Art of Problem Solving
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

2011 AIME II Problems/Problem 2: Difference between revisions

← Older editNewer edit →
Pkustar (talk | contribs)
87 edits
No edit summary
Pkustar (talk | contribs)
87 edits
No edit summary
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 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), $sqrt(x^2+(x/3)^2)=900$.

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