Does our approach to teaching math fail even the smartest kids ?

Quotes from that article "According to research from the University of California, Los Angeles, as many as 60 percent of all college students who intend to study a STEM (science, technology, engineering, math) subject end up transferring out. In an era when politicians and educators are beside themselves with worry over American students’ lagging math and science scores compared to the whiz kids of Shanghai and Japan, this attrition trend so troubles experts it has spawned an entire field of research on “STEM drop-out,” citing reasons from gender and race to GPAs and peer relationships."

A Skill for the 21st Century: Problem Solving by Richard Rusczyk, founder of "Art of Problem Solving".

Top 10 Skills We Wish Were Taught at School, But Usually Aren't

from Lifehacker

## Wednesday, June 14, 2017

## 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.

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,

#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 =

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 =

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 =

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

**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

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 dollars**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**.## Monday, April 10, 2017

### The Largest Rectangle Inscribed in Any Triangle

From Mathcounts Mini : Maximum area of inscribed rectangles and triangles

\(\Delta EHI\sim\Delta EFG\) \(\rightarrow\) \(\dfrac {a} {c}=\dfrac {d-b} {d}\)\(\rightarrow\) \(a=\dfrac {c\left( d-b\right) } {d}=\dfrac {-c\left( b-d\right) } {d}\)

We are going to find out what the largest area of a rectangle is with the side length a and b.

It can be shown that by substituting the side length "a" with the previous equation + completing the square that the largest area is half of the area of the triangle the rectangle is embedded.

\(a\times b=\dfrac {-c\left( b-d\right) \times b} {d}=\dfrac {-c\left( b^{2}-bd\right)} {d}= \dfrac {-c\left( b-\dfrac {1} {2}d\right) ^{2}+\dfrac {1} {4}dc} {d}\).

From there, you know that when \(b= \dfrac {1} {2}d\), it will give you the largest area, which is \(\dfrac {1} {4}dc\).

\(a=\dfrac {-c\left( b-d\right) } {d}= \dfrac {-c\left( \dfrac {1} {2}d-d\right) } {d}=\dfrac {c\left( d-\dfrac {1} {2}d\right) } {d}=\dfrac {1} {2}c\).

Thus, the maximum rectangle area occurs when the midpoints of two of the sides of the triangle were joined to make a side of the rectangle and its area is thus 50% or half of the area of the triangle or 1/4 of the base times height.

\(\Delta EHI\sim\Delta EFG\) \(\rightarrow\) \(\dfrac {a} {c}=\dfrac {d-b} {d}\)\(\rightarrow\) \(a=\dfrac {c\left( d-b\right) } {d}=\dfrac {-c\left( b-d\right) } {d}\)

We are going to find out what the largest area of a rectangle is with the side length a and b.

It can be shown that by substituting the side length "a" with the previous equation + completing the square that the largest area is half of the area of the triangle the rectangle is embedded.

\(a\times b=\dfrac {-c\left( b-d\right) \times b} {d}=\dfrac {-c\left( b^{2}-bd\right)} {d}= \dfrac {-c\left( b-\dfrac {1} {2}d\right) ^{2}+\dfrac {1} {4}dc} {d}\).

From there, you know that when \(b= \dfrac {1} {2}d\), it will give you the largest area, which is \(\dfrac {1} {4}dc\).

\(a=\dfrac {-c\left( b-d\right) } {d}= \dfrac {-c\left( \dfrac {1} {2}d-d\right) } {d}=\dfrac {c\left( d-\dfrac {1} {2}d\right) } {d}=\dfrac {1} {2}c\).

Thus, the maximum rectangle area occurs when the midpoints of two of the sides of the triangle were joined to make a side of the rectangle and its area is thus 50% or half of the area of the triangle or 1/4 of the base times height.

## Thursday, March 9, 2017

### For Mainly High School Students : Quest for USAJMO or USAMO

Good luck on your quest :)

AMC forum from AoPS

AHSME Problems and Solutions

AMC 10 Problems and Solutions

AMC 12 Problems and Solutions

AIME problems and solutions

2016AMC-10 A Math Jam

2016 AMC-10 B Math Jam

2016 AIME I Math Jam

2016 AIME II Math Jam

2015 AMC-10 A Math Jam

2015 AMC-10 B Math Jam

2015 AIME I Math Jam

2015 AIME II Math Jam

2014 AMC-10 A Math Jam

2014 AMC-10 B Math Jam

2014 AIME I Math Jam

2014 AIME II Math Jam

2012 AMC-10 A Math Jam

2012 AMC-10 B Math Jam

2012 AIME I Math Jam

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2011 AMC-10/12 A Math Jam

2011 AMC-10/12 B Math Jam

2011 AIME I Math I Jam

2011 AIME II Math Jam

2010 AMC-10/12 A Math Jam

2010 AMC-10/12 B Math Jam

2010 AIME I Math Jam

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Coach Monks's High School Playbook

AMC forum from AoPS

AHSME Problems and Solutions

AMC 12 Problems and Solutions

AIME problems and solutions

**AMC, AIME videos from AoPS resources****AMC-10/12 A, B and AIME Math Jam Transcripts from AoPS**

2016 AMC-10 B Math Jam

2016 AIME I Math Jam

2016 AIME II Math Jam

2015 AMC-10 A Math Jam

2015 AMC-10 B Math Jam

2015 AIME I Math Jam

2015 AIME II Math Jam

2014 AMC-10 B Math Jam

2014 AIME I Math Jam

2014 AIME II Math Jam

2012 AMC-10 B Math Jam

2012 AIME I Math Jam

2012 AIME II Math Jam

2011 AMC-10/12 A Math Jam

2011 AMC-10/12 B Math Jam

2011 AIME I Math I Jam

2011 AIME II Math Jam

2010 AMC-10/12 A Math Jam

2010 AMC-10/12 B Math Jam

2010 AIME I Math Jam

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Labels:
AIME I,
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Math Jams

## Tuesday, February 14, 2017

### 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)

789000 divided by the LCM of 7, 8 and 9, which is 504 = 1565.47...

Try 504 * 1566 = 789264 (it works)

The answer is

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.

Then solve.

The key is to see 2

2

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 =

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

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.

Vieta's Formula and the Identity Theory

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.

\(11\equiv -2\left ( mod13\right)\) ; \((-2)^{12}\equiv 4096\equiv 1\left ( mod13\right)\)

Or use Fermat's Little Theorem (Thanks, Spencer !!)

\(11^{13-1}\equiv 11^{12}\equiv1 (mod 13)\)

My students should get a virtual bump if they got this question wrong.

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 proportionCross 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

All we have left is to add the two bases. The total area of thebases is . So our final answer is

Using dimensional change and ratio, proportion.

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)ional change and ratio, proportion.

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)

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 2

^{10 }is 1024 or about 10^{3}2

^{30}= ( 2^{10 })^{3 }or about (10^{3 })^{3}about 10^{9}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 proportionCross 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

**Solution II :**Using dimensional change and ratio, proportion.

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)ional change and ratio, proportion.

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]

\(\dfrac {1} {2}\left( 2\times 10\pi + 2\times 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\pi \right)\)* \(\left( 10-10\times \dfrac {\sqrt [3] {2}} {\sqrt [3]{3}}\right)\)

### 2013 Mathcounts State Harder Problems

You can download this year's Mathcounts state competition questions here.

Trickier 2013 Mathcounts State Sprint Round questions :

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

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:

2,4,6,3 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,

\(\overline {ZY}:\overline {WV}=5:8\) -- 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

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 6,3,2,42,4,6,3 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

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.

Here is a proof to demonstrate that the largest area of a rectangle inscribed in a triangle is

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

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.

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.

\(\dfrac {7000} {x}+\dfrac {10000} {2x}=3000\) \(\rightarrow\)

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

\(\rightarrow\)\(\overline {ED}= \overline {CD}\times \dfrac {24} {24+25}\) = \(\dfrac {7\times 24} {25}\times \dfrac {24} {24+25}\) =

\(\dfrac {x} {21}=\dfrac {8-y} {8}\)

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

\(\rightarrow\)\(\overline {ED}= \overline {CD}\times \dfrac {24} {24+25}\) = \(\dfrac {7\times 24} {25}\times \dfrac {24} {24+25}\) =

**\(\dfrac {576} {175}\)**## Thursday, January 12, 2017

### 2017 Mathcounts Competition Preparation Strategies

**Hola, meet you at the state or/ and the Nationals. :D**

**Stay focused (hang in there), sleep well, eat healthy, exercise and please, please don't get sick...**

**Relaxed and Zenlike... yet focused (how...hmmm... that is a GREAT question)**

**Check out this video, see if it works and tells you something.**

**https://www.youtube.com/watch?v=T5rnD6u5EtY**

**Write on the comments and let me know if that really helps : Yup -- for the love of learning and have the chance to meet the other kindred spirits**

**Lacking sleep will fog your brain. Not skipping questions will cost you.**

**Most important, best of luck and have fun.**

It's the Mathcounts chapter/state/AMCs season. Kids are more motivated.

YEAH !!!!! :)

Hi, Thanks for visiting my blog.

E-mail me at

**thelinscorner@gmail.com**if you want to join my groups.
Currently I'm running different levels of problem solving group lessons, and it's lots of fun learning along with students from different states.

My most advanced group of students are just AMAZING !! to say the least.

Ha ha, we are using AMC-10, 12 questions as countdown round practices and some can solve the first few AIME problems in less than a minute. Oh dear !!

So many students are not learning smart.

Problem solving is really fun (and a lot of the times very hard, yes).

Good questions are intriguing and delicious, so come join our vibrant community and have the pleasure of finding things out on your own.

Problem solving is really fun (and a lot of the times very hard, yes).

Good questions are intriguing and delicious, so come join our vibrant community and have the pleasure of finding things out on your own.

**There is no overnight success.**

**My other blogs :**

thelinscorner : Standardized test preps, books, links/videos for life-time learning

Take care and have fun learning.

**Don't forget other equally interesting activities/contests, which engage your creativity and imagination.**

**Some also require team work. Go for those and have fun !!**

**Don't just do math.**

**In Chinese :**《專題報導》亞裔尖子生 職場難出頭？

Before going full throttle mode for competition math, please spend some time reading this

well- thought-out article from BOGTRO at AoPS "Learn How to Learn".

It will save you tons of time and numerous, unnecessary hours without a clear goal, better method in mind.

Less is more. My best students make steady, very satisfactory progress in much less time than those

counterparts who spent double, triple, or even more multiple times of prep with little to show.

It's all about "deliberate practices", "tenacity", and most of all, "the pleasure of finding things out on your own".

Take care and have fun problem solving.

I have been coaching students for many years. By now, I know to achieve stellar performance you need :

Grit (from TED talk), not only that but self-awareness (so you can fairly evaluate your own progress) and a nurturing-caring environment. (Parents need to be engaged as well.)

Thanks a lot !! Mrs. Lin

"Work Smart !!" , "Deliberate practices that target your weakness ", " Relax and get fully rested.", "Pace your time well", "Every point is the same so let go of some questions first; you can always go back to them if time permits."

"It's tremendous efforts preparing for a major event on top of mounting homework and if you are the ones who want to try that, not your parents and you work diligently towards your goal, good for you !!"

"Have fun, Mathcounts changes lives, because at middle school level at least, it's one of those rare occasions that the challenges are hard, especially at the state and national level."

**Now, here are the links to get you started:**

Of course use my blog. Whenever I have time I analyze students' errors and try to find better ways (the most elegant solutions or the Harvey method I hope) to tackle a problem. Use the search button to help you target your weakness area.

Last year's Mathcounts competition problems and answer key

This year's handbook questions.

Near the end of the handbook, there is a page called

**problem index (page 82 and 83 for 2013-2014 handbook)**.

For state/national prep, find your weakness and work on the problems backwards, from the hardest to the easiest.

Here are some other links/sites that are

**the best.**

Mathcounts Mini : At the very least, finish watching and understanding most of the questions from 2010 till now and work on the follow-up sheets, since detailed solutions are provided along with some more challenging problems.

For those who are aiming for the state/national competition, you can skip the warm-up and go directly to "The Problems" used on the video as well as work on the harder problems afterward.

Art of Problem Solving

The best place to ask for help on challenging math problems. Some of the best students/coaches/teachers are there to help you better your problem solving skills.

Register for Alcumus and start using the great tool to practice deliberately.

Change the setting based on the levels of your proficiency of different topics. Do Not Rush !!

**Awesome site!!**

For concepts reviewing, try the following three links.

Mathcounts Bible

Mathcounts Toolbox

Coach Monks's Mathcounts Playbook

You really need to understand how each concept works for the review sheets to be useful.

To my exasperation, I have kids who mix up the formulas without gaining a true understanding and appreciation of how an elegant, seemingly simple formula can answer myriads of questions.

You don't need a lot of formulas, handbook questions, or test questions to excel.

You simply need to know how the concepts work and apply that knowledge to different problems/situations.

Hope this is helpful!!

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