AIME 2014 I · 第 10 题

Solution 1

Let $F$ be the new tangency point of the two disks. The smaller disk rolled along minor arc $\overset{\frown}{AF}$ on the larger disk.

Let $\alpha = \angle AEF$, in radians. The smaller disk must then have rolled along an arc of length $5\alpha$, since the larger disk has a radius of $5$. Since all of the points on major arc $\overset{\frown}{BF}$ on the smaller disk have come into contact with the larger disk at some point during the rolling, and none of the other points on the smaller disk did,

$$\overset{\frown}{BF}=\overset{\frown}{AF}=5\alpha$$ .

Since $\overline{AC} || \overline{BD}$,

$$\angle BDF \cong \angle FEA$$ so the angles of minor arc $\overset{\frown}{BF}$ and minor arc $\overset{\frown}{AF}$ are equal, so minor arc $\overset{\frown}{BF}$ has an angle of $\alpha$. Since the smaller disk has a radius of $1$, the length of minor arc $\overset{\frown}{BF}$ is $\alpha$. This means that $5\alpha + \alpha$ equals the circumference of the smaller disk, so $6\alpha = 2\pi$, or $\alpha = \frac{\pi}{3}$.

Now, to find $\sin^2{\angle BEA}$, we construct $\triangle BDE$. Also, drop a perpendicular from $D$ to $\overline{EA}$, and call this point $X$. Since $\alpha = \frac{\pi}{3}$ and $\angle DXE$ is right,

$$DE = 6$$ $$EX = 3$$ and

$$DX = 3\sqrt{3}$$ Now drop a perpendicular from $B$ to $\overline{EA}$, and call this point $Y$. Since $\overline{BD} || \overline{EA}$,

$$XY = BD = 1$$ and

$$BY = DX = 3\sqrt{3}$$ Thus, we know that

$$EY = EX - XY = 3 - 1 = 2$$ and by using the Pythagorean Theorem on $\triangle BEY$, we get that

$$BE = \sqrt{31}$$ Thus,

$$\sin{\angle BEA} = \frac{\sqrt{27}}{\sqrt{31}}$$ so

$$\sin^2{\angle BEA} = \frac{27}{31}$$ and our answer is $27 + 31 = \boxed{058}$.

Solution 2

First, we determine how far the small circle goes. For the small circle to rotate completely around the circumference, it must rotate $5$ times (the circumference of the small circle is $2\pi$ while the larger one has a circumference of $10\pi$) plus the extra rotation the circle gets for rotating around the circle, for a total of $6$ times. Therefore, one rotation will bring point $D$ $60^\circ$ from $C$.

Now, draw $\triangle DBE$, and call

$$\angle BED = x^{\circ}$$ We know that $\overline{ED}$ is 6, and $\overline{BD}$ is 1. Since $EC || BD$,

$$\angle BDE = 60^\circ$$ By the Law of Cosines,

$$\overline{BE}^2=36+1-2\times 6\times 1\times \cos{60^\circ} = 36+1-6=31$$ and since lengths are positive,

$$\overline{BE}=\sqrt{31}$$ By the Law of Sines, we know that

$$\frac{1}{\sin{x}}=\frac{\sqrt{31}}{\sin{60^\circ}}$$ so

$$\sin{x} = \frac{\sin{60^\circ}}{\sqrt{31}} = \frac{\sqrt{93}}{62}$$ As $x$ is clearly between $0$ and $90^\circ$, $\cos{x}$ is positive. As $\cos{x}=\sqrt{1-\sin^2{x}}$,

$$\cos{x} = \frac{11\sqrt{31}}{62}$$ Now we use the angle sum formula to find the sine of $\angle BEA$:

$$\sin 60^\circ\cos x + \cos 60^\circ\sin x = \frac{\sqrt{3}}{2}\frac{11\sqrt{31}}{62}+\frac{1}{2}\frac{\sqrt{93}}{62}$$ $$= \frac{11\sqrt{93}+\sqrt{93}}{124} = \frac{12\sqrt{93}}{124} = \frac{3\sqrt{93}}{31} = \frac{3\sqrt{31}\sqrt{3}}{31} = \frac{3\sqrt{3}}{\sqrt{31}}$$ Finally, we square this to get

$$\frac{9\times 3}{31}=\frac{27}{31}$$ so our answer is $27+31=\boxed{058}$.

Solution 3

We could use coordbash to solve this problem. Assume that $E$ is the origin, and $E, A, C$ are collinear points on the x-axis. Then it's trivial to see that $D=(6\cos(60^\circ),6\sin(60^\circ))$, and since $BD$ is parallel to $AC$, $B=(6\cos(60^\circ)-1,6\sin(60^\circ))$. After simplifying, we get $B=(2,3\sqrt{3})$.

Assume there is a point $F$ such that $F=(0, 3\sqrt{3})$. Then $BF$ is parallel to $EA$, meaning $\angle FBE=\angle BEA$. $\sin(\angle FBE)$ is just $\frac{BF}{BE}=\frac{3\sqrt{3}}{\sqrt{31}}$, and squaring the fraction gives us the answer $\frac{27}{31}=\boxed{058}$.

-Marchk26