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Difference between revisions of "2011 AIME II Problems/Problem 2"
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Problem:
 
Problem:
  
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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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On square ABCD, point E lies on side AD and point F lies on side BC, so that BE=EF=FD=30. Find the area of the square ABCD.
  
 
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Solution:
 
Solution:
  
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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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Drawing the square and examining the given lengths, you find that the three segments cut the square into three equal horizontal sections (I dont know how to make a diagram so somebody please insert one)
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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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Therefore, (x being the side length), <math>sqrt(x^2+(x/3)^2)=900</math>.
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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 21:26, 30 March 2011

Problem:

On square ABCD, point E lies on side AD and point F lies on side BC, so that BE=EF=FD=30. Find the area of the square ABCD.

Solution:

Drawing the square and examining the given lengths, you find that the three segments cut the square into three equal horizontal sections (I dont know how to make a diagram so somebody please insert one) Therefore, (x being the side length), $sqrt(x^2+(x/3)^2)=900$.

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