Difference between revisions of "2000 AIME I Problems/Problem 14"
m |
|||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
+ | In triangle <math>ABC,</math> it is given that angles <math>B</math> and <math>C</math> are congruent. Points <math>P</math> and <math>Q</math> lie on <math>\overline{AC}</math> and <math>\overline{AB},</math> respectively, so that <math>AP = PQ = QB = BC.</math> Angle <math>ACB</math> is <math>r</math> times as large as angle <math>APQ,</math> where <math>r</math> is a positive real number. Find the greatest integer that does not exceed <math>1000r</math>. | ||
== Solution == | == Solution == | ||
+ | {{solution}} | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=2000|n=I|num-b=13|num-a=15}} |
Revision as of 18:42, 11 November 2007
Problem
In triangle it is given that angles and are congruent. Points and lie on and respectively, so that Angle is times as large as angle where is a positive real number. Find the greatest integer that does not exceed .
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |