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2011 AIME II Problems/Problem 3: Difference between revisions

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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.
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)
Set up an equation where ''x'' is the measure of the smallest angle, and ''y'' is the increase in angle measure.
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
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.

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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