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I accidentally stumbled upon this, I left my RUclips on autoplay and woke up in the middle of the night to hear some guy solving math problems and I was sooo confused 😭 epic video tho

(24:45 for possibly the worst silly of my competition career) If you want more high school competition prep resources, please subscribe and turn on notifications!

What do you think the dhr score will be?

starts 3:40

I actually realized aime problems are not that hard

thanks a lot for bringing up this series,it is really helpful for my amc 10 test

pretty cool problem. What question on the amc 10 do you think this would be at?

Thanks for the help

How did you draw the, I don't know, figure? Do you call it that? Like, the point is I don't know how to draw it, and it seems pretty much easier once you draw the diagram.

Good problem

no explanation?

Very popular series I see...

Another way to find the area of the triangle AMC is to find the area of UMC (12 * 8 / 2) and then double it which is 12 * 8 .

isn’t it strictly just less than instead of less than or equal to

So this is what bashing looks like.

Ironically, during the 10B, I also got 144 due to messing up #19 for a different reason. I multiplied by 1/2 instead of 1/4 at the ends because I assumed you could choose from North or South.

I am not sure whether 560 should be included or not. Reversing it makes 065. Can it be included?

any idea why F is on the circle? I have no clue

quick path , very nice!

This solution is a bit too convoluted. There are easier solutions which require much less hard calculations.

I think that if you are well trained in complex numbers, finding a root of every answer is simple. cis40 is a root in answer E.

You could also use the fact that any polynomial P(x)=Q(x)+r(x)/P(x), where P(x) is the polynomial, Q(x) is the quotient, and r(x) is the remainder. You get same answer.

ᵖʳᵒᵐᵒˢᵐ 🤣

By Vieta, the sum of the pairwise product of the roots (z1z2+z1z3+... +z3z4) is 3, so we take conjugates of both sides. I hope this helps.

confusing when you convert the conjugate of z1z2+z1z3+....... to 3.........

Very well explained! It makes perfect sense to take a side cross-section, but I would have never thought of taking a top cross-section. I will definitely consider that when doing other 3D geo problems!