AIME 1993 · 第 12 题

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