2013 Mathcounts State Prep: Harder State Questions

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2004 Mathcounts State Sprint #19: The points (x, y) represented in this table lie on a straight line. The point (13, q) lies on the same line. What is the value of p + q? Express your answer as a decimal to the nearest tenth. 

 

Solution: 

#19:  Look at the table and you'll see each time x + 2, y would -3. 

-5 to -14 is (-9), three times (-3) so p = 2 + 3 x 2 = 8

p would = 13 when 2 + 5.5 * 2 = 13 so q = -5  +  (5.5) * (-3) = -21.5

p + q = -13.5

2004 Mathcounts State Sprint #24: The terms x, x + 2, x + 4, ..., x + 2n form an arithmetic sequence, with x an integer. If each term of the sequence is cubed, the sum of the cubes is - 1197. What is the value of n if n > 3?

Solution: 

The common difference in that arithmetic sequence is 2 and the sum of the cubes is -1197, which means that these numbers are all odd numbers. (cubes of odd numbers are odd and the sum of odd terms of odd numbers is odd. )

(-5)³ + (-7)³ + (-9)³ = -1197  However, n is larger than 3 (given) so the sequence will look like this:

 (-9)³+ (-7)³+ (-5)³+ (-3)³+ (-1)³ + (1)³ + (3)³ = -1197

 x = - 9 and x + 2n = 3, plug in and you get -9 + 2n = 3;  n = 6

2013 Mathcounts Basic Concept Review: Some Common Sums/numbers

These are some common sums that appear on Mathcounts often.

\(1+2+3+\ldots +n=\dfrac {n\left( n+1\right) } {2}\)

 \(2+4+6+\ldots .2n=n\left( n+1\right)\)

\(1+3+5+\ldots .\left( 2n-1\right) =n^{2}\)

The above are all arithmetic sequences.

The sum of any arithmetic sequence is average times terms (how many numbers).
Besides, the mean and median are the same in any arithmetic sequence.
Combining these knowledge, along with distributive rules some times
(case in point, sum of multiples of n, etc...) will expedite the calculation.

The "nth" triangular number is \(\dfrac {n\left( n+1\right) } {2}\)

The sum of the first n triangular numbers is \(\dfrac {n\left( n+1\right) \left( n+2\right) } {6}\).

 \(1^{2}+2^{2}+3^{2}+\ldots +n^{2}\) = \(\dfrac {n\left( n+1\right) \left( 2n+1\right) } {6}\)

\(1^{3}+2^{3}+3^{3}+\ldots +n^{3}=\)\(\left[ \dfrac {n\left( n+1\right) } {2}\right] ^{2}\)

Harder Geometry Question from National Mathcounts

Question: Similar to a 96 National Mathcounts question: Circles of the same colors are congruent and tangent to each other. What is the ratio of the area of the largest circle to the area of the smallest circle?

Solution :
The most difficult part might be to find the leg that is "2R - r".
R and r being the radius of the median and the smallest circle.

Using Pythagorean theorem, you have \(\left( R+r\right) ^{2}= R^{2}+\left( 2R-r\right) ^{2}\)
\(R^{2}+2Rr+r^{2}=R^{2}+4R^{2}-4Rr+r^{2}\)
Consolidate and you have 6Rr = \(4R^{2}\)\(\rightarrow\) 3r = 2R \(\rightarrow\) r =\(\dfrac {2R} {3}\) The ratio of the area of the largest circle to the area of the smallest circle is thus \(\rightarrow\)\(\dfrac {\left( 2R\right) ^{2}} {\left( \dfrac {2R} {3}\right) ^{2}}\) = 9 .

2013 Mathcounts State Prep: Counting and Probability

Question #1: Rolling two dice, what is the probability that the product is a multiple of 3?
Solution I :
As long as one of the two numbers turn up as 3 or multiple of 3 (in this case, a "6"), the product of the two numbers will be a multiple of 3.
There are 6 * 6 = 36 ways to get 2 numbers. Out of the 36 pairs, you can have
3 - 1
3 - 2
3 - 3
3 - 4
3 - 5
3 - 6, 6 ways.
However, there are only 5 ways left if the other die has a 3 since 3 - 3 only counts as once.
1 - 3
2 - 3
4 - 3
5 - 3
6 - 3
Next we look at "at least one number is "6".
6 - 1
6 - 2
6 - 4 (We already used 6 - 3)
6 - 5
6 - 6 so 5 ways.
The other way around, we have
1 - 6
2 - 6
4 - 6
5 - 6  Total 4 ways, so the answer is \(\frac{\Large{( 6 + 5 + 5 + 4 )}}{\Large{36}}\) = \(\frac{\Large{ 20}}{\Large{36}}\) = \(\frac{\Large{5}}{\Large{9}}\).

Solution II: 
The easiest way to solve this problem is to use complementary counting, which is 1 (100% or total possible way) - none of the the multiples of 3 showing up, so 1 - \(\frac{\Large{4}}{\Large{6}}\) *\(\frac{\Large{4}}{\Large{6}}\) = 1 - \(\frac{\Large{2}}{\Large{3}}\) * \(\frac{\Large{2}}{\Large{3}}\)=  \(\frac{\Large{5}}{\Large{9}}\)

Question #2: [2002 AMC-12B #16] Juan rolls a fair regular eight-sided die. Then Amal rolls a fair regular six-sided die. What isthe probability that the product of the two rolls is a multiple of 3?
Solution:
Using complementary counting (see solution II of the previous question), 1 - \(\frac{\Large{6}}{\Large{8}}\) *\(\frac{\Large{4}}{\Large{6}}\) = 1 - \(\frac{\Large{3}}{\Large{4}}\) * \(\frac{\Large{2}}{\Large{3}}\)=  \(\frac{\Large{1}}{\Large{2}}\)

2013 Mathcounts State Prep: Similar Triangles and Height to the Hypotenuse

There are many concepts you can learn from this image, which cover numerous similar right triangles, ratio/proportion/dimensional change and the height to the hypotenuse.

Question:
#1: Δ ABC is a 3-4-5 right triangle. What is the height to the hypotenuse? 
Solution: 
Use the area of a triangle to get the height to the hypotenuse. 
Let the height to the hypotenuse be "h"
The area of Δ ABC is  \(\Large\frac{3*4}{2}\)= \(\Large\frac{5*h}{2}\)
Both sides times 2 and consolidate: h =  \(\Large\frac{3*4}{5}\) = \(\Large\frac{12}{5}\)

Practice: What is the height to the hypotenuse?

Question:
#2: How many similar triangles can you spot?
Solution: 
There are 4 and most students have difficulty comparing the largest one with the other smaller ones.
Δ ABC is similar to Δ ADE, Δ FBD, ΔGEC. Make sure you really understand this and can apply this to numerous similar triangle questions. 

Question: 
#3: What is the area of  □ DEGF if \(\overline{BF}\) = 9 and \(\overline{GC}\) = 4
Solution: 
Using the two similar triangles Δ FBD and  ΔGEC (I found using symbols to find the corresponding legs
 to be much easier than using the lines.), you have \(\frac{\Large{\overline{BF}}}{\Large{\overline{FD}}}\) = \(\frac{\Large{\overline{GE}}}{\Large{\overline{GC}}}\).
s (side length of the square) = \({\overline{GE}}\) =  \({\overline{FD}}\)
Plug in the given and you have 9 * 4 = s² so the area of □ DEGF is 36 square units. (each side then is square root of 36 or 6)

Question: 
#4: Δ ABC is a 9-12-15 right triangle. What is the side length of the square? 
Solution :
The height to the hypotenuse is\(\frac{\Large{9*12}}{\Large{15}}\) = \(\frac{\Large{36}}{\Large{5}}\)
Δ ABC is similar to Δ ADE. Using base and height similarities, you have \(\frac{\Large{\overline{BC}}}{\Large{\overline{DE}}}\) = \(\frac{\Large{15}}{\Large{S}}\) = \(\frac{\frac{\Large{36}}{\Large{5}}}{\frac{\Large{36}}{\Large{5}} - \Large{S}}\)
Cross multiply and you have 108 - 15*S = \(\frac{\Large{36}}{\Large{5}}\) *S
S =\(\frac{\Large{180}}{\Large{37}}\)