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Difference between revisions of "1989 AIME Problems/Problem 6"

(→‎Solution: Writing a new solution to this problem.)
(→‎Solution 2)
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Then, we have <math>PX = BY</math>. Since we know that <math>AY + YB = AB</math>, we get <math>4t + t = 100</math>. Solving for <math>t</math>, we get <math>t = 20</math>. Our answer is then equivalent to <math>8t</math>. Thus, <math>8(20) = \boxed{160}</math> meters is the solution.
 
Then, we have <math>PX = BY</math>. Since we know that <math>AY + YB = AB</math>, we get <math>4t + t = 100</math>. Solving for <math>t</math>, we get <math>t = 20</math>. Our answer is then equivalent to <math>8t</math>. Thus, <math>8(20) = \boxed{160}</math> meters is the solution.
 +
- Spacesam
  
 
== See also ==
 
== See also ==

Revision as of 23:08, 9 January 2019

Problem

Two skaters, Allie and Billie, are at points $A$ and $B$, respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB$. At the same time Allie leaves $A$, Billie leaves $B$ at a speed of $7$ meters per second and follows the straight path that produces the earliest possible meeting of the two skaters, given their speeds. How many meters does Allie skate before meeting Billie?

[asy] pointpen=black; pathpen=black+linewidth(0.7); pair A=(0,0),B=(10,0),C=6*expi(pi/3); D(B--A); D(A--C,EndArrow); MP("A",A,SW);MP("B",B,SE);MP("60^{\circ}",A+(0.3,0),NE);MP("100",(A+B)/2); [/asy]

Solution

Label the point of intersection as $C$. Since $d = rt$, $AC = 8t$ and $BC = 7t$. According to the law of cosines,

[asy] pointpen=black; pathpen=black+linewidth(0.7); pair A=(0,0),B=(10,0),C=16*expi(pi/3); D(B--A); D(A--C); D(B--C,dashed); MP("A",A,SW);MP("B",B,SE);MP("C",C,N);MP("60^{\circ}",A+(0.3,0),NE);MP("100",(A+B)/2);MP("8t",(A+C)/2,NW);MP("7t",(B+C)/2,NE); [/asy]

\begin{align*}(7t)^2 &= (8t)^2 + 100^2 - 2 \cdot 8t \cdot 100 \cdot \cos 60^\circ\\ 0 &= 15t^2 - 800t + 10000 = 3t^2 - 160t + 2000\\ t &= \frac{160 \pm \sqrt{160^2 - 4\cdot 3 \cdot 2000}}{6} = 20, \frac{100}{3}.\end{align*}

Since we are looking for the earliest possible intersection, $20$ seconds are needed. Thus, $8 \cdot 20 = \boxed{160}$ meters is the solution.

Alternatively, we can drop an altitude from $C$ and arrive at the same answer.

Solution 2

Can someone please help me make an asymptote diagram for this?

Let the point of intersection be $P$. We can draw a line that goes through $P$ and is parallel to $\overline{AB}$. Letting this line be the $x$ axis, we can reflect $B$ over the $x$ axis to get $B'$. Note that as reflections preserve length, $B'X = XB$.

We then draw lines $BB'$ and $PB'$. We can let the foot of the perpendicular from $P$ to $BB'$ be $X$, and we can let the foot of the perpendicular from $P$ to $AB$ be $Y$. In doing so, we have constructed rectangle $PXBY$.

By $d=rt$, we have $AP = 8t$ and $PB' = 7t$. Furthermore, we have $30-60-90$ triangle $PAY$, so $AY = 4t$ and $PY = 4t\sqrt{3}$. Since we have $PY = XB = B'X$, $B'X = 4t\sqrt{3}$. By Pythagoras, $PX = t$.

Then, we have $PX = BY$. Since we know that $AY + YB = AB$, we get $4t + t = 100$. Solving for $t$, we get $t = 20$. Our answer is then equivalent to $8t$. Thus, $8(20) = \boxed{160}$ meters is the solution. - Spacesam

See also

1989 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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