Some of the harder/hardest questions at Mathcounts can be tackled at ease using mass point geometry
so spend some time understanding it.
Basics
2014-15 Mathcounts handbook Mass Point Geometry Stretch
from page 39 to page 40
(Talking about motivation, yes, there are students already almost finish
this year's Mathcounts' handbook harder problems.)
From Wikipedia
From AoPS
Mass Point Geometry by Tom Rike
Another useful notes
Videos on Mass Point :
Mass Points Geometry Part I
Mass Points Geometry : Split Masses Part II
Mass Points Geometry : Part III
other videos from Youtube on Mass Points
It's much more important to fully understand how it works, the easier questions the weights align
very nicely.
The harder problems the weights are messier, not aligning nicely, so you need to find ways to may them integers (LCM) for easier solving.
Let me know if you have questions. I love to help (:D) if you've tried.
Showing posts with label 2015 Mathcounts state/nationals prep. Show all posts
Showing posts with label 2015 Mathcounts state/nationals prep. Show all posts
Sunday, March 6, 2022
Tuesday, February 14, 2017
2013 Mathcounts State Harder Problems
You can download this year's Mathcounts state competition questions here.
Trickier 2013 Mathcounts State Sprint Round questions :
Sprint #14:
From Varun:
Assume the term "everything" refers to all terms in the given set.
1 is a divisor of everything, so it must be first.
Everything is a divisor of 12, so it must be last.
The remaining numbers left are 2, 3, 4, and 6.
2 and 3 must come before 6, and 2 must come before 4.
Therefore, we can list out the possibilities for the middle four digits:
2,3,4,6
2,3,6,4
3,2,4,6
3,2,6,4
2,4,3,6
There are 5 ways--therefore 5 is the answer.
From Vinjai:
Here's how I did #14:
First, notice that 1 must be the first element of the set and 12 must be the last one.
So that leaves only 2,3,4,6 to arrange.
We can quickly list them out.
The restrictions are that 2 must be before 4, 3 must be before 6, and 2 must be before 6:
2,3,4,6 3,2,4,6 4,2,3,6 6,2,3,4
2,3,6,4 3,2,6,4 4,2,6,3 6,2,4,3
2,4,3,6 3,4,2,6 4,3,2,6 6,3,2,4
2,4,6,3 3,4,6,2 4,3,6,2 6,3,4,2
2,6,3,4 3,6,2,4 4,6,2,3 6,4,2,3
2,6,4,3 3,6,4,2 4,6,3,2 6,4,3,2
Only the bold ones work. So, the answer is 5.
#17: Common dimensional change problem
\(\overline {ZY}:\overline {WV}=5:8\) -- line ratio
The volume ratio of the smaller cone to the larger cone is thus \(5^{3}: 8^{3}\).
The volume of the frustum is the volume of the larger cone minus the volume of the smaller cone
= \(\dfrac {8^{3}-5^{3}} {8^{3}}\times \dfrac {1} {3}\times 8^{2}\times 32\times \pi\) = 516\(\pi\)
More problems to practice from Mathcounts Mini
#24: The answer is \(\dfrac {1} {21}\).
#28: Hats off to students who can get this in time !! Wow!!
From Vinjai:
Trickier 2013 Mathcounts State Sprint Round questions :
Sprint #14:
From Varun:
Assume the term "everything" refers to all terms in the given set.
1 is a divisor of everything, so it must be first.
Everything is a divisor of 12, so it must be last.
The remaining numbers left are 2, 3, 4, and 6.
2 and 3 must come before 6, and 2 must come before 4.
Therefore, we can list out the possibilities for the middle four digits:
2,3,4,6
2,3,6,4
3,2,4,6
3,2,6,4
2,4,3,6
There are 5 ways--therefore 5 is the answer.
From Vinjai:
Here's how I did #14:
First, notice that 1 must be the first element of the set and 12 must be the last one.
So that leaves only 2,3,4,6 to arrange.
We can quickly list them out.
The restrictions are that 2 must be before 4, 3 must be before 6, and 2 must be before 6:
2,3,4,6 3,2,4,6 4,2,3,6 6,2,3,4
2,3,6,4 3,2,6,4 4,2,6,3 6,2,4,3
2,4,3,6 3,4,2,6 4,3,2,6 6,3,2,4
2,4,6,3 3,4,6,2 4,3,6,2 6,3,4,2
2,6,3,4 3,6,2,4 4,6,2,3 6,4,2,3
2,6,4,3 3,6,4,2 4,6,3,2 6,4,3,2
Only the bold ones work. So, the answer is 5.
#17: Common dimensional change problem
\(\overline {ZY}:\overline {WV}=5:8\) -- line ratio
The volume ratio of the smaller cone to the larger cone is thus \(5^{3}: 8^{3}\).
The volume of the frustum is the volume of the larger cone minus the volume of the smaller cone
= \(\dfrac {8^{3}-5^{3}} {8^{3}}\times \dfrac {1} {3}\times 8^{2}\times 32\times \pi\) = 516\(\pi\)
More problems to practice from Mathcounts Mini
#24: The answer is \(\dfrac {1} {21}\).
#28: Hats off to students who can get this in time !! Wow!!
From Vinjai:
For
#28, there might be a nicer way but here's how I did it when I took the sprint
round:
#
4's # 3's # 2's # 1's
# ways
1 2 0
0 3
1 1 1
1 24
1 1 0
3 20
1 0 3
0 4
1 0 2
2 30
1 0 1
4 30
0 2 2
0 6
0 2 1
2 30
0 2 0
4 15
0 1 3
1 20
0 1 2
3 60
0 0 3
4 35
TOTAL:
277
#29:
\(\Delta ADE\) is similar to \(\Delta ABC\)
Let the two sides of the rectangle be x and y (see image on the left)
\(\dfrac {x} {21}=\dfrac {8-y} {8}\)
x =\(\dfrac {21\left( 8-y\right) } {8}\)
xy = \(\dfrac {21\left( 8-y\right) } {8}\) * y = \(\dfrac {-21y\left( y-8\right) } {8}\) =
\(\dfrac {-2l\left( y-4\right) ^{2}+21\times 16} {8}\)
From the previous equation you know when y = 4, the area \(\dfrac {21\times 16} {8}\)is the largest. The answer is 42.
Here is a proof to demonstrate that the largest area of a rectangle inscribed in a triangle is
half of the area of that triangle.
#30:
Solution I :
If (x, y) are the coordinates of the center of rotational points, it will be equal distance from A and A' as well as from B and B'.
Use distance formula, consolidate/simplify and solve the two equations, you'll get the answer (4, 1).
Solution II:
How to find the center of rotation from Youtube.
From AoPS using the same question
To sum up:
First, connect the corresponding points, in this case A to A' and B to B'.
Second, find the equation of the perpendicular bisector of line \(\overline {AA'}\), which is
y = - x + 5
and \(\overline {BB'}\), which is y = 5x - 19
The interception of the two lines is the center of rotation.
The answer is (4, 1).
2013 Mathcounts Target :
#3:
RT = D, unit conversions and different rates are tested here:
Make Joy's rate (speed) uphill be x m/s, his downhill speed be 2x m/s.
It takes Greg 3000 seconds (time) to reach the starting point and that is also what it takes Joy to
ride up to the hill and down to the same point.
\(\dfrac {7000} {x}+\dfrac {10000} {2x}=3000\) \(\rightarrow\) x = 4 m/s
#8:
Using "finding the height to the hypotenuse".( click to review)
you get \(\overline {CD}=\dfrac {7\times 24} {25}\).
Using similar triangles ACB and ADC, you get \(\overline {AD}=\dfrac {576} {25}\).
[\(\dfrac {24} {x}=\dfrac {25} {24}\)]
Using angle bisector (click to review),
you have \(\overline {AC}:\overline {AD}=\overline {CE}:\overline {ED}= 24: \dfrac {576} {25}\) = 600 : 576 = 25 : 24
\(\rightarrow\)\(\overline {ED}= \overline {CD}\times \dfrac {24} {24+25}\) = \(\dfrac {7\times 24} {25}\times \dfrac {24} {24+25}\) = \(\dfrac {576} {175}\)
\(\dfrac {x} {21}=\dfrac {8-y} {8}\)
x =\(\dfrac {21\left( 8-y\right) } {8}\)
xy = \(\dfrac {21\left( 8-y\right) } {8}\) * y = \(\dfrac {-21y\left( y-8\right) } {8}\) =
\(\dfrac {-2l\left( y-4\right) ^{2}+21\times 16} {8}\)
From the previous equation you know when y = 4, the area \(\dfrac {21\times 16} {8}\)is the largest. The answer is 42.
Here is a proof to demonstrate that the largest area of a rectangle inscribed in a triangle is
half of the area of that triangle.
#30:
Solution I :
If (x, y) are the coordinates of the center of rotational points, it will be equal distance from A and A' as well as from B and B'.
Use distance formula, consolidate/simplify and solve the two equations, you'll get the answer (4, 1).
Solution II:
How to find the center of rotation from Youtube.
From AoPS using the same question
To sum up:
First, connect the corresponding points, in this case A to A' and B to B'.
Second, find the equation of the perpendicular bisector of line \(\overline {AA'}\), which is
y = - x + 5
and \(\overline {BB'}\), which is y = 5x - 19
The interception of the two lines is the center of rotation.
The answer is (4, 1).
2013 Mathcounts Target :
#3:
RT = D, unit conversions and different rates are tested here:
Make Joy's rate (speed) uphill be x m/s, his downhill speed be 2x m/s.
It takes Greg 3000 seconds (time) to reach the starting point and that is also what it takes Joy to
ride up to the hill and down to the same point.
\(\dfrac {7000} {x}+\dfrac {10000} {2x}=3000\) \(\rightarrow\) x = 4 m/s
#8:
you get \(\overline {CD}=\dfrac {7\times 24} {25}\).
Using similar triangles ACB and ADC, you get \(\overline {AD}=\dfrac {576} {25}\).
[\(\dfrac {24} {x}=\dfrac {25} {24}\)]
Using angle bisector (click to review),
you have \(\overline {AC}:\overline {AD}=\overline {CE}:\overline {ED}= 24: \dfrac {576} {25}\) = 600 : 576 = 25 : 24
\(\rightarrow\)\(\overline {ED}= \overline {CD}\times \dfrac {24} {24+25}\) = \(\dfrac {7\times 24} {25}\times \dfrac {24} {24+25}\) = \(\dfrac {576} {175}\)
Monday, January 26, 2015
2015 Mathcounts State/National Prep
Harder concepts from Mathcounts Mini :
Also try the follow-up problems with detailed solutions. Don't do every question. Just the ones you think is hard.
Geometry :
#27 : Area and Volume
#30: Similar Triangles and Proportional Reasoning
#34: Circles and Right Triangles
#35: Using Similarity to Solve Geometry Problems
#41: Analytic Geometry : Center of Rotation
More questions to practice from my blog post.
From Mathcounts Mini : Video tutorials on counting and probability for Mathcounts state/national prep concepts are in the of difficulty.
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.
From Mathcounts Mink : Center of rotation, equal distance from a point to other two points or two lines, angle bisector (last problem on the follow-up worksheet)
Mathcounts Mini : #41 - Analytic Geometry
Also try the follow-up problems with detailed solutions. Don't do every question. Just the ones you think is hard.
Geometry :
#27 : Area and Volume
#30: Similar Triangles and Proportional Reasoning
#34: Circles and Right Triangles
#35: Using Similarity to Solve Geometry Problems
#41: Analytic Geometry : Center of Rotation
More questions to practice from my blog post.
From Mathcounts Mini : Video tutorials on counting and probability for Mathcounts state/national prep concepts are in the of difficulty.
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.
From Mathcounts Mink : Center of rotation, equal distance from a point to other two points or two lines, angle bisector (last problem on the follow-up worksheet)
Mathcounts Mini : #41 - Analytic Geometry
Subscribe to:
Posts (Atom)
