The Easiest Problem on the Hardest Test

This Putnam Series was given on the Putnam exam in 1997 (Problem A1). Let's figure out a way to deal with this Putnam problem!

For those that don't know, the Putnam math competition features some super interesting and challenging problems!

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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.

Putnam 1997 A1
Putnam Exam 1997 A1

#math #brithemathguy #putnam

[BriTheMathGuy]

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[karunk]

Now we just need “the easiest question on the easiest test” and “the hardest question on the easiest test” and we will complete the dynasty!!

[Juuueeel]

Like the new animations!

There's a small mistake starting 2:24. The right side of the upper right equation should be 5^2 + (11 + x)^2, not 5^2 + (11 + x^2). However the calculations are done correctly, so it's just a display error.

[BriTheMathGuy]

Tried a bunch of new animations (and sounds) in this one. Feedback?

Thanks so much for viewing! If you liked it, great! If you didn't like it leave a dislike and tell me why! :)

[razzka]

You actually explained this so well. Right now I’m an undergrad student doing math, but I’ve never really taken much geometry. So the fact that I could understand this should make you proud

[RisetotheEquation]

1. I was going to do an A1 on area between curves using literally the same title. You beat me to it, so back to the drawing board.
2. The animations, sounds, conceptual layout, and geometric renderings were spot on. I know how time consuming these videos can be, - especially geometry - so your work is greatly appreciated!

[pocarski]

I actually used an entirely different solution, based on the Euler line. First, let's construct line AM. M is the midpoint of BC, so AM is a median and therefore passes through the centroid. For any triangle the orthocenter, centroid and circumcenter lie on the same line, therefore the centroid will be the point of intersection of AM and HO. We will call this point X. The Euler line's property that 2 XO = XH means that we can determine XH to be 22/3, since HO is 11. We can now use congruence of AMF and AXH to find that AH = 10. This allows to find AO using Pythagoras through AH and HO, which turns out to be sqrt(221). AO is equal to BO because radius, therefore we can once again find BM using Pythagoras through BO and OM. BM = sqrt(221-25) = 14, therefore BC = 28.

[manucitomx]

I quite like the animations, the sounds are a little distracting.
As per usual, great explanations. Loving this channel.

[Woodchuckler]

why i felt the need to watch this in year 9 i have no idea

[deepjyoti]

As we know o and H we know it's euler line that means we can able to find vertexA also we know radii of Circumcircle as radi of nine point circle is half so we can find all three vertex by contructing bigger circle(circumcircle),

[vatrqx]

I need to solve this on my own 😎 thank you for the motivation 💙

[sicapanjesis]

Great video...I always hate geometry due to all of its constructions, but how you did it, now I am starting to realize how to approach a problem, like spotting the right triangles etc...Keep on

[ThAlEdison]

Do exploring this problem, I came across the 9-point circle. The center of the 9-point circle is N, which is midway along HO. And it has some interesting properties.
The points F and M are both on the 9-point circle, as is a point halfway along the line segment AH. Let's call that point P, so because of how circles work FH is the same length as HP, and because P is halfway along AH, AH is twice the length of FH. Further because of intersecting chords we can say AH^2=(r-11)(r+11) where r is the radius of the circumcircle, and 11 is the length OH. r=sqrt(221).
BM^2=r^2-OM^2=221-25=196.
BM=CM and BM+CM=BC=28

[Aiden-xnwo]

Here I provide a faster solution to solve this problem (provided that you know the necessary results):
Since HFMO is a rectangle, the Euler line OH is parallel to BC. The centroid G lies on OH and AM, so AH/HF=AG/GM=2. This implies AH=10.
Applying Pythag twice, AH^2+OH^2=AO^2=BO^2=BM^2+MO^2, which gives BM=14 so BC=28.

[citruslime]

Wouldn't you also be able to solve this using calculus? Since O is the radius, and you just figure out the rate of change of the diameter as O moves down 5 units?

[nabla_mat]

Thinking in an “Euclidean” way, I would rather demonstrate that ΔCHF ~ ΔABF, which means CF/FH = AF/FB, namely
(x+22)/5 = (y+5)/x

[nikolatopalov]

There's a theorem that says that AH=2OM. We can easily calculate the radius of the circle via Pythagorean theorem for triangle HOA and calculate BM thereafter.

[mesory]

Whenever im solving geometrical problems, i never come across the idea to use slopes to find an unknown value, so this is new to me

[RealCrate]

I see you changed geometry problem to just problem to clickbait the geometry haters xD

[jorgecasanova]

Very nice video. I personally would had try to avoid the analytic techniques with slopes using the clasical triangle similarity (ABF and HCF), getting the same equation.