1993 AIME Problems/Problem 12

Revision as of 09:16, 10 May 2008 by Dgreenb801 (talk | contribs)

Problem

The vertices of $\triangle ABC$ are $A = (0,0)\,$, $B = (0,420)\,$, and $C = (560,0)\,$. The six faces of a die are labeled with two $A\,$'s, two $B\,$'s, and two $C\,$'s. Point $P_1 = (k,m)\,$ is chosen in the interior of $\triangle ABC$, and points $P_2\,$, $P_3\,$, $P_4, \dots$ are generated by rolling the die repeatedly and applying the rule: If the die shows label $L\,$, where $L \in \{A, B, C\}$, and $P_n\,$ is the most recently obtained point, then $P_{n + 1}^{}$ is the midpoint of $\overline{P_n L}$. Given that $P_7 = (14,92)\,$, what is $k + m\,$?

Solution

If we have points (p,q) and (r,s) and we want to find (u,v) so (r,s) is the midpoint of (u,v) and (p,q), then u=2r-p and v=2s-q. So we start with the point they gave us and work backwards. We make sure all the coordinates stay within the triangle. We have: $P_7=(14,92) P_6=(2\cdot14-0, 2\cdot92-0)=(28,184) P_5=(2\cdot28-0, 2\cdot 184-0)=(56,368) P_4=(2\cdot56-0, 2\codt368-420)=(112,316) P_3=(2\cdot112-0, 2\cdot316-420)=(224,212) P_2=(2\cdot224-0, 2\cdot212-420)=(448,4) P_1=(2\cdot448-560, 2\cdot4-0)=(336,8)$ (Error compiling LaTeX. Unknown error_msg) So the answer is 344.

See also

1993 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions