Author Topic: 50 AMC Lectures Problems Book 2  (Read 17280 times)

yongcheng3315

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50 AMC Lectures Problems Book 2
« on: January 08, 2013, 06:31:08 PM »
50 AMC Lectures Problems Book 2 (PDF FILE)

$14.99

http://www.mymathcounts.com/50-AMC-Lectures-Problems-Book-2.php


This book contains over 500 problems (with solutions) accompanying the lectures 26 through 50 of our 50 Lectures for American Mathematics Competitions books covering the following topics:

Counting                        
How to solve inequality   
Polynomial                        
Geometric sequence      
Binomial theorem      
Gaussian function   
Geometry Parallelograms
Logarithms      
 Word problems (distance and speed)   
Geometry Trapezoids      
Substitution method   
Analytic Geometry Circles and lines   
Probability   
System of equations      
Diophantine equations   
Geometry Circles      
Geometry Menelaus and Ceva Theorems    
Cauchy inequality      
Congruence      
Series and sequences      
Recursion method      
Trigonometry applications   
Analytic Geometry Ellipse   
Analytic Geometry Hyperbola      
Complex numbers      
« Last Edit: January 16, 2013, 07:58:11 PM by yongcheng3315 »

dmz

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Re: 50 AMC Lectures Problems Book 2
« Reply #1 on: December 09, 2013, 10:56:34 AM »
Could you please check problem 12 on Chapter 35. My son says the problem might be incorrectly stated.

Thanks,

Deming

yongcheng3315

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Re: 50 AMC Lectures Problems Book 2
« Reply #2 on: December 09, 2013, 03:48:17 PM »
Thanks. I checked it should be DC = 4. Good thing is that we can verify this because we have the solution. Thanks again.

dmz

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Re: 50 AMC Lectures Problems Book 2
« Reply #3 on: December 10, 2013, 04:09:15 PM »
Yes, DC=4 works, my son says. Many thanks for your prompt reply!

dmz

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Re: 50 AMC Lectures Problems Book 2
« Reply #4 on: December 10, 2013, 08:04:59 PM »
Yes, DC=4 works, my son says. Many thanks for your prompt reply!
He also came up with a simpler method to solve the problem:

Let the intersection point of the two diagonals be E. Triangles CDE and ABE are similar. If we let DE = x, EB = 6-x. By similar triangles, x/6-x = 4/8=1/2. Solve for x=2. Thus EB = 4. Triangle ABE is right and AB = 8, EB = 4. This is a 30-60-90 triangle because of the side lengths. Therefore angle CAB = 30 degrees.

yongcheng3315

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Re: 50 AMC Lectures Problems Book 2
« Reply #5 on: December 12, 2013, 07:57:11 AM »
Thanks a lot. This is a good solution. Do you mind if we add this solution when we revise the book?