2013 Mathcounts Nationals Sprint #29 and similar problem hints

Mathcounts problems : state/nationals level

2013 Mathcounts Natinals Sprint # 28

2013 Mathcounts National Sprint #28 : In right triangle ABC, shown here,  line AC = 5 units and line BC = 12 units. Points D and E lie on  line AB and line BC respectively, so that line CD is perpendicular to line AB and E is the midpoint of line BC. Segments AE and CD intercept at point F. What is the ratio of AF to FE ? Express your answer as a common fraction.

                                                     Solution I : Using similar triangles

                                                        Solution II : Use Mass Point Geometry

2013 Mathcounts School and Chapter Harder Problems

You can now download and discuss with your friends this year's school and chapter problems.
Here is the link to the official Mathcounts website.

Some more challenging problems from this year's Mathcounts school/or chapter problems.

2013 school team #10 : Three concepts are testing here :
Hint: 
a. If you get rid of the remainder, the numbers will be evenly divided into 192, so you are looking at
those factors of 192 - 12 = 180

b. To leave a remainder of 12, those factors of 180 that are included in the Set must be smaller than 12, otherwise, you can further divide it.

c. To find the median, make sure to line up the numbers from the smallest to the largest and find the middle numbers. If there are even numbers of factors larger than 12, average the middle two. Otherwise, the middle number is the answer.

\(180=2^{2}\times 3^{2}\times 5 \) so there are (2 + 1) (2 + 1) (1 + 1) = 18 factors

The list on the left side gives you the first 9 and if you times those numbers with "5", you get 9 other factors,which are 5, 10, 20, 15, 18, 60, 45, 90 and 180.

Discard the factors that are smaller or equal to 12 and list all the other factors in order and find the median.

The answer is "36".



2013 Chapter Sprint:
#21: Dimensional change problem : The height of the top pyramid is \(\dfrac {2} {3}\) of the larger
pyramid so its volume is \(\left( \dfrac {2} {3}\right) ^{3}\) of the larger pyramid.

\(\left( \dfrac {2} {3}\right) ^{3}\times \dfrac {1} {3}\times \left( \dfrac {36} {4}\right) ^{2}\times 12
= \) \(96  cm^{3}\)

# 24:  According to the given:   \(xyz=720\)   and   \(2( xy+yz+zx)= 484 \) so
\(( xy+yz+zx )= 242\)

Since x, y and z are all integers, you factor 720 and see if it will come up with the same x, y and z values
for the second condition.

Problem writer(s) are very smart using this number because the numbers "6", "10", "12" would give you
a surface area of 252. (not right)

The three corrrect numbers are "8", "9", and "10" so the answer is \(\sqrt {8^{2}+9^{2}+10^{2}}=\) \(7\sqrt {5}\)

#25: Geometric probability: Explanations to similar questions and more practices below. 

Probability with geometry representations form Aops.

Geometric probability from "Cut the Knots".

#26: This one is similar to 2002 AMC-10B #21, so try that question to get more practices. 
2002 AMC-10B #21 link 

#27:
\(\dfrac {1} {A}+\dfrac {1} {B}=\dfrac {1} {2}\)
\(\dfrac {1} {B}+\dfrac {1} {C}=\dfrac {1} {3}\)
\(\dfrac {1} {C}+\dfrac {1} {A}=\dfrac {1} {4}\)
Add them up and you have  \(2 * (\dfrac {1} {A}+\dfrac {1} {B}+ \dfrac {1} {C})=\dfrac {13} {12}\)

\((\dfrac {1} {A}+\dfrac {1} {B}+ \dfrac {1} {C})=\dfrac {13} {24}\)

\(\dfrac {1} {\dfrac {1} {A}+\dfrac {1} {B}+\dfrac {1} {C}} = \)\(\dfrac {{24}} {13}\) hours

#28: Hint : the nth triangular number is the sum of the first "n" natural numbers and \(\dfrac {n\left( n+1\right) } {2}\) is how you use to find the sum.
From there, you should be able to find how many numbers will be evenly divided by "7".

#29 : Circle questions are very tricky so make sure to find more problems to practice accuracy.

#30 :  
Solution I: 
Read the solution that is provided by Mathcounts.org here.
Solution II:
Case 1 : \(x-1 > 0\rightarrow x > 1\) Times ( x - 1) on both sides and you have
\(x^{2}-1>8\) so x > 3 or x < -3 (discard)

Case 2: \(x-1 < 0\) so \(x < 1\) \(\rightarrow x^{2}-1 < 8\) [You need to change the sign since it's negative.]-3 < x < 3. Combined with x < 1 you have the range as -3 < x < 1
The answer is 60%.

                                               2013 Mathcounts Target #7 and 8: 

Target question #8 is very similar to 2011 chapter team #10. 
It just asks differently.   
Read the explanations provided on the Mathcounts official website.
They are explained very well.
Let me know if there are other easier ways to tackle those problems.

Hope this is helpful !! Thanks a lot !! Good luck on Mathcounts state.

Geometry : Harder Chapter Level Quesitons

Question #1 : The area ratio of two equilateral triangles are 4 to 9 and the sum of their perimeter is 30 √ 3 . What is the area of the a. smaller triangle   b. larger triangle?  

Solution: 
If the area ratio of two similar polygons is 4 to 9, their corresponding line ratio would be √ 4  to √ 9  
or 2 to 3.[Make sure you know why.]
The perimeter of the two equilateral triangles is 30 √ 3 so the smaller triangle has a perimeter of
2/5 *  30√ 3 or 12 √ 3. One side is 4√ 3 . Using the formula of finding the area of an equilateral triangle \(\dfrac{\sqrt3*s^2}{4}\) , you get the area to be 12√ 3.

Use the same method to get the area of the larger triangle as 27√ 3.
You can also use ratio relationship to get the area of the larger triangle by 
multiply 12√ 3 by 9/4.


2007 Mathcounts Chapter Sprint #30: In parallelogram ABCD, AB = 16 cm, DA = 32 cm, and sides AB and DA form a 45-degree interior angle. In isosceles trapezoid WXYZ with WX ≠ YZ, segment WX is the longer parallel side and has length 16 cm, and two interior angles each have a measure of 45 degrees. Trapezoid WXYZ has the same area as parallelogram ABCD. What is the length of segment YZ?

Solution I:
Make sure you know how to get the unknown leg fast. The height of the parallelogram is 8√2, so the area of the parallelogram is 48 square units. [Check out the special right triangle section here if you can't get the height fast.]

Let YZ of the trapezoid be x and draw the height. Using 45-45-90 degree angle ratio, you'll get the height. (See image above.)
Area of the trapezoid is average of the two bases time height. WX = 16 (given)
\(\dfrac{(16+x)* (16-x)}{4}\) = 48 ; 256 - x² = 192 ;  - x² = - 64;  x = 8 = YZ

Solution II: 

Make the y be the height of the trapezoid. YZ = 16 - 2y.  \(\dfrac{(16-2y + 16)}{2}\) * y = 48
\({(16 -y)* y = 48}\)\(\rightarrow\) \({16y -y^2 = 48}\) \(\rightarrow\) \({y^2 - 16y + 48 = 0}\) \(\rightarrow\) \({(y -12)(y -4)=0}\) \(\rightarrow\) \({y = 4}\) or \({y = 12}\)(doesn't work)
 YZ = 16 - 2y. Plug in y = 4 and you have  YZ = 8

Solution III: Let the height be y and you have \(\dfrac{(\overline{YZ}+ 16)* y}{2}"\) = 48 ; ( YZ + 16) * y = 96
When there are some numbers multiply together equal another number, it's a factoring question.
32 * 3 = 96, YZ = 16 (doesn't work)
24 * 4 = 96, YZ =8

2017 Mathcounts State Prep: Some Counting and Probability Questions on Dot Grids

Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here. (It's free.)

#5 1993 Mathcounts National Target : Find the probability that four randomly selected points on the geoboard below will be the vertices of a square? Express your answer as a common fraction.

#5 2004 AMC 10A: A set of three points is chosen randomly from the grid shown. Each three-point (same image as the below question) set has the same probability of being chosen. What is the probability that the points lie on the same straight line?

2007 Mathcounts Chapter Sprint #29 : The points of this 3-by-3 grid are equally spaced
horizontally and vertically. How many different sets of three points of this grid can be the three
vertices of an isosceles triangle?

Solution:

#5 National Target: There are 16C4 = \(\dfrac {16\times 15\times 14\times 13} {4\times 3\times 2\times 1}\)= 1820 ways to select 4 points on the geoboard.

There are 3 x 3 = 9  one by one squares and 2 x 2 = 4 two by two squares and 1 x 1 = 1 three by three squares. (Do you see the pattern?)

                                                         

There are 4 other squares that have side length of √ 2
and 2 other larger squares that have side length of √ 5.

9 + 4 + 1 + 4 + 2 = 20 and \(\dfrac {20} {1820}=\dfrac {1} {91}\)

#5: Solution:
AMC-10A: There are 9C3 = \(\dfrac {9\times 8\times 7} {3\times 2\times 1}\)= 84 ways to chose the three dots and 8 of the lines connecting the three dots will form straight lines. (Three verticals, three horizontals and two diagonals.) so 

\(\dfrac {8} {84}=\dfrac {2} {21}\)

#29: Solution: 
Use the length of the two congruent legs to solve this problem systematically. 

 There are 16   1 - 1 - \(\sqrt {2}\)    isosceles triangles.
There are 8    \(\sqrt {2}\)  by  \(\sqrt {2}\) by 2 isosceles triangles. (See that ?)

There are 4     2 - 2 - \(2\sqrt {2}\)  isosceles triangles.
There are 4    \(\sqrt {5}\)  by  \(\sqrt {5}\) by 2 isosceles triangles.
Finally, there are 4  \(\sqrt {5}\)  by  \(\sqrt {5}\) by \(\sqrt {2}\) isosceles triangles.
Add them up and the answer is 36.