I thought that the exam was somewhat easier than high-school level math competitions in the US (i.e., AIME-level competitions). I was never much good at those, thought; the times in parentheses indicate how long it took me to arrive at the solution, but without writing a neat proof.
1. I didn't do this using any clever technique; I just substituted (e^ix+e^-ix)/2 for cos(x) and solved the resulting equation. (~10 minutes)
2. One of the conditions tells you that if f(x)=0, then x is odd; the other tells you that x is even. (~3 minutes)
3. The first one can be easily done by L'Hopital's rule; the second can be done by looking at the limit of 1/x as x goes to infinity (you get xe^x) rather than the limit of x as x goes to 0. (~5 minutes)
4. This is obvious; given a particular gap between the numbers, there are only a finite amount of squares which are less than that distance apart. (~1 minute)
5. I probably did it the messy way, but I applied Fermat's little theorem and then did some computations. (~10 minutes)
6. This is just crying out for induction to be used, but I got bogged down in calculations until I saw a simpler way. So, by induction, suppose that the statement holds true up to n. Then for n+1, you have that 1-1/n!+(n+1)/[(n-1)!+n!+(n+1)] should equal 1-1/(n+1)!. The proof becomes easier if you notice that (n-1)!+n!+(n+1)! = (n-1)!(n+1)^2, so that the numerator of the second fraction cancels out. Then just take -1/n! and 1/(n-1)!(n+1) to the same denominator to get the result. (~15 minutes)
7. Ugh, I hate problems involving logs, whether physical or mathematical. I didn't even try this one. (~infinity minutes)
8. Draw a circle of radius 0.5 cm around the center of the circle; at most one point can lie in this region. Hence at least 7 points must lie in the outer annulus. However, this annulus has area 3pi/4. For each point, draw a circle of radius 0.5 cm around it; then 7 such points have area 7pi/4. However, their centers must also lie within the annulus, which is impossible (although this is not easy to prove; there's probably an easier way). (not completed, ~15 minutes)
9. The derivative of the expression is the same as the expression without the first term. Hence, if there is a repeated root, then both the initial expression and its derivative must be 0, so the first term must be 0. But this implies that x = 0, so the entire expression is 1, a contradiction. (~2 minutes)
10. I have no idea. (~infinity minutes)
11. Square the expression to get that sqrt(21)+3sqrt(7)+7sqrt(3) is rational; subtract the original expression from this to get that sqrt(7)+3sqrt(3) is rational. Square this to get that sqrt(21) is rational, which is impossible. (~3 minutes)
12. This can be solved by a system of equations; in drawing the top of the rhombus, you create a smaller isosceles triangle that is similar to the initial one. If you call the bottom of this triangle (i.e., the top of the rhombus) x, and one of its sides a, then you have that a/[sqrt(5)/2] = x/1 and that a+x=sqrt(5)/2. This can be solved to find x. (~10 minutes)
13. First note that (b+i/b)^4 = b^4+4b^2i-4i/b^2-6+1/b^4 = 4i(b^2-1/b^2). Square this to get -16(b^4-2+1/b^4)=-16(4)=-64. Then square this once more to get 4096 as the answer. (~3 minutes)
1. I read jbuniii's answer before trying this.
3. a. This just requires finding out what 10 000 is in base 6.
b. We have 36a+6b+c=81c+9b+a, so b = 5/3(7a-16c). Thus, 7a-16c=3, so b = 5. Trial-and-error yields that a=5 and b =2, so the number is 552. (~15 minutes)
4 a. Suppose a^2+b^2=c^2. If a or b is divisible by 2, then ab/2 is also divisible by 2. If neither a nor b is divisible by 2, then looking at a^2+b^2 mod 4 shows that c^2 is divisible by 2 but not by 4, which is impossible. So the area is divisible by 2. If neither a nor b is divisible by 3, then a^2+b^2 is equal to 2 mod 3; but c^2 cannot be equal to 2 mod 3. So either a or b is divisible by 3 and the area is divisible by 3. So the area is divisible by 6.
b. Drop some perpendicular bisector, l, which divides some side of the triangle into p and q. Then l(p+q) is rational by assumption, and p+q is rational by assumption, so l must be rational. Also, p = sqrt(a^2-l^2) and q = (b^2-l^2) = r-p, where a and b are the other sides of the triangle. So if p is irrational, then sqrt(b^2-l^2)=k-sqrt(a^2-l^2), so b^2-l^2=k^2-2sqrt(a^2-l^2)+a^2-l^2, which implies that sqrt(a^2-l^2) is rational, a contradiction. Hence p and q are both rational, and we have the desired partition. (~15 minutes)
7. This is obvious; if a and b are the common differences in the arithmetic progressions, then a line is given by (a1, b1)+t(a, b). (~1 minute)
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