Archive - Jan 7, 2008

Date

Suffolk Mathelete 2007-2008 4th meet questions

Posted January 7th, 2008 by Mgccl
in

Another Mathelete meet yesterday, I record all the questions, they are extremely simple compare to the questions in national contests(I guess it's below the AMC 10 level, but I might be wrong, never took one.). It's still abusive questions to my school. SWRHS really sucks at math, we don't even provide AP Calculus BC courses. I'm listing the question and solution for the sake to show other matheletes in my school.
Friday will be the Suffolk math tournament. Last year, the prize and "lunch" really doesn't cover the cost of our teams. (Ace the tournament = A Rubik's cube).

4-1
Write down 3 three-digit perfect square, when combined, it contain every non-zero digit once.

In matheletes, there are only one possible answer, it can help to eliminate some answers.
102 to 312 are the possible candidates, total of 21 numbers.
100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961
Remove ones with repeating digits
169, 196, 256, 289, 324, 361, 529, 576, 625, 729, 784, 841, 961
Now, because there are only one possible answer, numbers with exact same repeating number can also be removed
289, 324, 361, 529, 576, 729, 784, 841,
Now is the part for some smart trial and error.
First of all, one of the number must be 324 and 361, since it must contain a 3. Pick one and start elimination, soon the right answer will turn out
361, 529, 784
Personally, I don't like questions require a lot of repetitive job.

4-2

Image created with The Geometer's Sketchpad. I got a Chinese student version from helpful Chinese friends.
3 circles packs on top of 2 circles. 2 circles pack on those 3 circles. Now, use a rubber band and bind all the circle together. How long is the rubber band if the radius of each circle is 1.


connect the center of the circles, 6 equilateral triangles with side of 2 can be created. each one have 2 side to the out side. translate those sides, and it will cover the straight part of the band. The rest of the band equal to the circumference of one circle. The result is 2*6 + 2π = 12+2π

4-3
\frac{x^{n+2} +x^n}{x^{n+4}+x^{n+2}}=x^k
find k

\begin{eqnarray}\frac{x^{n+2} +x^n}{x^{n+4}+x^{n+2}} &=&\frac{x^2x^n+x^n}{x^4x^n+x^2x^n}\\ &=&\frac{(x^2+1)x^n}{(x^2+1)x^{n+2}}\\ &=&\frac{1}{x^{2}}\\ \end{eqnarray}
k=-2

4-4
Semi-circle with diameter AB, AB = 1. C locates on the arc. AC and BC are perpendicular. find the ∡CAB=θ to the nearest degree if Area of triangle ABC is half of the area of the semi-circle.

area of semi-circle is π/8
area of triangle is π/16
solve
\begin{eqnarray}\frac{sin(\theta)cos(\theta)}{2}&=&\frac{\pi}{16}\\ sin(\theta)cos(\theta)&=&\frac{\pi}{8}\\ \frac{sin(2\theta)}{2}&=&\frac{\pi}{8}\\ sin(2\theta)&=&\frac{\pi}{4}\\ 2\theta&=&sin^{-1}(\frac{\pi}{4})\\ \theta &\approx& 26^\circ\\ \end{eqnarray}

4-5
(x!)^2 - 7x! + 6 = 0
Find all x

y = x!
y^2 - 7y + 6 = 0
use quadratic equation and find y = {1, 6}
x = {0,1,3}

4-6
if an is the nth number larger than 1000 and has 4 different digits, find a111

The last digit always have 7 variations, 15*7+1 = 106.
The 2nd last digit always have 8 variations.
The smallest number after 1023 with the 2nd digit replaced with another number is 1203.
After 15-8 = 7
the third digit must be the 7th smallest digit, which would be 9
the last digit must be the 111-106, 5th smallest digit possible, which is 7
1297

I wonder who come up with those questions.

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