Difference between revisions of "2012 AIME II Problems/Problem 4"
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== Solution == | == Solution == | ||
+ | Let P, Q, and R be the east-west distance of the field, the north-south distance, and the distance from the southeast corner to point D, respectively. | ||
+ | |||
+ | Ana's distance to point D = <math>P + Q + (P - R) = 2P + Q - R</math> | ||
+ | |||
+ | Bob's distance to point D = <math>Q + R</math> | ||
+ | |||
+ | Cao's distance to point D = <math>\sqrt{Q^2 + R^2}</math> | ||
+ | |||
+ | Since they arrive at the same time, their distance/speed ratios are equal, so: | ||
+ | |||
+ | <cmath>\frac{2P + Q - R}{8.6} = \frac{Q + R}{6.2} = \frac{\sqrt{Q^2 + R^2}}{5}</cmath> | ||
+ | |||
+ | <cmath>\frac{2P + Q - R}{43} = \frac{Q + R}{31} = \frac{\sqrt{Q^2 + R^2}}{25}</cmath> | ||
+ | |||
+ | Looking at the last two parts of the equation: | ||
+ | |||
+ | <cmath>\frac{Q + R}{31} = \frac{\sqrt{Q^2 + R^2}}{25}</cmath> | ||
+ | |||
+ | <cmath>25 (Q + R) = 31 \sqrt{Q^2 + R^2}</cmath> | ||
+ | |||
+ | <cmath>625 Q^2 + 1250 QR + 625 R^2 = 961 Q^2 + 961 R^2</cmath> | ||
+ | |||
+ | <cmath>336 Q^2 - 1250 QR + 336 R^2 = 0</cmath> | ||
+ | |||
+ | <cmath>168 (\frac{Q}{R})^2 - 625 \frac{Q}{R}+ 168 = 0</cmath> | ||
+ | |||
+ | <cmath>\frac{Q}{R} = \frac{625 \pm \sqrt{625^2 - 4 \cdot 168^2}}{336}</cmath> | ||
+ | |||
+ | <cmath>\frac{Q}{R} = 24/7\;or\;7/24</cmath> | ||
+ | |||
+ | Looking at the first two parts of the equation above: | ||
+ | |||
+ | <cmath>\frac{2 P + Q - R}{43} = \frac{Q + R}{31}</cmath> | ||
+ | |||
+ | <cmath>62 P + 31 Q - 31 R = 43 Q + 43 R</cmath> | ||
+ | |||
+ | <cmath>P = \frac{6}{31}Q + \frac{37}{31} R</cmath> | ||
+ | |||
+ | If <math>\frac{Q}{R} = \frac{24}{27}</math>: | ||
+ | |||
+ | <cmath>R = \frac{7}{24} Q</cmath> | ||
+ | |||
+ | <cmath>P = \frac{6}{31} Q + \frac{37}{31} \cdot \frac{7}{24} Q = \frac{13}{24} Q</cmath> | ||
+ | |||
+ | However, this makes P < Q, but we are given that P > Q. | ||
+ | Therefore, <math>\frac{Q}{R} = \frac{7}{24}</math>, and: | ||
+ | |||
+ | <cmath>R = \frac{24}{7} Q</cmath> | ||
+ | |||
+ | <cmath>P = \frac{6}{31} Q + \frac{37}{31} \cdot \frac{24}{7} Q = \frac{30}{7} Q</cmath> | ||
+ | |||
+ | <cmath>P : Q : R = 30 : 7 : 24</cmath> | ||
+ | |||
+ | The solution is <math>P + Q + R = \framebox{061}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2012|n=II|num-b=3|num-a=5}} | {{AIME box|year=2012|n=II|num-b=3|num-a=5}} |
Revision as of 13:11, 3 April 2012
Problem 4
Ana, Bob, and Cao bike at constant rates of meters per second,
meters per second, and
meters per second, respectively. They all begin biking at the same time from the northeast corner of a rectangular field whose longer side runs due west. Ana starts biking along the edge of the field, initially heading west, Bob starts biking along the edge of the field, initially heading south, and Cao bikes in a straight line across the field to a point
on the south edge of the field. Cao arrives at point
at the same time that Ana and Bob arrive at
for the first time. The ratio of the field's length to the field's width to the distance from point
to the southeast corner of the field can be represented as
, where
,
, and
are positive integers with
and
relatively prime. Find
.
Solution
Let P, Q, and R be the east-west distance of the field, the north-south distance, and the distance from the southeast corner to point D, respectively.
Ana's distance to point D =
Bob's distance to point D =
Cao's distance to point D =
Since they arrive at the same time, their distance/speed ratios are equal, so:
Looking at the last two parts of the equation:
Looking at the first two parts of the equation above:
If :
However, this makes P < Q, but we are given that P > Q.
Therefore, , and:
The solution is .
See also
2012 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |