AIME 1989 · 第 6 题

Solution

Label the point of intersection as $C$. Since $d = rt$, $AC = 8t$ and $BC = 7t$. According to the law of cosines,

$$\begin{aligned}(7t)^2 &= (8t)^2 + 100^2 - 2 \cdot 8t \cdot 100 \cdot \cos 60^\circ\\ 0 &= 15t^2 - 800t + 10000 = 3t^2 - 160t + 2000\\ t &= \frac{160 \pm \sqrt{160^2 - 4\cdot 3 \cdot 2000}}{6} = 20, \frac{100}{3}.\end{aligned}$$ 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

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:

Now, if we drop an altitude from point$M$, we get :

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

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