## Monday, June 10, 2019

### 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

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)

#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

#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)

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 usinthe Pythagorean  is $\sqrt{10^2-5^2}=5\sqrt{3}$.
Usingthediagram below, let $r$ be the radius of the top cone and let $h$ be the height of the topcone.
Let $s=\sqrt{r^2+h^2}$ 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 $$\dfrac{r}{5}=\dfrac{s}{10}=\dfrac{\sqrt{r^2+h^2}}{10}.$$Cross multiplying yields $$10r=5\sqrt{r^2+h^2}\implies 100r^2=25r^2+25h^2\implies 75r^2=25h^2\implies 3r^2=h^2\implies h=r\sqrt{3}.$$This is what we need.

Next, the volume of the original cone is simply $\dfrac{\pi\times 25\times 5\sqrt{3}}{3}=\dfrac{125\sqrt{3}}{3}$.

The volume of the top cone is $\dfrac{\pi\times r^2h}{3}$.
From the given information, we know that $$\dfrac{125\sqrt{3}}{3}-\dfrac{\pi\times r^2h}{3}=\dfrac{125\sqrt{3}}{9}\implies 125\sqrt{3}-r^2h=\dfrac{125\sqrt{3}}{3}\implies r^2h=\dfrac{250\sqrt{3}}{3}.$$We simply substitute the value of $h=r\sqrt{3}$ from above to yield $$r^3\sqrt{3}=\dfrac{250\sqrt{3}}{3}\implies r=\sqrt{\frac{250}{3}}.$$We will leave it as is for now so the decimals don't get messy.

We get $h=r\sqrt{3}\approx 7.56543$ and $s=\sqrt{r^2+h^2}\approx 8.7358$.

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 $5\times 10\times \pi=50\pi$.
The surface area of the top cone is $\pi\times r\times s\approx 119.874$.
So our lateral surface area is

All we have left is to add the two bases. The total area of thebases is $25\pi+\pi\cdot r^2\approx 138.477$. So our final answer is $$37.207+138.477=175.684\approx\boxed{176}.$$
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 $$\dfrac {\sqrt  {2}} {\sqrt  {3}}$$.

Using this ratio, we can get the radius of the smaller circle as 10 * $$\dfrac {\sqrt  {2}} {\sqrt  {3}}$$ and the radius of the top circle of the frustum as 5 * $$\dfrac {\sqrt  {2}} {\sqrt  {3}}$$.

Now we can solve this :

$$\dfrac {1 } {2 }$$$$\left[ 10^{2}\pi -\left( 10\times \dfrac {\sqrt  {2}} {\sqrt  {3}}\right) ^{2}\pi \right]$$ + $$5^{2}\pi +\left( 5\times \dfrac {\sqrt  {2}} {\sqrt {3}}\right) ^{2}\pi$$ = 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 $$\dfrac {\sqrt  {2}} {\sqrt  {3}}$$.

Using this ratio, we can get the radius of the smaller circle as 10 * $$\dfrac {\sqrt  {2}} {\sqrt  {3}}$$ and the radius of the top circle of the frustum as 5 * $$\dfrac {\sqrt  {2}} {\sqrt  {3}}$$.

Now we can solve this :

$$\dfrac {1 } {2 }$$$$\left[ 10^{2}\pi -\left( 10\times \dfrac {\sqrt  {2}} {\sqrt  {3}}\right) ^{2}\pi \right]$$ + $$5^{2}\pi +\left( 5\times \dfrac {\sqrt  {2}} {\sqrt {3}}\right) ^{2}\pi$$ = 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]
$$\dfrac {1} {2}\left( 2\times 10\pi + 2\times 10\times \dfrac {\sqrt  {2}} {\sqrt  {3}}\pi \right)$$* $$\left( 10-10\times \dfrac {\sqrt  {2}} {\sqrt {3}}\right)$$