AIME 1993 · 第 13 题

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

$$\begin{aligned}y&=-\frac{100}{t}x+200-\frac{5000}{t}\\50^2&=x^2+y^2\end{aligned}.$$ 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.

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

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

$$\begin{aligned} \dfrac{OQ}{DS}=\dfrac{RO}{BD}&\implies\dfrac{50}{200}=\dfrac{RO}{\sqrt{200^2+4t^2}}\\ &\implies RO=\dfrac{1}{4}\left(\sqrt{200^2+4t^2}\right)\\ &\implies RO=\dfrac{1}{2}\left(\sqrt{100^2+t^2}\right) \end{aligned}$$ And we obtain

$$\begin{aligned} PR-OP&=RO\\ 2t-50&=\dfrac{1}{2}\left(\sqrt{100^2+t^2}\right)\\ 4t-100&=\sqrt{100^2+t^2}\\ (4t-100)^2&=\left(\sqrt{100^2+t^2}\right)^2\\ 16t^2-800t+100^2&=t^2+100^2\\ 15t^2&=800t\\ t&=\dfrac{800}{15} \end{aligned}$$ so we have $t=\frac{160}{3}$, and our answer is thus $160+3=\boxed{163}$.

Solution 4

We can use areas to find the answer. Since Jenny and Kenny are 200 feet apart, we know that they are side by side, and that the line connecting the two of them is tangent to the circular building.

We know that the radius of the circle is 50, and that $\overline{AJ} = x$, $\overline{BK} = 3x$.

Illustration:

By areas, $[OJK] + [AJOF] + [OFBK] = [ABKJ]$. Having right trapezoids, $[AFOJ] = \frac{x+50}{2} \cdot 100$. The other areas of right trapezoids can be calculated in the same way. We just need to find $[OJK]$ in terms of $x$.

If we bring $\overline{AB}$ up to where the point J is, we have by the Pythagorean Theorem, $\overline{JK} = 2\sqrt{x^2+10000} \Rightarrow [OJK] = \frac{1}{2} \overline{JK} \cdot \overline{OD}$.

Now we have everything to solve for $x$.

$$[OJK] + [AJOF] + [OFBK] = [ABKJ]$$ $$50 \sqrt{x^2 + 10000} + \frac{x+50}{2} \cdot 100 + \frac{3x+50}{2} \cdot 100 = \frac{x+3x}{2} \cdot 200$$ After isolating the radical, dividing by 50, and squaring, we obtain: $15x^2 - 800x = 0 \Rightarrow x = \frac{160}{3}$.

Since Jenny walks $\frac{160}{3}$ feet at 1 foot/s, our answer is $160+3 = \fbox{163}$.

Solution 5

Consider our diagram here, where Jenny goes from $A$ to $D$ and Kenny goes from $B$ to $C$. Really what we are asking is the length of $AD$, knowing that $AB$ and $CD$ are tangents and $AB$ is perpendicular to the parallel lines. Draw lines $EF$ and $GH$ as shown such that they are parallel to $BC$ and $AD$ and are tangent to the circle. Additionally, let $MN$ be the median of trapezoid $ABCD$. Notice how $AH=MH=50$ and $BE=ME=50$ as well, so $EF$ and $GH$ are medians of trapezoids $MBCN$ and $AMND$, respectively. If we set $AD=x$ and $BC=3x$, then $MN=2x$ so $HG=\frac{3}{2}x$ and $EF=\frac{5}{2}x$.

Now we will find $FG$ in two different ways. For the first way, notice that $EFGH$ has its inscribed circle, so $EF+GH=EH+FG$. This means that $FG=\frac{3}{2}x+\frac{5}{2}x-100=4x-100$. However, if we let $P$ be the altitude from $G$ to $EF$, then notice how $PF=x$. Since $PG=100$, then by the Pythagorean Theorem, we also have that $FG=\sqrt{x^2+10000}$. After that, then we have an equation in $x$, which is

$$\sqrt{x^2+10000}=4x-100.$$ Squaring both sides, it expands to

$$x^2+10000=16x^2-800x+10000,$$ and since $x$ is nonzero, then once we cancel $10000$ from both sides, we can divide $x$ and get the new equation

$$x=16x-800\implies x=\frac{160}{3}.$$ Because Jenny travels this at $1$ foot per second, then it will take $\frac{160}{3}$ seconds, so our answer is $\boxed{163}$.

~~ethanzhang1001

Solution 6

Let the point where Jenny started be $A = (0,0)$. Then let $D = (t,0)$ the point Jenny is when she can see Kenny. Similarly, let $B = (0,200)$ be where Kenny started, and let $C = (3t,200)$ the point Kenny is when she can see Jenny. Then the equation of the line passing through $C, D$ is $100x-ty-100t=0$. Now, note that the center of the building is at $O = (50,100)$, and the distance from $O$ to $CD$ is $50$, so, using the distance from a point to a line formula, we get

$$\frac{|100\cdot50-100a-100a|}{\sqrt{10000+a^2}} = 50,$$ which simplifies to $15a^2-800a = 0$ from clearing denominators, squaring to remove absolute value and radical, then cancelling and rearranging. But this is equivalent to $a = \frac{160}{3}$, so our answer is $\boxed{163}$.

~Yiyj1