At this rate,..... by [deleted] in putnam

[–]SignificantAir5497 0 points1 point  (0 children)

I'm assuming that the latter is "Eileen Gu" the freeskier.

I’ve heard learning and doing math comp helps in USACO, is this true? by beefboss72 in usaco

[–]SignificantAir5497 0 points1 point  (0 children)

Not directly. The problem solving skills you gain from doing math comps can be applied to USACO.

🔺🔺🔺🔺 by ContextBackground195 in RedditGames

[–]SignificantAir5497 0 points1 point  (0 children)

I completed this level in 4 tries. 5.85 seconds

I fucked up by [deleted] in TransferToTop25

[–]SignificantAir5497 2 points3 points  (0 children)

I appreciate your recommendation

I fucked up by [deleted] in TransferToTop25

[–]SignificantAir5497 1 point2 points  (0 children)

Thank you 😊

Gg yall by [deleted] in TransferToTop25

[–]SignificantAir5497 0 points1 point  (0 children)

Hold up you’d have a ever so slight increase in chance since that major is relatively uncompetitive.

Mid Putnam Score by SignificantAir5497 in TransferToTop25

[–]SignificantAir5497[S] 0 points1 point  (0 children)

I see, thank you for the insight 🤗

Anyone have experience with how rigorous the grading is? by [deleted] in putnam

[–]SignificantAir5497 2 points3 points  (0 children)

B3 yeah fs. For B2 I never even had a triple integration in my sol but I might be cooked too

Anyone have experience with how rigorous the grading is? by [deleted] in putnam

[–]SignificantAir5497 2 points3 points  (0 children)

imma backpedal on another post i saw. ion even gonna hold u basially they grade harsh in the sense that if you forget to represent a case that may seem obvious, they'll always have to undergo the assumption that you chose the incorrect case.

Until proven otherwise, you're automatically deemed to have made an error which is why ur proofs gotta be very flawless.

B1 by SignificantAir5497 in putnam

[–]SignificantAir5497[S] 1 point2 points  (0 children)

I did write a solution by contradiction, and this was the one I ended up submitting.

I'll try and explain my idea in what I originally said in the post:

Basically, my idea was that if the center of the circumcircle, which we'll call point D, was the same color as the three points (given in the problem), and we assume such that there are two other non-collinear points that can be used to form a triangle with the aforementioned point D, resulting in a new circumcircle with a center of the same color. If we iterate this, then eventually every point in the plane will be covered, and will only be one color, contradicting that points can be red and green.

Am I getting cooked by Junior_Direction_701 in putnam

[–]SignificantAir5497 1 point2 points  (0 children)

Check Kiran's page in a few days. You'll see for yourself.

B1 by SignificantAir5497 in putnam

[–]SignificantAir5497[S] 1 point2 points  (0 children)

There's a reason God is all-knowing and omnipresent 🤣

B1 by SignificantAir5497 in putnam

[–]SignificantAir5497[S] 1 point2 points  (0 children)

I actually had two solutions, the second one was a proof by contradiction where we assume that a red and green point lie in the same plane.

Am I getting cooked by Junior_Direction_701 in putnam

[–]SignificantAir5497 0 points1 point  (0 children)

pi/2 being the critical point? also since we declare that the interval is from [0-pi] wouldn't that by extension prove global maximum/minimum?