Question

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 $P_2,$ $P_3,$ $P_4,$ $\ldots$ 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 $\overline{P_nL}.$ Given that $P_7=(14,92),$ what is $k+m?$

Answer

Answer: 344

Concepts:coordinate geometry power of 2 recursion

Difficulty rating: 2600

Small Hint:

Reverse the six midpoint operations by multiplying the equation for $P_7$ by $64$

Big Hint:

The six rolled vertices receive the distinct weights $1,$ $2,$ $4,$ $8,$ $16,$ and $32$ ; use the $x$ -coordinate first

Solution:

Let $X$ and $Y$ be the sums of the weights $1,$ $2,$ $4,$ $8,$ $16,$ and $32$ assigned to rolls of $C$ and $B,$ respectively. Iterating the midpoint rule gives $\begin{aligned}64P_7&=P_1+X(560,0)\\&\quad+Y(0,420).\end{aligned}$ Hence $k=896-560X.$ Because $P_1$ is interior, $0<k<560,$ forcing $X=1$ and $k=336.$ The triangle inequality for its coordinates then gives $0<m<168.$ Since $m=5888-420Y,$ the only possible integer $Y$ is $14,$ giving $m=8.$ Therefore $k+m=344.$

1993 AIME Problem 12

Answer: 344

Difficulty rating: 2600

Small Hint:

Big Hint:

Solution:

Problem 12 in Other Years