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

So the answer is\(\dfrac{20* 3 *12}{6^4}\) = \(\dfrac{5}{9}\)

Notes to 2018 Mathcounts chapter more interesting problems

2018 Mathcounts Chapter Spring problems : solutions down below 
Thanks to a boy mathlete who tried these problems and e-mail me for feedback. 

Please try these problems first before reading the explanations. :D

#25 : Three employees split a bonus valued at some number of dollars. Arman first receives $10 more than one third of the total amount. Bernardo then receives $3 more than one half of what was left. Carson receives the remaining $25. What is the total dollar value of the bonus?

#27: For a particular list of four distinct integers the mean, median and range have the same value. If the least integer in the list is 10, what is the greatest value for an integer in the list?

#29: There are two values of x such that \( |\dfrac {x-2018} {x-2019}|=\dfrac {1} {6}\). . What is the absolute difference between these two values of x? Express your answer as a common fraction.

Target #8 : Four congruent circles of radius 2 cm intersect with their centers at intersection points as shown. What is the area of the shaded region? Express your answer in terms of π.

#25 : It's easier if you go backward and use inverse operations to solve this question. 
(25 + 3) *2 = 56 and 56 + 10 = 66, which is \( \dfrac {2} {3}\) of the original bonus value, or 
what is left after \( \dfrac {1} {3}\) was given out. 

\( \dfrac {2} {3}\) of bonus is 66 dollars, so the answer is 99. 

#27: Let the average be x and the three other numbers be a, b, c and \( a < b < c \).
The least number is 10 (given), so 
\( 10+a+b+c \) = \( 4x \)---> equation 1 
\( \dfrac {a+b} {2}\) = \( x \) (how to find the median), so \( a+b\) = \( 2x \) --- (2)
\(c-10 = x\) (given because it's the range), \(c= x + 10\) ---(3)
Substitute (2) and (3) to equation one and you have \(10 + 2x + x + 10 = 4x\), so \(x = 20 \)
\(C = 20 + 10 = 30\), the answer 

#28 : Let \(x - 2018 = y\) , then \(x - 2019 = y -1\)
We then have either 
\( |\dfrac {y} {y -1}|=\dfrac {1} {6}\)  \(\rightarrow\) \(y\) = \( \dfrac {-1} {5}\)

or \( |\dfrac {y} {y-1}|=\dfrac {-1} {6}\) \(\rightarrow\) \(y\) = \( \dfrac {1} {7}\)

Their positive difference is \( \dfrac {12} {35}\) , the answer. 


Target #8 : Thanks to a 5th grader girl mathlete's solution: 

If you move parts around, you'll see the answer is exactly a semi-circle with a radius 2, so the answer is \(2\pi\), the answer. 






or check out another solution from me: