Half of a deathly area...
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- Опубликовано: 9 сен 2024
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![The most beautiful equation in math, explained visually [Euler’s Formula]](http://i.ytimg.com/vi/f8CXG7dS-D0/mqdefault.jpg)
Advanced college geometry and we’re still calling sectors pizza slices. Never change math.
That has changed. Chapatti is becoming more and more popular in maths.
@@donaldbesong8853 right
Nahhh tht's high school level
Lol, this is high school level at most
Me after a long day of studying for finals: "let's get my daily dose of RUclips to de-stress."
RUclips algorithm: here's some math completely unrelated to your degree
Me: "hmmm yes, I'd like to know what that shaded area is too"
I have my second of three 8 hour medical licensing exams in the morning, yet here I am.
@@riceagainst1 a bit late but did you pass the exams? lol
@@riceagainst1 RESPOND
Literally my case 😂
@@riceagainst1 how was it then
"AND THATS A GOOD PLACE TO STOP", the fourth brother said calmly.
What a scam wow it just Re direct me to payment and actually i lost my 653$ From my account and the password Given was Wrong wow what a scam salute to u guys.
@@theandroidguy6032 no way you fell for that
@@theandroidguy6032 InstaPwn mean instaPayWeNow
lmao you guys are dumb enough for falling to that?
@@Lightwar49 I think that Makai and Angelo are the same person. Makai makes the claim and Angelo says it works so people will trust Makai. Android Guy must trying to destroy their credibility by saying that it's a scam (it definitely is, I've seen this comment everywhere), even though he probably didn't fall for for it.
As a wise man once said: "I wish I were high on potenuse"
Mr. Jackson, that is enough!
Dumbest joke I've ever heard, yet I find myself smiling.
@@hypercodedOld But, I said it first!
Gabriel Iglesias approves
Obama wants to know your location
Me who's terrible at math: "Ah yes, of course! Triangles have THREE sides... It's so obvious now!"
😂
ah yes, finally, a cube
what? it isnt?
This is comedy for me. Me laughing at my own ignorance.
Is this the start of real life Breaking Bad but instead of chemistry, Walter White does calculus?
@@justanothernick3984 Did we watch two different versions of Breaking Bad?
now shift the triangles altitude line 1% away from the center of the circle, rotate it clockwise 1 degrees and do it again
Great idea!, (*I have no idea what he is telling*)
Some people just like to see the world burn :D
This is exactly the thing stopping me from studying. They are all scripted problems in school, why should I bother learning them when the actual problems are mostly solved using computer aided stuff these days?
@@pubsvm7355 Without the ability to at least comprehend the principles behind these problems, you can't solve a more complex version. If you're asked to analyze a system that isn't given to you in the form of a problem, you wouldn't even be able to tell enough to get it to where you'd be able to use a computer to solve it. I know this because I'm an engineer and a tutor.
@VK It's entirely possible to solve that problem by hand, you'd just use calculus, not geometry.
I managed to follow this perfectly, right up until to the point he said "Hello there...".
- Did you get that Max?
- Not quite, Chief.
- Well, which part didn't you get?
- The part after you said: now listen carefully.
General Kenobi
I think this solution complicated things a bit. When you find the first Pi/6, you will already know the center angle is Pi/6(similar triangle or simply calculate the shared pi/3 angle), and the edge is 1 by 4+2sqrt(3) - 3+2sqrt(3)
I did that rn on my own too, I only struggled finding radius. Once I saw how to get hypotenuse, yes, its a lot easier our method
Before seeing the video, I was planning to do it by calculus:
1. Set the bottom left point as the origin
2. Make equation for the circle: (y+r)^2+x^2=r^2
3. Make equation for the line: y=mx+b, where m=-(3+2.sqrt3)/(2+sqrt3) and b=2+sqrt3.
4. Find radius by differentiating the circle and setting it equal to to the slope, m.
5. Set the two equations equal to get the x component of the point of intersection (point of tangency)
6. Integrate the difference between the equation of the line and the equation of the circle over the range of 0 to the X value that we just found, the result should be the area.
You can do that and if you use a computer it is way quicker
The faster method is by vector calculus using the double integral: int_0^(sqrt(3)/2) int_(sqrt(3)+sqrt(3-x^2))^(-(2+sqrt(3))/(3+2sqrt(3))x+2+sqrt(3)) dy dx and computing it into wolfram alpha
I would have done the same, had I not been in the bathroom when I had the idea 🤣
I'm trying integration atm, but it's 5am, and I'm tired lol
@@dimaryk11
I did with simple 10 standard geometry just after waitching thumbnail in 5min
The fact that the shaded area was found without the aid of integral calculus is a bit surprising.
It's implicit
How the fuck is it surprising? There's a triangle and a circle, why on freaking Earth would you use integral calculus, you dummy?!?
@@aaaab384 just because you have not the foggiest of ideas how to solve it with integral calculus doesn't mean it cannot be done that way. I am here to learn new things, not to ridicule anyone who uses a different approach than mine.
@@aaaab384 with integral calculus it's very much easier to solve this problem
@@leandromonteiro8613 do go on
There is a much easier solution. The key is to see that the triangle formed by connecting the origin of the circle to the point of tangent is similar to the larger triangle. This allows you to solve for R. Also, since the ratio of the legs of the large triangle is SQRT(3) (after rationalizing the denominator), the triangle is a 30-60-90 triangle. You can then use R and the angle to obtain the area.
Right, I came here to write this :-)
Nice
lmao
@@steveshaff8356 It is a fun trip, also killing a fly with a sledgehammer would be quite spectacular, i'd watch a youtube video about it.
@@tatomar001 Someone is working on that right now.
Once you found the third side, I quickly realized that the triangle was a 30-60-90 triangle. 30-60-90 triangles have the property such that the longer leg of the triangle is sqrt(3) times longer than the short leg, and the hypotenuse is 2x as long as the shorter leg.
He used that to find the smaller leg of the triangle that includes the shaded area. Although, yes, he did not realize the pi/6 instantly with that property applied, and instead used the sine theorem.
@@SyRose901 Youre right, i was just hoping to share a tip so everyone can know when to apply that fact. It's good to remember trig values 👍
The happiness when you figure it out correctly on your own and it turns out to be a simpler method.
can you describe the method you used ?
@@ouadii1427 use pythagorean identities and you can see the base triangle is a 30 - 69 - 90 triangle
@@williampeng2962 69?
@@klaumbazswampdorf1764 He must be from the sex dimension.
@@klaumbazswampdorf1764 60 my bad, typo
Hey Michael, I am from India. In my school days I was taught to do a perfect square factorization under a square root sign. It is the same method you used ( taking terms and equating to a and b) but it's done in shorter steps which may be a difficult for beginners or students who lack practice.
What we do is this -
2ab√3 = 16√3
ab =8√3
a^2 + b^2 = 28
We got two equations. Now start by selecting values for a and b
( a =√3 , b = 8, a^2 + b^2 >28
a = 2√3 , b = 4, a^2 + b^2 = 28....
So a=2√3 and b=4
The feeling of pride solving it and having the same answers but different solutions.
ditto :)
YES
me too, but i only did it in theory, going through the counts I got stuck at first hypotenuse. and still have no idea what he did there lol
@@GasNobili we could do it too!!
@@FrazerOR yes but using several concepts in solving these kind of problems is fun!
Me: Who has no idea what he's doing
Also me: Yep, seems about right.
The most colorful Michael Penn video so far.
There is a simpler method on 6:21 , which is by simply splitting 28 such that it forms the addition of two squares i.e. 4 and 2 root 3. I would say it's simpler as it is a direct method and doesn't require many steps. Anyways, nice video.
how do you think in such a way to spot that? Im convinced it's pretty easy with just a^2 and b^2 but the multiplying by 3 would screw me up... And where does the notation of "root 3" come from anyways? This comment has lead me down quite the math's hole, thankyou.
@@thepower7803 oh well the way that you would think like that in this situation is by making the expression within the square root be a copy of the expression a^2+b^2+2ab so that you could simplify the expression and remove the square root, as for the method shown by Mr. Penn as he mentioned it only works in certain situations and not all.
@@thepower7803 One of the most consistent way is to separate the x*sqrt(3) in the initial expression, into multipliers that match the a and b you are trying to find, so 2ab. It's trial and error at that point.
Love this. Make it a perect square if u can. Boom!~
This has to be first use of the half angle formula I’ve ever seen.
Hope it won't be your last :)
More well known is tan(2x) = 2tanx/(1-(tanx)^2)
Do you live in a cave?
ikr
If tan(2x)=p, tan(x)=t
p=2t/(1-t²), p(1-t²)=2t
pt²+2t-p=0,
t=(-1+-_/(1+p²))/p=
(-1+-1/|cos(2t)|)/(sin(2t)/cos(2t))
=(-|cos(2t)|+-1)/(sgn(cos(2t))sin(2t))
If cos(2t)0, it could be
(-cos(2t)+-1)/(sin(2t))
Now we check those and see
(1-cos(2t))/sin(2t)=
(1-(1-2sin²t))/(2sin(t)cos(t))
=2sin²(t)/(2sin(t)cos(t)),
So if sin(t)0 , we can cancel sin(t),
So that is sin(t)/cos(t)=tan(t)
It's 3 AM on a sunday.
I haven't gone to school in 6 years orso.
Tomorrow I have a D&D session I still need to prep for.
And before this video I was watching Clips from the Castlevania netflix series.
My brain: "Mmm indeed, How DOES one calculate that area?"
RUclips really is something else.
Dude...
Its 2 am on a wednesday night.
I havent gone to school in 9 years.
Was watching d&d videos on youtube.
My brain. Yes! Remember them maths you used to like but never had to use in the past 9 years... Lets get that brain cracking again hahaha!
I like the parallel of our brain!
O and how did your session go!?
Math is beautiful
I dont even speak english, but I understood everything
says the guy who types english?
@@jaybeevee6994 good point
@@jaybeevee6994 thats a point, então vou falar em português mesmo
me who speaks English 👁️👄👁️
Dont lie to yourself bro
Great video! I was taken by surprise by your "Good place to stop", cause I thought you would make a common denominator for the solution.
I had concluded my graduation in mechanical engineering 11 years ago, and Google still suggesting mathematics subjects at 1:00Pm.
- Congratulations by clear explanation 👏🏽👏🏽👏🏽👏🏽👏🏽
and then in the corner of the paper, you see:
*NOT DRAWN ACCURATELY*
I'm physicist and this channel kept my attention with many interesting problems, thank you Michael, great stuff!
Eae kkkkkkkk
Condense way:
Step 1) You should know the two congruent triangle by tangent property
Step 2) Get the angle (pi/12) and the hence the radius of circle by trigonometry
Step 3) Final Area = Big triangle - Area of 2 congruent triangles - Area of half circle + 2 * Sector area with angle(pi/2 - pi/12)
I don't know much math, but from looking at the still shot of the video this is what I hoped he would describe. Thank you
To find the length 1 at 13:10 you could also use the theorem that tangent lines at a circle are equal. 4+2sqrt3-3-2sqrt3=1
Oh nice! Didn't know that!!!
It doesn't need all that effort it's a common right angle triangle 📐 that has lengths of ( 1, √3 , 2) or you could say ( k , k√3 , 2k)
He also already established SSS congruence, but the tangent thing works, too
"okay so we are going to this this in a couple of steps"
Ahh here we go again
Just a comment on style - I would’ve noted that the triangle is similar to the standard 1-2-sqrt3 special triangle, done all computations in terms of those simpler quantities, and then multiplied the answer by the factor of 2+sqrt3
my thoughts tooooo
Very astute observation which negates the ‘a+/3b’ manipulation and solving simultaneous non-linear equations.
9:47 There is a more easy way to find the radius. The small triangle at the top is also rectangular and the top corner is pi/3. tan(pi/3) = r / (4 + 2 * sqrt(3) - 3 - 2 * sqrt(3)) = r / 1 = r. tan(pi / 3) = sqrt(3) ==> r = sqrt(3). You original way is over engineered.
What do you mean by the top small triangle is rectangular? I don't understand...
@@thelazynarwhal Triangle of the following three dots: the top point, the center of the circle point and the touch point of the circle and the hypotenuse of the big triangle.
The main problem is calculate the “radius”
And the “radius” could be calculated much more easily !!
Each tangent leg /line to the circle is exactly of same length ( by definition) = 3+2√3
Since hypotenuse (h) = 4+2√3
Then, h - 3+2√3 = ( 4+2√3 ) - ( 3+2√3 ) = 1
*“That 1”* is the length ( of one of the leg) of the small “Top Triangle “ ( *which is congruent with the big triangle* )
The other Leg is “r “ ( radius)
So
1/r = ( 2 + √3 ) / ( 3+2√3 )
Then r = ( 3+2√3 ) / ( 2 + √3 ) = √3
=========
*Note*
If we see The small “Top Triangle “ is the half of an Equilateral Triangle with each side = 2 and “ height” = √3
So each angle is 60º
Then the top angle of the small figure is 60ª ( 1/6 of the 360º)
Then, everything is very EASY ! and you don't need trigonometry !!
===========
Good moment to remember Einstein : *Imagination is more important than knowledge*
*KISS principle* ¡!
well done!
... "Corporate want you to find the difference between these two pictures"...
Excellent observation!
I used this same method
Haha he is a beast on math but no in trigonometry
Me: Thinks I'm okay at math and want to proceed with something surrounding math in the future.
Michael Penn: Crushes my dreams
I never thought of placing a nested root equal to a + b√n and solving for a and b. Factoring an integer (like -4) and solving for each factor as a system of equations was also a new idea. Quite interesting techniques.
That I loved more than the actual problem.
It is, I would have tried to complete the square under the square root
@@penniesshillings Same here.
I would have used a calculator
That was some crazy math! So simple with Trig, algebra, definitions, and geometry, and yet so complicated! Thank you Professor Penn!
To find the radius r and the angle alpha, use the similarity between the left-top triangle and the big green triangle (AA) would be simpler. :)
Yes, more than half of the calculations are completely unnecessary, First, calculate angle (2+√3)/(3+2√3)=tan(ϴ)=1/√3 =π/6, Hipotenuse = 2*(2+√3)=(4+2√3). Now you can calculate in your head that one side of the upper triangle is 1 (4+2√3)-(3+2√3), then you can calculate all sides because two angles are the same in similar triangles. Hypotenuse of small triange = 1*2=2 , r =(2+√3)-2 = √3. After that is trivial. Everything without square roots, factorizations, tan half-angle formula, and Pitagoras theorem. Only similar triangles and sin/tan of π/6(30ˇ). To be honest I watch these videos just to see how this guy can make things more complicated than they already are.
You're absolutely right. One thing I learned is that whenever you get a circle in your problem, the first thing you should do is to draw radii to all the interesting points on the circle. Then some relationships just pop out. When I labeled the lengths of the smaller 30-60-90 triangle in terms of r, the radius of the circle, I was able to solve for it without anything fancy like half-angle formulas. Once you know r=√3 you're all set.
For alpha it is even simpler: since we already have theta, 180-90-theta gives the opper angle (clearly 60°). With 180-90-60 gives you alpha (again 30°)
@@Isitar09 even simpler this is a problem for gifted 8th graders with no knowledge of trigonometry, you can use the height of a regular triangle for solving it.
Exactly what I came here to say. As usual, the solution the teacher gives is unnecessarily complicated. I guess this is because they want to teach advanced formulas, but there are very few examples where they would be useful. I solved this in much simpler way than in the video and reached the same result.
generalized formula to find the area (with a being the vertical length of the triangle and b being the horizontal length)
ab/2-(b(b(tan(tan^-1(a/b)))-(((tan^-1(a/b))/360)(b(tan(tan^-1(a/b))2))
General solution for triangle with sides a,b,c; side a being the side going through the circle:
T = ((r^2)/2)((a/b)-arctan(a/b))
My reaction seeing the problem:
"I see no obvious trick.... I am going to parametrize the **** out of this thing !"
A brutal solution. But hey, it works !
My math teacher always told us that we can use tricks, but before we do so, we have to learn HOW it works and WHY. Only then we can apply formulas to problems that can be solved with one. He actually made very slight changes to problems in tests so they looked like a formula could be applied, but that wasn't really possible. Still some people applied the formula and got the wrong result in turn. I'm quite thankful for his teaching methodology, as after being 20 years out of school now I still remember how to tackle many math problems, even if I don't do it regularly.
The first several minutes - I immediately noticed that the long leg is just root3 times the short leg. That made this a lot easier. Also after 10:00 - the circle has two tangents meeting at the pi/6 vertex. They must be collinear. Thus as one is 3 + 2root2, the other must be, so there's an excess on the left of the point of tangency of 1. That's when this got really easy.
I have to admit, my math education only went up to Calculus 4 and differential equations in college, and after going into medicine none of this makes sense anymore. Definitely a good career move!
13:51 You could find alpha by noticing that alpha plus the other two angles is a straight line and you already know the other two angles are both pi/4 - pi/12 = pi/6. So alpha = pi/2 - pi/3 = pi/6.
13:50 : one can find triangle area by heron formula and alpha is trivially equal to theta, pi/6 so the arc area can be computed.
This is the first video I’ve seen of you. And my mind exploded, my eyes widened! I love seeing how everything comes together. I am astonished, this makes me more interested in geometry. I loved this video, it made me really happy! :)
Just a thought, if you are already doing the guesswork on value of a and b by assuming common integers, instead of factorizing the polynomial you maybe could just say since ab=8 then there are only a handful of pairs to try out, especially since both a and b are positive.
tan(θ) = (2+√3)/(3+2√3) = √3/3 -> θ = pi/6. That saves 5 minutes.😃
You can get a solution after finding the radius by drawing a horizontal line from the circle center, which ends up drawing a triangle with two of the desired area plus a quarter of the circle
1. From a/b = tan(x) find x = pi/6. Then: c = 2a.
2. Reflect the triangle on its height to get big triangle c-c-2b with circle inscribed.
Area of this big triangle:
A = a×2b/2 = (c+c+2b)× r/2 and get r.
3. Notice that upper corner has the
angle pi/3, thus corresponding sector is pi/6.
4. S = (a - r) × r × sin(pi/6)/2 - pi × r^2/12
🙃👍
I saw the thumbnail, and I wondered if we'd be paramaterizing the problem for a general solution amd started w/o watching. I've now watched for some insights bc I knew Michael knew much more what he was doing than I would know myself. The closer here is way better than what I started to do. I was going to compute a general scalene triangle formed by chords from the original right vertex to the tangent of the hypotenuse of the triangle and from said tangent to the diameter of the circle where the circumference crosses the leg it's symmetric about. Which was part one, then add the adjoined chord area at the end of this pseudo sector so ultimately I could subtract the sum from the encompassing triangle which includes the similar right triangle and true sector that I'll use now. Man, that just makes too much sense- thanks for that👍
Is NO ONE going to comment on the brilliant Harry Potter reference made in the title?! I guess I have to.
What exactly is the reference is about?
@@johnjordan3552 The deathly hallows is from harry potter, in fact, its the name of the last book and movie(HP and the deathly hallows) It's basically the elder wand(pretty much a very powerful wand) which is the straight line then there is the resurrection stone(the circle) and the invisibility cloak(the triangle) which all come together to form the deathly hallows and its symbol and they play a big part of the final book/2 movies. So, in the video the triangle and all that looks like half of the deathly hallows so it's just referencing that. Hope it's clear. Sorry if I rambled I love HP lol
@@krettzy3540 OMG thanks ^^
Awesome bro I wouldn't have identified it myself awesome 😎 👏 😀 🙌 👌 😄 😎 👏 😀 🙌 👌 😄
The Venn Diagram I'M imagining is the one of A = Those who are intelligent...and B = Those who like the whole harry potter nonsense. I'm guessing A n B is such a tiny sliver of almost-non-existence as to make no difference.
I think every math teacher should watch this.. just so that they can feel how their students feels after explaining something to them.
a very clear solution that's well explained?
Don't project your confusion onto the rest of the world.
Even though this is a good explanation, I’m a teacher and I understand what you meant!😊
him: find the shaded area
me: *points at the shaded area* FOUND IT!
The hypotenuse can be found much easier by taking 2+sqrt(3) as a and you get the sides to be a and sqrt(3)*a. Hence the third side is 2a. From there on we get, pi/6 directly for theta. After that it should be straightforward to get the area of the triangle-area of sector
My man really spent the first 6 minutes to add a one to the bottom leg of that triangle 😂😂😂
but he did it in such a clear way that it made sense to even me. I rather him take the time
If you are watching on a tablet, double tap on the screen to advance 10 seconds. I skipped large portions of this video.
@@draugami if youre on a laptop or computer use the right/left arrow keys. if you do do SHIFT+? you will see the keyboard shortcut menu
This is a pretty good explanation and arguably a very long one ,
I would suggest, instead of calculating the hypotenuse, we can use
tan ø = 2 + √3 / 3 + √2 => which would have given us π/6 as the angle ,
and with that angle we can find the radius of the circle with
tan (π/12) = radius / 3 + √2 => this will give us √3
Now , we will be left with 1 big triangle , 1 Circle, 2 similar small triangles inside the big triangle
1) Calculate the area of big triangle
2) Calculate the area of circle => and make it to half ( as one side of triangle runs through the center of circle )
3) Calculate the area of 1small triangle and multiply it by 2 ( as we have 2 of them )
Area of shaded region = Area of big triangle - ( Area of semi circle ) - ( Area of 2 similar triangles )
I'm not sure whether this is a proper way of solving this problem , but this seems to yield the result
Yes, all those complicating steps calculating the hypotenuse was not necessary.
This helped in revising my concepts of geometry & Trigonometry.
Very good video. After 12:30 it was just to calculate the area of the triangle and then subtract the area of the semicircle
The right corner point of the triangle, connected to the center of the circle, forms the bisector of the interior angle (the one on the right obviously) of the triangle.
I couldn’t ...not say it...sorry.
But you could have shown it in a much easier way: considering the fact that radii of the same circle are equal, it follows that the center is equidistant from both sides of the angle, thus the center belongs to the bisector of the angle.
Another way is that the two sides of the angle are tangents to the circle. So the bottom side of the triangle and the segment on the right (formed by the right corner point and the point of tangency) have equal lengths.. because ..that’s what tangents do.. lol (Actually, 2 tangents issued from the same exterior point to the same circle are equal). And the line joining this exterior point and the center of the circle is an axis of symmetry of the figure (just the circle and the 2 tangents with no other elements). So by this symmetry, this axis is a bisector of the angle.
Also...
I prefer to “de-nest” the expressions this way:
Let x=radical(28+16radical3) and y=radical(28-16radical3).
Then (x+y)^2=x^2+y^2+2xy=64 (just pure calculation)
and (x-y)^2=x^2+y^2-2xy=48
which yields that x+y=8 (-8 is rejected since x>y>0) and also x-y=4radical3 (again, -4radical3 is rejected since x>y so x-y>0)
adding the above 2 equations gives us 2x=8+4radical3 or x=4+2radical3.
I think using the conjugate is better, leaves less room for guess work or trial and error.
This gives a really nice idea for a geometry problem for my students. Thanks a lot my friend. Great content.
@Michael Penn -- more easily to calculate that length=1 line segment at the top: the top tangent segments the hypothenuse into that top small segment, and the longer segment, which you proved it already equals the triangle's base. So, it's hypotenuse MINUS base (4+2sqrt3 - (3+2sqrt3)) = 1
I love these videos. I love how I remember bits and pieces of algebra and trig from high school along the way and also learn new things I never was exposed to in school. Please keep it up!
13:40 You can also use similar triangles to figure out the angle
7:33 i saw that peek michael, seems like not even the best of us know the unit circle by heart. don't worry i won't tell anyone ;).
I tracked as he went along, and it's very educational for the methods, but I noticed as the video went on a couple more "intuitive" ways to solve for some of the same answers, which require very little algebra or trigonometric work in comparison:
1. The Hypotenuse - Why not first check if we already are familiar with the triangle? This is a 1-2-√3 triangle, as it is a right triangle with 30º and 60º. If you even look at the initial sides, they have similar aspects, and if you just run a check by simple division, the bottom side is equal to the leftmost one multiplied by √3. Therefore, the Hypotenuse is equal to the left side * 2.
2. The Radius - By extension, If we look at the triangles created by bisecting the rightmost angle, a section of the hypotenuse must equal to the length of the adjacent bottom side, with the remaining length equal to exactly 1 (Hypotenuse - Bottom). And since the top angle is 60º, the smallest triangle at the top must be an exact 1-2-√3 triangle! So it's hypotenuse is 2, which is a shared length with the left side, so the remaining length of the left side (the Radius) must be √3.
3. The Slice - As mentioned above, we know just by comparisons and congruency that the angle of the slice is 30º. And of course the angle of the whole circle is 360º. So the area of the pizza slice must be 1/12 that of the circle (30/360). The circle's area being easily calculable as 3 * π, the Slice's area is 3/12 * π, or π/4.
4. Final - So the area we are looking for, without any tangent or cosine formulas, without any factors of -4 to unnest radicals, is [√(3)/2] - (π/4)
Of course, I just personally remember that particular common triangle better than several of the equations he brought up. And this won't always be applicable, and you'll have to use methods similar to his, but why not exemplify that with a triangle that isn't as nice? Anyways, good video and excellent education all the same!
Went from watching Key and Peele to this, not sure how it happened, but it did.
Same ahahah
13:15 The base of the main triangle (3+2sqr(3) ) is the same distance between the tangent and the pi/6 vertice angle, so, the distance of the small triangle to find is (4+2sqr(3) - (3+2sqr(3) ) = 1. No need pitagoras for this segment.
You made that much more complicated than necessary.
Start by examining theta. The tangent of theta is the ratio of our given sides, which easily simplifies to root(3)/3, (multiply the ratio by [(3 - 2*root(3))/(3 - 2*root(3)] )making our triangle similar to a simple right triangle with sides 1, root(3) and 2, in other words, half of the equilateral triangle. That means the hypotenuse of the original triangle is going to have a length that is twice the length of the smaller side or 2*[2 + root(3)] = 4 + 2*root(3), as you discovered with your complicated technique of “un-nesting” the square root.
After drawing the radius perpendicular to our original triangle and determining the two congruent right triangles created by joining the center of the circle with the 30 degree angle, you could see that the hypotenuse of our original triangle is now divided into two segments, and one of the segments is 3 + 2*root(3), so that means the other segment is
[4 + 2*root(3)] - [3 + 2*root(3)] = 1. Working with Pi/12 was never necessary. The circular section has an angle of Pi/6 because the small triangle is similar to our original one (they are both right triangles that share the angle at the top). And that quickly reveals the other side and hypotenuse of the small triangle that has side =1 to be root(3) and 2, which shows that the radius of the circle is root(3). We can confirm the hypotenuse of the small triangle is 2 since the original side is
2 + root(3) and the radius is root(3). No complicated trigonometric formula was required.
The calculation of the area is then straightforward, just as you describe it.
no theta necessary. Only Pythagoras for phuck sake
You choosing not to just solve the simultaneous equation by substitution was killing me lmao. So much more reliable
HOMEWORK : This is from the Guts Round of the 4th Annual Harvard-MIT November Tournament.
Let S be a set of consecutive positive integers such that for any integer n in S, the sum of the digits of n is not a multiple of 11. Determine the largest possible number of elements of S.
@@zombiekiller7101 I checked with desmos (dear lord, the digit function is a pain in the ass to input, i mean there’s a ceiling of a log on the top of a sigma!!!) and it seems to be true, and it seems to repeat but i’m not sure, no proof here yet
@@zombiekiller7101 No
SOLUTION
We claim that the answer is *38*
This can be achieved by taking the smallest integer in the set to be 999981. Then, our sums of digits of the integers in the set are 45, ... , 53, 45, ... , 54, 1, ... ,10, 2, ... , 10, none of which are divisible by 11.
Suppose now that we can find a larger set S: then we can then take a 39-element subset of S which has the same property. Note that this implies that there are consecutive integers a−1, a, a+1 for which 10b, ... , 10b+9 are all in S for b=a−1, a, a+1. Now, let 10a have sum of digits N. Then, the sums of digits of 10a+1, 10a+2 , ... , 10a+9 are N+1, N+2, ..., N+9, respectively, and it follows that n≡1 (mod 11).
If the tens digit of 10a is not 9, note that 10(a+1)+9 has sum of digits N+10, which is divisible by 11, a contradiction. On the other hand, if the tens digit of 10a is 9, the sum of digits of 10(a−1) is N−1, which is also divisible by 11. Thus, S has at most 38 elements.
Motivation: We want to focus on subsets of S of the form {10a, ..., 10a+ 9}, since the sum of digits goes up by 1 most of the time. If the tens digit of 10a is anything other than 0 or 9, we see that S can at most contain the integers between 10a−8 and 10a+18, inclusive. However, we can attempt to make 10(a−1)+9 have sum of digits congruent to N+9 modulo 11, as to be able to add as many integers to the beginning as possible, which can be achieved by making 10(a−1)+9 end in the appropriate number of nines. We see that we want to take 10(a−1) + 9 = 999999 so that the sum of digits uponadding 1 goes down by 53≡9 (mod 11), giving the example we constructed previously.
May I get some help for this question below?
If p is prime and n is any natural number.
Prove that (n+1)^p - n^p - 1 is divisible by p.
I think that may have something to do with Fermat's little theorem but I don't know what to do with it.
@@user-cr4fc3nj3i you can see that (something)^p= (something) x ( (something)^[p-1] )
and using fermats little theorem you have (something)^[p-1] congruent to 1 when p is prime using congruences you can easily do the rest of the problem
I am not good with trig identities and did not feel comfortable working with a pi/12 so my solution diverged around the 10 minute mark. At that point, you can see the top right leg of the little triangle encompassing the area is equal to one and also that the little triangle is similar to the big triangle. That also gives you the radius and therefore the shaded area.
Did you know? You are an awesome teacher ❤️
this was so interesting, I hadn't tought about separating the angle and creating another triangle. My head just went boooom
me too.
5 am, and for some random reason i am watching this...looked so hard in the beggining, thx for making it look easy at the end
to find base of above triangle we can also use congruency of triangle formula then it'll be (4+2√3 -(3+2√3)) = 1.
Me as an English major, seeing the first 3 minutes of this…
Okay I’ve seen too much RUclips, need to get back reading my books. 😅
Me too, lol. What type of literature are you studying?
Me too..I used to love to work calculus problems. I wasn't brilliant, but it was fun..
I havent watch the whole video but checked my answer in the end. The half-angle equation is abundant here. Since the 30 degree is out, the other angle is 60 degree, hence the angels of the smaller triangle are out. Applying the tan ratio could easily get the number for R. Then the area of the shadow could be calculated, which may save half of your time
Ignore the nit-pickers. It is interesting to see the many ways we could potentially navigate such a problem. Often for mathematicians, knowing the vast array of tools you have at your disposal is the most important thing. Knowing that one thing can be expressed another way, that you can reveal information about A by doing B first, that the more you dissect the given information you have, the more you have to work with, which makes it easier to solve the problem, is ultimately what this channel is about.
I always try these before watching the video -- here, I didn't see anything but the circle and triangle, so I did the general solution for any angle in the bottom right of the triangle (which, with the height of the triangle, drives everything).
I looked for the area of the little triangle (peak - tangent point - center of circle) and subtracted the area of the sector - the angle in radians times the radius...
So - 1/2 * r * (h-r) sin a (where a is the angle in the bottom right corner). The sector is then ra.
r = (cos a - cos^2 a)/sin^2 a, so know a and h, and you know everything!
Fun fact: you can do this with the help of co-ordinate geometry .
How so?
@@jeevithapatel94 take the the perpendicular sides as X and y axis and calculate the equation of the hypotenuse line. Take a point on y axis say (0,k) and find the perpendicular distance to the hypotenuse line and Equate with the given condition i.e. both X axis and hypotenuse line are tangent to the circle. You got the radius . Find the area of small triangle containing the required area formed by making a perpendicular to the hypotenuse line from the centre of circle and VOILA! You are more than half done . All that is left is calculate the area of section of circle and substract.
Nice
To be fair, the vast majority of geometry problems can be bashed with coordinate geometry.
@@swapnamoy6134 i don't want to be a fun destroyer but isn't it "Voilà" that you meant instead of wallah ? Or is it some vocabulary I, a simple-minded French people, have not in my bag ?
I saw that 1 from when you got e hypotenuse. The bottom line is 3+2root3, the length of the hypotenuse from the bottom right corner to the tangent point is also 3+2root3, thus making that last little bit 1
This video should be renamed "How to needlessly overuse trigonometry to solve an elementary geometric problem".
13:05
The pythagorean theorem is not actually needed. It's just the length of the hypothenuse of the big right triangle minus the length of the base of the big right triangle.
(4+2sqrt3)-(3+2sqrt3)=1
This is because of the two triangles that were proven to he congruent using the SSS theorem.
This is beautiful
At 5:20, when you're talking about factoring -4 and its possible values, I did it for -4 and 1, as you said. In the value of x, calculating the result, gave a slight difference.
- Gave 7.46410161514 in your case;
- Gave 7.83012701892 in the case I calculated.
I thought it had to do with the calculator I was using and its precision, but it seems that the value changes a little even based on the "a" and "b" that we find by one method or the other.
Using 2 and -2 (reversing what you calculated) drops to negative values, so I discarded it as absurd.
I just saw this while solving it, so I'm not sure what happened, but I'll try to check further.
Can u do olympiad geometry problems michael?
28+ 16 căn 3 =4 (7+4 căn 3)=4 (2^2 + 2.2. căn 3+ 3)=4. (2+ căn 3)^2. Bạn thấy cách này vs cách của Penn cách nào hay hơn nhỉ
Mark the center of the circle and draw a radius to the point of tangency to the hypotenuse of the triangle. There are a couple of 45-45-90 triangles now and thus the radius of the circle can be found without trig...
When you literally have nothing to do so you watch someone do math
Find hypotenuse of the traingle. Draw the radius to tangent.. The upper angle is 60 and the area of the smaller upper traingle can be found.. Subtract the upper smaller sector.. Done
That´s the kind of problem you solve in a test only if you have memorized the solution the day before.
13:00 This unknown length can be calculated even easier. We know that big hypotenuse is 4+2sqr(3), but earlier we saw that part of this hypotenuse is equal to bottom cathetus 3+2sqr(3), so we can subtract it and get 1.
yeah thats a better approach. I saw it too!
If processing were interesting, this would be interesting. He glosses over the points of math conceptual connection which turns math into a tic-tac-toe practice.
a^2 + 3b^2 = 28.....(equation1)
2ab = 16.....(equation 2)
2ab = 16=> ab=8 => a=8/b.
Put a=8/b in equation 1 and search the possible value according to the equation.
15:42
Lol
you are really early😂
I appreciate it
is that you ? :D
@@matzew6462 I'm not Michael Penn. Just a memer that is committed to that joke for too long lol
13:04 why use the pythagorean theorum? The side at the bottom is the same length as the larger portion of the hypotenuse, so just subtract 3 + 2sqrt3 from 4 + 2sqrt3
My teacher: Calculate the electromanetic field on an infinite line of electrons on any point.
Me: get me an infinite line of electrons and i will
this appeared randomly on my reccomended and it reminded me how much i love geometry and math, this was fun to understand
Why did you make this so much harder than it had to be.
No, he made the explanation pretty easy. It was not for people who are already expert in maths. It is for everyone. Also I like the identity that
sqrt(x + y*sqrt(3)) = a + b*sqrt(3)
I didn't know it earlier.
P.S. I am not a maths major.
(2 + sqrt(3) ) : (3 + 2sqrt(3) ) = 1 : sqrt(3), so ratio of three sides = 1 : sqrt(3) : 2, which is easy to get the hypotenuse = 2 * (2 + sqrt(3)) and the angle from sin = 1/2.
"Now we can look at that and it might seem kinda tricky, but it will actually be helpful to multiply this equation by -2 and then add the two equations"
I was lost way before this buddy. Why arent the a and b sides of the triangle just set to an easy numbers like 1000 or something so we dont have worry about what 3+2√3 is without a calculator. thats not the end goal of this video. the end goal is that shaded bit
the goal is a way of finding out what that bit is with basics of trigonometry. Any kind of value is irrelevant if you know how to do math.
@@timo3681 i agree completely. but instead of just laying out the steps this video focus's for several minutes at a time solving out math problems that i think the majority of people who stumble across this video will find dificult to understand. Not only are we showing the pythagoreans step, lets make it annoying to understand if you dont remember this from grade school when you're 30. Not only are we going to use triginometry now to find the angle or the hyptotenuse, we are going to show that angle as a fraction of π rather than a degree (if i was ever tought this in school i dont remember it even remotely, knowing my teenage self I was probably trying to actively forget it).
Im sure hes dumbing down the question as far as he can. But as a carpenter who occasionally has to rely on my calculator to figure out areas of objects, the choices of math terms makes it an effort to pay attention to the steps in the recipe to find the area of that little "deathly" area during this video. So yes i have saved this video to my "how to do math stupid" playlist, but in this 16 minute video it takes him 6:13 to through the pythagoreans portion. Hell I get annoyed with a blueprint when the slope of a roof is presented as a % rather than a degree or fraction (I now have to drawn things out on a piece of plywood rather than just pull out my speed square).
Pretty sure im rambling now.... ill stop
The pi's and the square roots are v. much necessary to make calculation easy in these trig problems; without them cancelling won't occur, and you basically won't be able to do shit without a calculator!
@13:23 you start to figure out the angle in the last / top right-angled triangle you've created, but you already know it's 30 degrees, because the large original triangle is 90-60-30 which means the top angle is 60 degrees in both triangles . . . which means the angle formed by the 2 radii from centre of circle must be 30 degrees.
I really liked your solution and explanation though.
I'll not pretend about how Indians are good at math, this problem was given to me in my 9th grade and the first time I couldn't solve it but now after 4 yrs when I saw the problem again it took me 30 sec to solve it.
When having two sides with roots in it, i would always getting rid of at least one of the roots, by scaling the task (typically safes some time):
Here a = 2+sqr(3), b = 3+2sqrt(3), so i would scale by (2-sqr(3)) or (2sqrt(3)-3); example for scaling by (2-sqr(3)) :
A = (2-sqr(3)) a = (2-sqr(3)) (2+sqr(3)) = 1
B = (2-sqr(3)) b = (2-sqr(3)) (3+2sqrt(3)) = sqrt(3)
C = (2-sqr(3)) sqrt(a^2 + b^2) = sqrt(A^2 + B^2) = 2
c = C / (2-sqr(3)) = C (2-sqr(3)) / ((2-sqr(3))(2+sqr(3))) = 4+2sqr(3)
Also helps (though you could see the factor of sqrt(3)) noticing that the triangle is half of an equilateral triangle (with the height beeing b).
Mirror the triangle on a to get a bigger triangel, where the circle is an inner circle and compute the radius r:
(2-sqr(3))r = R = 2 Area / perimeter = 2 0.5 2B A / (C + C + 2B) = 2 0.5 2sqrt(3) 1 / (2 + 2 + 2sqrt(3)) = sqrt(3) / (2 + sqrt(3)) = 2sqrt(3)-3
r = sqrt(3)
Draw a line from the center of the circle perpendicular to c to get a similar triangle to the task triangle (so angle at center is 30°), let Y to be part of C (analog y):
R/B = Y/A
Y = R A/B
(2-sqrt(3)) y = (2sqrt(3)-3) 1 / sqrt(3)
(2-sqrt(3)) y = 2-sqrt(3)
y = 1
Area_shaded = 0.5 r y - PI r^2 30°/360° = 0.5 sqrt(3) 1 - PI sqrt(3)^2 1/12 = 0.5 sqrt(3) - 0.25 PI