2020 AIME I Problems/Problem 8

Revision as of 16:23, 12 March 2020 by Jhawk0224 (talk | contribs) (Solution 2 (Complex))

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Problem

Solution 1 (Coordinates)

We plot this on the coordinate grid with point $O$ as the origin. We will keep a tally of the x-coordinate and y-coordinate separately.

First move: The ant moves right $5$. Second move: We use properties of a $30-60-90$ triangle to get $\frac{5}{4}$ right, $\frac{5\sqrt{3}}{4}$ up. Third move: $\frac{5}{8}$ left, $\frac{5\sqrt{3}}{8}$ up. Fourth move: $\frac{5}{8}$ left. Fifth move: $\frac{5}{32}$ left, $\frac{5\sqrt{3}}{32}$ down. Sixth move: $\frac{5}{64}$ right, $\frac{5\sqrt{3}}{64}$ down.

Total of x-coordinate: $5 + \frac{5}{4} - \frac{5}{8} - \frac{5}{8}  - \frac{5}{32} + \frac{5}{64} = \frac{315}{64}$. Total of y-coordinate: $0 + \frac{5\sqrt{3}}{4} + \frac{5\sqrt{3}}{8} + 0 - \frac{5\sqrt{3}}{32} - \frac{5\sqrt{3}}{64} = \frac{105\sqrt{3}}{64}$.

After this cycle of six moves, all moves repeat with a factor of $(\frac{1}{2})^6 = \frac{1}{64}$. Using the formula for a geometric series, multiplying each sequence by $\frac{1}{1-\frac{1}{64}} = \frac{64}{63}$ will give us the point $P$.

$\frac{315}{64} \cdot \frac{64}{63} = 5$, $\frac{105\sqrt{3}}{64} \cdot \frac{64}{63} = \frac{5\sqrt{3}}{3}$. Therefore, the coordinates of point $P$ are $(5,\frac{5\sqrt{3}}{3})$, so using the Pythagorean Theorem, $OP^2 = \frac{100}{3}$, for an answer of $\boxed{103}$.

-molocyxu

Solution 2 (Complex)

We put the ant in the complex plane, with its first move going in the positive real direction. Take \[|\sum_{k=0}^{\infty} (5\frac{e^{k\pi i / 3}}{2^k})|^2\] and this is an infinite geometric series. Summing using $\frac{a}{1-r}$ gives $\boxed{103}.$ ~awang11

See Also

2020 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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