Maths Olympiad Prep

Library / /45 of 120

, 2012

Geometry Difficulty 5.0 AIME, harder Prove it Saudi Arabia

Three equal circles of radius RR are given such that each one passes through the centers of the other two. Find the area of the common region.

Solution

Figure 1
Let O1O_1, O2O_2, O3O_3 be the centers of the three circles and SS the area of the common region. The three sectors with centers O1O_1, O2O_2, O3O_3 which subtend the arcs O2O3O_2O_3, O1O3O_1O_3, O2O1O_2O_1, respectively, cover the surface of area SS and twice more the surface of triangle O1O2O3O_1O_2O_3.

The area of triangle O1O2O3O_1O_2O_3 is R234\frac{R^2\sqrt{3}}{4}.

On the other hand, the area of each of these three circular sectors equals 13\frac{1}{3} the area of a semicircle of radius RR, hence it is 16πR2\frac{1}{6}\pi R^2.

Hence
12πR2=S+2R234. \frac{1}{2}\pi R^2 = S + 2 \cdot \frac{R^2\sqrt{3}}{4}.
Therefore S=12(π3)R2. \text{Therefore } S = \frac{1}{2}(\pi - \sqrt{3})R^2.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.