Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1993 AIME Problems/Problem 13

Revision as of 10:10, 20 July 2009 by Azjps (talk | contribs) (fmt)

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 $t\,$ be the amount of time, in seconds, before Jenny and Kenny can see each other again. If $t\,$ 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 $(-50,100)$ and $(-50,-100).$ Then at time $t$, they end up at points $(-50+t,100)$ and $(-50+3t,-100).$ The equation of the line connecting these points and the equation of the circle are \begin{align}y&=-\frac{100}{t}x+200-\frac{5000}{t}\\50^2&=x^2+y^2\end{align}. When they see each other again, the line connecting the two points will be tangent to the circle at the point $(x,y).$ Since the radius is perpendicular to the tangent we get \[-\frac{x}{y}=-\frac{100}{t}\] or $xt=100y.$ Now substitute \[y= \frac{xt}{100}\] into $(2)$ and get \[x=\frac{5000}{\sqrt{100^2+t^2}}.\] Now substitute this and \[y=\frac{xt}{100}\] into $(1)$ and solve for $t$ to get \[t=\frac{160}{3}.\] Finally, the sum of the numerator and denominator is $160+3=\boxed{163}.$

Solution 2

Let $A$ and $B$ be Kenny's initial and final points respectively and define $C$ and $D$ similarly for Jenny. Let $O$ be the center of the building. Also, let $X$ be the intersection of $AC$ and $BD$. Finaly, let $P$ and $Q$ be the points of tangency of circle $O$ to $AC$ and $BD$ 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]

From the problem statement, $AB=3t$, and $CD=t$. Since $\Delta ABX \sim \Delta CDX$, $CX=AC\cdot\left(\frac{CD}{AB-CD}\right)=200\cdot\left(\frac{t}{3t-t}\right)=100$.

Since $PC=100$, $PX=200$. So, $\tan(\angle OXP)=\frac{OP}{PX}=\frac{50}{200}=\frac{1}{4}$.

Since circle $O$ is tangent to $BX$ and $AX$, $OX$ is the angle bisector of $\angle BXA$.

Thus, $\tan(\angle BXA)=\tan(2\angle OXP)=\frac{2\tan(\angle OXP)}{1- \tan^2(\angle OXP)} = \frac{2\cdot \left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)^2}=\frac{8}{15}$.

Therefore, $t = CD = CX\cdot\tan(\angle BXA) = 100 \cdot \frac{8}{15} = \frac{160}{3}$, and the answer is $\boxed{163}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1993_AIME_Problems/Problem_13&oldid=32428"