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1993 AIME Problems/Problem 12

Revision as of 16:38, 8 November 2010 by Andrew Kim (talk | contribs) (→‎Solution 1)

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

Solution 1

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: $P_7=(14,92) P_6=(2\cdot14-0, 2\cdot92-0)=(28,184) P_5=(2\cdot28-0, 2\cdot 184-0)=(56,368) P_4=(2\cdot56-0, 2\cdot368-420)=(112,316) P_3=(2\cdot112-0, 2\cdot316-420)=(224,212) P_2=(2\cdot224-0, 2\cdot212-420)=(448,4) P_1=(2\cdot448-560, 2\cdot4-0)=(336,8)$ So the answer is 344.

Solution 2

Let $L_1$ be the $n^{th}$ roll that directly influences $P_{n + 1}$. Note that $P_7 = \cfrac{\cfrac{\cfrac{P_1 + L_1}2 + L_2}2 + \cdots}{2\ldots} = \frac {(k,m)}{64} + \frac {L_1}{64} + \frac {L_2}{32} + \frac {L_3}{16} + \frac {L_4}8 + \frac {L_5}4 + \frac {L_6}2 = (14,92)$. Then quickly checking each addend from the right to the left, we have the following information (remembering that if a point must be $(0,0)$, we can just ignore it!): for $\frac {L_6}2,\frac {L_5}4$, since all addends are nonnegative, a non-$(0,0)$ value will result in a $x$ or $y$ value greater than $14$ or $92$, respectively, and we can ignore them, for $\frac {L_4}8,\frac {L_3}{16},\frac {L_2}{32}$ in a similar way, $(0,0)$ and $(0,420)$ are the only possibilities, and for $\frac {L_1}{64}$, all three work. Also, to be in the triangle, $0\le k\le560$ and $0\le m\le420$. Since $L_1$ is the only point that can possibly influence the $x$ coordinate other than $P_1$, we look at that first. If $L_2 = (0,0)$, then $k = 2^6\cdot14 = 64\cdot14 > 40\cdot14 = 560$, so it can only be that $L_2 = (560,0)$, and $k + 560 = 2^6\cdot14\implies k = 64\cdot14 - 40\cdot14 = 24\cdot14 = 6\cdot56 = 336$. Now, considering the $y$ coordinate, note that if any of $L_3,L_4,L_5$ are $(0,0)$ ($L_3$ would influence the least, so we test that), then $\frac {L_3}{32} + \frac {L_4}{16} + \frac {L_5}8 < \frac {420}{16} + \frac {420}8 = 79\pm\epsilon < 80$, which would mean that $P_1 > 2^6\cdot(92 - 80) = 64\cdot12 > 42\cdot10 = 420\ge m$, so $L_3,L_4,L_5 = (0,420)$, and now $\frac {P_1}{64} + \frac {420}{2^5} + \frac {420}{2^4} + \frac {420}{2^3} = 92\implies P_1 = 64\cdot92 - 420(2 + 4 + 8) = 64\cdot92 - 420\cdot14$ $= 64(100 - 8) - 14^2\cdot30 = 6400 - 512 - (200 - 4)\cdot30 = 6400 - 512 - 6000 + 120$ $= - 112 + 120 = 8$, and finally, $k + m = 336 + 8 = \boxed{344}$.

See also

1993 AIME (Problems • Answer Key • Resources)
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
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