Difference between revisions of "2014 AIME I Problems/Problem 10"

Revision as of 13:37, 15 March 2014

Problem 10

A disk with radius $1$ is externally tangent to a disk with radius $5$ . Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\circ$ . That is, if the center of the smaller disk has moved to the point $D$ , and the point on the smaller disk that began at $A$ has now moved to point $B$ , then $\overline{AC}$ is parallel to $\overline{BD}$ . Then $\sin^2(\angle BEA)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

[diagram needed]

Let $F$ be the new tangency point of the two disks. The smaller disk rolled along minor arc $\overarc{AF}$ 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 it has a radius of $5$ . Since all of the points on major arc $\overarc{BF}$ 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, the length of major arc $\overarc{BF}$ equals the length of minor arc $\overarc{AF}$ , or $5\alpha$ . Since $\overline{AC} || \overline{BD}$ , $\angle BDF \cong \angle FEA$ , so the angles of minor arc $\overarc{BF}$ and minor arc $\overarc{AF}$ are equal, so minor arc $\overarc{BF}$ has an angle of $\alpha$ . Since the smaller disk has a radius of $1$ , the length of minor arc $\overarc{BF}$ 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 $\alpha = \frac{\pi}{3}$ , $\angle DXE$ is right, and $DE = 6$ , $EX = 3$ 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 $27 + 31 = \boxed{058}$ .

@@ Line 4: / Line 4: @@
 == Solution ==
-<asy>
-pair E = (0,0);
-pair C = (6,0);
-pair X = (3,0);
-pair Y = (2,0);
-pair A = (5,0);
-pair D = (3,sqrt(27))
-pair B = (2,sqrt(27))
-pair F = (5/3,5sqrt(27)/6)
-draw(E--C);
-draw(C--D);
-draw(E--D);
-draw(B--E);
-draw(B--D);
-draw(D--X);
-draw(B--Y);
-dotfactor = 3;
-dot("$A$",A,dir(135));
-dot("$B$",B,dir(215));
-dot("$D$",D,dir(305));
-dot("$F$",F,dir(45));
-dot("$E$",E,dir(135));
-dot("$C$",C,dir(215));
-dot("$X$",X,dir(305));
-dot("$Y$",Y,dir(45));
-</asy>
 [diagram needed]

2014 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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Difference between revisions of "2014 AIME I Problems/Problem 10"

Revision as of 13:37, 15 March 2014

Problem 10

Solution

See also