Difference between revisions of "1989 AIME Problems/Problem 6"
Flyhawkeye (talk | contribs) (→Solution) |
(→Solution: Writing a new solution to this problem.) |
||
Line 23: | Line 23: | ||
Alternatively, we can drop an altitude from <math>C</math> and arrive at the same answer. | Alternatively, we can drop an altitude from <math>C</math> 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 <math>P</math>. We can draw a line that goes through <math>P</math> and is parallel to <math>\overline{AB}</math>. Letting this line be the <math>x</math> axis, we can reflect <math>B</math> over the <math>x</math> axis to get <math>B'</math>. Note that as reflections preserve length, <math>B'X = XB</math>. | ||
+ | |||
+ | We then draw lines <math>BB'</math> and <math>PB'</math>. We can let the foot of the perpendicular from <math>P</math> to <math>BB'</math> be <math>X</math>, and we can let the foot of the perpendicular from <math>P</math> to <math>AB</math> be <math>Y</math>. In doing so, we have constructed rectangle <math>PXBY</math>. | ||
+ | |||
+ | By <math>d=rt</math>, we have <math>AP = 8t</math> and <math>PB' = 7t</math>. Furthermore, we have <math>30-60-90</math> triangle <math>PAY</math>, so <math>AY = 4t</math> and <math>PY = 4t\sqrt{3}</math>. Since we have <math>PY = XB = B'X</math>, <math>B'X = 4t\sqrt{3}</math>. By Pythagoras, <math>PX = t</math>. | ||
+ | |||
+ | 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. | ||
== See also == | == See also == |
Revision as of 23:08, 9 January 2019
Contents
Problem
Two skaters, Allie and Billie, are at points and
, respectively, on a flat, frozen lake. The distance between
and
is
meters. Allie leaves
and skates at a speed of
meters per second on a straight line that makes a
angle with
. At the same time Allie leaves
, Billie leaves
at a speed of
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]](http://latex.artofproblemsolving.com/a/3/0/a30f95a4bc471ad519d35b683f2f319118167660.png)
Solution
Label the point of intersection as . Since
,
and
. 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]](http://latex.artofproblemsolving.com/a/e/d/aedca2f1d05dc487c293ffbc289231edb6ca2cb9.png)
Since we are looking for the earliest possible intersection, seconds are needed. Thus,
meters is the solution.
Alternatively, we can drop an altitude from 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 . We can draw a line that goes through
and is parallel to
. Letting this line be the
axis, we can reflect
over the
axis to get
. Note that as reflections preserve length,
.
We then draw lines and
. We can let the foot of the perpendicular from
to
be
, and we can let the foot of the perpendicular from
to
be
. In doing so, we have constructed rectangle
.
By , we have
and
. Furthermore, we have
triangle
, so
and
. Since we have
,
. By Pythagoras,
.
Then, we have . Since we know that
, we get
. Solving for
, we get
. Our answer is then equivalent to
. Thus,
meters is the solution.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.