How did everyone find it? I didn't get q5 or 6, and was wondering if people had any insights?

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Original post by Wormscarf67

How did everyone find it? I didn't get q5 or 6, and was wondering if people had any insights?

(i don't think we're allowed to discuss olympiads yet cuz china still have to do it tomorrow)

Original post by Wormscarf67

How did everyone find it? I didn't get q5 or 6, and was wondering if people had any insights?

I struggled with the first few questions. I’d appreciate some help understanding how to do them?

Original post by Cridplbn

Is the exam over now because I would like to see which ones I got right.

Id guess so. They generally release paper/solutions the following day which would be Monday in this case, but the exam is over so ...

I don't want to discuss it in detail yet until the paper and sols are out, but my solution for Q5 was not exactly inaccessible but it was pretty obscure (involving an interesting property of squares) - I assume there are more obvious ways of solving part (b), but I couldn't think of any alternative solutions. Q6 was easier imo but still a little tricky

edit: pretty sure its over so for q5 you can use the fact that b^2 and the two discriminants form an arithmetic sequence and for q6 a construction extending a line does the job

edit: pretty sure its over so for q5 you can use the fact that b^2 and the two discriminants form an arithmetic sequence and for q6 a construction extending a line does the job

(edited 1 year ago)

Original post by Jam.123

What about for Q1? Me and the other person who did it from my school are both pretty sure it was impossible because the m% of water and losing (m-5)% of this to reach 50% water just doesn't work in our opinion

try m=80

4 parts water to 1 part fruit becomes 1:1

(this comes from the quadratic m^2 - 205m + 10000 btw)

(edited 1 year ago)

Original post by Jam.123

So m=80%:

80% - 80%(75%) = 20%

I feel like I'm just being stupid though

80% - 80%(75%) = 20%

I feel like I'm just being stupid though

20% was the original weight of the non-water mass, so the new ratio is 1:1

Original post by _wxyz_

20% was the original weight of the non-water mass, so the new ratio is 1:1

Ohhhh, I feel so stupid now because of forgetting to account for mass being lost altogether.

Thanks though

how did you get that quadratic?Personally, I got m^2 - 105m 5000 which yielded no results so i was unable to solve the problem

Original post by cardtricks

how did you get that quadratic?Personally, I got m^2 - 105m 5000 which yielded no results so i was unable to solve the problem

sry m^2 - 105m + 5000

20m^2-41m+20=0 is the final quadratic you'd need

Original post by cardtricks

sry m^2 - 105m + 5000

Original post by Cridplbn

what was the explanation behind 6, I only had a couple of minutes to comprehend it and the only instinct I had was to use the abc conjecture. Any explanations

extend DE hit the circle again at P and then triangles CEP and BPC are similar, which gives you that EC/CP = 1/2 and then since triangles DBE and CPE are similar so DE/DB = 1/2

(edited 1 year ago)

Original post by Jam.123

20m^2-41m+20=0 is the final quadratic you'd need

because m-m(m-0.05)=1-m

Original post by Cridplbn

For question 2 we’re the answers 0 and 1

can you remind me what q2 was?

Original post by Cridplbn

Took a quick snap

yeah, pretty sure you're right (i assume you used parity proof)

(edited 1 year ago)

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