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.

## Monday, June 10, 2019

## 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)

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]

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

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