^{2}.

Expanding, {\left( 2+\sqrt{3}\right)}^{2}=4+4\sqrt{3}+3=7+4\sqrt{3}.

The ratio is 7+4\sqrt{3}:1, the third option.

Shortcut. Since 2-\sqrt{3} is the reciprocal of 2+\sqrt{3}, the ratio is simply {\left( 2+\sqrt{3}\right)}^{2} — and the options being 2+\sqrt{3} and its square makes the intended trap obvious.