2014 Mathcounts State Prep : Folding Paper Questions

Folding paper questions are not too bad so here are two examples:
 

Question #1:  The two side lengths of rectangle ABCD is "a" and "b". If you fold along EF and the point B now converges on point D, what is the length of \(\overline {EF}\)?

Solution I :
Let the length of \(\overline {FC}\)  be x and the length of \(\overline {FD}\)  is thus a - x.

Using Pythagorean theorem you'll easily get x (the \( x^{2}\) part cancel each other out) and from there get the length of a - x.

\(\overline {HD}\) = \(\dfrac {1} {2}\) of the hypotenuse. (Use Pythagorean theorem or Pythagorean triples to get that length,)

Again, using Pythagorean theorem you'll get the length of \(\overline {HF}\). Times 2 to get \(\overline {EF}\).

Solution II : 
After you find the length of x, use \(\overline {EG}\), which is a - 2x and b as two legs of the right triangle EGF, you can easily get \(\overline {EF}\). (Pythagorean theorem)

Question #2: 

What about this time you fold B to touch the other side.
 What is the length of EF? 

This one is not too bad.

Do you see there are two similar triangles?

Just make sure you use the same corresponding sides to get the desired
length.

Answer to one mathleague quesiton from AoPS

Question is here.

You have two congruent triangles. 17-same angle and- x (SAS)
Using distance formula, the two green lines are of the same length.
\(\left( a-25\right) ^{2}+\left( 20-15\right) ^{2}=a^{2}+20^{2}\)
a = 5

Use another distance formula to get x -- the blue line.
\(\sqrt {\left( 5-17\right) ^{2}+20^{2}}=\sqrt {544}= 4\sqrt {34}\)

Sum and Product of roots : Vieta -- > Questions from 2010-2011 Mathcounts Super Stretch

Questions: (detailed solutions below)

#1 : What is the sum of the solutions of 6x² + 5x - 4 = 0 ? Express your answer as a common fraction. 

#2 : A quadratic equation of the form x² + kx + m = 0 has solutions x = 3 + 2√ 2  and 3 - 2√ 2 
What is the value of k + m? 

#3 : What is the sum of the reciprocals of the solutions of 4x² - 13x + 3 = 0 ? Express your answer as a common fraction. 













Solutions : 

#1:  6x² + 5x -4 = 0 divided the whole equation by 6 and you have x² + (5/6) x - 4/6 = 0, which means that the sum of the solutions is - 5/6. 

#2: The two roots are 3 + 2√ 2  and 3 - 2√ 2 , which means that -k = 3 + 2√ 2 + 3 - 2√ 2 
k = -6;  m = (3 + 2√ 2 ) (3 - 2√ 2 ) = 9 - 8 = 1 so m + k = -6 + 1 = -5

#3:
Solution I:
4x² - 13x + 3 = 0; divided the whole equation by 4 and you have  x² - (13/4)x + 3/4 = 0,
which means that the sum of the two roots, if they are x and y, are 13/4 and their product is 3/4.

1/x + 1/y = (x + y) / xy = 13/4  divided by 3/4 = 13/3

Solution II: Tom shows Rob and Rob shows me how to solve this using another method.

The original equation is 4x² -13x + 3 = 0 To find the sum and product of the reciprocals, you flip the equation so it becomes 3x² - 13x + 4 = 0

Using the same way you find the sum of the two roots,the answer is 13/3.

Find the area of the petal, or the football shape.

Find the area of the football shape, or the petal shape.
The below Mathcounts mini presents two methods.

Circle and area revisited from Mathcounts mini

The first question is exactly the same as this one.
Besides the two methods on the videos, you can also use the following methods.

Solution III:
You can also look at this as a Venn Diagram question.
One quarter circle is A and the other is B, and both are congruent. (center at opposite corner vertexes)

The overlapping part is C.

A + B - C = 6^2 so C = A + B - 36 or 18pi - 36





                                                                                               

Solution IV:

If you use the area of the rectangle,
which is 6 x 12 minus, the area of the half circle with a radius 6, you get the two white spots that are un-shaded.

Use the area of the square minus that will again give you the answer.
\(6^{2}-\left( 6*12-\dfrac {6^{2}\pi } {2}\right)\)
= 18pi - 36




Similar triangles and triangles that share the same vertexes/or/and trapezoid

Another link from my blog

Similar triangles, dimensional change questions are all over the place so make sure you really
understand them.

Take care and happy problem solving !!

Counting I : Ways to Avoid Over Counting or Under Counting

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

Question #1: How many ways to count a triple of natural numbers whose sum adds up to 12 
if  A. order doesn't matter? B. if order matters?

Solution I:
A. Find a systematic way to solve this type of problem in an organized manner.
We can start with the smallest natural number "1".
(1, 1, 10), (1, 2, 9), (1, 3, 8), (1, 4, 7), (1, 5, 6)  Since (1, 5, 6) is the same as (1, 6, 5) so stop.
(2, 2, 8), (2, 3, 7), (2, 4, 6), (2, 5, 5)
(3, 3, 6), (3, 4, 5)
(4, 4, 4) so total there are 12 ways.

B. Since in this case order matters so if you add up all the possible ways, for example, there are "3" ways to arrange (1, 1, 10) -- 3!/2! = 3
This is similar to "how many ways to arrange the letters 'odd'.
If all letters/numbers are different, you have 3! ways to arrange them; however, in this case, the two numbers 1, 1 are indistinguishable and there are 2! ways to arrange them, thus 3!/2! = 3

There are 3! or 6 ways to arrange triples such as (1, 2, 9).
Add all the possible ways and there are total 55 ways.

Solution II:
B.There is a much easier way to tackle this question, using bars and stars or sticks and stones method.
There are 11 spaces between 12 stones. @__@__@__@__@__@__@__@__@__@__@__@
If you places the two sticks on two of the spaces, you'll split the stones into three groups.
For example, if you have @ | @@@@@ | @@@@@@
The triples are 1, 5, and 6.
If it's @@@@@@@@ | @@ | @@ , the triples are 8, 2, and 2.
So 11C2 = 55 ways

Let me know if you are still confused.  Have fun problem solving !!