Difference between revisions of "1993 AIME Problems/Problem 13"
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Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let <math>t\,</math> be the amount of time, in seconds, before Jenny and Kenny can see each other again. If <math>t\,</math> is written as a fraction in lowest terms, what is the sum of the numerator and denominator? | Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let <math>t\,</math> be the amount of time, in seconds, before Jenny and Kenny can see each other again. If <math>t\,</math> is written as a fraction in lowest terms, what is the sum of the numerator and denominator? | ||
− | == Solution 1== | + | __TOC__ |
+ | == Solution == | ||
+ | === Solution 1=== | ||
Consider the unit cicle of radius 50. Assume that they start at points <math>(-50,100)</math> and <math>(-50,-100).</math> Then at time <math>t</math>, they end up at points <math>(-50+t,100)</math> and <math>(-50+3t,-100).</math> | Consider the unit cicle of radius 50. Assume that they start at points <math>(-50,100)</math> and <math>(-50,-100).</math> Then at time <math>t</math>, they end up at points <math>(-50+t,100)</math> and <math>(-50+3t,-100).</math> | ||
The equation of the line connecting these points and the equation of the circle are <cmath>\begin{align}y&=-\frac{100}{t}x+200-\frac{5000}{t}\\50^2&=x^2+y^2\end{align}.</cmath> When they see each other again, the line connecting the two points will be tangent to the circle at the point <math>(x,y).</math> Since the radius is perpendicular to the tangent we get <cmath>-\frac{x}{y}=-\frac{100}{t}</cmath> or <math>xt=100y.</math> Now substitute <cmath>y= \frac{xt}{100}</cmath> into <math>(2)</math> and get <cmath>x=\frac{5000}{\sqrt{100^2+t^2}}.</cmath> Now substitute this and <cmath>y=\frac{xt}{100}</cmath> into <math>(1)</math> and solve for <math>t</math> to get <cmath>t=\frac{160}{3}.</cmath> Finally, the sum of the numerator and denominator is <math>160+3=\boxed{163}.</math> | The equation of the line connecting these points and the equation of the circle are <cmath>\begin{align}y&=-\frac{100}{t}x+200-\frac{5000}{t}\\50^2&=x^2+y^2\end{align}.</cmath> When they see each other again, the line connecting the two points will be tangent to the circle at the point <math>(x,y).</math> Since the radius is perpendicular to the tangent we get <cmath>-\frac{x}{y}=-\frac{100}{t}</cmath> or <math>xt=100y.</math> Now substitute <cmath>y= \frac{xt}{100}</cmath> into <math>(2)</math> and get <cmath>x=\frac{5000}{\sqrt{100^2+t^2}}.</cmath> Now substitute this and <cmath>y=\frac{xt}{100}</cmath> into <math>(1)</math> and solve for <math>t</math> to get <cmath>t=\frac{160}{3}.</cmath> Finally, the sum of the numerator and denominator is <math>160+3=\boxed{163}.</math> | ||
− | ==Solution 2== | + | ===Solution 2=== |
Let <math>A</math> and <math>B</math> be Kenny's initial and final points respectively and define <math>C</math> and <math>D</math> similarly for Jenny. Let <math>O</math> be the center of the building. Also, let <math>X</math> be the intersection of <math>AC</math> and <math>BD</math>. Finaly, let <math>P</math> and <math>Q</math> be the points of tangency of circle <math>O</math> to <math>AC</math> and <math>BD</math> respectively. | Let <math>A</math> and <math>B</math> be Kenny's initial and final points respectively and define <math>C</math> and <math>D</math> similarly for Jenny. Let <math>O</math> be the center of the building. Also, let <math>X</math> be the intersection of <math>AC</math> and <math>BD</math>. Finaly, let <math>P</math> and <math>Q</math> be the points of tangency of circle <math>O</math> to <math>AC</math> and <math>BD</math> respectively. | ||
− | <asy> | + | <center><asy> |
− | size(8cm); | + | size(8cm); defaultpen(linewidth(0.7)); |
pair A,B,C,D,P,Q,O,X; | pair A,B,C,D,P,Q,O,X; | ||
A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); X=(300,0); | A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); X=(300,0); | ||
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draw(Circle(O,50)); | draw(Circle(O,50)); | ||
label("$A$",A,SW); label("$B$",B,NNW); label("$C$",C,S); label("$D$",D,NE); label("$P$",P,S); label("$Q$",Q,NE); label("$O$",O,W); label("$X$",X,ESE); | label("$A$",A,SW); label("$B$",B,NNW); label("$C$",C,S); label("$D$",D,NE); label("$P$",P,S); label("$Q$",Q,NE); label("$O$",O,W); label("$X$",X,ESE); | ||
− | </asy> | + | </asy></center> |
From the problem statement, <math>AB=3t</math>, and <math>CD=t</math>. Since <math>\Delta ABX \sim \Delta CDX</math>, <math>CX=AC\cdot\left(\frac{CD}{AB-CD}\right)=200\cdot\left(\frac{t}{3t-t}\right)=100</math>. | From the problem statement, <math>AB=3t</math>, and <math>CD=t</math>. Since <math>\Delta ABX \sim \Delta CDX</math>, <math>CX=AC\cdot\left(\frac{CD}{AB-CD}\right)=200\cdot\left(\frac{t}{3t-t}\right)=100</math>. | ||
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Thus, <math>\tan(\angle BXA)=\tan(2\angle OXP)=\frac{2\tan(\angle OXP)}{1- \tan^2(\angle OXP)} = \frac{2\cdot \left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)^2}=\frac{8}{15}</math>. | Thus, <math>\tan(\angle BXA)=\tan(2\angle OXP)=\frac{2\tan(\angle OXP)}{1- \tan^2(\angle OXP)} = \frac{2\cdot \left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)^2}=\frac{8}{15}</math>. | ||
− | Therefore, <math>t = CD = CX\cdot\tan(\angle BXA) = 100 \cdot \frac{8}{15} = \frac{160}{3}</math>, and the answer is <math>\boxed{163}</math>. | + | Therefore, <math>t = CD = CX\cdot\tan(\angle BXA) = 100 \cdot \frac{8}{15} = \frac{160}{3}</math>, and the answer is <math>\boxed{163}</math>. |
+ | |||
== See also == | == See also == | ||
{{AIME box|year=1993|num-b=12|num-a=14}} | {{AIME box|year=1993|num-b=12|num-a=14}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 11:10, 20 July 2009
Problem
Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let be the amount of time, in seconds, before Jenny and Kenny can see each other again. If
is written as a fraction in lowest terms, what is the sum of the numerator and denominator?
Solution
Solution 1
Consider the unit cicle of radius 50. Assume that they start at points and
Then at time
, they end up at points
and
The equation of the line connecting these points and the equation of the circle are
When they see each other again, the line connecting the two points will be tangent to the circle at the point
Since the radius is perpendicular to the tangent we get
or
Now substitute
into
and get
Now substitute this and
into
and solve for
to get
Finally, the sum of the numerator and denominator is
Solution 2
Let and
be Kenny's initial and final points respectively and define
and
similarly for Jenny. Let
be the center of the building. Also, let
be the intersection of
and
. Finaly, let
and
be the points of tangency of circle
to
and
respectively.
![[asy] size(8cm); defaultpen(linewidth(0.7)); pair A,B,C,D,P,Q,O,X; A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); X=(300,0); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(X); draw(A--B--X--cycle); draw(C--D); draw(P--O--Q); draw(O--X); draw(Circle(O,50)); label("$A$",A,SW); label("$B$",B,NNW); label("$C$",C,S); label("$D$",D,NE); label("$P$",P,S); label("$Q$",Q,NE); label("$O$",O,W); label("$X$",X,ESE); [/asy]](http://latex.artofproblemsolving.com/e/8/5/e854091993a5c5a12a513b87e411d392cda1b333.png)
From the problem statement, , and
. Since
,
.
Since ,
. So,
.
Since circle is tangent to
and
,
is the angle bisector of
.
Thus, .
Therefore, , and the answer is
.
See also
1993 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |