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: 

Geometric mean of a right triangle

Geometric mean of a right triangle :

Video tutorial 

Notes

Practice I 

Practice II : focus on the height (or altitude)

Practice III : focus on the legs 

The Largest Rectangle Inscribed in Any Triangle

From Mathcounts Mini : Maximum area of inscribed rectangles and triangles

\(\Delta EHI\sim\Delta EFG\) \(\rightarrow\) \(\dfrac {a} {c}=\dfrac {d-b} {d}\)\(\rightarrow\) \(a=\dfrac {c\left( d-b\right) } {d}=\dfrac {-c\left( b-d\right) } {d}\)

We are going to find out what the largest area of a rectangle is with the side length a and b.
It can be shown that by substituting the side length "a" with the previous equation + completing the square that the largest area is half of the area of the triangle the rectangle is embedded.

\(a\times b=\dfrac {-c\left( b-d\right) \times b} {d}=\dfrac {-c\left( b^{2}-bd\right)} {d}= \dfrac {-c\left( b-\dfrac {1} {2}d\right) ^{2}+\dfrac {1} {4}dc} {d}\).

From there, you know that when \(b= \dfrac {1} {2}d\), it will give you the largest area, which is \(\dfrac {1} {4}dc\).

\(a=\dfrac {-c\left( b-d\right) } {d}= \dfrac {-c\left( \dfrac {1} {2}d-d\right) } {d}=\dfrac {c\left( d-\dfrac {1} {2}d\right) } {d}=\dfrac {1} {2}c\).

Thus, the maximum rectangle area occurs when the midpoints of two of the sides of the triangle were joined to make a side of the rectangle and its area is thus 50% or half of the area of the triangle or 1/4 of the base times height.

Proof without words from Mr. Rusczyk 

Try using different types of triangles to experiment and see for yourself.
Paper folding is fun !!!!!
It's very cool :D

Dimensional Change questions I:

Questions written by Willie, a volunteer.  Answer key and detailed solutions below.

1a. There is a regular cylinder, which has a height equal to its radius. If the radius and height are both increased by 50%, by what % does the total volume of the cylinder increase?

1b. If the radius and height are both decreased by 10%, by what % does the total volume of the cylinder decrease?

1c. If the radius is increased by 20% and the height is decreased by 40%, what % of the volume of the original cylinder does the volume of the new cylinder represent?

1d. If the radius is increased by 40% and the height is decreased by 20%, what % of the volume of the original cylinder does the volume of the new cylinder represent?

1e. If the height is increased by 125%, what % does the radius need to be decreased by for the volume to remain the same?

2. If the side of a cube is increased by 50%, by what % does the total surface area of the cube increase?

3a. If the volume of a cube increases by 72.8%, by what % does the total surface area of the cube increase?

3b. By what % did the side length of the cube increase?

4. You have a collection of cylinders, all having a radius of 5. The first cylinder has a height of 2, the second has a height of 4, the third a height of 6, etc. The last cylinder has a height of 50. What is the sum of the volumes of all the cylinders (express your answer in terms of pi)?

Answer key: (Each question should not take you more than 30 seconds to solve if you really understand the concepts involved.)

 

 

1a.  The volume of a cylinder is πr²x h (height). The radius itself will be squared and the height stays at constant ratio. The volume will increased thus (1.5)³ - 1³ -- the original 100% of the volume = 2.375

=237.5%

1b.  Like the previous question: 1³ - 0.9³ [when it's discount/percentage decrease, you use the 100% or 1 - the discount/decrease percentage] = 0.271 =  27.1% decrease

1c.  1.2² [100% + 20% increase = 1.2] x 0.6 [100% -40% = 0.6] = 0.864  or  

86.4% of the original volume

1d.  1.4² [100% + 40% increase = 1.4] x 0.8 [100% -20% = 0.8] = 1.568 = 156.8% of the original volume

1e.  When the height of a cylinder is increased 125%, the total volume is is 225% of the original cylinder, or 9/4.

Since the radius is used two times (or squared), it has to decrease 4/9^1/2 = 2/3 for the new cylinder to have the same volume as the old one. [9/4 times 4/9 = 1 or the original volume.]

1 - (2/3) = 1/3 = 0.3 = 33.3%

2. Surface area is 2-D so 1.5² - 1 = 1.25 = 125% increase

3a. If a volume of a cube is increased by 72.8 percent, it's 172.8% or 1.728 of the original volume. Now you are going from 3-D (volume) to 2-D (surface area). 1.728^2/3 = 1.44 or 44% increase. [Don't forget to minus 1 (the original volume) since it is asking you the percentage increase.]

3b. From surface area, you can get the side increase by using 1.44^1/2 = 1.2, so 20% increase.

Or you can also use 1.728^1/3 = 1.2;  1.2 - 1 = 20%

4. The volume of a cylinder is πr²x h . (2 + 4 + 6 + ...50) x 5²π = (25 x 26) x 25π =16250π