## Monday, October 10, 2016

### 2012 Harder Mathcounts State Target Questions

Check out Mathcounts here -- the best competition math program for middle schoolers up to the
state and national level.

# 6: A semicircle and a circle are placed inside a square with sides of length 4 cm, as shown. The circle is tangent to two adjacent sides of the square and to the semicircle. The diameter of the semicircle is a side of the square. In centimeters, what is the radius of the circle? Express your answer as a decimal to the nearest hundredth. [2012 Mathcounts State Target #6]

#6:  Solution:
Using Pythagorean theory: (2 + r)2 = (4-r)2 + ( 2- r)2
4 + 4r + r2 = 16 - 8r + r2 + 4 - 4r + r2
r2 - 16 r + 16 = 0
Using the quadratic formula You have 8 ± 4√ 3
Only 8 - 4 3 = 1.07 works

There is a Mathcounts Mini #34 on the same question. Check that out !!

The above question looks very similar to this year's AMC-10 B #22, so try that one.
(cover the answer choices so it's more like Mathcounts)

2014 AMC-10 B problem #22

#8: In one roll of four standard, six-sided dice, what is the probability of rolling exactly three different numbers? Express your answer as a common fraction. [2012 Mathcounts State Target #8]

Solution I : Permutation method
If order matters, there are 6 * 5 * 4 * 1 ways to choose the number, 1 being the same number as one of the previous one.
Let's say if you choose 3 1 4 1.

Now for the placement of those 4 numbers on the 4 different dice. There are 4C2 ways to place where the two "1" will
be positioned so the answer is : $$\dfrac {6\times 5\times 4\times 1\times 4C2} {6^{4}}$$ = $$\dfrac{5}{9}$$
Solution II:  Combination method

There are 6C3 = 20 ways to choose the three numbers.

There are 3 ways that the number can be repeated. [For example: If you choose 1, 2, and 3, the fourth number could be 1, 2 or 3.]

There are $$\dfrac {4!} {2!}$$ =12
ways to arrange the chosen 4 numbers.[same method when you arrange AABC]
So the answer is$$\dfrac{20* 3 *12}{6^4}$$ = $$\dfrac{5}{9}$$