Geometry puzzle + Tour through my university

Рет қаралды 11,402

6 жыл бұрын

Help me create more free content! =)
/ mathable
Thought I might show you where I'm studying =) And also, don't forget to solve the puzzle :3
Twitter: / flammablemaths
Visit my website! =)
mathable.me/

Пікірлер: 143

@atrimandal4324 6 жыл бұрын

8:14 "Chemistry is bad Physics" - Papa Flammy😎

@garvett6660 4 жыл бұрын

Flammy: “Chemistry is bad Physics” Chemists: “Poison is bad for your health”

@yahav897 6 жыл бұрын

I'm glad I found this channel, keep up the great work and dedication! :D

@diegotejada55 6 жыл бұрын

1:43 that's wrong! g = 10 pi^2 = 3^2 = 9 Some "student"

@yashkrishnatery9082 3 жыл бұрын

But π^2=10

@diegotejada55 3 жыл бұрын

@@yashkrishnatery9082 Correct. But π^2 = e^2 = 4

@timothyaugustine7093 6 жыл бұрын

I like how clean it is there!

@bryandl6446 6 жыл бұрын

You've rustled your engineer bois, I got cube root 2 but that's like 1.3

@TheReptileDragons 6 жыл бұрын

Happy sqrt (g) day!

@2003kunju 5 жыл бұрын

got cube root of 2 by finding special (equilateral and right) triangles, using mass points to find another distance (sqrt(3)/(1+x), precisely) and solving polynomial obtained through pythagorean theorem on right triangle with legs 1 and sqrt(3)/(1+x) and hypotenuse x

@GermanSnipe14 6 жыл бұрын

1:32 CHRIST JENS THEY HAVE FAMILIES

@RetsamX 6 жыл бұрын

0:09 fast getroffen ! Nice video boi :D

@Absilicon 6 жыл бұрын

Holy shit boi! Dissing every shitty subject like a true mathematician. #respect

@yuval2be 6 жыл бұрын

Flammable Maths, will you make a video about how to solve it correctly? I'm sure there are couple solutions, and I'm very curious to see how you solve it!

@revooshnoj4078 6 жыл бұрын

I got cube root of 2 Upper left triangle is equilateral and lower left triangle is isosceles by solving for some angles we get that upper right triangle is a right triangle we then let the top right right angle =y we get that bottom right angle = 60-y then we use law of sines on the big triangle on the right knowing that cosy=1/x and siny=sqrt(x^2-1)/x we get sin(60-y)/1=sin120/x+1 which simplifies to the equation 0=x^4+2x^3-2x-4 which simplifies to 0=(x+2)(x^3-2) which yields cube root of two

@IlTrojo 6 жыл бұрын

At the very end I got 0 =x^4 *-* 2x^3 *+* 2x - 4 which simplifies to 0=(x-2)(x^3-2). This would leave to disallow x = 2, which is easily done by noting that would force y = pi/3 and those two lines would be parallel. But I might as well have been duped by some minus sign somewhere.

@Hello-fb7sp 6 жыл бұрын

I found the cube root of 2, did I do well papa flammy ?

@E1Luch 5 жыл бұрын

2^(1/3), it can be solved without any trig whatsoever with just similar triangles and pythagorean. Lets the side with length 1+x be named AD and the equilateral triangle be named ABC, such that C lies in between B and D. Draw a perpendicular AH to BC. AH = sqrt(3)/2, HC = 1/2 so AH^2 + (HC + CD)^2 = 1+x, so x(x+2) = CD^2 + CD. Then let E be point on AD such that DE = 1. Extend CE and draw a perpendicular DF = CD/2 to it. Then we can see that ACE and DEF are similar, DE/AE = 1/x = DF/AC = CD/2, so CD = 2/x, so x(x+2) = 2(x+2)/x^2, so sorry for me showing off

@PandaWaffle3 6 жыл бұрын

HAHAHAHAH.I actually had to stop the video to laugh, because of the engineering joke. I love the channel! Have you thought about doing videos on statistics?

@ChakraaAAAA 6 жыл бұрын

Oh ho Mr Avenger (trick shot)💂💂💣

@Soundillusions94xyz 6 жыл бұрын

Hey, I'm an engineering student!

@_neophyte 3 жыл бұрын

1:33

@rishabhdhiman9422 6 жыл бұрын

I think the answer is cube root of 2. Seems unnatural though. If that's the case then this configuration can not be created solely using compass and straight-edge.

@kenhaley4 6 жыл бұрын

cube root of 2 looks right! How did you get this?

@eduinst3342 6 жыл бұрын

Answer is "x" because x is still unknown

@gnikola2013 6 жыл бұрын

1:30 I freaking lol'd

@sarpkaplan4449 6 жыл бұрын

blitzkrieg at the start lul

@anhthiensaigon 5 жыл бұрын

Lul troo german uni campus looks like a damn military camp

@rishabhdhiman9422 6 жыл бұрын

Here's my proof. I'll refer to this image and the naming is according to that, drive.google.com/file/d/1zw6OyBlhgtbgDFXlNnji8wv9Z5QXe7Pu/view?usp=drivesdk. AE = x. Note that AB = BC = CA = BD = 1. Therefore, ACD lie on a circle with diameter AD and center B which implies that ∠ACD=90° which in turn implies that ∠ACE=90°. By Pythagora's, CE^2 = x^2 - 1 and CD^2 = 2^2-1^2=3. Since, B, C, F are collinear. By menelaus theorem, (DB*AF*CE)/(AB*EF*CD) = 1 => (x+1)√(x^2-1) = √3 It is simple enough to prove that 2^(1/3) is the only positive real solution of this equation.

@rishabhdhiman9422 6 жыл бұрын

Thank you!

@andresxj1 6 жыл бұрын

Solving for the final equation sqrt(3)=(x+1)sqrt(x² -1), I end up with (x³ -2)(x+2)=0, and there you have the real solutions x=-2 and x=cuberoot(3). But then I found out that there are two complex solutions, and we don't obtain them unless we factorize (x³ -2). Why is that? I thought it wouldn't be necessary factorization once I can state x³ -2=0. Can anybody explain it to me? Thanks in advance!

@rishabhdhiman9422 6 жыл бұрын

Andy Arteaga, if you are familiar with euler's identity, it would become easier. We can write x^3 = 2 as, x^3 = e^(2 pi i) 2, which gives the second solution of x = e^(2/3 pi i) * cbrt(2) = cbrt(2) ( cos(2/3 pi)+i sin(2/3 pi)) = cbrt(2)(-1/2+sqrt(3)/2) It can also be written as x^3 = e^(4 pi i) 2 => x = e^(4/3 pi i)*cbrt(2) = cbrt(2)(-1/2-sqrt(3)/2). If we write it as 2e^(6 pi i) we simply get cbrt(2)e^(2 pi i), it can be seen by the periodicity of cosine and sine or by that fact e^(2 pi i) = 1 or that a cubic equation has only three roots that these are the only three solutions of the equation x^3-2=0. I'd recommend you to read about roots of unity as it a beautiful topic bringing together polynomials, complex numbers, trigonometry and number theory all under one big roof with allcthe pieces falling together perfectly. Also, x = -2 is not a solution, put it in the original equation to see what happens (;

@andresxj1 6 жыл бұрын

Rishabh Dhiman Oh, I see sqrt(3)=/=-sqrt(3), but it appeared as a solution since I squared both sides of the equation and the minus vanished. Thank you for the explanation, and the suggestion!

@revooshnoj4078 6 жыл бұрын

You mean B, C and F are collinear. Anyways that was a nice application of menalaus theorem

@dalitas 6 жыл бұрын

Isnt a Hörsahl an aula IE audiotorium?

@knexator_ 6 жыл бұрын

Quality content

@noobkilla3 5 жыл бұрын

so u're saying you only like the physics and math buildings in your campus. sounds perfectly reasonable to me!

@bramlentjes 6 жыл бұрын

As a mathematician, we should make meme about 1:35!

@GrandMoffTarkinsTeaDispenser 6 жыл бұрын

DID YOU JUST TRASHTALK ENGINEERS SON

@gammaknife167 6 жыл бұрын

I've got 1.99 3sf. I used cosine rule twice. Is there an exact form?

@sansamman4619 6 жыл бұрын

Flammable Maths, Google 3sf it's rounding to 3 significant figures. But, I still don't understand what he's saying :(

@gammaknife167 6 жыл бұрын

What is there not to understand??? Anyway I made a mistake, probably have still made a mistake (:/) but I have a new answer of 1.043679947... . 3sf is shorhand for 3 significant figures, I'm surprised you didn't know the shorthand :O Does the actual answer involve a surd?

@rishabhdhiman9422 6 жыл бұрын

Rohan Skatedude its cube root of 2.

@gammaknife167 6 жыл бұрын

SPOILER ALERT and my construction doesn't agree with that result - are you super sure about it? Please don't explain the answer, I want to figure it out for myself.

@gammaknife167 6 жыл бұрын

You're bad and you should feel bad. Can you use ceva/menalaus? EDIT: It works! I have a quartic and cube root 2 works, but how would I solve the quartic??

@46pi26 6 жыл бұрын

6:58 pure sciences vs. applied sciences yet again

@46pi26 6 жыл бұрын

And again at 8:21

@hi_pd Жыл бұрын

Golm campus!!! Cheers 🍻

@ChakraaAAAA 6 жыл бұрын

I think blackpenredpen is also with you

@blackpenredpen 6 жыл бұрын

MATHEIO ICON INFINITY TO MATHS He is way too far! I can only watch his videos lol

@ChakraaAAAA 6 жыл бұрын

yes lol hahaha

@dalitas 6 жыл бұрын

Isn't the cantina called "mensa" at German universities?

@dalitas 6 жыл бұрын

Flammable Maths which feels weird as a Swede since in Swedish it would mean "to period" (the biological one not the maths one)

@XanderGouws 6 жыл бұрын

7:47 I spy with my little eye... something beginning with "Jens' girlfriend"

@lisafehlau1547 6 жыл бұрын

Hi :D

@Hello-fb7sp 6 жыл бұрын

I found x=sqrt(2) , did I do well papa flammy ?

@46pi26 6 жыл бұрын

Here's the real mathematical puzzle: given that the mysterious figure at 7:48 is in fact Jens' girlfriend, differentiate (with respect to pr0n) the integral of the Jens sexiness function with the Lisa sexiness function using the famed sexyboi integral to find the sexiness difference between Mr. and Ms. Flammable. Of course, the Jens sexiness function is limit as pr0n--> infinity of phi^pr0n^e (phi here is the golden ratio, not azimuthal angle), and the Lisa sexiness function is of course just the limit as pr0n--> infinity of e^pr0n^phi. Note that the sexyboi integral is of course the volume integral from minimum to maximum height, or from feet to head.

@Arthur-so2cd 3 жыл бұрын

What the fuck

@danielescotece7144 6 жыл бұрын

Why 9:59 stretch it by 2 more seconds and double the adds my boi! More cpm for you!

@danielescotece7144 6 жыл бұрын

Btw I am an engineering student ! I can relate with the trash lecture room! It's the same here in Milan!

@user-ux9rc3gj8v 6 жыл бұрын

is it Sqrt[ (1/Sqrt[3] - 2^(1/3)/Sqrt[3] + 2^(2/3)/Sqrt[3] )^2 + 1] ，approximately 1.25992？

@user-ux9rc3gj8v 6 жыл бұрын

i forget to simplify this ,Sqrt[ (1/Sqrt[3] - 2^(1/3)/Sqrt[3] + 2^(2/3)/Sqrt[3] )^2 + 1] =Sqrt[ ( 1 + 2^(2/3) + 2^(4/3) - 2*2^(1/3) + 2*2^(2/3) - 2*2 ) / 3 + 1 ]=Sqrt[ ( 3 * 2^(2/3) - 3 ) / 3 + 1] = 2^(1/3).

@user-ux9rc3gj8v 6 жыл бұрын

are you on vacation? why is there no one in your university?

@user-ux9rc3gj8v 6 жыл бұрын

@einsteingonzalez4336 4 жыл бұрын

#doublingthecube Congratulations! You just discovered the neusis construction (yes, it’s on Wikipedia) and a way to double the cube!

@KOmBos2011 6 жыл бұрын

"I hate reading books" LUL

@takaharrue 6 жыл бұрын

are you a physics student?

@tropixMw2 6 жыл бұрын

nerd

@tropixMw2 6 жыл бұрын

Do you even Pi groups?

@paulroler2124 6 жыл бұрын

Ist x die Wurzel aus 1+1/9 ?

@paulroler2124 6 жыл бұрын

OMG... Ich meine Wurzel 1 + 1/3 ?!!

@rockybond42 6 жыл бұрын

3:36 you forgot what Buero translates to... nice recovery, though!

@ShadowZZZ 6 жыл бұрын

ich liebe deine ignoranz den anderen studiengängen gegenüber. irgendwie müssen wir armseeligen uns ja motivieren xD

@bemolio2956 4 жыл бұрын

nice looking campus german boi, seems like it has very clean bathrooms cheers from my letrina

@williamfeasey1486 6 жыл бұрын

64-36*sqrt(3) = 1.65 (3sf)

@amineelfardi4311 6 жыл бұрын

😂😂Dr. Sheldon Cooper is proud of you In some way I got X=2

@unknown360ful 6 жыл бұрын

#NotificationSquadBois Where are you?

@im_Smitty 6 жыл бұрын

Fukn died at 1:35

@Jodabomb24 6 жыл бұрын

> really huge > 200-300 people Here at McGill we have one lecture hall which can hold upwards of 700 people. O.o

@Vedvart1 6 жыл бұрын

I got x = 2/sqrt(3). For reference, I'll call the triangles A thru D, left to right. Triangle AB is simply A U B. By using law of cosines twice on triangles A and AB, I worked out that the leftmost angle is 30 degrees (and the unknown side of A is sqrt(3)). With this information, you can find that triangle C is actually a right triangle with hypotenuse x. Using a variation of Pythagorean Theorem (derived-ish in a 3Blue1Brown video linked below at 7:18), with a=2, b=x, h=1, you can solve to get x = 2/sqrt(3). 3Blue1Brown video: kzfaq.info/get/bejne/mpOfZshzntnKnKc.htmlm18s

@lweyhacker5557 6 жыл бұрын

Vedvart To use the identity presented in the 3blue1brown video , the angle between sides a and b has to be 90° degrees , and this is not actually the case in this problem

@user-vg2qf3iw5v 6 жыл бұрын

x^3=2?

@user-vg2qf3iw5v 6 жыл бұрын

Flammable Maths I think I am the third

@FernandoVinny 6 жыл бұрын

Do you know about IMPA? Instituto de Matemática Pura e Aplicada, in Brazil. A center of reference, they make extremely high quality math. Search 'IMPA' on KZfaq, there's a lot of free courses in english

@quantumchill5237 6 жыл бұрын

The pedagogical building looks like a correctional facility. But the physics building 💦💦💦😍

@hgaclark9031 6 жыл бұрын

(2root3)/2?

@hgaclark9031 6 жыл бұрын

I feel like it’s a circle theorem question but maybe not probably why I might be wrong