AIME 2004 I · 第 14 题

Solution

Looking from an overhead view, call the center of the circle $O$, the tether point to the unicorn $A$ and the last point where the rope touches the tower $B$. $\triangle OAB$ is a right triangle because $OB$ is a radius and $BA$ is a tangent line at point $B$. We use the Pythagorean Theorem to find the horizontal component of $AB$ has length $4\sqrt{5}$.

Now look at a side view and "unroll" the cylinder to be a flat surface. Let $C$ be the bottom tether of the rope, let $D$ be the point on the ground below $A$, and let $E$ be the point directly below $B$. Triangles $\triangle CDA$ and $\triangle CEB$ are similar right triangles. By the Pythagorean Theorem $CD=8\cdot\sqrt{6}$.

Let $x$ be the length of $CB$.

$$\frac{CA}{CD}=\frac{CB}{CE}\implies \frac{20}{8\sqrt{6}}=\frac{x}{8\sqrt{6}-4\sqrt{5}}\implies x=\frac{60-\sqrt{750}}{3}$$ Therefore $a=60, b=750, c=3, a+b+c=\boxed{813}$.

Solution 2

Note that by Power of a Point, the point the unicorn is at has power $4 \cdot 20 = 80$ which implies that the tangent from that point to the tower is of length $\sqrt{80}=4\sqrt{5},$ however this is length of the rope projected into 2-D. If we let $\theta$ be the angle between the horizontal and the rope, we have that $\cos\theta=\frac{1}{5}$ which implies that $\sin\theta=\frac{2\sqrt{6}}{5}.$ Note that the portion of rope not on the tower is $4\sqrt{5} \cdot \frac{5}{2\sqrt{6}}= \frac{5\sqrt{30}}{3},$ the requested length of rope is $20-\frac{5\sqrt{30}}{3}=\frac{60-\sqrt{750}}{3}$ thus the requested sum is $\boxed{813}.$

~ Dhillonr25