Difference between revisions of "2011 AIME II Problems/Problem 3"

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Problem:
 
Problem:
  
In triangle ABC, AB=(20/11)AC. The angle bisector of angle A intersects BC at point D, and point M is the midpoint of AD. Let P be the point of intersection of AC and the line BM. The ratio of CP to PA can be expresses in the form m/n, where m and n are relatively prime positive integers. Find m+n.
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The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle.
  
 
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Solution:
 
Solution:
  
I have no idea, but can be solved with relatively simple geometry. (I ran out of time to solve this one)
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Set up an equation where ''x'' is the measure of the smallest angle, and ''y'' is the increase in angle measure.
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You get 18''x''+153''y''=2880, because (x+0y)+(x+y)+(x+2y)+...(x+17y)=18''x''+153''y''=the total angle measures of all of the angles in an 18-gon=2880
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Solving the equation for integer values (or a formula that I don't know) you get ''x''=7, and ''y''=18
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The smallest angle is therefore 7.

Revision as of 21:21, 30 March 2011

Problem:

The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle.


Solution:

Set up an equation where x is the measure of the smallest angle, and y is the increase in angle measure. You get 18x+153y=2880, because (x+0y)+(x+y)+(x+2y)+...(x+17y)=18x+153y=the total angle measures of all of the angles in an 18-gon=2880 Solving the equation for integer values (or a formula that I don't know) you get x=7, and y=18 The smallest angle is therefore 7.

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