Difference between revisions of "1993 AIME Problems/Problem 12"
m |
|||
| Line 1: | Line 1: | ||
== Problem == | == 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 == | ||
| + | {{solution}} | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=1993|num-b=11|num-a=13}} | |
Revision as of 00:22, 26 March 2007
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
This problem needs a solution. If you have a solution for it, please help us out by adding it.
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 | ||
