# Solution 72: Hexagon Inside a Nine-Point Circle

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• Published on Dec 27, 2018
• Let's use a property of the Nine-Point Circle to compare the areas of an inscribed hexagon and the triangle.
Congratulations to attyfarbuckle, Luis Perez, Eric Schneider, Farzad Saeidi, Devansh Sehta, and Henry M. for successfully solving this math challenge question! attyfarbuckle was the first person to solve the question.
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ME: I bet, that the hexagon's area will be half the triangle's area!
ME: *watching video*
ME:
ME: Indeed
I watched too many mathematic riddles not to expect a ridiculously simple ratio ;)

• Rahul 5 months ago

On the real line place an object at 1.After every flip of a fair coin ,move the object to the right by 1unit if the outcome is head and to the left by 1 unit if the outcome is tail. Let N b a fixed positive integer .Game ends when the object reaches either 0 or N . Let P(N) denote the probability of the object reaching N.Find the formula for P(N) for any positive integer N.

• Rahul 5 months ago

In the complex plane ,let u,v be two distinct solutions of z^2019 -1=0. Find the probability that |u+v|>_ 1 .

• Rahul 5 months ago

Find all polynomials p(x) such that (p(x))^2 = 1+ xp(1+x) for all real numbers x

• Lewis Tran 5 months ago

I'm loving these videos, even as a non-mathematics major. What hardware and software do you use for your videos?

• Omar Fahmy 5 months ago

Can anyone help me in this problem : "find with proof all prime p and q for which p
^3 + 19q^3 + 2018 is the cube of a prime" ?

• Radhika Shenoy 5 months ago

Wow you are intelligent in solving difficult maths problem.

• Himansh Negi 5 months ago +1

I have a tip for you,
Just start solving IIT-Jee integration problems and get flooded with views and subscribers.
Thank me later. You deserve more subs.❤️

• adandap 5 months ago +9

Having never laid ears on the nine-pint circle before, this problem was hard work. I ended up just brute forcing it with coordinate geometry. Thankfully Mathematica looked after the tedious algebra.

• Felipe Lorenzzon 5 months ago

I didn't post a solution because I was late. I did it by connecting the points shown on the circle to its center and finding all angles by using parallel lines and triangle similarities. Then, trigonometry gave me the length of every segment and I found the area. A bunch of work

• Sudheer Thunga 5 months ago

We could also use the properties of the altitude being dropped on the side(a=bcosC+ccosB) to get it directly.......

• UbuntuLinux 5 months ago +2

Extra: CIB is a 30-60-90 triangle, so BC = 2BI. D is mid of BC so BD = 1/2BC = BI. Triangle BID has a 60° angle and two equal/congurent(?) side, so it's equiterial

• Arbitrary Renaissance 5 months ago

Holy cow. Yeah, I wasn't going to get this one. xD

• Gergő Dénes 5 months ago

Our teacher talked about the Euler-line and the Nine-point Circle (In our language it's mentioned as Feuerbach-circle), but she only showed us the proof for the Euler-line :/
Overall great video, and shows how much impact knowing a small fact can have.

• an–bn 5 months ago

❤️

• Johannes H 5 months ago

Good thing I didn't try that one
Or maybe not, but I don't know many geometric tricks

• James Wilson 5 months ago +5

Very cool! That is yet another interesting thing you can do to a triangle! I may have never known that had I not watched this video.

• Alexander Mendez 5 months ago +1

I would have used analytic geometry, but your solution is easier.

• Andrés Robles 5 months ago

Donde puedo ver esos challenge?

• Anto15 5 months ago

usclip.net/video/YN9Keh2HSSg/video.html