Difference between revisions of "1993 AIME Problems/Problem 12"

Revision as of 17:01, 26 March 2009

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: $P_7=(14,92) P_6=(2\cdot14-0, 2\cdot92-0)=(28,184) P_5=(2\cdot28-0, 2\cdot 184-0)=(56,368) P_4=(2\cdot56-0, 2\codt368-420)=(112,316) P_3=(2\cdot112-0, 2\cdot316-420)=(224,212) P_2=(2\cdot224-0, 2\cdot212-420)=(448,4) P_1=(2\cdot448-560, 2\cdot4-0)=(336,8)$ (Error compiling LaTeX. Unknown error_msg) So the answer is 344.

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, $k + m = 336 + 8 = \boxed{344}$ .

@@ Line 3: / Line 3: @@
 == 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:
 <math>P_7=(14,92)
@@ Line 12: / Line 13: @@
 P_1=(2\cdot448-560, 2\cdot4-0)=(336,8)</math>
 So the answer is 344.
+===Solution 2===
+Let <math>L_1</math> be the <math>n^{th}</math> roll that directly influences <math>P_{n + 1}</math>. Note that <math>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)</math>. Then quickly checking each addend from the right to the left, we have the following information (remembering that if a point must be <math>(0,0)</math>, we can just ignore it!): for <math>\frac {L_6}2,\frac {L_5}4</math>, since all addends are nonnegative, a non-<math>(0,0)</math> value will result in a <math>x</math> or <math>y</math> value greater than <math>14</math> or <math>92</math>, respectively, and we can ignore them, for <math>\frac {L_4}8,\frac {L_3}{16},\frac {L_2}{32}</math> in a similar way, <math>(0,0)</math> and <math>(0,420)</math> are the only possibilities, and for <math>\frac {L_1}{64}</math>, all three work. Also, to be in the triangle, <math>0\le k\le560</math> and <math>0\le m\le420</math>. Since <math>L_1</math> is the only point that can possibly influence the <math>x</math> coordinate other than <math>P_1</math>, we look at that first. If <math>L_2 = (0,0)</math>, then <math>k = 2^6\cdot14 = 64\cdot14 > 40\cdot14 = 560</math>, so it can only be that <math>L_2 = (560,0)</math>, and <math>k + 560 = 2^6\cdot14\implies k = 64\cdot14 - 40\cdot14 = 24\cdot14 = 6\cdot56 = 336</math>. Now, considering the <math>y</math> coordinate, note that if any of <math>L_3,L_4,L_5</math> are <math>(0,0)</math> (<math>L_3</math> would influence the least, so we test that), then <math>\frac {L_3}{32} + \frac {L_4}{16} + \frac {L_5}8 < \frac {420}{16} + \frac {420}8 = 79\pm\epsilon < 80</math>, which would mean that <math>P_1 > 2^6\cdot(92 - 80) = 64\cdot12 > 42\cdot10 = 420\ge m</math>, so <math>L_3,L_4,L_5 = (0,420)</math>, and now <math>\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</math>
+<math>= 64(100 - 8) - 14^2\cdot30 = 6400 - 512 - (200 - 4)\cdot30 = 6400 - 512 - 6000 + 120</math>
+<math>= - 112 + 120 = 8</math>, and finally, <math>k + m = 336 + 8 = \boxed{344}</math>.
 == See also ==
 {{AIME box|year=1993|num-b=11|num-a=13}}

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