Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Find the answer

If an arc of 6060^{\circ} in circle I has the same length as an arc of 4545^{\circ} in circle II, find the ratio between the area of circle I and the area of circle II.
(a) 169\frac{16}{9}
(b) 916\frac{9}{16}
(c) 43\frac{4}{3}
(d) 34\frac{3}{4}
(e) 69\frac{6}{9}

Multiple choice: answer with the letter of the option you want.

Solution

The correct option is (b).

Since the 6060^{\circ} arc of circle I has the same length as the 4545^{\circ} arc in circle II, we conclude that the radius of circle I is smaller than that of circle II. Let's denote the radii of circles I and II by rr and RR, respectively.

In circle I, the length of the 6060^{\circ} arc is equal to 1/61 / 6 of its circumference, i.e., 2πr/62 \pi r / 6. Similarly, in circle II, the length of the 4545^{\circ} arc is equal to 1/81 / 8 of its circumference, i.e., 2πR/82 \pi R / 8. Therefore, 2πr/6=2πR/82 \pi r / 6 = 2 \pi R / 8, or r/R=6/8=3/4r / R = 6 / 8 = 3 / 4. Finally, we have

 area of circle I  area of circle II =πr2πR2=(rR)2=(34)2=916 \frac{\text { area of circle I }}{\text { area of circle II }}=\frac{\pi r^{2}}{\pi R^{2}}=\left(\frac{r}{R}\right)^{2}=\left(\frac{3}{4}\right)^{2}=\frac{9}{16}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.