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 "2005 AMC 10A Problems/Problem 23"
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
<math>BCDE</math> is a [[square (geometry) | square]]. [[Point]] <math>A</math> is chosen outside of <math>BCDE</math> such that [[angle]] <math>BAC= 120^\circ</math> and <math>AB=AC</math>. Point <math>F</math> is chosen inside <math>BCDE</math> such that the [[triangle]]s <math>ABC</math> and <math>FCD</math> are [[congruent (geometry) | congruent]]. If <math>AF=20</math>, compute the [[area]] of <math>BCDE</math>.
+
Let <math>AB</math> be a diameter of a circle and let <math>C</math> be a point on <math>AB</math> with <math>2\cdot AC=BC</math>. Let <math>D</math> and <math>E</math> be points on the circle such that <math>DC \perp AB</math> and <math>DE</math> is a second diameter. What is the ratio of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math>?
  
 
<math> \mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3} </math>
 
<math> \mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3} </math>
  
 
==Solution==
 
==Solution==
−
http://img501.imageshack.us/img501/5625/tempym3.png
+
http://img443.imageshack.us/img443/8034/circlenc1.png
  
−
Let a side of <math>BCDE</math> be <math>x</math>.
+
{{solution}}
−
 
 
−
 
 
−
<math>AS=ST+AT=\frac{x}{2}+\frac{x\sqrt{3}}{6}</math>
 
−
 
 
−
<math>SF=ST-AT=\frac{x}{2}-\frac{x\sqrt{3}}{6}</math>
 
−
 
 
−
<math>\left(\frac{x}{2}+\frac{x\sqrt{3}}{6}\right)^2+\left(\frac{x}{2}-\frac{x\sqrt{3}}{6}\right)^2=20^2</math>
 
−
 
 
−
<math>x=10\sqrt{6}</math>
 
−
 
 
−
 
 
−
The area of <math>BCDE</math> is <math>600</math>.
 
  
 
==See also==
 
==See also==

Revision as of 12:10, 6 May 2007

Problem

Let $AB$ be a diameter of a circle and let $C$ be a point on $AB$ with $2\cdot AC=BC$. Let $D$ and $E$ be points on the circle such that $DC \perp AB$ and $DE$ is a second diameter. What is the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$?

$\mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3}$

Solution

http://img443.imageshack.us/img443/8034/circlenc1.png

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_23&oldid=14841"