2013 Mathcounts State Prep : Angle Bisect and Trisect Questions

Proof : 
2y = 2x + b (exterior angle = the sum of the other two interior angles)
--- equation I

y = x + a (same reasoning as above)
--- equation II

Plug in the first equation and you have
2y = 2x + 2a = 2x + b

2a = b

  
Here is the link to the Angle Bisector Theorem, including the proof and one example.

Angle ABC and ACB are both trisected into three congruent angles of x and y respectively. 
If given angle "a" value, find angle c and angle b.  

Solution: 3x + 3y = 180 - a

From there, it's very easy to find the value of x + y
and get angle c, using 180 - (x + y).

Also, once you get 2x + 2y, you can use the same method -- 180 - (2x + 2y) to get angle b

Counting II : Practice Counting Systematically

Counting Coins 
Lots of similar questions appear on Mathcounts tests. Be careful when there are limits, for example, the sum of the coins do not exceed ___ or you have to have at least one for each type, etc...

Partition 

The Hockey Stick Identity from Art of Problem Solving
Same as "Sticks and Stones", or "Stars and Bars" methods

Applicable question: Mathcounts 2008 Chapter #9--During football season, 25 teams are ranked by three reporters (Alice, Bob and Cecil). Each reporter assigned all 25 integers (1 through 25) when ranking the twenty-five teams. A team earns 25 points for each first-place ranking, 24 points for each second-place ranking, and so on, getting one point for a 25th place ranking. The Hedgehogs earned 27 total points from the three reporters. How many different ways could the three reporters have assigned their rankings for the Hedgehogs? One such way to be included is Alice - 14th place, Bob - 17th place and Cecil - 20th place.

Solution I :
Let's see how it could be ranked for Hedgehogs to get 27 points from the three reporters.

A          B         C

1           1        25     1 way for C to get 25 points and the other two combined to get 2 points

1           2        24

2           1        24    2 ways for C to get 24 points and the other two combined to get 3 points.

1           3        23

3           1        23

2           2        23    3 ways for C to get 23 and the other two combined to get 4 points.

.

.

25        1         1     25 ways for C to get 1 point and the other two combined to get 26 points.

1 + 2 + 3 + ...25 = \(\dfrac {25\times 26} {2}=325\)

Solution II:

Use 26C2. Look at this questions as A + B + C = 27 and A, B C are natural numbers. To split the objects into three groups (for Alice, Bob, and Cecil), we must put 2 dividers between the 27 objects. (You can't grant "0" point.) There are 26 places to put the dividers, so 26C2 and the answer is \(\dfrac {26\times 25} {2}=325\)

2013 Mathcounts State Prep: Partition Questions

#24 2001 Mathcounts Sate Sprint Round: The number 4 can be written as a sum of one or more natural numbers in exactly five ways: 4, 3+1, 2 + 1 + 1, 2 +2 and 1 + 1 + 1 + 1; and so 4 is said to have five partitions. What is the number of partitions for the number 7?

#2: Extra: Try partition the number 5 and the number 8. 

Solution: 

#24: You can solve this problem using the same technique as counting coins:

Counting coins questions

7     6    5    4    3    2    1

1                                           1 way

       1                                    1 way

             1                 1           2 ways ( 5 + 2 or 5 + 1 + 1)

                   1    1                  1 way

                   1           1           2 ways ( 4 + 2 + 1 or 4 + 1 + 1 + 1)

                         2     0   1      1 ways

                         1     2           3 ways ( 3 + 2 + 2, 3 + 2 + 1 + 1 and 3 + 1 + 1 + 1 + 1)

                                3           4 ways (2 + 2 + 2 + 1, 2 + 2 + 3 ones, 2 + 5 ones and 7 ones.)

Total 15 ways.

The partitions of 5 are listed below (There are 7 ways total.):

5   4   3   2   1

1                           1 way

     1                      1 way

          1    1           2 ways  (3 + 2 and 3 + 1 + 1)

                2           3 ways  (2 + 2 + 1, 2 + 1 + 1 + 1 and 1 + 1 + 1 + 1 + 1)

There are 22 ways to partition the number 8.