**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 + r

^{2}= 16 - 8r + r

^{2}+ 4 - 4r + r

^{2}

r

^{2}- 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 \(\dfrac {4!} {2!}\) =12ways to arrange the chosen 4 numbers.[same method when you arrange AABC]