1993 AIME Problems/Problem 12

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 $\boxed{344}$ .

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

1993 AIME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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1993 AIME Problems/Problem 12

Contents

Problem

Solution

Solution 1

Solution 2

See also