Three Pole Problems : Similar triangles

Question: If you know the length of x and y, and the whole length of \(\overline {AB}\),

A: what is the ratio of a to b and 

B: what is the length of z.






 Solution for question A:
\(\Delta\)ABC and \(\Delta\)AFE are similar so \(\dfrac {z} {x}=\dfrac {b} {a+b}\). -- equation 1
Cross multiply and you have z ( a + b ) = bx

\(\Delta\)BAD and \(\Delta\)BFE are similar so \(\dfrac {z} {y}=\dfrac {a} {a+b}\). -- equation 2
 Cross multiply and you have z ( a + b ) = ay

bx = ay so \(\dfrac {x} {y}=\dfrac {a} {b}\)  same ratio

Solution for question B: 
Continue with the previous two equations, if you add equation 1 and equation 2, you have:
\(\dfrac {z} {x}+\dfrac {z} {y}=\dfrac {b} {a+b}+\dfrac {a} {a+b}\)
\(\dfrac {zy+zx } {xy}=1\) \(\rightarrow\) z = \(\dfrac {xy} {x+y}\)

Applicable question: 

\(\overline {CD}=15\) and you know \(\overline {DB}:\overline {BC}=20:30=2:3\) 
 so  \(\overline {DB}=6\) and \(\overline {BC}=9\) 

\(\overline {AB}=\dfrac {20\times 30} {\left( 20+30\right) }\) = 12

2014 Mathcounts State/National/AMC Prep : More Counting and Probability Problems from Mathcounts Mini / AoPS

From Mathcounts Mini : Video tutorials on counting and probability for Mathcounts state prep

Counting the Number of Subsets of a Set

Constructive Counting

More Constructive Counting 

Probability and Counting

Probability with Geometry Representations  : Oh dear, the second half part is hilarious.

Probability with Geometry Representations : solution to the second half problem from previous video

Try this one from 1998 AIME #9. It's not too bad.

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.