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

Latest revision as of 12:59, 15 July 2023

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 $P_{n-1}=(x_{n-1},y_{n-1}) = (2x_n\bmod{560},\ 2y_n\bmod{420})$ Then $P_7=(14,92)$ , so $x_7=14$ and $y_7=92$ , and we get $\begin{array}{c||ccccccc} n & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\ \hline\hline x_n & 14 & 28 & 56 & 112 & 224 & 448 & 336 \\ \hline y_n & 92 & 184 & 368 & 316 & 212 & 4 & 8 \end{array}$

So the answer is $\boxed{344}$ .

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 $L_1 = (0,0)$ , then $k = 2^6\cdot14 = 64\cdot14 > 40\cdot14 = 560$ ,

so it can only be that $L_1 = (560,0)$ , 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 $L_2,L_3,L_4$ are $(0,0)$ ( $L_2$ would influence the least, so we test that),

then $\frac {L_2}{32} + \frac {L_3}{16} + \frac {L_4}8 < \frac {420}{16} + \frac {420}8 = 79\pm\epsilon < 80$ ,

which would mean that $P_1 > 2^6\cdot(92 - 80) = 64\cdot12 > 42\cdot10 = 420\ge m$ , so $L_2,L_3,L_4 = (0,420)$ ,

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 1: / Line 1: @@
 == Problem ==
+The vertices of <math>\triangle ABC</math> are <math>A = (0,0)\,</math>, <math>B = (0,420)\,</math>, and <math>C = (560,0)\,</math>.  The six faces of a die are labeled with two <math>A\,</math>'s, two <math>B\,</math>'s, and two <math>C\,</math>'s.  Point <math>P_1 = (k,m)\,</math> is chosen in the interior of <math>\triangle ABC</math>, and points <math>P_2\,</math>, <math>P_3\,</math>, <math>P_4, \dots</math> are generated by rolling the die repeatedly and applying the rule: If the die shows label <math>L\,</math>, where <math>L \in \{A, B, C\}</math>, and <math>P_n\,</math> is the most recently obtained point, then <math>P_{n + 1}^{}</math> is the midpoint of <math>\overline{P_n L}</math>.  Given that <math>P_7 = (14,92)\,</math>, what is <math>k + m\,</math>?
 == Solution ==
+===Solution 1===
+If we have points <math>(p,q)</math> and <math>(r,s)</math> and we want to find <math>(u,v)</math> so <math>(r,s)</math> is the midpoint of <math>(u,v)</math> and <math>(p,q)</math>, then <math>u=2r-p</math> and <math>v=2s-q</math>. So we start with the point they gave us and work backwards. We make sure all the coordinates stay within the triangle. We have <cmath>P_{n-1}=(x_{n-1},y_{n-1}) = (2x_n\bmod{560},\ 2y_n\bmod{420})</cmath>
+Then <math>P_7=(14,92)</math>, so <math>x_7=14</math> and <math>y_7=92</math>, and we get <cmath>\begin{array}{c||ccccccc}
+n & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\
+\hline\hline
+x_n & 14 & 28 & 56 & 112 & 224 & 448 & 336 \\
+\hline
+y_n & 92 & 184 & 368 & 316 & 212 & 4 & 8
+\end{array}</cmath>
+So the answer is <math>\boxed{344}</math>.
+===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_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</math>
+<math>\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_2,L_3,L_4</math> are <math>(0,0)</math> (<math>L_2</math> would influence the least, so we test that),
+then <math>\frac {L_2}{32} + \frac {L_3}{16} + \frac {L_4}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_2,L_3,L_4 = (0,420)</math>,
+and now <math>\frac {P_1}{64} + \frac {420}{2^5} + \frac {420}{2^4} + \frac {420}{2^3} = 92</math>
+<math>\implies P_1</math>
+<math> = 64\cdot92 - 420(2 + 4 + 8)</math>
+<math> = 64\cdot92 - 420\cdot14= 64(100 - 8) - 14^2\cdot30 </math>
+<math>= 6400 - 512 - (200 - 4)\cdot30</math>
+<math> = 6400 - 512 - 6000 + 120</math>
+<math>= - 112 + 120</math>
+<math> = 8</math>,
+and finally, <math>k + m = 336 + 8 = \boxed{344}</math>.
 == See also ==
-* [[1993 AIME Problems]]
+{{AIME box|year=1993|num-b=11|num-a=13}}
+{{MAA Notice}}

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

Page

Toolbox

Search