Solution for probability Problem for area

Thanks everyone for trying the problem and posting in the comments, and congrats for those of you who got the right answer!


Problem:
Suppose a and b are single-digit positive integers chosen independently and at random. What is the probability that the point (a,b) lies above the parabola y=ax^2-bx.


Solution:
Because the point (a,b) needs to be above the parabola ax^2-bx, the value of y has to be greater than the function itself. Therefore we get y>ax^2-bx. Next, we're finding the probability that point (a,b) is above the parabola, so we just plug in a for x and b for y. Thus, we get b>a*a^2-b*a = a^3-ab. Now we can solve for b in terms of a.


b>a^3-ab 
b+ab>a^3
b(1+a)>a^3
b>(a^3)/(1+a) (We can divide by (1+a) because in the problem it states a is positive)


Now we have the inequality b>(a^3)/(1+a). From this you can see that (a^3)/(1+a) is an increasing function, since a^3 grows faster than 1+a. From this, you see that the largest integer you can plug in for a is 3 because plugging in a 4 for a requires b to be a double digit integer. We have restricted a to 1≤a≤3. Now, we plug in values of a and see what positive single digit integers of b can make the inequality true.


If we plug in 1 for a, we get b>1/4. All numbers 1-9 can be plugged in for b for the inequality to hold. Which gives 9 possible ordered pairs. 
If we plug in 2 for a, we get b>8/3. Numbers 3-9 can be plugged in for b which gives 7 ordered pairs.
If we plug in 3 for a, we get b>27/4. Number 7-9 can be plugged in for b which gives 3 ordered pairs.


In total, the number of possible ordered pairs is 9+7+3=19. This is out of the total number of 9*9=81 ordered pairs. Hence the probability is 19/81.


Here's the next problem:
Problem:
A square of side length 1 and a circle of radius (√3)/3 share the same center. What is the area inside the circle, but outside the square?


Good luck! and remember calculators are not allowed. 

AMC12A 2011 Problem #14

I've gotten several comments stating that the problems I've been doing have been too simple, and that they want more of a challenge. For this reason, I have decided to skip some of the simpler problems on the test, and move on to a more challenging section.
Another change I will be making, is I will now be posting the next day's question at the bottom of each post, for those of you who want to try and do the problems by yourself.


Problem:
Suppose a and b are single-digit positive integers chosen independently and at random. What is the probability that the point (a,b) lies above the parabola y=ax^2-bx.


Good luck! and remember, calculators are not allowed.

AMC12A 2011 Problem #2

I don't think anyone will have trouble doing this problem, but I wrote a solution anyway.


Problem:
There are 5 coins placed flat on a table according to the figure. What is the order of the coins from top to bottom.

Solution:
From looking at this picture, you can easily tell that coin C is the coin at the very top because it overlaps coins D, E, and A, and coins A and D are overlapping coin B. Coin D is overlapping coin A and B and overlapped by coin E. That makes coin E the second in the order of top to bottom and coin D the third. Because coin A overlaps coin B, coin A is going to be the 4th and coin B is the 5th. Therefore the order is (C,E,D,A,B).

AMC12A 2011 Problem #1

I'm going to do math problems from various tests such as the AMC or the SAT on this blog, and hopefully it can help people out with tests. If there are any requests to do specific problems just leave them in the comments.
I'm starting off with problems from AMC12A that we had earlier this year. 


Problem:
A cell phone plans costs $20 each month, plus 5 cents per text message sent, plus 10 cents for each minute used over 30 hours. In January Michelle sent 100 text messages and talked for 30.5 hours. How much did she have to pay?


Solution:
$20.00+($0.05*100)+($0.10*30)=$28.00


Because the cell phone plan costs $20 each month, we can start out the cost at $20. Michelle sent 100 text messages in that month, and the cost of text messaging is 5 cents per text message. Thus, the extra cost from text messaging is $0.05*100 which is equal to $5.00. Now the total cost is $25. The problem states that there is an additional cost of 10 cents for each minute used over 30 hours. Michelle talked on the phone for 30.5 hours which means she spent an additional .5 hours or 30 minutes on the phone. Because the cost of an extra minute on the phone is 10 cents, we multiply $0.10*30 which equals $3.00. Now we add that to the total cost and we get $28.00.