Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2011 AIME II Problems/Problem 2"
Line 1: Line 1:
 
Problem:
 
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.
+
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:
 
Solution:
  
−
Set up an equation where ''x'' is the measure of the smallest angle, and ''y'' is the increase in angle measure.
+
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)
−
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
+
Therefore, (x being the side length), <math>sqrt(x^2+(x/3)^2)=900</math>.
−
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 22:26, 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), $sqrt(x^2+(x/3)^2)=900$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_2&oldid=37882"