Difference between revisions of "2012 AIME II Problems/Problem 4"

Revision as of 14:08, 4 July 2013

Problem 4

Ana, Bob, and Cao bike at constant rates of $8.6$ meters per second, $6.2$ meters per second, and $5$ 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 $D$ on the south edge of the field. Cao arrives at point $D$ at the same time that Ana and Bob arrive at $D$ for the first time. The ratio of the field's length to the field's width to the distance from point $D$ to the southeast corner of the field can be represented as $p : q : r$ , where $p$ , $q$ , and $r$ are positive integers with $p$ and $q$ relatively prime. Find $p+q+r$ .

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 = $P + Q + (P - R) = 2P + Q - R$

Bob's distance to point D = $Q + R$

Cao's distance to point D = $\sqrt{Q^2 + R^2}$

Since they arrive at the same time, their distance/speed ratios are equal, so:

$\frac{2P + Q - R}{8.6} = \frac{Q + R}{6.2} = \frac{\sqrt{Q^2 + R^2}}{5}$

$\frac{2P + Q - R}{43} = \frac{Q + R}{31} = \frac{\sqrt{Q^2 + R^2}}{25}$

Looking at the last two parts of the equation:

$\frac{Q + R}{31} = \frac{\sqrt{Q^2 + R^2}}{25}$

$25 (Q + R) = 31 \sqrt{Q^2 + R^2}$

$625 Q^2 + 1250 QR + 625 R^2 = 961 Q^2 + 961 R^2$

$336 Q^2 - 1250 QR + 336 R^2 = 0$

$168 (\frac{Q}{R})^2 - 625 \frac{Q}{R}+ 168 = 0$

$\frac{Q}{R} = \frac{625 \pm \sqrt{625^2 - 4 \cdot 168^2}}{336}$

$\frac{Q}{R} = 24/7\;or\;7/24$

Looking at the first two parts of the equation above:

$\frac{2 P + Q - R}{43} = \frac{Q + R}{31}$

$62 P + 31 Q - 31 R = 43 Q + 43 R$

$P = \frac{6}{31}Q + \frac{37}{31} R$

If $\frac{Q}{R} = \frac{24}{7}$ :

$R = \frac{7}{24} Q$

$P = \frac{6}{31} Q + \frac{37}{31} \cdot \frac{7}{24} Q = \frac{13}{24} Q$

However, this makes P < Q, but we are given that P > Q. Therefore, $\frac{Q}{R} = \frac{7}{24}$ , and:

$R = \frac{24}{7} Q$

$P = \frac{6}{31} Q + \frac{37}{31} \cdot \frac{24}{7} Q = \frac{30}{7} Q$

$P : Q : R = 30 : 7 : 24$

The solution is $P + Q + R = \framebox{061}$ .

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 == See Also ==
 {{AIME box|year=2012|n=II|num-b=3|num-a=5}}
+{{MAA Notice}}

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Difference between revisions of "2012 AIME II Problems/Problem 4"

Revision as of 14:08, 4 July 2013

Problem 4

Solution

See Also