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Difference between revisions of "1993 AIME Problems/Problem 13"

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We see that <math>m\angle SBD=m\angle ORQ</math> since they are corresponding angles, and thus by angle-angle similarity, <math>\triangle QOR~\triangle SDB</math>.
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We see that <math>m\angle SBD=m\angle ORQ</math> since they are corresponding angles, and thus by angle-angle similarity, <math>\triangle QOR\sim\triangle SDB</math>.
  
  

Revision as of 11:53, 6 November 2016

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 $\boxed{163}$.

Solution 3

[asy] size(8cm); defaultpen(linewidth(0.7)); pair A,B,C,D,P,Q,O,R,S; A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); R=(100,106.667); S=(0,53.333); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(R); dot(S); draw(A--B--D--C--cycle); draw(P--O); draw(D--S); draw(O--Q--R--cycle); draw(Circle(O,50)); label("$A$",A,SW); label("$B$",B,NNW); label("$C$",(200,-205),S); label("$D$",D,NE); label("$P$",(100,-205),S); label("$Q$",Q,NE); label("$O$",O,SW); label("$R$",R,NE); label("$S$",S,W); [/asy]

Let $t$ be the time they walk. Then $CD=t$ and $AB=3t$.


Draw a line from point $O$ to $Q$ such that $OQ$ is perpendicular to $BD$. Further, draw a line passing through points $O$ and $P$, so $OP$ is parallel to $AB$ and $CD$ and is midway between those two lines. Then $PR=\dfrac{AB+CD}{2}=\dfrac{3t+t}{2}=2t$. Draw another line passing through point $D$ and parallel to $AC$, and call the point of intersection of this line with $AB$ as $S$. Then $SB=AB-AS=3t-t=2t$.


We see that $m\angle SBD=m\angle ORQ$ since they are corresponding angles, and thus by angle-angle similarity, $\triangle QOR\sim\triangle SDB$.


Then \begin{align*} \dfrac{OQ}{DS}=\dfrac{OR}{BD}&\implies\dfrac{50}{200}=\dfrac{OR}{\sqrt{200^2+4t^2}}\\ &\implies OR=\dfrac{1}{4}\left(\sqrt{200^2+4t^2}\right)\\ &\implies OR=\dfrac{1}{2}\left(\sqrt{100^2+t^2}\right) \end{align*}


And we obtain \begin{align*} PR&=OP+OR\\ 2t&=50+\dfrac{1}{2}\left(\sqrt{100^2+t^2}\right)\\ 4t-100&=\sqrt{100^2+t^2}\\ (4t-100)^2&=\left(\sqrt{100^2+t^2}\right)^2\\ 16t^2-800t+100^2&=t^2+100^2\\ 15t^2&=800t\\ t&=\dfrac{800}{15} \end{align*}

so we have $t=\frac{160}{3}$, and our answer is thus $160+3=\boxed{163}$.

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

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