Difference between revisions of "2020 AIME I Problems/Problem 8"

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== Problem ==
 
== Problem ==
  
== Solution ==
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== Solution 1 (Coordinates) ==
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We plot this on the coordinate grid with point <math>O</math> as the origin. We will keep a tally of the x-coordinate and y-coordinate separately.
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First move: The ant moves right <math>5</math>.
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Second move: We use properties of a <math>30-60-90</math> triangle to get <math>\frac{5}{4}</math> right, <math>\frac{5\sqrt{3}}{4}</math> up.
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Third move: <math>\frac{5}{8}</math> left, <math>\frac{5\sqrt{3}}{8}</math> up.
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Fourth move: <math>\frac{5}{8}</math> left.
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Fifth move: <math>\frac{5}{32}</math> left, <math>\frac{5\sqrt{3}}{32}</math> down.
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Sixth move: <math>\frac{5}{64}</math> right, <math>\frac{5\sqrt{3}}{64}</math> down.
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Total of x-coordinate: <math>5 + \frac{5}{4} - \frac{5}{8} - \frac{5}{8}  - \frac{5}{32} + \frac{5}{64} = \frac{315}{64}</math>.
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Total of y-coordinate: <math>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}</math>.
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After this cycle of six moves, all moves repeat with a factor of <math>(\frac{1}{2})^6 = \frac{1}{64}</math>. Using the formula for a geometric series, multiplying each sequence by <math>\frac{1}{1-\frac{1}{64}} = \frac{64}{63}</math> will give us the point <math>P</math>.
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<math>\frac{315}{64} \cdot \frac{64}{63} = 5</math>, <math>\frac{105\sqrt{3}}{64} \cdot \frac{64}{63} = \frac{5\sqrt{3}}{3}</math>.
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Therefore, the coordinates of point <math>P</math> are <math>(5,\frac{5\sqrt{3}}{3})</math>, so using the Pythagorean Theorem, <math>OP^2 = \frac{100}{3}</math>, for an answer of <math>\boxed{103}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 16:14, 12 March 2020

Note: Please do not post problems here until after the AIME.

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 $\boxed{103}$.

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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