Can you solve this ALMOST IMPOSSIBLE third grade geometry puzzle?!
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- Опубліковано 19 гру 2024
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Hey, my name is Presh Talwhatsoever and I hope you are going to enjoy this clickbait titled video! A huge shoutout to all the people who managed to solve it! =)
Twitter: / flammablemaths
Visit my website! =)
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So this third grader must have had calculus in his 4th grade, abstract algebra in 5th grade, topology in 6th grade and differential geometry in 7th grade. He would definitely have defended his thesis in 10th grade.
Then he would proove Riemann Hypothesis at age 25, then reject all professorship offers from hundreds of universities and tell that he is dissapointed in maths. He would then spend the reat of his life living reclusively with his mother.
Pretty sure he proved E8 was the meaning of life by the time he was in 11th
@@ty6339 hmmm sounds like Grigori Perelman
? All u need is law of sines
@@yashuppot3214 Yeah you're right. Law of sines is 3rd grade math
/s
when I was in third grade I used trigonometry to solve problems too
law of sins=3rd grade math
With Menelaus' theorem you can completely avoid all the sins, and just solve x^4+2x^3-2x-4=0, which the result is kind of obvious if you know how to factor the polynomial. So actually a 3rd grade should be able to do it with 1 simple geometry theorem and basic algebra.
illusiwind what 3rd graded knows how to solve a 4th degree polynomial?
@@ethanJ496 Wow you must be a real genius if you solve 4th degree polynomial in 3rd grade :p
@@That_One_Guy...
Well, maybe I am, if being the top 0.01% is within the range of "genius".
I don't see myself fit in, though. I've been collecting the real genius' website, e.g Stephen Wolfram, Terence Tao, John Baez etc, which are actually enlightening.
@@ethanJ496
The cringe is too strooooooo-
“180 degrees, so Pi/2.” -Papa Flammy
Yes
Sometimes his genius is almost... frightening
@@anitagofradump5195 It seems you didn't get it.
Bro just divide by zero and it's legit.
/s
He's clearly a Tao person I guess
I love how he started by thoroughly explaining how there are 180 degrees in a triangle then zipped off to do some fast trigonometry.
“Did you figure it out?”
Hi, my name is *PRESSURE LOCKER*
I thought of 1
There is no way in hell this would stump Terrance Tao. He'd sneeze and his snot would some how produce the answer.
To be fair, Terrance Tao is an incredible mathematician (probably even an understatement), in analysis and NT, as far as I’m aware, though his greatest weakness as he said was the geometric side of maths (he said in his interview with numberphile that it was something similar). So saying that he might struggle with it is similar to saying that a professional sprinter would struggle with tennis.
Top notch clickbait from a top notch shitposter. I can't complain.
My name is fresh toewalker :)
LOL
omg im dying
My name is pressure talker
presh facking talwalkar
His actual name is *pressure cooker*
LOL HIS NAME IS PARESH TALWAKAR!
fun fact: this is exactly the construction for making a length of 2^(1/3) unit long (aka for doubling the cube), by using a compass, a straightedge, and a marked ruler
Shorter path without using trig:
1. identify all angles as in video
2. identify right triangle as in video, mark side length as sqrt(x^2-1)
3. the 30° and 60° next to the right angle form another right angle, in particular they form a 30-60-90 triangle with sides of length 2 and 1 given, hence we mark the long side as sqrt(3)
4. apply Menelaus's theorem to find that
(1+x)/1 * sqrt(x^2-1)/sqrt(3) * 1/1 = 1
which simplifies to
(1+x)sqrt(x^2-1)=sqrt(3)
which may be solved as in the video.
My geometric solution in 5min: first spend 15secs figuring out the right angle and 30deg angle and then lengthen the vertical middle line downwards for b units and draw a line from the bottom right point perpendicular to the lengthened middle line. Then through similar triangles we have 1/(1+b)=x/(1+x) => b=1/x. And in the bigger right triangle we created we have (1+b)^2+(sqrt(3)b)^2=(1+x)^2 => 4b^2 +2b=x^2+2x. Finally because b=1/x we can get the exact same equation from the video and factor out x+2 we get x=2^(1/3)
“Presh Talwhatsoever” as soon as I read that I started dying haha
Liked because you roast MindYourDecisions.
Alexander Sanchez ooooh this clickbaiter i hate him
How can i block this man
My issue with those kinds of titles is that they draw people to the problem for the wrong reason: don't do a math problem to be better than somebody else--do it because the problem is beautiful or interesting.
CarmeleH There's an extension called VideoBlocker. I use it on multiple channels, like BuzzFeed, WatchMojo, Danger Dolan, etc. It works like a charm.
Yoav Carmel so what, it is still a good chanel
6/10
Good maths, but not enough timers and animations in powerpoint.
I found an answer before watching this video and I was able to do it without any trigonometry at all. I did use similar triangles and the Pythagorean Theorem to get that same polynomial x^4 + 2x^3 - 2x - 4 = 0 which I had to factor. So, this could be done with 9th grade math.
This will be a lot easier to describe if I assign some labels to points on the graph. Call the point at the very top of the diagram "a". Call the point at the center "b", the point at the upper right "c" and the point at the far lower right "d". Once you realize that x is the hypotenuse of a right triangle abc, you can construct two more right triangles. To get the first one, extend ab straight down and locate on that line the point "e" which is horizontal to point d, so when we connect e to d we see that aed is a right triangle. To get the second right triangle, extend line segment bc to the right until you reach the point "f" which is vertical to d, so dfc is a right triangle. Notice that angle bac and angle fdc are alternate interior angles, an therefore are congruent, so dfc and abc are similar triangles. For abc, observe the ratio 1 over x. The corresponding parts of triangle dfc are an unknown length of line segment df and 1. Therefore, the length of df is 1/x. Notice that bedf is a rectangle, so 1/x is also the length of be. Notice that triangle bed is a 30-60-90 triangle, therefore the length of ed is sqrt(3) times 1/x. Now apply the Pythagorean Theorem to triangle aed. The length of ae is 1+1/x. The length of ad is x+1. The length of ed is is sqrt(3)/x. By PT, (x+1)^2 = 3/x^2 + (1+1/x)^2. Expand to get x^2 + 2x + 1 = 3/x^2 + 1 + 2/x + 1/x^2. Collect like terms and subtract 1 from both sides. Then multiply both sides by x^2 to get x^4 + 2x^3 = 4 + 2x, hence x^4 + 2x^3 - 2x - 4 = 0. Then factor the polynomial as (x^3 - 2)(x + 2) = 0 and the only positive real solution is x = cuberoot(2).
When I first saw my answer, I said to myself, "That doesn't look right." I wasn't expecting a cube root to show up in a diagram full of right triangles with square roots all over the place. But I couldn't find a mistake in my reasoning, so I gave up and watched the video, only to discover that by golly I had the right answer after all. Huh. Well, that explains why I couldn't find my mistake; there wasn't one. It just surprised me.
Idk man. I looked at this and it was pretty obvious that it would be 2^(1/3)
You need to open your minds eye and see Plato's paradise.
I don't understand ... What does this have to do with Plato's Republic?
top notch clickbait skills
I totally agree!!
blackpenredpen hi🙂
Too little red arrows
9 out of 16 simillar videos are by mindyourdecisions. Boi you really confuzed youtube with that title
BTW how can i block this man from showing up in my recommended videos?
Flammable Maths lol thanks anyway
Using the right angled triangle to get the missing side √(x² - 1) and then the sine rule to get two definitions for sin(angle on far right) solve for x. You get -2, two complex and ³√2 as solutions. Since x ISA length the last one is the solution to this problem. 😀
Every time you say law of sines, I hear "law of science" xD
I feel stupid now because I actually thought he said science
@@erikkonstas Fundamental theorem of engineering is probably another way you can solve a thrid grade problem.
@@MusicalInquisit if only I had known the F. T. O. E. in elementary school, I would have been able to solve e:th degree polynomials already in πth grade.
Oh yeah, I remember how when I was in the third grade, we did this stuff every day ;-)
5:21 that θ looks hella weird.
Yep
I think he called it vheta(v theta), probably his own made up symbols
@@That_One_Guy... Theta
@@ryanjagpal9457 vheta
@@That_One_Guy... That’s how he pronounced is it right?
I found a very nice solution using mass points technique. Law of cosines gives the third side of bottom left triangle to be root 3. Set top point and bottom left point to each be mass 1, because the ratio of the segments is 1. This makes the mass of the bottom right point to be x. Middle point on the right becomes mass x+1, so we can use the mass ratios again to find that that missing side of the right triangle is sqrt(3)/(x+1). Pythagorean theorem finishes it off because we have all three sides in terms of x.
The obvious question is now: Does that mean that we can double the unit cube with compass and straightedge?
Nice video! I solved it by also using the Sine Law, but got a shorter solution, I guess. On a triangle we have
(x+1)/sin(120°) = 1/sin(v).
on a smaller triangle, we get
1/sin(30°) = a/sin(v).
Then,
a(x+1)/sin(120°) = 1/sin(30°).
Squaring both sides and taking into account that a^2=x^2-1, you arrive to
(x^2-1)(x+1)^2 = 3,
which can be rewritten as
(x^3-2)(x+2)=0.
Great solution! I think I also found an easier way to solve the problem by extending the side with length 1 (adjacent to the 90 degree angle drawn at 3:00) downward. By doing so, you create two similar right triangles allowing you to calculate their height, and use simple trig (no variables, just using 30-60-90 principles) to calculate their base all in terms of x. Then, you get (1 + 1/x)^2 + (sqr3 / x)^2 = (x+1)^2 which yields the same equation in the video and cube root of 2 as an answer.
Gotta love that description though, it's top tier. Awesome solution too, I got fairly close but ended up giving up after having gotten the answer to the fourth root of 5 like 4 times in a row. I just realized that I completely messed up expanding the sin function the right way hah. Great video :P
11:28 Shouldn't x = rad(3)/rad(x^2 - 1) - 1 turn into (x + 1)(rad[x^2 - 1]) = rad(3)?
Dan you're right 12:00
Using Euclidian geometry to find a cube root? This is amazing. I've never seen this done before. They all said it couldn't be done.
The imposible thing is to find a cube root using ruler and compass.
Can you solve this preschool level geometry puzzle that probably even stumped some random infant on the street?!
I find it extremely confident on your part to write equivalent that fast with no double take or second thought... if you are right it's great... I'd love to be as confident as you.. I carefully go one direction first, look left and right in the zebra crossing then come back whence I came....
Totally not ripping off MindYourDecisions titles xD
Because all third graders know the law of sines...
And quantum mechanics as well.
poor Terence Tao :
Presh Talwakar xD
This is way more challenging then problems from Press toewalker
I decided to do this problem without your solution and wasted 10 mins using 3rd grade tricks until I read the video description... Lol u got me
Kasîm I was simply trying to figure out all the angles like in the 3rd grade then i read the description about the clickbait and the sarcasm directed at another channel. The question seems very easy as you can figure out many angles easily then you have to apply sine formula which is not 3rd grade obviously
"Third grade" being third YEAR university
Love the humour here. “And that is actually the solution, we won’t care about complex solutions…”
Pre-watch: Clearly not drawn to scale, but after many dead-ends, I get x = ∛2.
Label the vertices, starting from lower-left, clockwise, going up to the apex, down the right side and ending at the "convex point," A[LL], B, C[peak], D, E[LR], F.
Then ∆BCF is equilateral, ∆ABF is isosceles, and from these two facts we quickly see that
∠ABF = 120º; ∠BAF = ∠BFA = ∠DFE = 30º; CF ⊥ AD
Call ∠FCD = α, so that ∠FDE = 90º+α; also call EF = y. Then in Rt ∆CFD, we have
x = 1/cosα
And by the Law of Sines in ∆DEF,
y/sin(90º+α) = 1/sin30º; i.e.,
y/cosα = 2; xy = 2
Now by the Law of Cosines in ∆CEF,
1 - 2y cos120º + y² = 1 + y + y² = (x+1)²
and multiplying across by x²,
x² + 2x + 4 = x²(x+1)² = x⁴ + 2x³ + x²
x⁴ + 2x³ - 2x - 4 = 0
(x + 2)(x³ - 2) = 0
and rejecting the one negative and two complex solutions, we conclude
x = ∛2
So, Flammy: How did you figure it out? Let's watch . . .
Pretty much the same!
Just goes to show that great minds think alike! ;-)
Fred
@@PapaFlammy69 Thanks!
Fred
Hello this is Preshtal Walker. Hold up...
presh talwarkar
@@adwaitT priston tarwanker
There's gotta be a cleaner solution than this, hm. Will have to take a look at this later but i'm really glad you do videos on this level too!
Let the equilateral triangle in the figure be ABC with AB extended to D such that AB=1 & BD=1. BC is extended to point E which is joined to A and DC meets EA in F. We are given EF=1. Now apply Menelaus's Theorem in triangle ADF with the line BCE as the transversal. Note that DC=√3 and CF=√(x²-1). We can then claim: (3/x²-1)^(1/2)*(1/(1+x))*(1/1)=1 which yields x=2^(1/3). That's all there's to it!
Hey flammable! Nice clickbait my homeomorphic friend!
A quicker way to get from 3:30 to 12:12 is: (x+1)/sin(90+30) = 1/sin(v), 1/sin(30) = a/sin(v), x^2 = 1 + a^2. Then solve the set without unpacking sin(v)
I was almost done solving it. I got to the same equation you did at the end, but then i thougth that since it is a 4th power equation that I wont be able to solve.I should have just sticked to it and done the factoring
How's this a 15 minute video? I just put my ruler against the screen and got 2 ⅓ immediately
I tried to solve this problem. I got blocked quickly, then saw in your video the law of sins, which i didn't know (i used Al-Kashi). Then i stopped watching and keep trying with new keys to solve this. After searching, i had to solve cos(pi/3 - sqrt(x²-1)/2)=1/x . At this point, i threw in the towel, saw an approximation of the solution (1.26) with my calculator. Finally i watched the video to the end and saw your beautiful solution. Wp !
Sweet problem. I'm glad I didn't decide to do it because it took more work than I expected.
x = 1.5, notice 1-1-1 equilateral triangle, reflect it leftwards, extend lines of length 1 and the another line on the far left such that they intersect, notice this second line is exactly half of 60 degrees that is 30 degrees, in order for a new line of length 1 to appear beyond length x there must be a line segment of length 1/2 (angle congruency) that is at least normal to the relatively horizontal line that touches the two equilateral triangles, so x = 1 + 1/2 = 1.5.
1:38 180° or pi/2 ???
Its so funny when people double comment
Its so funny when people double comment
Ranjan Bhat do you realize what you just did there?
Ranjan Bhat do you realize what you just did there?
Kappa ccino Yeah that was the joke. Hipocracy
You almost got me confused at the 11:30 haha. Such an interesting problem for pi grade students
:)
I can't fault you for the wonky proportions, since the cube root makes this non-constructible
I saw this question yesterday and solved it with an easier way that people here might want to know. Just an option. First apply Menelaus theorem starting from the side where x is.(if one doesnt know what this theorem is about, nothing to worry, when you use the similarity of triangles there, that will give us the same thing ) And use Pythagoras theorem on the right triangle where x is the hypotenuse. In that right angle, let the unknown side be of length y. That will give you the quartic equation faster and some factorization will do the job.
I solved the Fermat problem in 3rd grade. My whole family did it a year after after I taught them how to do it (sarcasm).
7:00 Sum of interior angle of triangle is 180,it looks like a triangle but isn't a triangle if you look closely at the angles they made, they dont have corresponding angles so its not a big triangle.
I'm having a harder time trying to figure out in what school they teach 3rd graders this
I got stuck at the point where I had cos (a)/sin (b) in my equation and didn't see the use of that inner triangle to make it one angle. Good job.
Flammable Maths be like:
Hi! My name is *stressed Paul Walker*
Just saw this, I think this is a lot easier using Menelaus' thm. You still need to find that 90 deg angle. your a= sqrt(x^2-1). Now for isosceles triangle you need to find the 3rd side length knowing the sides and angles = sqrt(3). using Menelaus' theorem :
1/(1+x) 1/1 sqrt(3)/sqrt(x^2-1) = 1
you get the 4th order polynomial to solve. When I saw it, that theorem came to mind right away. But needed to calculate a two lengths to be able to use it.
i remember getting this problem when I was in 3rd grade
Asian Spotted
I mean.... I remember _having learned_ this shut- but 23 years without having needed to do any of it, it's not exactly a fresh memory.
Using t instead of theta, this is a short way:1/sin(pi/6)=a/sin(t), then a=2sin(t): x^2=1+a^2=1+(2sin(t))^2, then 2sin(t)=sqrt(x^2-1). In the other triangle (x+1)/sin(pi/3)=1/sin(t), then x+1=sqrt(3)/2sin(t)=sqrt(3)/sqrt(x^2-1) and it is solved.
More geometryproblems please :)
1:37
"That angle right here is 180°, π/2"
This is precisely why we should use τ, no confusion.
3rd grade? I don't think so
Cubi Cardi youre right its second grade geometry
@@solidsinek14 Are you kidding? It's Preschool geometry at best.
Hey mate, loved the problem and your solution, I noticed a short cut though - at about 7min, instead of using the big triangle and the double angle formula, you can use the smaller triangle with sides of length 1 and x+1, and angles theta and 90+30 = 120 degrees, and another application of the sine rule, so x+1 = sin(120)/sin(theta), this allows you to get to the quartic for x a little faster. Cheers.
His accent is so strong that UA-cam's auto-generated subtitles think he's speaking German
Dude I actually thought this was a third grade problem and thought you were being over the top and then I read the comments. xD
:'D
I asked a 3rd grader for help and he shoved me in a locker while saying, "Only idiots don't know the Law of Sines." So I'm only here to see the cat at the end of the video.
Lord Flam has created an unsolvable 3.14159265th grade problem. Fresh can't cope with vat.
dX :)
is anyone else hearing minecraft sounds in this vid
you know you are on the right path when the coefficients cancel each other
1:34 How do you know thats a straight line? none of the rest of that diagram is draw to scale so why assume a straight line when theres a point half way along it?
A third grade geometry problem!?! I didn't learn this in third grade.
I have to admit that at first I thought you were joking. But it is really not easy! 3:35 made me so curious about he law of science!
* law of sines :)
@@samuelbarham8483 law of signs*
+ and + makes +, + and - makes -, etc.
im a common man , i got mindyourdecisions in thumbnail , i clicked it
After you figured out the 90deg angle you could've said a=sqrt(x^2-1) (and the line segment to the left sqrt(3)) and then used the Menelaus's theorem to instantly get an equation in x.
How did you get the cube root of two. Isn't it impossible to draw using only a straightedge and a compass.
Trigonometry should be taught in grade 3, more useful than learning how to multiply 4 and 3.
Menelaus gets you to that quartic equation much faster.
It's not 3rd grade papa, it is 11th grade math
At 11:39, isn't it supposed to be (x+1)(sqrt[x^2 - 1])?
oh nevermind im stupid
Oh God that brings back childhood memories...
Since the length x = cbrt(2) and we know that cbrt(2) is not a constructible number (see the problem of doubling the cube), this whole construction of triangles is not possible by using just a compass and straightedge.
Incredible how a single variable can be solved using cotagents and square roots and so on!!!
Flammable Maths Would you say this is as complicated as it gets when solving for a single length in relation to just triangles?
Hahahha I learned law of sines in precalculus. Good thing I took precal in the third grade.
I am watching from Turkey your videos are awesome thanks man!
thank you for the aneurysm of looking at this diagram
I solved this one using Menelaus' theorem... It fits almost perfectly... You still have to solve for the roots of that 4th degree polynomial, but it was easier
If we know the equilateral triangle isn't drawn accurately, why are we allowed to assume the straight line ("point a to b" @ 1:29) is in fact supposed to represent a straight line?
Tbh the key to this is probably Menelaus' theorem. In other words, a third grader may be able to solve it, provided they have read about Menelaus' theorem before. The rest of the problem (e.g. law of sine) may be circumvented. But someone who has never seen Menelaus' theorem may not be able to solve it even if they're much older and great at math.
I done it much the same way as *FM* , simply using three clear facts:
Right angle △(1,a,x) gives a=√(x²-1)
Small △(1,a, 30°, _∨_ ) sin-rule gives a= 2 sin( _∨_ )
Larger △(1,x+1, 60°, _∨_ ) sin-rule gives x+1=√3 /2sin( _∨_ )
then eliminate sin( _∨_ ) giving same quartic:
(x+1)²(x²-1)-3=0
no need for double-angle trig formula just the sin-rule!
_PS I thought about △ similarity and/or cyclic quadrilaterals but that did not seem to work here_ ☹
I'm hoping you can help me with a problem. Similar to the Leibniz rule, can you prove (or possibly disprove) that the divergence of the integral of a vector field over a volume is equivalent to the integral of the divergence of the vector field over the volume?
Have been hunting for where I saw this problem for a few weeks and then it shows up in my recommends again
Do a digital reconstruction on how it should look
Menelaos hilft aus dem Chaos: Satz von Menelaos+Cosinussatz liefert einen 3-Zeiler.
Fresh Tallwater moment.
11:36 why is it x-1 and not x+1 ?
How out of proportions do the angles have to be so that you stop trusting straight lines, at least when separated?
I don't believe I had geometry math in my third grade math class.