2013 Mathcounts State Prep : Inscribed Circle Radius and Circumscribed Circle Radius of a right triangle

Question: \(\Delta\) ABC is a right triangle and a, b, c are three sides, c being the hypotenuse.
What is a. the radius of the inscribed circle and
b. the radius of the circumscribed circle? 

Solution a :
Area of the right \(\Delta\)ABC =  \(\dfrac {ab} {2}\) = \(\dfrac {\left( a+b+c\right) \times r} {2}\)
r =\(\dfrac {ab} {a+b+c}\)

Solution b: 
In any right triangle, the circumscribed diameter is the same as the hypotenuse, so the circumscribed radius is\(\dfrac {1} {2}\) of the hypotenuse, in this case \(\dfrac {1} {2}\) of c or \(\dfrac {1} {2}\) of \(\overline {AC}\)



Some other observations: 
A. If you only know what the three vertices of the right triangle are on a Cartesian plane, you can use distance formula to get each side length and from there find the radius.

B.In right \(\Delta\)ABC , \(\overline {AC}\) is the hypotenuse.
If you connect B to the median of \(\overline {AC}\), then \(\overline {BD}\) = \(\overline {AD}\) = \(\overline {CD}\) = radius of the circumscribed circle

2013 Mathcounts State Prep: Counting Problems

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#1: 2006 Mathcounts state : My three-digit code is 023. Reckha can’t choose a code that is the same as mine in two or more of the three digit-positions, nor that is the same as mine except for switching the positions of two digits (so 320 and 203, for example, are forbidden, but 302 is fine). Reckha can otherwise choose any three-digit code where each digit is in the set {0, 1, 2, ..., 9}. How many codes are available for Reckha?

Solution:
Do complementary counting. Use total possible ways minus those that are not allowed. 
 
You can't use two or more of the numbers that are at the same position (given) as 203, which means that you can't have 0 __ 3, __ 23, or 02__.

For each of the __, you can use 10 digits (from 0, 1, 2 ... to 9) so 10 + 10 + 10 = 30.

However, you repeat 023 three times in each case so you need to minus 2 back so not to over count.
30-2 = 28

Also, you can't just switch two digits, which means 320, 203 and 032 are not allowed.  { but 302 and 230 are allowed since you are switching all the digits }

There are 10 x 10 x 10 = 1000 digits total and 1000 - 28 - 3 = 969   The answer

#2: 2011 AMC-8 # 23: How many 4-digit positive integers have four different digits where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?

Solution: 
For the integer to be a multiple of 5, there are two cases: 

Case I: The unit digit is 5 :  __ __ __ 5

There are 4 numbers to choose for the thousandth digit [since 5 is the largest digit and you can't have "0" for the leading digit so there are 4 numbers 1, 2, 3, 4 that you can use], 4 numbers to choose for the hundredth

digit (0 and one of the remaining 3 numbers that are not the same number as the one in the thousandth digit) and 3 numbers to choose for the tenth digit (the remaining 3 numbers) so total 4 x 4 x 3 = 48 ways

Case II:  The unit digit is 0: __ __ __ 0

One of the remaining three numbers has to be 5, and for the remaining 2 numbers, there are 4C2 = 6 ways

to choose the 2 numbers from the numbers 1, 2, 3 or 4.
There are 3! arrangements for the three numbers so 6 x 3! = 36

48 + 36 = 84 ways

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.