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