Ok, since the welfare rep has kindly requested me to do so, and I'm getting too sick of mugging, and the 2nd round is coming, it pays to jian2 tao3 a bit before I try to get a gold for the last time...
1) 12560, verified with GC (YAY! I'm correct)
cos 30 you can evaluate yourself, then you multiply the expression by cos 10 divided by cos 10, then you realize a lot of stuff you can use double angle formula to simplify and cancel out! =)
2) 54 (correct again!)
I call the centre of circle K! K for 6K! haha =) Then angle a1a3a5 = 0.5 * angle a1Ka5 = angle a1Ka6 = 180 degrees - angle a2Ka3 - angle a3ka4 (because of all the equal arcs and stuff) = 180 degrees - angle a2Ka4 = 180 degrees - 2 angle a2a6a4 = 72 degrees. So the answer is 54 degrees.
3) 2009 (you might have guessed it!)
This 1 is plain steward's theorem, which states that APi^2 + PiB*PiC = AB*AC=49 for i=1,2,..., 41. So the sum is 49*41 = (guess what?) 2009.
4) 8, verified with GC (hehheh!)
Erm... note that (2x^2+6x-56) - (x^2-11x+24) = (x^2+17x-80). So you can let (x^2-11x+24) = A, (x^2+17x-80) = B, (2x^2+6x-56) = A+B, then you get stuff like A + A+B = B, then you play around with whether A is negative of B is negative and A>B and god knows what and your solution should be 8. Btw the graphic calculator tells me that equality holds for 0<=x<=8 then it's a strictly increasing function after that.
And as I type this LA has just beaten the crap out of Orlando at Amway Arena... AGAIN... for their 15th championship title... SIGH...
5) 1742, verified with graphmatica (Argh read question wrongly...)
Translate f(x) 2006 units to the left and call it g(x). Then g(x) is a power 5 polynomial whose x^5 coefficient is also 7. AND, g(-2) = 72, g(-1) = -30, g(0) = 32, g(1) = -24, g(2) = 24. Then introduce h(x) = g(x) - 7x^5. Then h(x) is a power 4 polynomial. AND, h(-2) = 296, h(-1) = -23, h(0) = 32, h(1) = -31, h(2) = -200. Then use silly silly "method of common defferences" and you get h(3) = 41, which means f(2009) is 1742.
6) 2, verified with GC (goddamnit i can't do this 1)
Double angle formula, triple angle formula, you name it, I use it.
(sin80/sin20) - [(sqrt3)/(2 sin80)]
=[cos 10/(2 sin10 cos10)] - (cos30/cos10)
=1/(2 sin10) - (4 cos^3 10 - 3 cos 10)/cos 10
=1/(2 sin10) - 4 cos^2 10 + 3
=(1 - 8 sin 10 cos^2 10 + 6 sin 10)/(2 sin10)
=[1 - 8 sin 10 (1 - sin^2 10) + 6 sin 10]/(2 sin10)
=(1 - 8 sin 10 + 8 sin^3 10 + 6 sin 10)/(2 sin10)
=(1 + 8 sin^3 10 - 2 sin 10)/(2 sin10)
=(2 sin 30 + 8 sin^3 10)/(2 sin10) - 1
=(6 sin 10 - 8 sin^3 10 + 8 sin^3 10)/(2 sin10) - 1
=3 - 1 = 2
7) 128? Anyone wants to check? By listing... *evil grin*
Compare the number formed when a is removed and when b is removed. Write them out in standard notation form (10 to the power of duno duno what). Then you get a=b (mod 7). Then you do the same for b and c. Then b=c (mod 7), with the help that 7 is a prime. Then you go on, and you get a=b=c=d=e=f=g=h (mod 7). In this scenario, any 6-digit number formed by a,...,h is divisible by 7. Just whack and see. So, say for the number abcdefg (h is removed) to be divisble by 7, since bcdefg is divisible by 7, a must be divisible by 7. SO, a=b=c=d=e=f=g=h=0 (mod 7). Then b,...,h = 0 or 7, and a = 7 since a cannot be 0. So total number of integers = 2^7 = 128.
8) 500, approximately verified by GC
Eh... Not sure what happened, I think i got the wrong side of the inequality using cauchy-schwarz, but the answer is still there... Anyway, i have (a-5b)(1-0.2)>=(sqrt a - sqrt b)^2, so a-5b>=500, inequality sign is the wrong way...
9) 224, approximately verifiesd by geometer sketchpad (OMG i guessed it)
Ya, I guessed it, so don't ask me for how I did it...
10) 2009 (Yes, again...)
I just denote cube root of 209 by k. Then x = 0.5 (k-1/k). Then x^2 = 0.25 (k-1/k)^2. Then 1 + x^2= 1 + 0.25 (k-1/k)^2 = 0.25 (k^2 + 1/k^2 - 2 + 4) = 0.25 (k+1/k)^2. Then sqrt (1+x^2) = 0.5 (k+1/k). The rest is easy.
11) 82215?
There are 3 scenarios: x < y < z < w, y < x < z < w, y < z < x < w. So for each scenario you can pick 4 elements from S, order does not matter cos it will be fixed by that inequality. So answer is 30C4 * 3 = 82215.
12) 38889? not sure if the final answer I put was this, i kept changing...
If you count the zeros in front of numbers that have less than 5 digits, then you have 100000 5-digit numbers, so there are 500000 digits, so by symmetry (i cheated) there are 50000 zeros. Then from 00000 to 09999 the 1st zero is not counted. So you minus off 10000. Then from 00000 to 00999 the 2nd zero is not counted, so you minus off 1000. Then you keep doing it, and you will get 50000-10000-1000-100-10-1 = 38889.
13) 96, should be confirmed... GRAHHHHHH
If k is a multiple of 3, since squares = 0 or 1 (mod 3), x^2-ky^2=8 has no integer solutions. So maximum value of k is 99, and k = 3 has no integer solutions. When k=1, x=3 ,y=1. When k=2, x=20,y=14. So minimum value of k is 3. So answer is 99-3 =96.
14) I'm guessing 21845, any other answers?
15) 4021, should be confirmed =)
Letting y=0 gives f(x)f(0) = f(3) + 6x.
Letting x=0 gives f(0)f(y) = f(3) + 3f(y) - 3f(0), following which letting y=x gives f(0)f(x) = f(3) + 3f(x) - 3f(0).
So 6x = 3f(x) - 3f(0), which gives f(x) = 2x + f(0), which gives f(3) = 6 + f(0),
Letting x=y=0 gives f(0)^2 = f(3).
So f(0)^2 - f(0) - 6 =0, which gives f(0) = 3 or -2, which gives f(x) = 2x + 3 or 2x - 2.
Sub back into original equation, f(x) = 2x - 2 does not work, so f(x) = 2x+3. So f(2009) = 4021.
16) 89970, should be confirmed =)
After some silly trial and error, we guess that A(n+1) = (2008+n)A(n). WHICH WORKS! So A993 = (2008+992)(2008+991)A991. So A993/(100A991) = 3000*2999/100 = 89970.
17) Anyone tried this 1?
18) 11440, tentatively confirmed (OMG I FORGOT TO DIVIDE 16 BY 4!!!!!)
Consider the act of putting 7 slots in between 15 balls, where extreme ends of the balls can also have slots placed there. After which, remove 1 ball from in between each slot, so that there are 9 balls left, with 7 slots in between. Let each digit be represented by no. of balls on the left side of each slot. So answer is 16C7=11440.
19) 127, should be confirmed =)
a^n - 1 can always be factorized into (a-1)(something something something). So for the number to be prime, a-1 must be = 1, so a = 2. Since 2^13 - 1 = 8191>5000, 2^12 - 1 = 4095 which is divisible by 5, 2^11 - 1 = 2047 which is 23*89, 2^10 - 1 = 1023 which is divisible by 3, 2^9 - 1 = 511 which is 7*73, 2^8 - 1 = 255 which is 3*5*17, and 2^7 - 1 = 127 which is a prime, the answer is 127.
20) 1005?
Cheat and whack, M looks like it's constant at 1004.5.
21) 56, tentatively confirmed (oh man, may the earth swallow me up, no time to finish this 1...)
Consider how many ways there are so that no 3 numbers are consecutive.
Case 1: 3 blocks of 2 numbers.
39 balls unpicked, 40 places to slot these 3 blocks, so no. of ways = 40C3 = 9880.
Case 2: 2 blocks of 2 numbers, 2 blocks of 1 number.
No. of ways = 40C2 * 38C2= 548340.
Case 3: 1 block of 2 numbers, 4 blocks of 1 number.
No. of ways = 40C1 * 39C4= 3290040.
Case 4: 6 blocks of 1 number.
No. of ways = 40C6= 3838380 (wow)
Total possible ways to pick 6 balls = 45C6 = 8145060
So p = 1 - (9880+548340+3290040+3838380)/8145060 = 0.0562819672....
So floor value of 1000p = 56.
22) 1, should be confirmed (I'll never ever forget that tan^-1 (infinity) is PI/2, not ZERO! ARGHHHHHHHHHHH)
tan^-1 [2/(2k+1)^2] = tan^-1 (2k+2) - tan^-1 (2k) (Try it yourself, it works). So sum of something something something = (2/pi) [tan^-1 (infinity) - tan^-1 (0)] = (2/pi) (pi/2 - 0) = 1.
23) 263, confirmed by GC (Ahhhhhh..... i just made a fool of myself... again...)
Don't know the general formula for this sum? Lazy to find out? Try... WHACKING!
11^5+10^5+9^5+8^5+7^5+6^5+5^5+4^5+3^5+2^5+1^5
= 161051+100000+59049+32768+16807+7776+3125+1024+243+32+1
= 381876 = 2^2 * 3 * 11 ^2 * 263.
24) 15, confirmed by GC (sigh...)
X(n+2) = 0.25 X(n+1) + 0.75 X(n) , ...................... So,
X(n+2) + X(n+1) +...+X(3) = 0.25 X(n+1) + X(n) + X(n-1) +...+ X(2) + 0.75 X(1).
So X(n+2) + 0.75 X(n+1) = X(2) + 0.75 X(1) = 24 + 3*3/4 = 105/4.
As n goes to infinity, X(n+2) approximates X(n+1) approximates X(n), so X(n) approaches (105/4) / (1 + 3/4) = 15.
25) Who wants to do this 1? Opportunity cost for getting the correct answer is 20 years of your life...
Hmm... that's about 12 or 13 correct... team award? As much chance as Newcastle playing in EPL next season *evil cackle to mister mugging for IChO*...
Yup, that's all!
DK
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At around 5.30pm, we eventually had to continue trekking so as to meet the class PUNCTUALLY AT 6PM (*coughs*) at Vivocity.... and of course to enjoy the aircon after spending about 6 hours in the sun... A pity Tze Han couldn't join us for dinner... but thanks Yong Lin, Yi Tyan and Huang Pei for joining us!!! As usual, after one hour of waiting (*coughs again*), dinner started. Another hour later, dinner ended and everyone tried to find something to do. Finally, after one more hour of slacking and doing nothing (or trying to test Yi Tyan more A level stuff), we decided to pack up and go home (to mug). But obviously, ppl like the President, Champion, DK and BlahKing didn't have to worry abt mugging since they happily went to the arcade at 9pm for more games!!!
btw, i think wear red underwear really brings u good luck...
