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"
(Remove extra problem section)
 
(11 intermediate revisions by 5 users not shown)
Line 1: Line 1:
−
Problem:
+
== Problem 2 ==
 +
On [[square]] <math>ABCD</math>, point <math>E</math> lies on side <math>AD</math> and point <math>F</math> lies on side <math>BC</math>, so that <math>BE=EF=FD=30</math>. Find the area of the square <math>ABCD</math>.
  
−
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 ==
 +
Drawing the square and examining the given lengths,
 +
<asy>
 +
size(2inch, 2inch);
 +
currentpen = fontsize(8pt);
 +
pair A = (0, 0); dot(A); label("$A$", A, plain.SW);
 +
pair B = (3, 0); dot(B); label("$B$", B, plain.SE);
 +
pair C = (3, 3); dot(C); label("$C$", C, plain.NE);
 +
pair D = (0, 3); dot(D); label("$D$", D, plain.NW);
 +
pair E = (0, 1); dot(E); label("$E$", E, plain.W);
 +
pair F = (3, 2); dot(F); label("$F$", F, plain.E);
 +
label("$\frac x3$", E--A);
 +
label("$\frac x3$", F--C);
 +
label("$x$", A--B);
 +
label("$x$", C--D);
 +
label("$\frac {2x}3$", B--F);
 +
label("$\frac {2x}3$", D--E);
 +
label("$30$", B--E);
 +
label("$30$", F--E);
 +
label("$30$", F--D);
 +
draw(B--C--D--F--E--B--A--D);
 +
</asy>
 +
you find that the three segments cut the square into three equal horizontal sections. Therefore, (<math>x</math> being the side length), <math>\sqrt{x^2+(x/3)^2}=30</math>, or <math>x^2+(x/3)^2=900</math>. Solving for <math>x</math>, we get <math>x=9\sqrt{10}</math>, and <math>x^2=810.</math>
  
−
----
+
Area of the square is <math>\fbox{810}</math>.
−
Solution:
 
  
−
Set up an equation where ''x'' is the measure of the smallest angle, and ''y'' is the increase in angle measure.
+
==See also==
−
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
+
{{AIME box|year=2011|n=II|num-b=1|num-a=3}}
−
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.
+
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 16:02, 9 August 2018

Problem 2

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, [asy] size(2inch, 2inch); currentpen = fontsize(8pt); pair A = (0, 0); dot(A); label("$A$", A, plain.SW); pair B = (3, 0); dot(B); label("$B$", B, plain.SE); pair C = (3, 3); dot(C); label("$C$", C, plain.NE); pair D = (0, 3); dot(D); label("$D$", D, plain.NW); pair E = (0, 1); dot(E); label("$E$", E, plain.W); pair F = (3, 2); dot(F); label("$F$", F, plain.E); label("$\frac x3$", E--A); label("$\frac x3$", F--C); label("$x$", A--B); label("$x$", C--D); label("$\frac {2x}3$", B--F); label("$\frac {2x}3$", D--E); label("$30$", B--E); label("$30$", F--E); label("$30$", F--D); draw(B--C--D--F--E--B--A--D); [/asy] you find that the three segments cut the square into three equal horizontal sections. Therefore, ($x$ being the side length), $\sqrt{x^2+(x/3)^2}=30$, or $x^2+(x/3)^2=900$. Solving for $x$, we get $x=9\sqrt{10}$, and $x^2=810.$

Area of the square is $\fbox{810}$.

See also

2011 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_2&oldid=97082"