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

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== Solution 2 ==
 
== Solution 2 ==
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Can someone please help me make an asymptote diagram for this?
 
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<center><asy>
 
<center><asy>
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pointpen=black; pathpen=black+linewidth(0.7);  
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draw((0,0)--(11,0)--(7,14)--cycle);
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pair A,B,P;
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draw((7,14)--(11,28));
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A=(0,0); B=(5,0); P=(3.5,7);
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draw((11,28)--(11,0));
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label("$A$",(-1,-1),N);
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label("$B$",(12,-1),N);
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label("$P$",(6,15),N);
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label("$B'$",(12,29),N);
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draw((-10,14)--(20,14));
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label("$X$",(12.5,15),N);
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draw((7,0)--(7,14),dashed);
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label("$Y$",(7,-4),N);
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draw((6,0)--(6,1));
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draw((6,1)--(7,1));
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draw((11,13)--(10,13));
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draw((10,13)--(10,14));
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label("$60^{\circ}$",(4,0.5),N);
 
</asy></center>
 
</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>.
  
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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>.
+
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>.
  
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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>.
+
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>.  
  
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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>.
+
Our answer, <math>AP</math>, is equivalent to <math>8t</math>. Thus, <math>AP = 8 \cdot 20 = \boxed{160}</math>.
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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.
 
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- Spacesam
 
  
 
==Solution 3==
 
==Solution 3==
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We know this from the <math>30-60-90</math> triangle that is formed. From this we get that:  
 
We know this from the <math>30-60-90</math> triangle that is formed. From this we get that:  
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<cmath>(7x)^2 = (4 \sqrt{3})^2 + (100-4x)^2</cmath>
+
<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 49x^2 - 48x^2 = x^2 = (100-4x)^2</cmath>
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~qwertysri987
 
~qwertysri987
 +
 +
== Solution 4 ==
 +
 +
<asy>
 +
draw((0,0)--(100,0)--(80,139)--cycle);
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label("8x",(0,0)--(80,139),NW);
 +
label("7x",(100,0)--(80,139),NE);
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label("100",(0,0)--(100,0),S);
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dot((0,0));
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label("A",(0,0),S);
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dot((100,0));
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label("B",(100,0),S);
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dot((80,139));
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label("M",(80,139),N);
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draw((80,139)--(80,0),dashed); // Altitude MP
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label("$P$",(80,0),S); // Label for point P
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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 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

See also

1989 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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All AIME Problems and Solutions

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