Difference between revisions of "1993 AIME Problems/Problem 12"
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===Solution 2=== | ===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> | + | 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_1 = (0,0)</math>, then <math>k = 2^6\cdot14 = 64\cdot14 > 40\cdot14 = 560</math>, so it can only be that <math>L_1 = (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>= 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>. | <math>= - 112 + 120 = 8</math>, and finally, <math>k + m = 336 + 8 = \boxed{344}</math>. |
Revision as of 21:18, 2 January 2019
Problem
The vertices of are , , and . The six faces of a die are labeled with two 's, two 's, and two 's. Point is chosen in the interior of , and points , , are generated by rolling the die repeatedly and applying the rule: If the die shows label , where , and is the most recently obtained point, then is the midpoint of . Given that , what is ?
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: So the answer is .
Solution 2
Let be the roll that directly influences . Note that . Then quickly checking each addend from the right to the left, we have the following information (remembering that if a point must be , we can just ignore it!): for , since all addends are nonnegative, a non- value will result in a or value greater than or , respectively, and we can ignore them, for in a similar way, and are the only possibilities, and for , all three work. Also, to be in the triangle, and . Since is the only point that can possibly influence the coordinate other than , we look at that first. If , then , so it can only be that , and . Now, considering the coordinate, note that if any of are ( would influence the least, so we test that), then , which would mean that , so , and now , and finally, .
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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