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.
Monday, January 6, 2014
Thursday, December 12, 2013
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.
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.
Tuesday, December 10, 2013
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.
\(\sqrt {\left( 5-17\right) ^{2}+20^{2}}=\sqrt {544}= 4\sqrt {34}\)
Thursday, December 5, 2013
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.
#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.
Monday, November 4, 2013
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 !!
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: 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 !!
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