You can
download this year's Mathcounts state competition questions here.
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 6,3,2,4
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,
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
#29:
\(\Delta ADE\) is similar to \(\Delta ABC\)
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.
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
\(\rightarrow\)\(\overline {ED}= \overline {CD}\times \dfrac {24} {24+25}\) = \(\dfrac {7\times 24} {25}\times \dfrac {24} {24+25}\) =
\(\dfrac {576} {175}\)
.
![\[\dfrac{r}{5}=\dfrac{s}{10}=\dfrac{\sqrt{r^2+h^2}}{10}.\]](https://latex.artofproblemsolving.com/2/0/5/205a8d45594c47d92393e07cffca02d3d3a5b415.png)
![\[10r=5\sqrt{r^2+h^2}\implies 100r^2=25r^2+25h^2\implies 75r^2=25h^2\implies 3r^2=h^2\implies h=r\sqrt{3}.\]](https://latex.artofproblemsolving.com/c/3/4/c34a4033b4515ae88c64fda9f900d84924094aa6.png)
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.![\[\dfrac{125\sqrt{3}}{3}-\dfrac{\pi\times r^2h}{3}=\dfrac{125\sqrt{3}}{9}\implies 125\sqrt{3}-r^2h=\dfrac{125\sqrt{3}}{3}\implies r^2h=\dfrac{250\sqrt{3}}{3}.\]](https://latex.artofproblemsolving.com/d/2/e/d2eb3f35bb46a600fd5db854308e1f7844964947.png)
![\[r^3\sqrt{3}=\dfrac{250\sqrt{3}}{3}\implies r=\sqrt[3]{\frac{250}{3}}.\]](https://latex.artofproblemsolving.com/a/d/9/ad9d39850b5a5aa905a1e879a6b5bfebc1804496.png)
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.![\[37.207+138.477=175.684\approx\boxed{176}.\]](https://latex.artofproblemsolving.com/9/b/9/9b9d7bd69023b24516b90a9c51dbce0f9fffbe08.png)
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