Tuesday, November 29, 2016

Geometry : Harder Chapter Level Quesitons

Question #1 : The area ratio of two equilateral triangles are 4 to 9 and the sum of their perimeter is 30 3 . What is the area of the a. smaller triangle   b. larger triangle?

Solution:
If the area ratio of two similar polygons is 4 to 9, their corresponding line ratio would be  4  to  9
or 2 to 3.[Make sure you know why.]
The perimeter of the two equilateral triangles is 30 3 so the smaller triangle has a perimeter of
2/5 *  30 3 or 12 3. One side is 4 3 . Using the formula of finding the area of an equilateral triangle $$\dfrac{\sqrt3*s^2}{4}$$ , you get the area to be 12 3.

Use the same method to get the area of the larger triangle as 27 3.
You can also use ratio relationship to get the area of the larger triangle by
multiply 12 3 by 9/4.

2007 Mathcounts Chapter Sprint #30: In parallelogram ABCD, AB = 16 cm, DA = 32 cm, and sides AB and DA form a 45-degree interior angle. In isosceles trapezoid WXYZ with WX ≠ YZ, segment WX is the longer parallel side and has length 16 cm, and two interior angles each have a measure of 45 degrees. Trapezoid WXYZ has the same area as parallelogram ABCD. What is the length of segment YZ?

Solution I:
Make sure you know how to get the unknown leg fast. The height of the parallelogram is 82, so the area of the parallelogram is 48 square units. [Check out the special right triangle section here if you can't get the height fast.]

Let YZ of the trapezoid be x and draw the height. Using 45-45-90 degree angle ratio, you'll get the height. (See image above.)
Area of the trapezoid is average of the two bases time height. WX = 16 (given)
$$\dfrac{(16+x)* (16-x)}{4}$$ = 48 ; 256 - x2 = 192 ;  - x2 = - 64;  x = 8 = YZ

Solution II:

Make the y be the height of the trapezoid. YZ = 16 - 2y.  $$\dfrac{(16-2y + 16)}{2}$$ * y = 48
$${(16 -y)* y = 48}$$$$\rightarrow$$ $${16y -y^2 = 48}$$ $$\rightarrow$$ $${y^2 - 16y + 48 = 0}$$ $$\rightarrow$$ $${(y -12)(y -4)=0}$$ $$\rightarrow$$ $${y = 4}$$ or $${y = 12}$$(doesn't work)
YZ = 16 - 2y. Plug in y = 4 and you have  YZ = 8

Solution III: Let the height be y and you have $$\dfrac{(\overline{YZ}+ 16)* y}{2}"$$ = 48 ; ( YZ + 16) * y = 96
When there are some numbers multiply together equal another number, it's a factoring question.
32 * 3 = 96, YZ = 16 (doesn't work)
24 * 4 = 96, YZ =8

Friday, November 25, 2016

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.

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: This question is very similar to 2012 chapter target #8.
See this link for similar 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 $$r^3\sqrt{3}=\dfrac{250\sqrt{3}}{3}\implies r=\sqrt[3]{\frac{250}{3}}.$$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 $$\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}$$.

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