Sequences are fun to learn and not really that difficult.
There are many similarities between arithmetic and geometric sequences, so
learn both together.
Enjoy !!!!!
From Mthcounts Mini: Sequences and Series
Easier concepts:
Sequences
Arithmetic sequence/determine the nth term
Arithmetic and geometric sequences
Mathcounts strategies : review some sums
Note : Don't just memorize, but really understand the concepts.
Harder concepts:
Sum and Average of An Evenly Space
Relationship between arithmetic sequences, mean and median
Sequences, series and patterns
Some Common Sums
Showing posts with label Mathcounts problems. Show all posts
Showing posts with label Mathcounts problems. Show all posts
Sunday, December 10, 2023
Friday, May 5, 2023
Pathfinder
From Mathcounts Mini :
Counting/Paths Along a Grid
From Art of Problem Solving
Counting Paths on a Grid
Math Principles : Paths on a Grid : Two Approaches
Question #1: How many ways to move the dominoes on a 6 by 6 checker board if you can only move the dominoes to the right or to the bottom starting from the upper left and you can't move the dominoes diagonally?
Solution :
You can move the dominoes 5 times to the right at most and 5 down to
the bottom at most, so the answer is \(\dfrac {\left( 5+5\right) !} {5! \times 5!}\) = 252 ways
Question # 2: How many ways can you move from A to B if you can only move downward and to right?
Solution : There are \(\dfrac {\left( 4+4\right) !} {4!\times 4!}\) * 2 * \(\dfrac {\left( 4+4\right) !} {4!\times 4!}\) = 9800 ways from A to B
Counting/Paths Along a Grid
From Art of Problem Solving
Counting Paths on a Grid
Math Principles : Paths on a Grid : Two Approaches
Question #1: How many ways to move the dominoes on a 6 by 6 checker board if you can only move the dominoes to the right or to the bottom starting from the upper left and you can't move the dominoes diagonally?
Solution :
You can move the dominoes 5 times to the right at most and 5 down to
the bottom at most, so the answer is \(\dfrac {\left( 5+5\right) !} {5! \times 5!}\) = 252 ways
Question # 2: How many ways can you move from A to B if you can only move downward and to right?
Solution : There are \(\dfrac {\left( 4+4\right) !} {4!\times 4!}\) * 2 * \(\dfrac {\left( 4+4\right) !} {4!\times 4!}\) = 9800 ways from A to B
Sunday, March 6, 2022
Mass Points Geometry
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.
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.
Monday, March 11, 2019
Hints/links or Solutions to 2014 Harder Mathcounts State Sprint and Target question
Links, notes, Hints or/and solutions to 2014 Mathcounts state harder problems.
2014, 2015 Mathcounts state are harder
Sprint round:
#14 :
Solution I :
(7 + 8 + 9) + (x + y + z) is divisible by 9, so the sum of the three variables could be 3, 12, or 21.
789120 (sum of 3 for the last three digits) works for 8 but not for 7.
21 is too big to distribute among x, y and z (all numbers are district),
thus only x + y + z = 12 works and z is an even number
__ __ 0 does't work (can't have 6 6 0 and the other pairs all have 7, 8 or 9)
264 works (789264 is the number)
Solution II :
789000 divided by the LCM of 7, 8 and 9, which is 504 = 1565.47...
Try 504 * 1566 = 789264 (it works)
The answer is 264.
#18:
Watch this video from Mathcounts mini and use the same method for the first question,
you'll be able to get the answer. It's still tricky, though.
#23 : Drop the heights of the two isosceles triangles and use similar triangles to get the length of FC.
Then solve.
#24:
The key is to see 210 is 1024 or about 103
230 = ( 210 )3 or about (103 )3about 109 so the answer is 10 digit.
#25:
As you can see, there are two Pythagorean Triples : 9-12-15 and 9-40-41.
Base (40-12) = 28 gives you the smallest area.
The answer is 28 * 18 = 504
#26 : Let there be A, B, C three winners. There are 4 cases to distribute the prizes.
A B C
1 1 5 There are 7C1 * 6C1 * \( \dfrac {3!} {2!}\) = 126 ways -- [you can skip the last part for C
because it's 5C5 = 1]
1 2 4 There are 7C1* 6C2 * 3! = 630
1 3 3 There are 7C1 * 6C3 * \( \dfrac {3!} {2!}\) = 420
2 2 3 There are 7C2 * 5C2 * 3 (same as above)
Add them up and the answer is 1806.
If you can't see why it's \( \dfrac {3!} {2!}\) when there is one repeat, try using easier case to help you understand.
What about A, B two winners and 4 prizes ?
There are 2 cases, 1 3 or 2 2, and you'll see how it's done.
#27 : Read this and you'll be able to solve this question at ease, just be careful with the sign change.
Vieta's Formula and the Identity Theory
#28: There are various methods to solve this question.
I use binomial expansion :
\(11^{12}=\left( 13-2\right)^{12}=12C0*13^{12}\)+ \(12C1*13^{11}*2^{1}\)+... \(12C11*13^{1}*2^{11}\)+ \(12C12*2^{12}\) Most of the terms will be evenly divided by 13 except the last term, which is \(2^{12}\) or 4096, which, when divided by 13, leaves a remainder of 1.
Solution II :
\(11\equiv -2\left ( mod13\right)\) ; \((-2)^{12}\equiv 4096\equiv 1\left ( mod13\right)\)
Solution III :
Or use Fermat's Little Theorem (Thanks, Spencer !!)
\(11^{13-1}\equiv 11^{12}\equiv1 (mod 13)\)
Cut the cone and observe the shape.
The circumference of the larger circle is 20pi (10 is the radius) and the base of
the cone circle circumference is 10pi (5 is the radius), which means that the cut-off cone shape is a half circle because it's \(\dfrac {10\pi } {20\pi }\) or \(\dfrac {1 } {2 }\) of the larger circle. (180 degrees)
To find the part that is the area of the frustum not including the top and bottom circles,
you use the area of the half circle minus the area of the smaller half circle.
Since the volume ratio of the smaller cone to larger cone = 2 to 3, the side ratio of the
two radius is \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).
Using this ratio, we can get the radius of the smaller circle as 10 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\) and the radius of the top circle of the frustum as 5 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).
Now we can solve this :
Cut the cone and observe the shape.
The circumference of the larger circle is 20pi (10 is the radius) and the base of
the cone circle circumference is 10pi (5 is the radius), which means that the cut-off cone shape is a half circle because it's \(\dfrac {10\pi } {20\pi }\) or \(\dfrac {1 } {2 }\) of the larger circle. (180 degrees)
To find the part that is the area of the frustum not including the top and bottom circles,
you use the area of the half circle minus the area of the smaller half circle.
Since the volume ratio of the smaller cone to larger cone = 2 to 3, the side ratio of the
two radius is \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).
Using this ratio, we can get the radius of the smaller circle as 10 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\) and the radius of the top circle of the frustum as 5 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).
Now we can solve this :
\(\dfrac {1 } {2 }\)\(\left[ 10^{2}\pi -\left( 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\right) ^{2}\pi \right] \) + \(5^{2}\pi +\left( 5\times \dfrac {\sqrt [3] {2}} {\sqrt {3}}\right) ^{2}\pi \) = about 176 (after you round up)
Solution III : Another way to find the surface area of the Frustum is :
median of the two half circle [same as median of the two bases] * the height [difference of the two radius]
\(\dfrac {1} {2}\left( 2\times 10\pi + 2\times 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\pi \right)\)* \(\left( 10-10\times \dfrac {\sqrt [3] {2}} {\sqrt [3]{3}}\right)\)
789120 (sum of 3 for the last three digits) works for 8 but not for 7.
21 is too big to distribute among x, y and z (all numbers are district),
thus only x + y + z = 12 works and z is an even number
__ __ 0 does't work (can't have 6 6 0 and the other pairs all have 7, 8 or 9)
264 works (789264 is the number)
Solution II :
789000 divided by the LCM of 7, 8 and 9, which is 504 = 1565.47...
Try 504 * 1566 = 789264 (it works)
The answer is 264.
#18:
Watch this video from Mathcounts mini and use the same method for the first question,
you'll be able to get the answer. It's still tricky, though.
#23 : Drop the heights of the two isosceles triangles and use similar triangles to get the length of FC.
Then solve.
#24:
The key is to see 210 is 1024 or about 103
230 = ( 210 )3 or about (103 )3about 109 so the answer is 10 digit.
#25:
As you can see, there are two Pythagorean Triples : 9-12-15 and 9-40-41.
Base (40-12) = 28 gives you the smallest area.
The answer is 28 * 18 = 504
#26 : Let there be A, B, C three winners. There are 4 cases to distribute the prizes.
A B C
1 1 5 There are 7C1 * 6C1 * \( \dfrac {3!} {2!}\) = 126 ways -- [you can skip the last part for C
because it's 5C5 = 1]
1 2 4 There are 7C1* 6C2 * 3! = 630
1 3 3 There are 7C1 * 6C3 * \( \dfrac {3!} {2!}\) = 420
2 2 3 There are 7C2 * 5C2 * 3 (same as above)
Add them up and the answer is 1806.
If you can't see why it's \( \dfrac {3!} {2!}\) when there is one repeat, try using easier case to help you understand.
What about A, B two winners and 4 prizes ?
There are 2 cases, 1 3 or 2 2, and you'll see how it's done.
#27 : Read this and you'll be able to solve this question at ease, just be careful with the sign change.
Vieta's Formula and the Identity Theory
#28: There are various methods to solve this question.
I use binomial expansion :
\(11^{12}=\left( 13-2\right)^{12}=12C0*13^{12}\)+ \(12C1*13^{11}*2^{1}\)+... \(12C11*13^{1}*2^{11}\)+ \(12C12*2^{12}\) Most of the terms will be evenly divided by 13 except the last term, which is \(2^{12}\) or 4096, which, when divided by 13, leaves a remainder of 1.
Solution II :
\(11\equiv -2\left ( mod13\right)\) ; \((-2)^{12}\equiv 4096\equiv 1\left ( mod13\right)\)
Solution III :
Or use Fermat's Little Theorem (Thanks, Spencer !!)
\(11^{13-1}\equiv 11^{12}\equiv1 (mod 13)\)
Target Round :
#3: Lune of Hippocrates : in seconds solved question.
^__^
#6: This question is very similar to this Mathcounts Mini.
My students should get a virtual bump if they got this question wrong.
#8: Solution I : by TMM (Thanks a bunch !!)
Using similar triangles and Pythagorean Theorem.
The height of the cone, which can be found using the Pythagorean is
.
Usingthediagram below, let
be the radius of the top cone and let 
be the height of the topcone.
Let
be the slant height of the top cone.
Drawing the radius as shown in the diagram, we have two right triangles. Since the bases of the top cone and the original cone are parallel, the two right triangles are similar. So we have the proportion![\[\dfrac{r}{5}=\dfrac{s}{10}=\dfrac{\sqrt{r^2+h^2}}{10}.\]](https://latex.artofproblemsolving.com/2/0/5/205a8d45594c47d92393e07cffca02d3d3a5b415.png)
Cross multiplying yields ![\[10r=5\sqrt{r^2+h^2}\implies 100r^2=25r^2+25h^2\implies 75r^2=25h^2\implies 3r^2=h^2\implies h=r\sqrt{3}.\]](https://latex.artofproblemsolving.com/c/3/4/c34a4033b4515ae88c64fda9f900d84924094aa6.png)
This is what we need.
Next, the volume of the original cone is simply
.
The volume of the top cone is
.
From the given information, we know that![\[\dfrac{125\sqrt{3}}{3}-\dfrac{\pi\times r^2h}{3}=\dfrac{125\sqrt{3}}{9}\implies 125\sqrt{3}-r^2h=\dfrac{125\sqrt{3}}{3}\implies r^2h=\dfrac{250\sqrt{3}}{3}.\]](https://latex.artofproblemsolving.com/d/2/e/d2eb3f35bb46a600fd5db854308e1f7844964947.png)
We simply substitute the value of 
from above to yield ![\[r^3\sqrt{3}=\dfrac{250\sqrt{3}}{3}\implies r=\sqrt[3]{\frac{250}{3}}.\]](https://latex.artofproblemsolving.com/a/d/9/ad9d39850b5a5aa905a1e879a6b5bfebc1804496.png)
We will leave it as is for now so the decimals don't get messy.
We get
and 
.
The lateral surface area of the frustum is equal to the lateral surface area of the original cone minus the lateral surface area of the top cone. The surface area of the original cone is simply
.
The surface area of the top cone is
.
So our lateral surface area is
All we have left is to add the two bases. The total area of thebases is
. So our final answer is ![\[37.207+138.477=175.684\approx\boxed{176}.\]](https://latex.artofproblemsolving.com/9/b/9/9b9d7bd69023b24516b90a9c51dbce0f9fffbe08.png)
Solution II : #3: Lune of Hippocrates : in seconds solved question.
^__^
#6: This question is very similar to this Mathcounts Mini.
My students should get a virtual bump if they got this question wrong.
#8: Solution I : by TMM (Thanks a bunch !!)
Using similar triangles and Pythagorean Theorem.
The height of the cone, which can be found using the Pythagorean is
. Usingthediagram below, let
be the radius of the top cone and let be the height of the topcone. Let
be the slant height of the top cone.Drawing the radius as shown in the diagram, we have two right triangles. Since the bases of the top cone and the original cone are parallel, the two right triangles are similar. So we have the proportion
Cross multiplying yields This is what we need.Next, the volume of the original cone is simply
. The volume of the top cone is
.From the given information, we know that
We simply substitute the value of from above to yield We will leave it as is for now so the decimals don't get messy.We get
and .The lateral surface area of the frustum is equal to the lateral surface area of the original cone minus the lateral surface area of the top cone. The surface area of the original cone is simply
. The surface area of the top cone is
. So our lateral surface area is
All we have left is to add the two bases. The total area of thebases is
. So our final answer is Cut the cone and observe the shape.
The circumference of the larger circle is 20pi (10 is the radius) and the base of
the cone circle circumference is 10pi (5 is the radius), which means that the cut-off cone shape is a half circle because it's \(\dfrac {10\pi } {20\pi }\) or \(\dfrac {1 } {2 }\) of the larger circle. (180 degrees)
To find the part that is the area of the frustum not including the top and bottom circles,
you use the area of the half circle minus the area of the smaller half circle.
Since the volume ratio of the smaller cone to larger cone = 2 to 3, the side ratio of the
two radius is \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).
Using this ratio, we can get the radius of the smaller circle as 10 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\) and the radius of the top circle of the frustum as 5 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).
Now we can solve this :
Cut the cone and observe the shape.
The circumference of the larger circle is 20pi (10 is the radius) and the base of
the cone circle circumference is 10pi (5 is the radius), which means that the cut-off cone shape is a half circle because it's \(\dfrac {10\pi } {20\pi }\) or \(\dfrac {1 } {2 }\) of the larger circle. (180 degrees)
To find the part that is the area of the frustum not including the top and bottom circles,
you use the area of the half circle minus the area of the smaller half circle.
Since the volume ratio of the smaller cone to larger cone = 2 to 3, the side ratio of the
two radius is \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).
Using this ratio, we can get the radius of the smaller circle as 10 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\) and the radius of the top circle of the frustum as 5 * \(\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\).
Now we can solve this :
\(\dfrac {1 } {2 }\)\(\left[ 10^{2}\pi -\left( 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\right) ^{2}\pi \right] \) + \(5^{2}\pi +\left( 5\times \dfrac {\sqrt [3] {2}} {\sqrt {3}}\right) ^{2}\pi \) = about 176 (after you round up)
Solution III : Another way to find the surface area of the Frustum is :
median of the two half circle [same as median of the two bases] * the height [difference of the two radius]
\(\dfrac {1} {2}\left( 2\times 10\pi + 2\times 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\pi \right)\)* \(\left( 10-10\times \dfrac {\sqrt [3] {2}} {\sqrt [3]{3}}\right)\)
Tuesday, September 18, 2018
Dimensional Change questions I:
Questions written by Willie, a volunteer. Answer key and detailed solutions below.
1a. There is a regular cylinder, which has a height equal to
its radius. If the radius and height are both increased by 50%, by what % does
the total volume of the cylinder increase?
1b. If the radius and height are both decreased by 10%, by
what % does the total volume of the cylinder decrease?
1c. If the radius is increased by 20% and the height is
decreased by 40%, what % of the volume of the original cylinder does the volume
of the new cylinder represent?
1d. If the radius is increased by 40% and the height is
decreased by 20%, what % of the volume of the original cylinder does the volume
of the new cylinder represent?
1e. If the height is increased by 125%, what % does the
radius need to be decreased by for the volume to remain the same?
2. If the side of a cube is increased by 50%, by what % does
the total surface area of the cube increase?
3a. If the volume of a cube increases by 72.8%, by what %
does the total surface area of the cube increase?
3b. By what % did the side length of the cube increase?
4. You have a collection of cylinders, all having a radius
of 5. The first cylinder has a height of 2, the second has a height of 4, the
third a height of 6, etc. The last cylinder has a height of 50. What is the sum
of the volumes of all the cylinders (express your answer in terms of pi)?
1b. Like the previous question: 13 - 0.93 [when it's discount/percentage decrease, you use the 100% or 1 - the discount/decrease percentage] = 0.271 = 27.1% decrease
1d. 1.42 [100% + 40% increase = 1.4] x 0.8 [100% -20% = 0.8] = 1.568 = 156.8% of the original volume
Answer key: (Each question should not take you more than 30 seconds to solve if you really understand the concepts involved.)
1a. The volume of a cylinder is πr2x h (height). The radius itself will be squared and the height stays at
constant ratio. The volume will increased thus (1.5)3 - 13 -- the original 100% of the volume = 2.375
=237.5%
1b. Like the previous question: 13 - 0.93 [when it's discount/percentage decrease, you use the 100% or 1 - the discount/decrease percentage] = 0.271 = 27.1% decrease
1c. 1.22 [100% + 20% increase = 1.2] x 0.6 [100% -40% = 0.6] = 0.864 or
86.4% of the original volume
1d. 1.42 [100% + 40% increase = 1.4] x 0.8 [100% -20% = 0.8] = 1.568 = 156.8% of the original volume
1e. When the height of a cylinder is increased 125%, the total volume is is 225% of the original cylinder, or 9/4.
Since the radius is used two times (or squared), it has to decrease 4/91/2 = 2/3 for the new cylinder to have the same volume as the old one. [9/4 times 4/9 = 1 or the original volume.]
1 - (2/3) = 1/3 = 0.3 = 33.3%
2. Surface area is 2-D so 1.52 - 1 = 1.25 = 125% increase
3a. If a volume of a cube is increased by 72.8 percent, it's 172.8% or 1.728 of the original volume. Now you are going from 3-D (volume) to 2-D (surface area). 1.7282/3 = 1.44 or 44% increase. [Don't forget to minus 1 (the original volume) since it is asking you the percentage increase.]
3b. From surface area, you can get the side increase by using 1.441/2 = 1.2, so 20% increase.
Or you can also use 1.7281/3 = 1.2; 1.2 - 1 = 20%
4. The volume of a cylinder is πr2x h . (2 + 4 + 6 + ...50) x 52π = (25 x 26) x 25π =16250π
Saturday, September 1, 2018
2011 Mathcounts Chapter Sprint Round solutions
#22: The answer is 2674.See left for explanations.
#23: Let the two consecutive positive integers be x and x + 1.
( x + 1 ) / x = 1.02, x + 1 = 1.02x, 0.02x = 1, x = 1 divided by 0.02 = 1 times 100/2 = 50
The two numbers are 50 and 51 and their sum is 50 + 51 = 101.
#24: The two x-intercepts when y is "0" are 10 or -10; the two y-intercepts when x is "0" are 5 or -5.
Area of a rhombus is D1 x D2 / 2 so the answer is [10-(-10)] x [5 -(-5)] = 100 square units.
#25: The area ratio of the two similar triangle is 150/6 so the line ratio is √ 150/6 or 5:1.
the length of the hypotenuse of the smaller triangle is 5 inches, so the other two legs are 3 and 4.
(a Pythagorean triple)
The sum of the lengths of the legs of the larger triangle is (3 + 4) * 5 = 35.
#26: To have same number of boys and girls, the committee needs to consist of 3 boys and 3 girls.
(6C3 x 4C3)/ 10C6 = 80/210 = 8/21
#27: When the point (3, 4) is reflected over the x-axis to B, B would = (3, -4).
When B is reflected over the line y = x to C, C would = (-4, 3).
The area of the triangle is [4 -(-4)] x [ 3 - (-4)]/ 2 = 28 square units.
#28: Tonisha is 45 miles ahead Sheila when Sheila leaves Maryville at 8: 15 a.m.
Each hour Sheila will be 15 miles closer to Tonisha. 45/15 = 3, which means that 3 hours
later Sheila will pass Tonisha.
8:15 + 3 hours = 11: 15 a.m.
#29: Using 30-60-90 degree angle ratio, you can make the radius be √ 3 and half of the side of the hexagon would be 1 so each side of the hexagonis 2.
The area of the hexagon is (√ 3/4) times 22 times 6 = 6√ 3.
The area of the circle is 3Î .
The fraction is 3Î /6√ 3 = √ 3 / 6
a = 3 and b = 6, ab = 18
Using Pythagorean theorem, you get the height to be 2√ 5 .
8 x 2√ 5 /2 = 8 √ 5 .
Sunday, April 23, 2017
Tricky Algebra Mathcounts National Questions: Counting Backwards
Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.
#24 1993 National Sprint: Bottle A contains more Diet Coke than Bottle B. Now do the following:
a. Pour from Bottle A into B as much Diet Coke as B already contains.
b. Pour from B into A as much Diet Coke as A now contains.
c. Pour from A into B as much Diet Coke as B now contains.
Both bottles now have 64 ounces. How many more ounces were in A than in B at the beginning?
#30: 1993 National Sprint: Auggie spent all of his money in 5 stores. In each store, he spent $1 more than one-half of what he had when he went in. How many dollars did Auggie have when he entered the first store?
#25: 1998 AMC-8 Three generous friends, each with some money, redistribute the money as follow: Amy gives enough money to Jan and Toy to double each amount has. Jan then gives enough to Amy and Toy to double their amounts. Finally, Toy gives enough to Amy and Jan to double their amounts. If Toy had 36 dollars at the beginning and 36 dollars at the end, what is the total amount that all three friends have?
Solution II: Work backwards
Since Auggie spent all his money at the 5th store. If there are x dollars left before he spent the money all at the 5th store. You can set up the equation such as this: x = 1 + 1/2 of x (according to the given)
So at the 5th store, he had 2 dollars.
Use the same strategy, if he had y dollars before he spent the money at the 4th store, he had
y = 1 + 1/2 of y + 2 ; y = 6
Use the same method, Aggie had 16 before he spent at the 3rd store, 30 before the 2nd store and finally,
62 dollars at the beginning.
#25: The total sum of what Amy, Jan, and Toy have stay constant so use Toy's amount to solve this problem.
Amy Jan Toy
? ? 36
First round Amy gave Jan and Toy double the amount of what each of them has, so
Amy Jan Toy
? ? 72
Second round Jan gave Amy and Toy double the amount of what each of them has, so
Amy Jan Toy
? ? 144
Third round Toy gave Amy and Jan double the amount of what each of them has an at the end Toy has 36 dollars Amy Jan 36
That means that at the second round, Amy + Jan = 144 - 36 = 108 dollars.
So they total have 108 + 144 = 252 dollars.
Download this year's Mathcounts handbook here.
#24 1993 National Sprint: Bottle A contains more Diet Coke than Bottle B. Now do the following:
a. Pour from Bottle A into B as much Diet Coke as B already contains.
b. Pour from B into A as much Diet Coke as A now contains.
c. Pour from A into B as much Diet Coke as B now contains.
Both bottles now have 64 ounces. How many more ounces were in A than in B at the beginning?
#30: 1993 National Sprint: Auggie spent all of his money in 5 stores. In each store, he spent $1 more than one-half of what he had when he went in. How many dollars did Auggie have when he entered the first store?
#25: 1998 AMC-8 Three generous friends, each with some money, redistribute the money as follow: Amy gives enough money to Jan and Toy to double each amount has. Jan then gives enough to Amy and Toy to double their amounts. Finally, Toy gives enough to Amy and Jan to double their amounts. If Toy had 36 dollars at the beginning and 36 dollars at the end, what is the total amount that all three friends have?
Solution I: Use Algebra:
#24: Let A contains x ounces and B contains y ounces and x > y (given).
After first pouring, A has (x - y) ounces left and B has 2y ounces (double the original amount)
After second pouring, A has ( 2x - 2y)(double the amount) ounces and B has (3y - x) ounces left.
After third pouring, A has (3x - 5y) ounces left and B has (6y - 2x) (double the amount)
3x - 5y = 64 times 2 for each terms 6x - 10y = 128 ----equation 3
6y - 2x = 64 times 3 for each terms 18y - 6x = 192 ---- equation 4
equation 3 + equation 4 and you have 8y = 320 and y = 40 ; Plug in any equation and you get x = 88
88 - 40 = 48 ounces
Solution II: Solving it backwards:
At the end,both A and B have 64 ounces, which is after same amount of Diet Coke being pour from A to B.
Thus before action C, A has 64 + half of 64 = 96 ounces and B has 32 ounces. [Make sure you understand this]
With the same reasoning, before action B, B has 32 + half of 96 = 80 oz. and A has 48 ounces.
Again, use the same strategy, you have before action A, A has 48 + half of 80 = 88 and 40.
The difference is 88 - 40 = 48 oz.
Solution I: Use Algebra
#30: Let Auggie had x dollars at the beginning. At the first store, he would spent 1 + (x/2) = (2+x)/2 and would have x - (2 +x)/2 = (x-2)/2 left
Solution II: Solving it backwards:
At the end,both A and B have 64 ounces, which is after same amount of Diet Coke being pour from A to B.
Thus before action C, A has 64 + half of 64 = 96 ounces and B has 32 ounces. [Make sure you understand this]
With the same reasoning, before action B, B has 32 + half of 96 = 80 oz. and A has 48 ounces.
Again, use the same strategy, you have before action A, A has 48 + half of 80 = 88 and 40.
The difference is 88 - 40 = 48 oz.
Solution I: Use Algebra
#30: Let Auggie had x dollars at the beginning. At the first store, he would spent 1 + (x/2) = (2+x)/2 and would have x - (2 +x)/2 = (x-2)/2 left
At the second store, he would spend 1 + (x-2)/4 and would have (x-2)/2 - 1 - (x-2)/4 or (x-6)/4 left
At the third store, he would spend 1 + (x-6)/8 and would have (x-14)/8 left
It looks like there's a pattern. At the fourth store, he would spend (x-30)/16
and at the 5th store he would spent (x-62)/32 = 0 so x - 62 = 0 and x = 62 dollarsSolution II: Work backwards
Since Auggie spent all his money at the 5th store. If there are x dollars left before he spent the money all at the 5th store. You can set up the equation such as this: x = 1 + 1/2 of x (according to the given)
So at the 5th store, he had 2 dollars.
Use the same strategy, if he had y dollars before he spent the money at the 4th store, he had
y = 1 + 1/2 of y + 2 ; y = 6
Use the same method, Aggie had 16 before he spent at the 3rd store, 30 before the 2nd store and finally,
62 dollars at the beginning.
#25: The total sum of what Amy, Jan, and Toy have stay constant so use Toy's amount to solve this problem.
Amy Jan Toy
? ? 36
First round Amy gave Jan and Toy double the amount of what each of them has, so
Amy Jan Toy
? ? 72
Second round Jan gave Amy and Toy double the amount of what each of them has, so
Amy Jan Toy
? ? 144
Third round Toy gave Amy and Jan double the amount of what each of them has an at the end Toy has 36 dollars Amy Jan 36
That means that at the second round, Amy + Jan = 144 - 36 = 108 dollars.
So they total have 108 + 144 = 252 dollars.
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
\(\rightarrow\)\(\overline {ED}= \overline {CD}\times \dfrac {24} {24+25}\) = \(\dfrac {7\times 24} {25}\times \dfrac {24} {24+25}\) = \(\dfrac {576} {175}\)
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