Does this red region have a name?? Let's find its area!

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Опубликовано:

14 дек 2021

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Комментарии : 106

@manucitomx 2 года назад

One of those rare problems I can solve by myself! Thank you, professor!

@squeekycheese 2 года назад

ditto

@theash4361 2 года назад

I used a sort of cheap method to solve this. Scale the whole picture so that the radius of the circle is 1. Then the area of the square is 4, and the area of the circle and everything else needs to add to 4. The tangent line on the circle has slope -1 and intersects at point (1/√2, 1/√2), so it's equation can be solved for to get y = √2 - x. Plugging in y=1 we find x=√2-1, so the length of the given triangle is 1 - (√2 - 1) = 2 - √2. Now we find the area of the scaled sliver as (4 - π - 2 * (2 - √2)^2)/8, and scale it by 1/(2-√2)^2 to obtain the area of the sliver in the given diagram. Once you simplify and all the dust settles you'll arrive at the same answer.

@Victor-sw4ne 2 года назад

you could usethe fact that two tangents to a circle beginning at the same point have the same lenght to find R a bit more easily as 1+sqrt(2)/2 :)

@alexey_burkov 2 года назад

I really like this kind of mathematical magic: 7:59. Very cool :D

@kimloantranthi6335 2 года назад

mathemagic for short, and yes, Michael is, to me, a mathemagician :>

@BikeArea Год назад

It's the famous magic knock. 😋 He could develop this into much fancier transitions if the editor of these videos is willing to go the extra mile for some additional eye candy. 😊

@udic01 2 года назад

3:37 since you know that the purple triangle is isoceles and you know that one of the sides is sqrt(2)/2 and that the sides are equal,, why use the pythagorian theorem?!

@goodplacetostop2973 2 года назад

8:20

@mustangjoe2071 2 года назад

I found r by simply adding 1 + sqrt2/2 The 1/8 piece can be split in half creating two congruent triangles which are mirrored. A bit less work.

@Actanonverba01 Год назад

This guy is GREAT. He makes the most complicated diagrams look simple. I usually ask myself, why didn't I think of that! lol

@wiseSYW 2 года назад

isn't this just a circle insribed in an octagon?

@karolakkolo123 2 года назад

Yes

@RexxSchneider 2 года назад

@@karolakkolo123 Indeed - and a regular octagon, in fact.

@teeweezeven 2 года назад

I noticed a regular octagon that the tangent lines make (the square and the small triangle four times). Since a regular octagon has all side lengths equal, it must be sqrt(2), the hypotenuse of the small triangle. Making the side length of the square 2+sqrt(2) and the radius of the circle half that. From there, it's just the area formula Michael wrote down

@Mystixor 2 года назад

Clever, I didn't see that

@charlessmith1931 2 года назад

I noticed the imaginary octagon immediately, as well to get the length of the diameter/side of the square. The area of the four corner triangles is 2, so the area of the octagon is the area of the square minus 2. The area of "our piece" is one-eighth the difference between the area of the octagon and the area of the circle. My answer looked like this: [ ( 4 - pi ) x ( 1 + root/2 )^2 - 2) ] / 8

@richardgurney1844 2 года назад

You have to know the octagon is regular before you start calculating lengths and areas though Now that we know the answer, it's clear that the octagon WAS regular - but how did you prove that beforehand?

@teeweezeven 2 года назад

@@richardgurney1844 copy the triangle to all four corners and extend their hypotenuses to get another square. The sides of the octagon are touching the circle at the midpoints (which isn't difficult to show) We can draw lines from the center to these 8 points. Rotating 45 degrees gives the same picture, hence all the side lengths must be equal. (Still not formal maybe but that's how I imagined it)

@charlessmith1931 2 года назад

@@richardgurney1844 By definition, the square is regular. The sides of the triangle(s) are equal thus must be 45-45-90 degrees giving a 135 degree internal angle for the octagon. The sides of the square and the hypotenuses of the triangles are tangent to the circle so they must be equidistant and similar.

@illusionist1872 2 года назад

At first seeing this problem, I questioned how it could be possible without knowing the radius of the circle. I was surprised how you managed to solve it so simply! Nothing but warm regards to you and your team.

@Mystixor 2 года назад

Even after getting into studying physics in university I still like these small puzzles. Solving these problems in your head feels like it will keep your mind fit very nicely

@user-er8ro4hv1e 2 года назад

so... does it have a name?

@alainbarnier1995 2 года назад

Day after day watching Michael's nice video, I solved this one by myself... great !

@IsaacKuo 2 года назад

Wow, so much use of the Pythagorian theorem when there are isoceles triangles all over the place so you can just do simple addition and subtraction to get r and such.

@ThAlEdison 2 года назад

I was worried, but my answer matched

@therealplnts8533 2 года назад

Michael is so smooth with the chalk

@user-vu1fj9hr1n 2 года назад

We can get the radius of the circle by just adding the length of tangent line segment(sqrt2/2) to the length given line segment(1). The Pythagorean theorem is too luxurious XD.

@abuzabid6473 2 года назад

Yeah that works but how did you figure that out. I mean what is the proof?

@user-vu1fj9hr1n 2 года назад

@@abuzabid6473 Oh, thank you for your comment. It's because of the properties of a circle and its tangent line. When we draw two tangent lines from a point which is outside of the circle, those two line segements have the same length because two triangles are congruent(right angle, common side, radius of the circle)

@jimskea224 2 года назад

I'll admit I was unable to guess the answer just looking at the diagram.

@yasinmohammadi8 2 года назад

Very nice 👌

@majidrazavi9467 2 года назад

Very interesting

@austinm271 2 года назад

Lighting gone rogue at 7:28!

@robshaw2639 2 года назад

I’d be interested to this one in N dimensions

@radadadadee 2 года назад

I was not even particularly interested in 2 dimensions if I'm being honest!

@assassin01620 2 года назад

@@radadadadee what about 1 dimension? :3

@MrRyanroberson1 2 года назад

I don't know if you're avoiding any particular technique this time, so I'll just do whatever works. Let t = sqrt(1/2). Center a square around the origin, with side length 2. Inscribe a circle tangent to all four sides. Draw the line with slope -1 tangent to the circle with pisitive x and y intercepts. The equation of this line is given through geometry with the circle: (y-t)=-(x-t); y=sqrt(2)-x. The line intersects the square at x or y = 1; respectively y or x = sqrt(2)-1. The distance from either to (1,1) is therefore 2-sqrt(2). The area in the square above the line is therefore the triangle with area 3-2sqrt(2) ~ 0.2. The area of the square in the upper right quadrant is 1. This can be broken up into four pieces: pi/4~0.4 from the circle, 3-2sqrt(2) from the triangle, and two same-area segments with area R. R = sqrt(2) - 1 -pi/8 ~ 0.02 reacale R such that the line segment (1,1):(1,sqrt(2)-1) measures length 1; that is: divide by twice the triangle's area. The result is (sqrt(2)-1-pi/8)(3/2+sqrt(2)) = sqrt(2)(1/2-pi/8)+1/2-3pi/16 ~ 0.4

@MrJdcirbo 2 года назад

I took a slightly different approach... I used the area of the square minus the area of the circle, divided by 4, minus the area of the isosceles triangle (1/2), all divided by two. I wound up with [r²(4-π)-2]/8. I didn't go the full distance to figure out what r was, though. It gave the same answer. I just didn't think of using the diagonal of the square to figure out r. Nice calculation, sir!

@dataflowgeometry 2 года назад

I would call it an "external-point barb of a circle" (since it takes on the shape of a barb). I'd solve it by area subtraction....an isoscoles triangle (external pt., and 2 pts. of tangency) minus a circular segment (part of that triangle that lies in the circle).

@txikitofandango 2 года назад

Perhaps I cheated a little, but I found the radius of the circle by setting the equation of the circle to (x - h)² + (y - h)² = h², plugging in (-½,-½), and solving for h. The distance from the center to either axis equals the radius of the circle, which equals h. The point (-½,-½) is the point of tangency between the triangle and the circle. At that point I was tempted to integrate between the two curves, but I thought better of it.

@txikitofandango 2 года назад

I did check my final answer in desmos against the integral between the curves. The decimals match!

@txikitofandango 2 года назад

tfw you check your irrational number answer and the decimals match

@thatkindcoder7510 2 года назад

@@txikitofandango Casually spends an infinite amount of time to check his irrational answer against his irrational answer

@xizar0rg 2 года назад

So much similar triangles. Definitely allows a different route to calculation.

@mariuserasmus1878 2 года назад

You can skip a few steps if you note that tangent lines that share a vertex are congruent. My students used the shortcut (I can't take any credit :)) to find the radius.

@OH-pc5jx 2 года назад

yeah babey, i’ve still got it!

@peterciccone620 2 года назад

To get r you can just use the fact that the side of the purple triangle and the segment from the purple triangle to the base of the big triangle are both tangent to the circle so they're equivalent, so r must be 1 + sqrt(2)/2

@ikocheratcr 2 года назад

I solved this in 2min, I love geometry problems, and Michael presents this as an 8min video. I got lost on the tensor video 1min in.... Not sure what to think. Also, and the name?

@thatkindcoder7510 2 года назад

You might firmly understand everything behind geometry, since that's what's built up during school, but tensor requires prerequisites that's not typically learnt during school, like set theory (and linear algebra, but that's sort of covered in school). I got lost when he introduced arbitrary operators between vector spaces, and showed properties about them, like you getting infinite dimensional spaces after what seems like something that should be similar to the cartesian product, but he's got tons of videos on the prerequisites that might make sense of what he did.

@RexxSchneider 2 года назад

Optionally, you might assume that Michael Penn's specialisations (check his Linked-In profile) include topics like vector spaces and tensors. Geometry and mental arithmetic, less so.

@mienzillaz 2 года назад

We can observe that slope of red line segment is -1 and it creates 45° angle with side of a square. It means that it is a side of octagon. From that we can calculate radius of octagon, its area and by having radius we will get area of inscribed circle as well. Take fraction of the difference of both areas and you're done. If that was concluded in video then sorry i haven't watched till the end.

@ahmadmazbouh 2 года назад

a:side length of the square r=½a(radius of the circle) d:the diagonal s:area of the square c:area of the circle x:the red area the triangle (T)sided by 1; 1; and the tangent (which =√2 u) has an area of T=½ u² so the triangle (t)sided by 1; the tangent and the diagonal has an area of t=½T=¼ u² ½d=½a√2 ½d=r+½√2 a=2+√2 s=a²=6+4√2 r=½(2+√2) c=πr²=π½(3+2√2) ⅛s=⅛c+x+¼ we have s and c so everybody knows the rest

@Timmmmartin 2 года назад

I love the way that Michael Penn doesn't feel the need to act like an entertainer when presenting his mathematical proofs. Maths is great fun on its own, without the teacher trying to make it even more fun.

@sldimaf 2 года назад

It's interesting that the result equals 1/16 (with more than 0,5% accuracy)

@pierre-henrilebrun36 2 года назад

No it’s not

@thatkindcoder7510 2 года назад

@@pierre-henrilebrun36 Engineers vs mathematicians be like

@GKinWor 2 года назад

@@thatkindcoder7510 its interesting that pi = 3 (less than 5% difference between the two)

@karolakkolo123 2 года назад

@@GKinWor it's interesting that x=sin x, with just abs(x-sin x)/x error!

@HoSza1 2 года назад

Guys, take a look at the value of exp(pi)-pi+9/10000. Why is it so close to 20?

@martintoilet5887 2 года назад

It seemed a little sus when he named the region "our region" and fill it with the color red.

@RexxSchneider 2 года назад

Perhaps the region voted for Trump.

@shaunbrowne9870 2 года назад

Looking at the thumbnail and assuming we're given a side length for the square: -cut square in half from upper righthand to lower lefthand. -the remaining triangle has a 45 degree angle in what was the corner and a 90 degree angle at the point of tangency--making it a 90-45-45 triangle whose non-hypotenuse sides are the length of the diagonal across the square minus the side length (because radius of circle is unchanging). -area of original square/2 - area of circle/8 - area of the triangle described above = red area. ...Turns out that's not what we're given. Alright, just use the fact that we already know what the hypotenuse of the triangle I described is to work our way backwards through the steps I listed and find the side length of the shape in the thumbnail, THEN do the things I said. (This last is a joke. But I mean, it WOULD work...)

@cernejr 2 года назад

In Wolfram Alpha: A = ((8+8*sqrt(2))-pi*(3+2*sqrt(2)))/16.0 r = sqrt(0.5)/(sqrt(2)-1) A = 0.06... r = 1.70711... A is approx. 0.5% of the area of the square, about 1/186 - surprisingly small. So small that I am suspicious that there is a mistake somewhere.

@MrSantiagoPrado 2 года назад

Around 79ish% of the area of the square is the circle (4/pi). You can check that the area of the for the four triangles is around 17ish% so it makes sense that what's left is 4ish% which is 8 times what you got.

@iesakhanji1995 2 года назад

You can also solve through integration. My only issue with this video is I’m having trouble understanding why the line from the ce yet of the circle to the corner bisects the hypotenuse created by the 45 45 triangle, making one side length of the shaded region root 2/2. I don’t know why you can assume that to be true. However I only get the right answer when I do assume this (1/16), making me wonder why this is so.

@pikcube 2 года назад

The short answer is symmetry. Draw our line from diagonal to diagonal. Since we are cutting a square through one of it's lines of symmetry (diagonal to diagonal) and cutting the circle through one of it's lines of symmetry (any diameter of the circle) the top of the line must be a reflection of the bottom of the line. Since the hypotenuse passes through this diagonal, the first part of the diagonal must be a reflection of the second part of the diagonal. Since reflection is a congruent transformation, the two parts are equal and the line must bisect the hypotenuse. This fact actually means that every right triangle we draw in this problem is isosceles as long as one side is parallel to a diagonal and one side is parallel to a side of the square. Given that the hypotenuse of a right isosceles triangle has a hypotenuse leg ratio of sqrt(2), this lets you figure out quite a few lengths in this problem extremely quickly (x must be 1/sqrt(2), the hypotenuse of the big triangle must be sqrt(2)*r, and that's enough to set up r + 1/sqrt(2) = sqrt(2)*r to find the radius).

@marcushendriksen8415 2 года назад

It's not assumed, it's a theorem. The tangent line segment can be thought of as a chord of the circle lying outside the circle, and any radius intersecting a chord at a right angle will bisect that chord, and its angle

@jmk6696 2 года назад

How about S= sqrt(2)-1- pi over 8 when theta = pi over 4? Isn't S=(1-cos (theta)) over sin(theta) - (theta) over 2? Am I wrong?

@moonwatcher2001 2 года назад

Eventually, what was the name of that region?

@juanramonvazquez3212 2 года назад

I do not know why, but my answer has the first term of the numerator as: 12+8sqrt(2); other than that i got the answer right, i guess good enough :) EDIT: now that i think through it again, i think Michael forgot to add the (sqrt(2))^2=2 after squaring the numerator in the R term with the 2^2=4; am i just not seeing something?

@NonTwinBrothers 2 года назад

What why'd he make the side lengths = 1??

@mohammadsalem5207 2 года назад

I was dying to see the area in decimal ...

@BikeArea Год назад

I second that. It feels uncomfortable. 😮

@skye_thenarrator 2 года назад

I’m at grade 9 and I literally hear: aghast cirkel square squid game half pizza cake with a lie

@thatkindcoder7510 2 года назад

I'm in year 12 and I hear the same thing

@antoniussugianto7973 2 года назад

Radius of the circle is...

@denisphelipon4695 2 года назад

So the lunule can be calculated .

@worldbfr3e263 2 года назад

Has anyone ever told you that you look like the music artist Caribou?

@markkinnard796 2 года назад

What is the name of the region?

@TheVanguard11 2 года назад

mn 4:32 how u know its the radius of the circle?

@marcozarantonello2180 2 года назад

Because the circle is inscribed to the square

@batchrocketproject4720 2 года назад

From inspection of the thumbnail, I set r=1 allowing 1 (area of square) - (√2-1)^2 (the corner triangle area) - π/4 (the sector area), all divided by 2 as the target is half of what's left. It evaluated to ~ 0.215 (x r^2 if a unit circle wasn't considered).

@ixajot 2 года назад

There is an easier way to get radius.

@TronSAHeroXYZ 2 года назад

Depends on the coefficient, and scaler tensors.

@BlackIntegral 2 года назад

BUT DOES IT HAVE A NAME!?!!

@titan1235813 2 года назад

Yes, it's called "Finding The Red Area" 🥰

@asdf8asdf8asdf8asdf 2 года назад

Mamikon Mnatsakanian would’ve been all over this… (Ask Tom Apostol)

@panyachunnanonda6274 2 года назад

It's a pretty Math. problem.

@JLvatron 2 года назад

Yes, the red area has a name, it's "Lawrence". Some people say "Larry", but that's just disrespectful.

@ThainaYu 2 года назад

I think this question would be more interesting to ask for maximum area of that shard or something more abstract

@dhanishsrinivas 2 года назад

Maximising area of the shard based on point of tangency is not intresting cuz it will just be the point where the square and circle meet

@ThainaYu 2 года назад

@@dhanishsrinivas Minimize then?

@antonvlaskin8496 2 года назад

@@ThainaYu Exactly the same. It's 0, and it's even easier to proof.

@ThainaYu 2 года назад

How about minimize the sum of both shard? Also in my idea the thing that move is not the circle but the line segment that cut the corner and also tangent to circle