Difference between revisions of "1989 AIME Problems/Problem 6"
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== Solution 2 == | == Solution 2 == | ||
− | + | <center><asy> | |
+ | draw((0,0)--(11,0)--(7,14)--cycle); | ||
+ | draw((7,14)--(11,28)); | ||
+ | draw((11,28)--(11,0)); | ||
+ | label("$A$",(-1,-1),N); | ||
+ | label("$B$",(12,-1),N); | ||
+ | label("$P$",(6,15),N); | ||
+ | label("$B'$",(12,29),N); | ||
+ | draw((-10,14)--(20,14)); | ||
+ | label("$X$",(12.5,15),N); | ||
+ | draw((7,0)--(7,14),dashed); | ||
+ | label("$Y$",(7,-4),N); | ||
+ | draw((6,0)--(6,1)); | ||
+ | draw((6,1)--(7,1)); | ||
+ | draw((11,13)--(10,13)); | ||
+ | draw((10,13)--(10,14)); | ||
+ | label("$60^{\circ}$",(4,0.5),N); | ||
+ | </asy></center> | ||
− | Let | + | Let <math>P</math> be the point of intersection between the skaters, Allie and Billie. 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>. 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>. | 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>. | + | By <math>d=rt</math>, we have <math>AP = 8t</math> and <math>PB = PB' = 7t</math>, where <math>t</math> is the number of seconds it takes the skaters to meet. 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>. |
− | + | As <math>PXBY</math> is a rectangle, <math>PX = YB</math>. Thus <math>AY + YB = AB \Rightarrow AY + PX = AB</math>, so we get <math>4t + t = 100</math>. Solving for <math>t</math>, we find <math>t = 20</math>. | |
− | + | ||
+ | Our answer, <math>AP</math>, is equivalent to <math>8t</math>. Thus, <math>AP = 8 \cdot 20 = \boxed{160}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | We can define <math>x</math> to be the time elapsed since both Allie and Billie moved away from points <math>A</math> and <math>B</math> respectfully. Also, set the point of intersection to be <math>M</math>. Then we can produce the following diagram: | ||
+ | |||
+ | <asy> | ||
+ | draw((0,0)--(100,0)--(80,139)--cycle); | ||
+ | label("8x",(0,0)--(80,139),NW); | ||
+ | label("7x",(100,0)--(80,139),NE); | ||
+ | label("100",(0,0)--(100,0),S); | ||
+ | dot((0,0)); | ||
+ | label("A",(0,0),S); | ||
+ | dot((100,0)); | ||
+ | label("B",(100,0),S); | ||
+ | dot((80,139)); | ||
+ | label("M",(80,139),N); | ||
+ | </asy> | ||
+ | |||
+ | Now, if we drop an altitude from point<math>M</math>, we get : | ||
+ | |||
+ | <asy> | ||
+ | size(300); draw((0,0)--(100,0)--(80,139)--cycle); | ||
+ | label("8x",(0,0)--(80,139),NW); | ||
+ | label("7x",(100,0)--(80,139),NE); | ||
+ | |||
+ | dot((0,0)); | ||
+ | label("A",(0,0),S); | ||
+ | dot((100,0)); | ||
+ | label("B",(100,0),SE); | ||
+ | dot((80,139)); | ||
+ | label("M",(80,139),N); | ||
+ | draw((80,139)--(80,0),dashed); | ||
+ | label("4x",(0,0)--(80,0),S); | ||
+ | label("100-4x",(80,0)--(100,0),S); | ||
+ | label("$4x \sqrt{3}$",(80,139)--(80,0),W); | ||
+ | label("$60^{\circ}$", (3,1), NE); | ||
+ | </asy> | ||
+ | |||
+ | We know this from the <math>30-60-90</math> triangle that is formed. From this we get that: | ||
+ | <cmath>(7x)^2 = (4 \sqrt{3} x)^2 + (100-4x)^2</cmath> | ||
+ | |||
+ | <cmath>\Longrightarrow 49x^2 - 48x^2 = x^2 = (100-4x)^2</cmath> | ||
+ | |||
+ | <cmath>\Longrightarrow 0=(100-4x)^2 - x^2 = (100-3x)(100-5x)</cmath>. | ||
+ | |||
+ | Therefore, we get that <math>x = \frac{100}{3}</math> or <math>x = 20</math>. Since <math> 20< \frac{100}{3}</math>, we have that <math>x=20</math> (since the problem asks for the quickest possible meeting point), so the distance Allie travels before meeting Billie would be | ||
+ | <math>8 \cdot x = 8 \cdot 20 = \boxed{160}</math> meters. | ||
+ | |||
+ | ~qwertysri987 | ||
+ | |||
+ | == Solution 4 == | ||
+ | |||
+ | <asy> | ||
+ | draw((0,0)--(100,0)--(80,139)--cycle); | ||
+ | label("8x",(0,0)--(80,139),NW); | ||
+ | label("7x",(100,0)--(80,139),NE); | ||
+ | label("100",(0,0)--(100,0),S); | ||
+ | dot((0,0)); | ||
+ | label("A",(0,0),S); | ||
+ | dot((100,0)); | ||
+ | label("B",(100,0),S); | ||
+ | dot((80,139)); | ||
+ | label("M",(80,139),N); | ||
+ | draw((80,139)--(80,0),dashed); // Altitude MP | ||
+ | label("$P$",(80,0),S); // Label for point P | ||
+ | draw(rightanglemark((80,0),(80,139),(100,0))); // Right angle mark | ||
+ | </asy> | ||
+ | |||
+ | Drop the altitude from <math>M</math> to <math>AB</math>, and call it <math>P</math>. <math>\triangle AMP</math> is a <math>30-60-90</math> triangle, so <math>AP = 4t</math> and <math>MP = 4\sqrt{3}t</math>, and by the Pythagorean theorem on <math>\triangle MPB</math>, <math>PB = t</math>. <math>AP + BP = AB</math>, so <math>t=20</math>. Therefore, <math>8t = \boxed{160}</math>. | ||
+ | |||
+ | ~~Disphenoid_lover | ||
== See also == | == See also == |
Latest revision as of 15:08, 14 May 2024
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
![[asy] draw((0,0)--(11,0)--(7,14)--cycle); draw((7,14)--(11,28)); draw((11,28)--(11,0)); label("$A$",(-1,-1),N); label("$B$",(12,-1),N); label("$P$",(6,15),N); label("$B'$",(12,29),N); draw((-10,14)--(20,14)); label("$X$",(12.5,15),N); draw((7,0)--(7,14),dashed); label("$Y$",(7,-4),N); draw((6,0)--(6,1)); draw((6,1)--(7,1)); draw((11,13)--(10,13)); draw((10,13)--(10,14)); label("$60^{\circ}$",(4,0.5),N); [/asy]](http://latex.artofproblemsolving.com/2/6/d/26d2715830064afebf909e7c7f4ee28b0cbbaaa8.png)
Let be the point of intersection between the skaters, Allie and Billie. 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
. 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
, where
is the number of seconds it takes the skaters to meet. Furthermore, we have
triangle
, so
, and
. Since we have
,
. By Pythagoras,
.
As is a rectangle,
. Thus
, so we get
. Solving for
, we find
.
Our answer, , is equivalent to
. Thus,
.
Solution 3
We can define to be the time elapsed since both Allie and Billie moved away from points
and
respectfully. Also, set the point of intersection to be
. Then we can produce the following diagram:
Now, if we drop an altitude from point, we get :
We know this from the triangle that is formed. From this we get that:
.
Therefore, we get that or
. Since
, we have that
(since the problem asks for the quickest possible meeting point), so the distance Allie travels before meeting Billie would be
meters.
~qwertysri987
Solution 4
Drop the altitude from to
, and call it
.
is a
triangle, so
and
, and by the Pythagorean theorem on
,
.
, so
. Therefore,
.
~~Disphenoid_lover
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.