SMO 2022 (Open Section) Answers
1. 4043
2. 1303
3. 100
4. 9
5. 11
6. 50
7. 144
8. 208
9. 18
10. 40
11. 544
12.
17
13. 3000
14. 11
15. 44
16. 99
17. 8
18. 121
19. 2
20. 4
21. 30
22. 200
23. 29
24. 27648
25. 2023
Click here for the review session today (2 Jun) at 6.30 pm.
27 comments:
for q8 i got 207 when x=3 and y=3+root3 can i check if im wrong or theres an error in the answers thanks
What’s the predicted cut off point for second round
What’s the estimated cut off for second round? Thanks for the answers :)
Is this year considered harder easier or the same difficulty as last year
Sketchy solutions by HCI
1. Pair terms
2. Sum
3. Everything is 1
4. Discriminant
5. Pythagoras
6. psle
7. Cauchy
8. Geometry
9. Directly bound w from the inequality given
10. Equilateral triangle
11. Subtract
12. Complete square
13. x>=4x^2/1+4x^2
14. Keep subbing into t, golden ratio
15. Concyclic with I
16. S(n)<=n
17. SFFT, each pair gives one solution by mod 4
18. x1 between 10 and 20, x2 between x1 and 20-x1
19. Bash
20. Discriminant
21. x-y,3x+3y+1,4x+4y+1 perfect squares, pell's equation
22. 345,156,246,123 gives 200, if one person has 4 then at least 1+100+100
23. q=2/3
24. Weighted amgm
25. x+1/x, sub y=1 and y=f(y)
I think easier
18/25 is enough right?
Less sketchy solutions from non-nush student(s)
1. Pair opposite terms and use 1/(x+1) + 1/(1/x + 1) = 1.
2. Last term is 2^(n+1)-2, so the answer is 1+2+...+510
3. 338350=1^2+2^2+...+100^2<=(x1)^2+(2x2)^2+...+(100x100)^2, so equality must hold everywhere, meaning all the xi are 1
4. Write as 4y+(y-m)^2=25, which has 2 positive solutions in y. Use discriminant, a=5 b=29/4
5. Drop heights from P to all 4 sides, by British Flag Theorem / Pythagoras we have 14^2+5^2=10^2+PC^2
6. Note EC/BE=FC/DF=1/2, so EFC= 1/2 x 1/3 x 1/3 x 180=10
7. A,B,C lie on x+z/2=6, and we want to minimize x^2+z^2, by cauchy (x^2+z^2)(1+1/4)>=(x+z/2)^2
8. We want to find the largest distance from the origin, which is obviously found by taking the point on the line through (0,0) and (2,3), so sqrt(S)=sqrt(13)+2
9. We have (w-3)(w-1)(w+1)(w+3)<=0 so -3<=w<=-1 or 1<=w<=3
10. Let O be the intersection of AC and BD. Construct point E on AC such that EOD is an equilateral triangle, since ECD and OBA are congruent AO=DE=DO, so ODA=30
11. Subtracting, (b-1)(c-a)=1. If b=0 then abc=0, else b=2 gives a=16, c=17
12. Completing the square gives |2root(x+1)-1|+|root(x+1)-1|. Split cases based on the sign of the terms, -8/9 is the result
13. Obviously all are positive. Note that x>=4x^2/1+4x^2, which is equivalent to x(2x-1)^2>=0. Thus, x>=y>=z>=x, so all are equal, and equality holds everywhere, so everything is 1/2
14. By repeatedly substituting t=root(1+root(1+...+t), we get t=root(1+root(1+... so t=root(1+t), which means t^2-t=1. Now t^4-t^3-t+10=t^2-t+10=11
15. Since ICD=IED=IBD=90, ICDBE is concyclic, so BDC=68. But BDC=90-A/2 so A=44
16. Note s(n)<=n so n^2-5n-9486<=0, which means n<100. n=99 works
17. (4x-93)(4y-19)=3x19x31. 4x-93 is 3 mod 4 and 4y-19 is 1 mod 4. Each pair of factors gives 1 "1 mod 4" and 1 "3 mod 4" factor, so each pair of factors give one solution. Note negative numbers allowed
18. obviously 10<=x_1<=20. For each x_1, there 20-x_1<=x_2<=x_1, so there are 1+3+5+...+21=121 solutions
19. BCE=135, BC=root(3)+1, BE=root(6)+root(@). By cosine rule CE=root(2), and by CH/HD=CE/BD we get DH=2.
20. Let r be the radius of the circle below, then x^2+(y-r)^2=r^2 and y=x^2/2 only has one solution, i.e. discriminant 0. This gives r=1. If t is the center of the circle above, x^2+(y-t)^2=(t-2)^2 and y=x^2/2 only has one solution, i.e. discriminant 0, so t=5, giving radius=3
21. Note (x-y)(3x+3y+1)=y^2 and (x-y)(4x+4y+1)=x^2. Since 3x+3y+1, 4x+4y+1 are relatively prime, they are both perfect squares, and so is x-y. Let x+y=t, then 3t+1, 4t+1 are both perfect squares, so they satisfy pell's eqn a^2=3b^2+1, which has root (2-root(3))(2+root(3)). Raising to power of 3 we find (a,b)=(26,15) as the smallest with a even, and this gives x+y=56. This gives x=30 y=26
22. Putting 50 ppl in (123) (156) (246) (345) we get 200. If all have <=3 this gives 600/3=200. If one has 4, say (1234), then we need 100 disjoint ppl in 5 and 6, so there are at least 1+100+100=201 ppl
23. Note q=2 has no solution and q=3 gives p=29
24. (12(a+b+c)/9)^9=((6a+6a+4b+4b+4b+3c+3c+3c+3c)/9)^9>=(6^2)(4^3)(3^4)a^2b^3c^4
25. Let P(x,y) be the assertion that f(y)=f(yf(x))+1/xyf(x)
P(x,1): f(f(x))=f(1)-1/xf(x)
P(x,f(y)): f(f(y))=f(f(y)f(x))+1/xf(y)f(x), so f(1)-1/yf(y)=f(f(y)f(x))+1/xf(y)f(x)
Swapping variables, 1/yf(y)+1/xf(y)f(x)=1/xf(x)+1/yf(y)f(x)
multiplying by xyf(x)f(y), xf(x)+y=yf(y)+x, which means xf(x)-x is a constant.
Let xf(x)-x=c, then f(x)=(c+x)/x=1+c/x
Substituting back into the original equation gives c=1, so the only solution is 1+1/x
can 13/25 get me to the second round
Do you cutoff is going to rise?
Can 11 get bronze? Thanks
What will 5 get
Estimation
HM: 6-8
Bronze: 9-11
Silver: >= 12
Rd 2: 13/14, probably.
Will silver remain at 10 or rise to 11 like in other years?
silver was at 11 in 2019-2020. not sure abt 2021, would be nice to get insights on 2021 cutoffs.
last year got 6/7 and got hm.
is 14 too high
2021 was 10 for silver. Not sure if this year is easier or harder tho wdyt
imo was easier than 2021, but comparable to 2020
is 24 enough for silver? I messed up on one qn and I'm worried
I think it’s very unlikely. Try again next year!
easier than last year
Round 2 selection is out?
yes, r2 list is out.
https://sms.math.nus.edu.sg/SMO/Open_R2_invitation.pdf
what is the cutoff?
I got 12 if I filled in my answers correctly and didn't get into round 2. Very sad now😭😭😭
same
can i ask if 14 makes the cutoff?
i get in with 16
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