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

Tuesday, September 27, 2016

2016 AMC-8 Prep

Interesting articles on math for this week:

How to Fall in Love With Math by Manil Suri from the New York Times

The Simpsons' secret formula : it's written by math geeks by Simon Singh from the Guardian

For this week's self studies (part I work):

2009 #25 : Review using the Harvey method. :D

2008 #25   Don't use the method on the link. Use the much faster method we talked about at our lesson.

2008 #24  Make a chart. Slow down on similar question such as this one. 
This type of problem is very easy to make mistake on under or overcounting. 
Skip first and definitely slow down and double, triple check. 

2007 #25 Read the solution if you don't get the method we talked about at our lesson.
It takes time to develop insights so you need to be patient.
If you understand the method, this question will be easy, right ?
Stay with this question longer.

2007 #24 
Aayush's method is faster.
(To get the sum of three digits that is a multiple of 3, you either get rid of 1 or 4 [do you see it jumps by 3, why?] ) , so the answer is 1/2.

2006 #25
I've seen other problems (AMC-10s) using the oddest prime, which is "2", the only prime number that is even.
Thus, make sure you understand this question.

2006#24  Taking out the factor question.
Also, learn "1001 = 7 x 11 x 13"
"23 x 29 = 667"

2005 #25 Venn diagram is your friend.

2005 #24 Working backwards is the way to go.

For part II work :
This week, work on the last 5 problems from AMC-8 year 2010, 2011, 2012, 1999 and 1998.
Here is the link from AoPs.

Sam' original question:
David has a bag of 8 different-colored six-sided dice. Their colors are red, blue, yellow, green, purple, orange, black, and white. What is the probability that David takes out a red die, rolls a 6, then takes out a purple die, and rolls another 6 without replacement?

The probability of rolling a 6 on a red die is 1/8 * 1/6 = 1/48. The probability of rolling a purple die and rolling a 6 after that, without replacement, is 1/7 * 1/6 = 1/42. Therefore, to get both events, 1/48 * 1/42 = 1/2016.

Evan's compiled question:
\(\sqrt {18+8\sqrt {2}}=a+b\sqrt {c}\)
a, b and c are positive integers. Find a + b + c.
Square both sides and you have \(18+8\sqrt{2}\) = \( a^2 + 2ab\sqrt {2} + b^2c\)
You can see ab = 4 = 4 x 1 and c = 2
a = 4, b = 1 and c = 2 so the sum is 7.

Sounak's problem:
A rhombus with sides 4 is drawn. It has an angle of 60 degrees. What is length of the longer diagonal?

Well first you have to draw the rhombus's height .The resulting triangle will be 30,60, 90 triangle.
We know the hypotenuse is 4 so now we know the rest of the sides are \(2\sqrt {3}\) and 2.
Now if we draw the diagonal we see that it makes another right angle triangle.
We know the legs of this triangle are the same as the previous lengths so then we know the diagonal is \(4\sqrt {3}\).