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

Latest revision as of 14:08, 14 May 2024

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

$[asy] draw((0,0)--(11,0)--(7,14)--cycle); draw((7,14)--(11,28)); draw((11,28)--(11,0)); label("$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]$

Let $P$ be the point of intersection between the skaters, Allie and Billie. 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 $B'$ . 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 $PB = PB' = 7t$ , where $t$ is the number of seconds it takes the skaters to meet. 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, $PX = t$ .

As $PXBY$ is a rectangle, $PX = YB$ . Thus $AY + YB = AB \Rightarrow AY + PX = AB$ , so we get $4t + t = 100$ . Solving for $t$ , we find $t = 20$ .

Our answer, $AP$ , is equivalent to $8t$ . Thus, $AP = 8 \cdot 20 = \boxed{160}$ .

Solution 3

We can define $x$ to be the time elapsed since both Allie and Billie moved away from points $A$ and $B$ respectfully. Also, set the point of intersection to be $M$ . 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 $M$ , 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 $30-60-90$ triangle that is formed. From this we get that: $(7x)^2 = (4 \sqrt{3} x)^2 + (100-4x)^2$

$\Longrightarrow 49x^2 - 48x^2 = x^2 = (100-4x)^2$

$\Longrightarrow 0=(100-4x)^2 - x^2 = (100-3x)(100-5x)$ .

Therefore, we get that $x = \frac{100}{3}$ or $x = 20$ . Since $20< \frac{100}{3}$ , we have that $x=20$ (since the problem asks for the quickest possible meeting point), so the distance Allie travels before meeting Billie would be $8 \cdot x = 8 \cdot 20 = \boxed{160}$ 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 $M$ to $AB$ , and call it $P$ . $\triangle AMP$ is a $30-60-90$ triangle, so $AP = 4t$ and $MP = 4\sqrt{3}t$ , and by the Pythagorean theorem on $\triangle MPB$ , $PB = t$ . $AP + BP = AB$ , so $t=20$ . Therefore, $8t = \boxed{160}$ .

~~Disphenoid_lover

@@ Line 1: / Line 1: @@
 == Problem ==
-Two skaters, Allie and Billie, are at [[point]]s <math>A^{}_{}</math> and <math>B^{}_{}</math>, respectively, on a flat, frozen lake. The [[distance]] between <math>A^{}_{}</math> and <math>B^{}_{}</math> is <math>100^{}_{}</math> meters. Allie leaves <math>A^{}_{}</math> and skates at a [[speed]] of <math>8^{}_{}</math> meters per second on a straight line that makes a <math>60^\circ</math> angle with <math>AB^{}_{}</math>. At the same time Allie leaves <math>A^{}_{}</math>, Billie leaves <math>B^{}_{}</math> at a speed of <math>7^{}_{}</math> 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?
+Two skaters, Allie and Billie, are at [[point]]s <math>A</math> and <math>B</math>, respectively, on a flat, frozen lake. The [[distance]] between <math>A</math> and <math>B</math> is <math>100</math> meters. Allie leaves <math>A</math> and skates at a [[speed]] of <math>8</math> meters per second on a straight line that makes a <math>60^\circ</math> angle with <math>AB</math>. At the same time Allie leaves <math>A</math>, Billie leaves <math>B</math> at a speed of <math>7</math> 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?
+<center><asy>
-[[Image:AIME_1989_Problem_6.png]]
+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></center><!-- Minsoen's image: [[Image:AIME_1989_Problem_6.png]] -->
 == Solution ==
 Label the point of [[intersection]] as <math>C</math>. Since <math>d = rt</math>, <math>AC = 8t</math> and <math>BC = 7t</math>. According to the [[law of cosines]],
-:<math>(7t)^2 = (8t)^2 + 100^2 - 2 \cdot 8t \cdot 100 \cdot \cos 60</math>
-:<math>0 = 15t^2 - 800t + 10000 = 3t^2 - 160t + 2000</math>
-:<math>t = \frac{160 \pm \sqrt{160^2 - 4\cdot 3 \cdot 2000}}{6} = 20, \frac{100}{3}</math>.
-Since we are looking for the earliest possible intersection, <math>20</math> seconds are needed. Thus, <math>8 \cdot 20 = 160</math> meters is the solution.
+<center><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></center>
+<cmath>\begin{align*}(7t)^2 &= (8t)^2 + 100^2 - 2 \cdot 8t \cdot 100 \cdot \cos 60^\circ\\
+&= 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*}</cmath>
+Since we are looking for the earliest possible intersection, <math>20</math> seconds are needed. Thus, <math>8 \cdot 20 = \boxed{160}</math> meters is the solution.
+Alternatively, we can drop an altitude from <math>C</math> and arrive at the same answer.
+== 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 <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>.
+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 ==
 {{AIME box|year=1989|num-b=5|num-a=7}}
+[[Category:Intermediate Geometry Problems]]
+{{MAA Notice}}

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

Page

Toolbox

Search