Difference between revisions of "2005 AMC 12B Problems/Problem 18"
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== Solution == | == Solution == | ||
| − | For angle <math>A</math> and <math>B</math> to be acute, <math>C</math> must be between the two lines that are perpendicular to <math>\ | + | For angle <math>A</math> and <math>B</math> to be acute, <math>C</math> must be between the two lines that are perpendicular to <math>\overline{AB}</math> and contain points <math>A</math> and <math>B</math>. For angle <math>C</math> to be acute, first draw a <math>45-45-90</math> triangle with <math>\bar{AB}</math> as the hypotenuse. Note <math>C</math> cannot be inside this triangle's circumscribed circle or else |
== See also == | == See also == | ||
* [[2005 AMC 12B Problems]] | * [[2005 AMC 12B Problems]] | ||
Revision as of 18:49, 12 September 2010
Problem
Let 
and 
be points in the plane. Define 
as the region in the first quadrant consisting of those points 
such that 
is an acute triangle. What is the closest integer to the area of the region ?
Solution
For angle 
and 
to be acute, must be between the two lines that are perpendicular to 
and contain points and . For angle to be acute, first draw a 
triangle with 
as the hypotenuse. Note cannot be inside this triangle's circumscribed circle or else
