SMO 2018 Open Rd 2 Namelist
Open Round 2 Cut-off: 13
SMO 2018 Open Rd 2 Questions
SMO 2018 Open Questions
SMO 2018 (Open Section) Answers
1. 3
2. 1540
3. 2
4. 1
5. 360
6. 112
7. 370
8. 8071
9. 1927
10. 2
11. 29
12. 2
13. 10
14. 33
15. 6
16. 270
17. 3
18. 26961
19. ? (*amended. Refer to **)
20. 2700
21. 2018
22. 81
23. 2520
24. 181
25. 70 (*amended)
**Checked with Lim Jeck Q19, he thinks the intended answer is 24 but said it is not attainable
48 comments:
First
Autist i was here before u
Q15 is 6 I think
14 is 31?
19 is 15 i think
I think so too
I got 3 for 13 odk
Q25 is 70, not 63, use Verrier's Lemma+similar triangles.
Q11 is 29, the graph is basically x+y=-1 and (1,1), which is the closest point.
Q15 is 6 I think, if I'm not wrong you trigo bash.
Yup 15 is 6
Q15 you can also menelaus’ theorem or stewart’s theorem
14 is not 33. If it was its its just like asking whats 6*11/2.
Q14 is 33.Just do chessboard coloring.
Thats like saying q14 is just 66/2
hi is this year more difficult than past yrs?
Yes hopefully r2 is like 10 or 11
can anyone estimate how many is bronze? thanks
can someone provide a solution for q15?
What's the ddifficulty of this year's paper? harder or easier
You see the thing is yes, it is
Q19 is not 15, I managed to find a counter example.
Let m=2 and solve for a1, you get a1^3-2a1^2+a1-1=0. Plug it into a cubic equation solver, you get 1.754877666246692 as one of the roots. Solve for 16a1-a3, you get somethung like 23.998 which is clearly more than 15, so 15 can't be the answer.
*something
how do you do qn13?
Is it harder than last year?
Last year was significantly easier i think
I got 24 for q19
pray for 5 as r2 cutoff tbh
i got 6 can i get to top 30?
lol got 6 too, careless for 2 questions
I got 11 correct... Hopefully can get in rd 2...
Can anyone estimate the cutoff point for entering second round?
maybe cutoff will be around 12 or 13 again
Will 9 points qualify for anything?
Q19 should be 24. First notice that the terms from a2 onwards are positive. If 01 and the rest of the terms are positive, so no solution. If a1 is negative, you have 16a1-a_{m+1} is negative, so we can ignore this case. Plug small values of m into WolframAlpha, we have 16a1-a_{m+1}=15, 23.9984, - 6.5..., - -1765...,.... This probably means 24 is the answer as 16a1-a_{m+1} decreases as m increases for m>2.
Believe 5 is cutoff point for r2.
c'mon that's absurd. 11 or 12 maybe more realistic
Q19 I also got 24. Manipulation of the recurrence relation can get 1/(a_{n})=1/(a_{n}-1)+1/(a_{n+1}-1). Plug into 1/a1+...+1/am =1, we can get 1/(a_{1}-1)-1/(a_{m+1}-1)=1 by method of difference. Then we can express a_{m+1} in terms of a1, which is 1/(2-a_{1}). Substitute it into 16a1-am+1 and differentiate. maximum value is reached when 2-a1=1/4. Sub in get 24.
I think this process is correct but a1=1.75 is clearly not correct. This is absurd... but 15 is clearly wrong given the counter-example
Q13 should be 47 with m=39 and n=8
I also got 47 for q13
whats verrier's lemma??
q19 is 24 when a1=7/4
why questions 23 is not 2523
@above because it can be -30 instead of 26
@above I mean 30 instaed of 29
Is the round 2 list out yet?
all the round 2 lists are on sms website; https://sms.math.nus.edu.sg/competitions/SMO2018/R2Open-2018.pdf
Hi for Q13 shouldn't sinB=sqrt(39)/8 so m+n=47?
I’m really dumb so I’m just gonna say
I got 4 for question 1, how you got 3?
So first it’s x=0,y=10.999999999
Then there’s x=3,y=7
So x difference is 3 y difference is 3.999999...
So since it’s right angled triangle using Pythagoras the area should be slightly under 5 which means floor will become 4
How you get 3?
U just sketch a quick x-y plan and mark out the areas where floor x = 0,1,2 (because 0 <= x <=3) then mark out the corresponding areas where floor y = 10, 9, 8. You'll get 3 1*1 squares so R = 1+1+1 = 3
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