a friendly triangle problem.

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Күн бұрын

A nice geometry problem from the 1983 "friendly" mathematics competition. Our solution uses elementary geometry and first semester calculus.
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Пікірлер: 166

@csodhi9969 4 жыл бұрын

If only both Michaels did a backflip...

@iridium8562 4 жыл бұрын

That’s an idea for the next video

@ThePharphis 3 жыл бұрын

collision risk

@akashthiagarajan4751 4 жыл бұрын

I see one Michael wasn’t enough.....

@stevejustin670 3 жыл бұрын

🤣

@tracyh5751 4 жыл бұрын

First the competition math, then the backflips, now the editing... Will Michael's flexing know no bounds??

@JalebJay 4 жыл бұрын

Guess we'll see how far his limit goes to.

@minh9545 3 жыл бұрын

@@JalebJay it will be very irrational to do so

@MatesMike 4 жыл бұрын

I love how 2 Barney Stinson teach us this kind of maths :D

@ezequielangelucci1263 3 жыл бұрын

pero hombreee, que haces aqui compañeroo

@blackpenredpen 4 жыл бұрын

I was hoping you said something to the other Prof Penn in the beginning. Lol

@Manuel-pd9kf 4 жыл бұрын

h e l l o

@aahaanchawla5393 3 жыл бұрын

want to like but its at 69 likes

@Marek-db8wl 2 жыл бұрын

I expected a high five and hoped for dual back flip though the divided screen (so he would both times end up in that part of image which is cut off) providing an illusion of both Profs colliding with each other in air head down and then disappearing.

@plislegalineu3005 Жыл бұрын

Just like Matt Parker in his videos sometimes?

@DragonKidPlaysMC 4 жыл бұрын

This is quickly becoming my favorite Math Channel!

@Green_Eclipse 4 жыл бұрын

There is a small mistake at the end. The radius should have 2pi in the denominator instead of pi.

@josanyotuve 3 жыл бұрын

Thanks God, I was trying to figure out what I did wrong,... but no :-) 2*pi, thanks

@iridium8562 4 жыл бұрын

I love the strange things he does in his intros, keep doing them! They’re slowly becoming the purpose of this channel xD

@user-rs4st3cj1z 4 жыл бұрын

r can be found MUCH more easier: We know that for a triangle s=pr, where s - the area of the triangle, p-half the perimeter, r - radius, the same as in the video. So for the smaller triangle (a-x)*(a-x)/2=(a-x+sqrt2*(a-x)/2)*r. So r=(a-x)/(2+sqrt2) or r=(2-sqrt2)*(a-x)/2

@euler73 4 жыл бұрын

Loved the playful start to this video. This is quickly becoming one of my favourite maths channels on KZfaq - loving the upload schedule as well ; keep up the great work. And again, thanks for the fantastic abstract algebra lectures as well.

@MrSmartdude11 4 жыл бұрын

Love the new style. Really the best Math channel out there! Keep up the good work.

@NoName-eh8fz 4 жыл бұрын

Good video and a very cool effect with the second Michael. I think it would be even cooler if the "old" one goes out of the picture instead of cutting if it is possible. :)

@stewartcopeland4950 4 жыл бұрын

This special splitting effect celebrates the 30k subscriber : congratulations and continue to share your passion with dynamism!

@cycklist 4 жыл бұрын

Fabulous editing. What a great educator you are.

@DragonKidPlaysMC 4 жыл бұрын

Love the new editing style!

@malawigw 4 жыл бұрын

b(a-x)^2 + x(a-x) can be factored as ( b(a-x) + x)*(a-x) = (ab - (1-b)x)*(a-x) this is 0 when x = a and x = ab/(1-b) and then we just use the fact that a quadratic function has its vertex on the x-value located at right between its zeroes.

@juanignaciodiaz28 3 жыл бұрын

It should be (ab - (b - 1)x) * (a - x) yielding x = ab / (b - 1) as the other 0.

@DavidSavinainen 4 жыл бұрын

What’s better than Michael? Two Michaels! Awesome editing there!

@ThePharphis 3 жыл бұрын

Great little problem! Thanks for sharing

@Aditya_V_R 4 жыл бұрын

I loved the geometry part. So neat

@ShinySwalot 3 жыл бұрын

The editing in the beginning was top budget TV levels, love this channel

@adityamohan7366 4 жыл бұрын

Man you are super-talented.An underappreciated polymath!

@rd-bz8dg 4 жыл бұрын

I don't know why this channel is not popular.

@stefanschroder4694 4 жыл бұрын

The best math channel ever

@ivoh2384 4 жыл бұрын

You don't even need calculus near the end. For a quadratic function f(x)=ax^2+bx+c, it has a min/max when x = -b/(2a), which means A(x) has a max when x = (2ab-a)/(b-1). No calculus needed.

@prabalbaishya6179 4 жыл бұрын

the fact that the extremum of a quadratic is obtained at x=-b/2a is itself obtained by the help of calculus. But obtaining the extemums of polynomials is so easy with calculus that people generally not tend to memorize specific results.

@chiquiflautro 4 жыл бұрын

but obviously the optimum/extremum x = -b/(2a) is derived from calculus

@ivoh2384 4 жыл бұрын

@Adam Romanov that's another way. Another one is to observe that the extremum is when x is the average of the roots due to the symmetry. The sum of the roots is -b/a by Viet's formula, so the extremum is at -b/a/2=b/(2a).

@prabalbaishya6179 4 жыл бұрын

@Adam Romanov Ya I know that for a quadratic, it's extremum can be obtained by Sridharacharya method. But I tried to generalize that for any general nth order polynomial, it's extremum can be found easily through calculus.

@mathissupereasy 4 жыл бұрын

Post more geometry. I love it.

@Jason-ot6jv 4 жыл бұрын

great video man, I enjoyed it a lot

@prabalbaishya6179 4 жыл бұрын

i like the editing sir ! also while finding the relation between r and x, we notice that we just need to find the radius of the incircle, and the radius of the incircle of any triangle is simply (its area)/(its semi-perimeter). so r=[0.5(a-x)^2]/[(a-x)(2+2^0.5)/2]. I think that this relation is pretty common and hence reduces our calculations :)

@jonathangrey6354 4 жыл бұрын

These new edits are so cool

@MarcoLiedekerken 4 жыл бұрын

You have to check if the found X falls in the range: 0

@geometrydashmega238 4 жыл бұрын

You're right. I've checked and it works. Approx. 0.32a

@malawigw 4 жыл бұрын

You have two zeroes to the the Area function b(a-x)^2 + x(a-x) one which is x = a which gives Area = 0. The other zero will give maximum just by logic.

@sosguitar92 4 жыл бұрын

It's trivial. x = a(2b-1)/(2b-2) that is obviously < a because 2b-1

@CTJ2619 2 жыл бұрын

I like these geometry problems

@morancium 3 жыл бұрын

A Classic Question!

@samrachkem7408 4 жыл бұрын

I thought the question ask to determine the dimensions of the geometry like in the plane or in space for example

@antoniopalacios8160 4 жыл бұрын

If we modify it to obtain (in addition to the rectangle), the largest isosceles triangle that is formed with all the remaining base and the lower part of the triangle itself, we obtain x=1/3a and h=1/3a, too (curiously). Thanks.

@gourabjitbiswas 4 жыл бұрын

❤ the edit...!

@felicote 2 жыл бұрын

The equations come out a bit cleaner if you define everything in terms of y (the height of the rectangle) instead of x.

@colinjava8447 2 жыл бұрын

I did it a similar way and got r = aT/[2(T^2-pi)] for T = 2+sqrt2, can't be arsed to check if its right though. I thought there may have been an interesting trick to solving it, but it was just bog standard differentiation.

@danishjuneja 4 жыл бұрын

I didn't know being a math professor gives you Quantum features.

@user-gc8cy5tx1w 4 жыл бұрын

It is getting out of hand : now there are two of them !

@mathematicalmonk1427 4 жыл бұрын

Try this one : Let P(n) be the product of all digits of a positive integer n. Can the sequence nk defined by nk+1 nk +P(nk) and initial term n1 ∈ N be unbounded for some n1 (AUO 1980).

@celineguler 3 жыл бұрын

this is very high quality, i feel lucky

@7177YT 4 жыл бұрын

lovely! thx!

@thayanithirk1784 4 жыл бұрын

Wow editing😍

@enricomartinez1212 4 жыл бұрын

we need 3 michaels for the next vid

@mathematicalminds8446 3 жыл бұрын

Sir, can you give solution of questions of higher mathematics specially group theory, real analysis etc special came in the entrence of TIFR (india)

@roberttelarket4934 4 жыл бұрын

Does anyone know the wording for a minimum distance problem I haven't seen from high school calculus in the late 1960 involving crossing a river and returning?

@duskhound2883 4 жыл бұрын

Burning tent problem?

@roberttelarket4934 4 жыл бұрын

@@duskhound2883: Thank you.

@heathledger8264 3 жыл бұрын

Starting line :here we gonna look at the problem ..same everytime

@jonathanhanon9372 4 жыл бұрын

You really should have switched x and a-x to simplify the problem by symmetry. Other than that good video.

@LucaIlarioCarbonini 4 жыл бұрын

Funny as mesmerizing video!

@kurt.dresner 3 жыл бұрын

Fancy editing! :D

@anon6514 2 жыл бұрын

3:13 My way was a bit longer lol I said the circle function was this: (x-h)(x-h) + (y-R)(y-R) = RR where h is the x-value of the circle's center. Substituting x for y, you get quadratic in x, whose roots are the intersections with y=x xx - (h+R)x + hh/2 = 0 But since it's a tangent intersection, the discriminant must be zero hh - 2Rh - RR = 0 h = R(1 +|- sqrt2) h is going to be positive, R is positive and 1 - sqrt2 is negative, so take the positive root. h = R(1 + sqrt2)

@alexanfung 4 жыл бұрын

Hi Michael, are you interested in those unsolved math problems? I do not mean to know how to solve it, but by any chance, will you explain why are they so hard to solve? For example the sofa problem and the Goldbach conjecture

@alexanfung 4 жыл бұрын

@@angelmendez-rivera351 thanks for your kind and deep explanation. And that's what I would like to know, say, what kind of theory do we lack of to prove/disprove Goldbach conjecture? Or Riemann Hypothesis? and why do we need such "tools" based on the properties of the problems? I really hope someone can explain.

@aviralchaudhary5477 3 жыл бұрын

At 3:25 how did the side came as underroot of 2r?? Please help

@kriswillems5661 4 жыл бұрын

Finally an easier question :)

@WilliametcCook 4 жыл бұрын

man didn't even let himself finish writing the problem before starting smh my head

@jameshenner5831 4 жыл бұрын

Graph and mathematics of solution if a=1, and the value of x is assigned to "Q". I couldn't call it "x" on Desmos. i.imgur.com/bbmcm9Q.jpg

@denisphelipon4695 3 жыл бұрын

The inside of a cube has to be projected on the faces of the same cube . That cube has nXn cubes on the faces , so it has (n-2 ) (n-2) small cubes inside . when is it possible ?

@vikaskalsariya9425 4 жыл бұрын

As time goes by, Michaels will multiply exponentially until they conquer the whole universe using maths and logic.

@ertyrtyui469 4 жыл бұрын

Presh Talwalkar stopped watching at 1:04

@tirlas 4 жыл бұрын

Gouguuūuûuuu Theorum

@shantanudhiman8194 3 жыл бұрын

Hahahaha, absolutely 😝🤣

@user-en5vj6vr2u 3 жыл бұрын

Fresh toadwalker

@titouant1936 2 жыл бұрын

I used a parameter t in [0, 1] to describe the situation. I defined v the vertical length of the rectangle to be: v = at from that I followed a similar path and got the horizontal length of the rectangle to be h = a(1-t) and the radius r=at*(1-sqrt(2)/2) The area A(t)=hv+pi*r^2 ends up as A(t)=a^2(t^2(pi(3/2 - sqrt(2)) - 1) + t) Then solving for t instead of h (called x in the video) in A's maximisation yields t0=1/2 * 1/(1-pi(3/2 - sqrt(2))) . From which we ca get r0, h0 and v0. But wait there is more, A(t0) = a^2 * 1/2 * t0 . So the ratio of the maximized area and the area of the triangle is A(t0)/(a^2 * 1/2) = t0. Here is a geogebra link to play with the setup: www.geogebra.org/classic/xn3gtgh6

@mariocortes1203 4 жыл бұрын

Okay great!

@AgentMCCityDE 3 жыл бұрын

what was that b for tho

@prnk1729 4 жыл бұрын

Instead, why not apply the maximization criteria to the two variable function A(x,r) :)

@justpaulo 3 жыл бұрын

I believe your final value for 'r' is not ok. 1st I noticed that it was numerically different from mine. 2nd I checked it using your 'b' value and a calculator and r should be ~ 0.2*a (the same I got). However your final expression for r is ~0.17*a.

@RickyKwokMath 4 жыл бұрын

Really thought you had a grad student do your work for you for a second

@solomou146 3 жыл бұрын

Goodafternoon. I think that in the denominator of the value of r, must be "2π" instead of "π". Isn't it?

@grn7797 4 жыл бұрын

how come there are two mikes in the beginning simultaneously?

@zehaannaik6918 4 жыл бұрын

The diagrams were bomb

@user-A168 4 жыл бұрын

Good

@AlexandreRibeiroXRV7 4 жыл бұрын

Michael has finally mastered the art of the Kage Bunshin no Jutsu. Jokes aside, great video!

@SyberMath 3 жыл бұрын

Hello Dr. Penn, you make great videos! Would you check out some of my geometry puzzles or algebra challenges?

@jpamado96 4 жыл бұрын

these are big editing skills

@temptemp563 2 жыл бұрын

I just like watching him move ...

@ThePingouin2ter 4 жыл бұрын

This is getting out of hand ! Now there are two of them !

@DeanCalhoun 4 жыл бұрын

“I understood that reference”

@IoT_ 4 жыл бұрын

@@DeanCalhoun I reference that understood)

@roberttelarket4934 4 жыл бұрын

At first I thought that was a student assistant on the left helping on the blackboard.

@pepeluchosykes6874 4 жыл бұрын

Es mas facil hallar todo en función del radio de la circunferencia ya que la expresión cuadratica sale sin termino independiente y es mas facil notar que el coeficiente principal es negativo Y para no aplicar parabola Formulon de vertice de parabola -DISCRIMINANTE/4a = k = MÁXIMO VALOR DE LA FUNCIÓN

@fym4x7 4 жыл бұрын

We need a Vsauce intro music on that one with double mike

@alivape 3 жыл бұрын

> friendly problem > requires two Michaels Can you see how this does not make sense?

@robertlynch7520 3 жыл бұрын

i started with 𝒂, 𝒓 and 𝒌, where 𝒌 = [0 .. 𝒂], or some between value. using very similar algebra, it becomes obvious that the total area is A = 𝒂𝒌 + (π𝒇² - 1) 𝒌², so taking the first derivative yields 𝒌' = 𝒂 + (2π𝒇² - 2)𝒌 Solving at zero yields 𝒌 = 𝒂 / ( 2 - 2π𝒇² ) Finally, 𝒇 = the corner-constant to transform a '𝒌' into an inscribed circle radius 𝒓 𝒇 = √2 / (2 + 2√2); such that 𝒓 = 𝒇𝒌 𝒌' zeroes out at 𝒌 = 0.684468; Working from the other end (computer geek side), an iterative solution finds the same answer in about 100 loops, to a lower precision. Differentiation is so much more valuable! ⋅-=≡ GoatGuy ✓ ≡=-⋅

@NarutoSSj6 4 жыл бұрын

Kage bunshin?

@memomariya2101 4 жыл бұрын

Cool

@MAREKROESEL 4 жыл бұрын

Choosing x for the vertical side makes the calculations much easier.

@georgecaplin9075 3 жыл бұрын

Friendly?! I would say diabolical.

@keepitplainsimple1466 4 жыл бұрын

1 Michael wasn't enough or what? Another one? Double fun! 😁😁

@keepitplainsimple1466 4 жыл бұрын

What editor did you use? VSDC? VSDC is awesome

@MichaelPennMath 4 жыл бұрын

I have been using Final Cut Pro X.

@ChazAllenUK 3 жыл бұрын

Oooh... that's not a square

@ChuckHenebry 4 жыл бұрын

I'm sad that you don't actually calculate the value of x at the end. If there weren't a circle involved-if we were just trying to maximize the area of the rectangle, then x=½a, and the rectangle is maximized by being a square. But once you put in a circle, optimizing its area forces the rectangle to become skinnier. But by how much? I get a value of x=.32a. That amounts to a 40% reduction in the short side of the rectangle, and a 13% reduction in its area.

@rezazom27 3 жыл бұрын

I got r = [a(2+sqrt92)]/2(6+4 Sqrt(2)-Pi), you are missing a 2 in front of Pi

@yossiyaari3760 3 жыл бұрын

I can follow this kind of a process. But for the life of me, I never could understand how I was supposed to come up with these kinds of solutions on my own. (even the easier ones)

@davidjames1684 4 жыл бұрын

You could have used both Mikes to parallel process and solve this problem about twice as quickly.

@tamirerez2547 3 жыл бұрын

9:09 You use the term "downward facing parabola". Sounds too long to say... In Israel math class we simply say "sad parabola" or " happy parabola" (some say "crying" / "smiling" parabola) You can easy guess which parabola we call "happy" 😌 and which we call "sad" 😟

@Marieadams.little.love.handles 4 жыл бұрын

We could have just chainruled

@klementhajrullaj1222 2 жыл бұрын

TWO IN ONE! 😀😉 ...

@oiuhnp98hp33 4 жыл бұрын

Distracting ping / in-the-way subscribe message

@simos11 3 жыл бұрын

A good place to stop the video could be to give us the numbers of x and r supposing a=1!

@rbdgr8370 4 жыл бұрын

At 2:58 Preshtal Walker be like "by Gogu Theorem" 😒

@mathissupereasy 4 жыл бұрын

I have never heard of that. I have been taught Pythagorean Theorem only.

@somebodyelse9130 4 жыл бұрын

He calls it that because the theorem arose first in China, so he calls it by what they call it in China, where it's called the Gougu theorem or the Shang Gao theorem. It's respectable, so there's no need to be rude.

@khalidf2403 4 жыл бұрын

I think there’s a mistake : r =a(2+sqr(2))/2(6+4sqr(2)-pi)

@benhetland576 4 жыл бұрын

@Khalid F : Yes, I believe you are right. He wrote the bottom part of the fraction as "12+8sqrt(2)-pi", but it probably should be 2×pi there like you have.

@rajdeepsrivastava3035 2 жыл бұрын

If you exploit the fact that tan theta will be 1 then it can be done without this many steps

@keinKlarname 4 жыл бұрын

Your twin brother seems to be a quiet person, right? Also mathematician?

@pattystomper1 3 жыл бұрын

4:00 I thought it was r + root 2 r Not 1+root 2 r. I paused to comment. I'm confused Either you made a mistake, or I'm missing something.

@dominickmancine6033 3 жыл бұрын

I think you're both right and confused. :) You're right that it's r + (root2)*r, but when you factor out the r you get r * (1 + root2). He writes it correctly, but doesn't say the parentheses (or "quantity") so it sounds like 1+root2 r.

@iljas275 2 жыл бұрын

the answer to the problem is given in crocodiles, like to measure the length of a snake in parrots))).

@szehoonglew1863 4 жыл бұрын

I try to solve by switching your "a-x" to y so that your "x" to a-y but the answer i get for radius is [ ( 1+ sqrt 2 ) / ( 8 + 6 sqrt 2 + pi sqrt 2 ) ] a ?

@stewartcopeland4950 4 жыл бұрын

r = (a/2) * (2 + 2^0.5)/((2 + 2^0.5)^2-pi)) --> about 0.2 * a