Art of Problem Solving: 2012 AIME I #14

Art of Problem Solving's Richard Rusczyk solves 2012 AIME I #14.

[OO-fgcr]

WOW, I just realized how amazing the advice "if you see a wacky condition, relax it" is.

[artix]

Love this dude, he makes mathematics so much fun

[ArtofProblemSolving]

Here, it's really the algebra that inspires the parallelogram once we have a+b+c=0 condition. That equation gives us a+b=-c, which means we have a very nice substitution for a+b, and a+b has a nice geometric representation as the diagonal of a parallelogram. And when you see that the right triangle consists of the other diagonals of the parallelograms, well, then you just know you're on the right track.

[ArtofProblemSolving]

We may experiment with that sometime, but I'm afraid it would consist of long stretches of time of me staring at the ceiling or mumbling to myself. It would need a lot of editing to be tolerable!

[timothygreen]

I don't know how you make such problems and methods looking so easy. However, because of your videos I'm almost done completing the intermediate algebra course in blue and doing amc 10/12 this year :). Thx for your help over the past year.

[csodhi]

3:10 24 isn't 250, even for very large values of 24.
I laughed my butt off after hearing this lol

[wontpower]

So for a general triangle (where the angle between p and q is x != 90), h^2=p^2+q^2-2pqcosx. So you if you want to substitute this in for p^2+q^2, you would get The extra 2pqcosx makes it impossible to get a constant ratio between h^2 and the sum of the squares of the magnitudes of a, b, and c. The only way 2pqcosx disappears is if cosx=0, which only happens if x=90 for x<180. Thus, the triangle in question has to be a right triangle in order to get that nice ratio.

[ArtofProblemSolving]

Any parallelogram can be modeled in complex plane as we did there. So, what we proved holds for any parallelogram. This was an example of how we can use complex numbers to prove general Euclidean geometry results.

[captainsnake]

So I know "the other guy" is supposed to be the viewer, but the other guy is also that guy on AOPS who always proceeds his solutions with "it's really trivial, I don't get why everyone's solutions are so complicated" and then comes up with the most black magic unintuitive solution you've ever seen.

[JoelFishel]

My bad, for some reason my video just cut off near the end whilst you were in the middle of the solution but i've refreshed and it's all fine. Thanks for the instructive video :)

[edgarwang]

My favourite solution is to translate the triangle until a is real, then b and c must be complex conjugates. We can easily solve the messy equation with the squares of magnitudes from there.

[takinrho]

Quick question. When working on geometry problems in general, don't complex numbers provide the same utility as vectors? Aren't they the same? BTW, great video. I'm glad I stayed till the end.

[msuper]

how do you know what additional lines to draw in geometry problems? like, what inspired you to go for these parallelograms? i always see solutions dropping seemingly random altitudes and stuff and i wonder how i would ever have thought of that...

[outofthedumps]

Who knew Sam Rockwell was so good at math.

[M_Chen]

4:14 Did he say "That's a great"?

[ArtofProblemSolving]

Not sure what you mean -- there's not a part 2 to this video.

[arghyadeepchatterjee]

I used vectors. Just used a+b+c = 0 and squared it. and used dot product for the perp sides of the triangle . The result came was
2a.c= -125-b^2 . So (c-a) ^2 is just a^2+b^2+c^2 + 125 = 375.

[WilliamLiu]

Does the parallelogram lemma at 15:58 apply to all parallelograms, or just ones in the complex plane?

[raymondwei]

Could have used the fact that the origin is the centroid of the right triangle, the median through b is 3/2 of |b| and it's half of the |h|

[זאבגלברד]

It is indeed interesting, your wish to understand the "behind" of the problem.