2 Circles 1 Square

[@MattBottomly]

I am 30+ years out of algebra, and just stumbled across this in my feed and I still watched the whole thing. That was a terrific explanation. Good teaching and very clear.

[@rnburley]

not going to lie, I have forgotten 90% of this.

[@bobbuilder3414]

Still i doesn't 😢 understand

[@BlackDreaded]

You were so bad at algebra that you failed to understand this is geometry 😋

[@doudi0101]

​@@bobbuilder3414 what exactly didn't you understand, I can explain to you

[@Joys_funclub]

same here

[@mikesmith-fw9nc]

I was never good at maths. But this was a delight to watch.

[@vitorstreetboys]

for me is because is not me doing the hard work

[@ast-og-losta]

If your following it and understand, your head and shoulders above most everyone else. Kudos to you;)

[@listerofsmeg884]

I didn't mind maths , but I could never get my head around rearranging equations

[@y.o.2478]

It's also extremely basic

[@Rchigo]

Mede it in 20 sec with my eye 😊

[@WALTERRIFIC]

You lost me after “hey guys”, but I watched the whole thing.

[@South_Van_Ilse_MTB]

holy shit, exact sme experience!!!!

[@Ambettertenu]

Same

[@LovelyMacawBird-mj7bi]

Why tho???

[@GhPuiu]

Well, now I know why this video was recommended to me

[@prasheinranaweera276]

same

[@rilijn]

I love the logic in math. When I saw the thumbnail, I couldn’t fathom a way to calculate that. 3 min and 34 seconds later it totally makes sense and it’s not even difficult math. Just logic and reasoning. Right on man!

[@qwelqwdjln123]

Just out of 7 years of engineering education and watching these videos reminded me of why I chose to follow this path at the first place. Don't forget guys we are not machines enjoy yourselves

[@imad7x]

Who's we here

[@iPlayDotaReligiously]

​@@imad7xengineer

[@anshikarana7476]

@@imad7x That's a valid question.

[@leightonrud66]

Engineers don't do this though. Calculators and a tape measure is all you'd need for any measurements.

[@saaheb085]

​@@imad7x, we = x

[@Frekspete]

Excellent work! I think another way of looking at it is to determine that the half lengths of the square is x/2, and by using the line you drew for the triangle you would quickly find that y actually is 5 - x/2, thus making the equation being: (5 - x/2)squared + (5-x) squared = 5 squared.

[@alifahmed9309]

Yap, this approach is much easier.

[@neopearl23]

Brilliant! Didn't think of it that way...

[@knotwilg3596]

That's how I solved it too: apply symmetry, then Pythagoras, then solve a quadratic equation.

[@shreyasmore07]

Yup I solved it like this itself😊

[@daoudsammy9716]

Yep ✨✨✨

[@mkbismuth829]

x=10 is actually a fairly reasonable result, it just refers to the square which is bounded by the upper arcs of the two circles (not by their lower arcs like the original case we are dealing with) ..

[@abhisheknanda85]

This is precisely what I love about math. The equations end up encoding all cases that fit the constraints used in the deductions.

[@vitec_dev3659]

You are blowing my mind

[@hammy2737]

sorry what do you mean bounded by upper arcs?

[@Wyvern07_]

​@@hammy2737I'm also wondering this

[@mkbismuth829]

@@hammy2737 I mean with x=2 , the square is in touch with the outer surfaces of the two circles but with x=10 the square is touching the inner surfaces of the circles

[@juhaszelodmusic]

I graduated in electrical engineering Bsc in 2021 and like algebra. But I always felt bad when had to solve at geometry problems. Your videos make me feel the happyness of understading and solving problems again. Thank you

[@NOTaYOUTUBERRR]

Me at 2am for no reason

[@muneebtahir4739]

I'm literally watching this at 2 am 💀

[@NOTaYOUTUBERRR]

Replying at 2am

[@jivvr]

caught me

[@CrashCrushh]

caught me

[@FaiajFida]

same here wtf

[@disparato]

I'm 44 years old, Maths was my subject in school and this came in my feed, I watch the whole thing and was genuinely rivetted. Andy you are a gift my friend

[@ratedaron]

Lair 🤮

[@youcefsiouda]

​@@ratedaron No, math is cool 😎

[@rawbeefking866]

@@ratedaronmath is the best subject

[@Zkflames]

@@ratedaron nice spelling bro

[@TestTubeBaba]

​​@@ratedaron Lair. Yup. Get inside the one you emerged from.

[@Dhairyasheel192]

This is why i always loved maths!!! Its logic always felt almost magical! Like anything can be explained and solved with Mathematics!!

[@SSchus87]

A lot of people do math videos. No one dows rhem as well as you do. Aside from perfect explanation, your personality is an addes bonus. I'm hooked!

[@Zandezz.]

Wow. I love how this used such simple math in combination to solve a bigger problem

[@raidzeromatt]

Yeah it was in my suggested videos and I just laughed watching him write the steps down Last time I did geometry like this was in 8th grade so it would've taken me a minute to put all those pieces together

[@PLF...]

thats pretty much the point of math

[@Flqmmable]

@@raidzeromatt oH mY. you are sooo special!

[@ZiyaB3ast]

​@@raidzeromatti knew how to do all the steps too, but you have to be creative and intuitive enough to discover what should be done here. You should of tried the problem on your own first if you're that smart

[@raidzeromatt]

@@ZiyaB3ast If you know Pythagorean and implicit functions it's just connect the dots Definitely would've taken me a minute to solve but it's not like you're being asked to find the rectangle with the largest area that will fit in that space Geometry doesn't suck until you learn derivatives

[@HoSza1]

Choose a coordinate system such that its origin is at the center of the bottom edge of the red square, y and x axes point towards up and to the right respectively. Since the top right corner (x,y) is on the circle we can write (x-r)²+(y-r)²=r² and if that point (and its mirror image on the other circle) would be moved a bit on the circle, it would distort the red square into a rectangle. The only way it will remain square is when y=2x. Substitute y into the equation of the circle and solve for x: (x-r)²+(2x-r)²=r². x²-2rx+r²+4x²-4rx+r²=r². 5x²-6rx+r²=0. Solutions: x=r and x=r/5. Reject the first and the area is A=2xy=4x²=4r²/25=4

[@MouhibBayounes]

That's how i found it 😂

[@TheGamingPalace123]

Bruhh😥😥my brain cells are gonna diyy💀💀💀

[@LEONLOVESMUSIC]

​@@TheGamingPalace123don't worry it's not that hard! Trust me.

[@sutirthjha4539]

​@@TheGamingPalace123maybe you haven't studied coordinate geometry yet, otherwise it is pretty easy

[@pikminman13]

While I didn’t get the answer, I was trying to do it in my head, and this was my approach. I just missed one step.

[@dipanshumaini]

I wish we had teachers like him in our school days . Simple explanation, to the point without any complexity 👏🏻

[@lpi3]

Every teacher is exactly like him. You was too small to understand

[@yureino]

Dont blame the teacher, you were just not paying attention.

[@Sharaa0]

he's explaining what he's doing in the most plain way possible... this is like the most baseline "teaching" ever, there's not much learning of things you can apply to other problems here (i.e strategies for setting up your solution)

[@juanospitia]

Dude, if your teacher actually explained like that, you'd DESPISE them

[@raghav3914]

You reminded me of my childhood. Those were some of the most tensed moment preparing for competitive exams.

[@marcusambler3543]

At 1:03 , we can define 5 + 5 = 2y + x , which gives us y = 5 - x/2 From here, you can use this result in Pythagoras, (5-x)^2 + (5 - x/2)^2 = 5^2 leading to a straightforward simplification of a quadratic giving the same results as you, 10 and 2. Personally I think this method is easier as you don’t have to deal with y, since you already know what it is, and it is quicker. Although, we all get the same answer in the end.

[@vincevd]

Used this solution aswell, think it's the most straight forward

[@varun9733]

Could you explain further please I didn't get it like how did you come up with 5+5 = 2y+x like did you took the radius and length of square or like what?

[@lordkedos]

@@varun97331:52

[@varun9733]

​@@lordkedos So it's the same theory alright.

[@craigauty8874]

My issue is when he gets to the quadratic equation he says just factor it, it's -10 and -2. That's fine in this simple example but just use the quadratic formula as your proof.

[@zmchenaustin]

Came here for the title, still waiting for the joke

[@martnava1661]

2 circles, one square walk into a bar. the square hits his head on the bar and says ouch the circle rolls around on the floor laughing

[@ProjectileGrommet]

Two circles go to the bar without a square He’s never a round

[@azDanqs]

2g 1c

[@phineasferb9128]

the fact that females are represented as circles in pedigrees makes this even funnier

[@AshwinSadh]

​@@phineasferb9128 We are slowly getting to the point 😁

[@nmaogs]

A simpler solution (what I did) is to draw a perpendicular from the base to the point where the circles meet, so y = 5 - x/2. Then the Pythagoras and quadratic method is just the same but there’s only one variable

[@nicholasfactor]

This is what i originally did in my head it’s so much simpler.

[@ValiTanasi]

Exactly how I thought of it too

[@sebastianmenconi6853]

That's how I solved by head yep

[@shoaibakhtar2476]

How bro ? 😊

[@karlacvalencia2409]

Could you explain it with more details? I didnt understand why /2

[@rjayme5]

For a general solution, using: "r" for Radius; "L" for lenght of the square side; the triangle drawn at 00:26, Use sin^2+cos^2=1. Sin=(r-L/2)/r ; Cos=(r-L)/r. Solving for L, you get a 2nd degree equation in which the solutions are 2r and 2r/5. As you know the side L is smaller than R, the only valid answer is 2r/5. For radius=5, L becomes 2, thus the Area (L²) = 4.

[@Vissou]

00:38 you mean + it’s good to precise for readers that you are talking about the angle at the center of the circle of this said right triangle

[@SH1N]

Ffs thank you, way shorter

[@maboh5622]

ok, but why didn’t you use x …lol

[@rjayme5]

I feel L and r are more representative in this case, but you can use any variable you like. However, if you must always use X when solving implicit problems, I have some bad news for you… 😅

[@johnnechaev1550]

@@rjayme5 😂duuude😂

[@youknowsahara]

I actually just finished this unit in math! Would never be able to solve it on my own but i could understood the steps and the procedure when you did it. Great video!

[@dielaughing73]

Im not sure how i feel about the title of this video

[@caseygreen1475]

bro 💀

[@gr4n737]

CAMER HERE TO FIND SOMEONE POINTING THIS OUT

[@deskasandreas4504]

😅😅😅😅

[@nayanverma6096]

😂

[@OmarFaruk-i9]

Bro intentionally did that

[@peterboneg]

To understand why it's a whole number, note that the triangle you formed first is the 5,4,3 pythagorean triple. We just require that the hypotenuse minus the shorter side is twice the hypotenuse minus the longer side, which of course it is. 5-3 gives the edge of the square.

[@adamesd3699]

Yeah, that is much faster.

[@doormat1]

That was my thought. As soon as he did the 3,4,5 right triangle I was expecting a quick resolution. Then he did polynomial math and I was like wait, why?

[@hellowill]

@@doormat1 hard same. Reminds me of people that don't realise the hypotenuse on a 45-45-90 triangle is just x*√2.

[@Pawn-Sac]

how do you know its a 3-4-5 triangle though

[@happy_labs]

@@doormat1 He hadn't established it was a 3,4,5 triangle though. It's not obvious to me why that would be the case, although I'd imagine there is a geometry theorem that could have been called on to show that

[@songthulethuy5839]

I love doing maths,especially geometry,when i watch this guy,i love his way to solving maths problem,it's make me understand faster.I can learn maths and English at once when i watch you =D

[@mrhankey20]

I haven't done this since high school over 20 years ago but I was able to keep up with you and it all made sense. Well done!

[@Schadock_Magpie]

You can also put the origin at the center of the bottom side of the red square and look at the coordinate of top right corner, we have: 1) y=2x from the square 2) (x-5)²+(y-5)²=5² from the right circle equation Putting 1 into 2, we get : (x-5)²+(2x-5)²=25 that give 5x² - 30x + 25 = 0 and then x² - 6x + 5 = 0. 1 is pretty obvious as a root, so it factor in (x-1)(x-5)=0 and since x=5m doesn't fit we have x=1 Right half of red square is 1*2=2m² so answer is 4m² It avoid a bunch of square roots and pythagore

[@gregjones4870]

Your mathematical reasoning is sound. Thank you!

[@pedro92073]

Exactly what I thought. I am surprised I can still easily do an exercise like this.

[@hentype]

I did the same starting with a cartesian coordinate and deriving y=2x.

[@Yue-vh5zg]

Why are you using coordinates in a problem solving problem? Also is it just me or coordinates are far too complicated for its simple concept?

[@asifmorshed9956]

​@@hentypeactually i didn't understand the Y=2X part.... Can you explain a bit?

[@mcc-us]

I always find the little tricks used to determine side lengths so fun. Good Job!

[@cy4n_knight_q8]

It's really weird feeling that you enjoy looking at math being solved while not the biggest fan of math. Kinda miss when I used to solve those in college

[@shabirparwaz]

This video made me feel good about myself. It has been more than 15 years since I graduated high school, but basic algebra still sticks with me and this is exactly the solution I would’ve gone with. Thanks for a delightful video.

[@PSYCH_IITR]

I'm 𝗰𝗵𝗮𝗹𝗹𝗲𝗻𝗴𝗶𝗻𝗴 you to solve this 𝗺𝗮𝘁𝗵 𝗽𝗿𝗼𝗯𝗹𝗲𝗺 and if you will be able to solve this i'll give you money as a reward. 𝗩𝗶𝗱𝗲𝗼 𝗹𝗶𝗻𝗸-:kzbin.info/www/bejne/n3exZXeDnZ2KqtEfeature=shared

[@notnullnotvoid]

There are a bunch of ways to solve this. An easier one is to recognize that the right triangle created at 0:40 has side lengths of `5 - x` and `5 - 0.5x` and a hypotenuse of `5`. Using the pythagorean theorem gives `(5 - x)^2 + (5 - 0.5x)^2 = 5^2` which can be expanded out and turned into a standard-form quadratic equation, then solved with the quadratic formula. No need to introduce an extra variable `y`.

[@sosven1115]

I like this reference in the title

[@Sonilotos]

Trauma

[@KatzRool]

no

[@PROTAEQUESO98]

Yes

[@arpit2859]

Anyone wants chocolate ice cream?

[@PROTAEQUESO98]

@@arpit2859 👯🥤🎱💩💩

[@solidUntilLiquidBeforeGas]

Really enjoyed this video, Andy - awesome! I had to roll up my sleeves, brush off engineering degree from decades ago and give it a shot! I happened to solve it with a much simpler approach. If the x-y axis is drawn right between the circles and at the bottom of the circles. Equation of the circle (x-5)^2 + (y-5)^2 = 25 must satisfy the point (a/2, a) on it, where "a" is the side of the square. Plugging this point into the equation for circle and solving for "a" will lead to same two values Andy reached (10 and 2). Bam!

[@thomasvanassche7108]

I did it like this as well, but it became painfully clear how long ago it was i last took a math course. About 20 highschool algebra mistakes later, i got there though ;)

[@bluecatlee1043]

There's another solution involving right triangles and trigonometry. 1st right triangle: Center top of circle, to center bottom of circle, to center bottom of square; with angles 26.57 degrees (using arctan(1/2)) and 90 degree corner. 2nd right triangle: Center top of circle, to center bottom of circle, to top left corner of square. Use trigonometry to determine the length of line from top left corner of square to center bottom of square, then pythagorean theorem for the sides of the square.

[@RunstarHomer]

I solved it by considering the horizontal line at the bottom to be the x-axis, and the vertical line between the two circles to be the y-axis, so that half of the red square is in the first quadrant. Then the top right vertex of the square can be the point (x,y). x is half the side length of the square and y is the whole side length, so we have y=2x. Substituting this into the equation of the right circle (x-5)^2+(y-5)^2 = 25 yields a quadratic that can be solved yielding x=1 and x=5, the former corresponding to a square with side length 2 and area 4.

[@bakersmileyface]

I just looked at it and thought "Yeah that looks about a 2"

[@ArtursZedins]

I wish there had been such videos 20 years ago. It is so easy to follow and understand how to get the result, step by step. 👍

[@hammy_and_britney]

Came for the title, stayed for the math

[@flovahkiin5783]

Only few will understand the title

[@gabrieliordache296]

Actually, if you consider the x=10 as a possible solution, you can see that it leads to another square, which has a side along the horizontal line to which both circles are tangent, but now the square is made by the whole distance between the tangent points, and the two diameters. If you imagine the two radii in each circle moving, the two solutions reflect the two possible perfect squares which can be built so that a side is along the tangent (horizontal line): one when the x is 2 and one when the x is 10.

[@jama211]

So cool!

[@outerlast]

oh nice

[@indiankid8601]

But then y will become 0. But we know that y is not zero. It is already defined so x cannot be 10. That's just a coincidential solution not real solution

[@Vaisaakh_Nagra]

Bro u are making a square under the table, that doesn't make sense, the square is above the table or horizontal

[@cithria583]

@indiankid8601 When they say solution, they are talking about solution to the quadratic equation, not the solution to the entire problem. You can use the exact same quadratic equation to solve two different problems, which is why it gives two different answers. They are only mentioning it because this explains why it results in both x=2 and x=10, not because it is complete nonsense, but because it is the solution to a different problem.

[@AnkitAye]

I did it through a bit of approximation. I drew an imaginary rectangle that encompasses both the circles from the outer side. Then I calculated the area of the rectangle which would be 200 meter squared, and the area of circles combined which would be π×(r)^2 = 158 (π value approx 3.14...taken so 157+a small value considered to be 1) , then we get the empty area in the rectangle without the circles which is 200-158= 42. Now if you have imagined the rectangle you already know that there are 6 different empty spaces in the figure and the ones in the corners are half of the ones in the middle. So in total there are 8 spaces as such and the ones we want are the two in the bottom centre, so 8/2 = 4, and 42/4 = 10.5 Then through extreme approximation just by looking at the figure and assuming that the square is surrounded by three triangles and a little excess curved area we can come to the conclusion that the square is about 2 triangles in area + 3 surrounding triangles + excess area, which would be 5 triangles in total and a little excess area(0.25 part ,not area in m) . So since we are concerned about only the square(2 triangles), 2/5.25 × 10.5 = 4 Now I got the answer to be exact 4(which as shown in the video is right), but if the values change in the question still I would have gotten a approx answer, which I could have rounded up to the closest whole number. And if there were options then it becomes even easier. (I was in the train and couldn't get my notebook so had to resort to approximation. Also the method shown in the video is very good, I'll remember that. Also I speculate that since I got the answer exact right maybe the ratio for the area between those two circles might be actually somewhere around 2/5.25 for all circles?, Idk but I'll look into it seems interesting.)

[@MrBlastZone3000]

I did the same thing. Very cool to know someone else tried this approach as well 👍

[@Wertyhappy27]

i did a guess style answer too, wasnt giving up on my answer, and once he said it and it mathced, I felt good

[@Prolute]

If you're going for an approximate answer I don't see why you'd make it so complicated. You can easily just visualize that the 5 meter line is 2 and a half red squares in length.

[@AnkitAye]

@@Prolute that's assuming that the figures are to scale. And if you actually know what you're doing then it hardly takes some minutes to figure out in your head. But I do see that it's a lot of words for explaining such an easy approach.

[@Prolute]

@@AnkitAye Pretty sure the figures are to scale, they have to be for this problem to be solvable at all lol

[@Crucialnas]

I thought this was a "two girls one cup" joke in my recommended. Glad I learned something different today.

[@puncherdavis9727]

You ever have That moment where your brain just turns into a flat line where nothing registers. Yeah that was about twelve seconds in

[@krabkrabkrab]

Try the equation for circle with origin at the lower point of a square that encapsulates the circle. The equation is (x-5)^2+(y-5)^2=25. But from the small square, you also know that at the point of contact of that square with the circle, y=2x. Simultaneous solution is x=1,y=2 or x=5,y=10. If x=1, then y=2, it is the square's side, so area is 4.

[@jamello29]

This is the solution I used, took me a minute to get there though!

[@_Nulupu]

Esattamente. 1 minuto. Cambio di riferimento nell'equazione del cerchio. Change of reference system. Change is always good😊

[@EinSofQuester]

That's the way I solved it. Very easy problem.

[@mylesmcvey1457]

Another way to solve this problem is to think about the dimensions of the square as two separate functions. The base of the square starts out at 2r and goes to 0 where the two circles touch. This width can be defined as 2r-2rsin(Θ). Similarly, the height of the square would start at 0, and work its way up to r. The height can be defined as r-rcos(Θ). From there this problem can be seen as the unique case where the width and height are the same, and so 2r-2rsin(Θ)=r-rcos(Θ). There's a little bit of trig involved from this point, but r cancels out almost immediately leaving just Θ to solve for, which should end up as 2*invtan(1/2) or roughly 53°. Plugging back into either the width or height equation gives a side length of 2 for the square.

[@thomasthuene3173]

right. And much more elegant.

[@sebastianalarcon1204]

Thats how I thought about it

[@ModestestRUST]

That was very pretty.

[@MatthewHolevinski]

I looked at it graphically and instinctively knew it was 4, because the r=5 setup, now had the r's been wildly different, ya probably would of done it that way.

[@a_39_yashraj48]

insta ID.. I need to understand this explanation

[@fuseboy08]

Every time I watch your videos, I feel energized and ready to crush my workout!

[@jothejoker6771]

Mathematician: let's use calculus to determine if the semi truck will fit between those two tipped over silos. Everyone else: *pulls out tape measure*

[@RhumpleOriginal]

Kids: * pull out measuring app on their tablet * GenZ: * Chat GPT wirelessly implants the answer into their brain *

[@thereddestsuninthesky]

This guy knew what he was doing with the title

[@gnatflatt]

in literally 4 videos, my interest in math has been reignited. so happy

[@rizwansmazum9476]

You suprise me in every video

[@getrudesiachiyako]

I love ❤

[@getrudesiachiyako]

Don't you Like

[@Kingsidthesloth]

This dude made it wayyyy more complicated than it needs to be

[@TheBiomedZed]

please share how you would do it simply

[@logitech4873]

​@@TheBiomedZedliterally just measure it with a ruler

[@Kingsidthesloth]

@@TheBiomedZed the square Is equal on all sides so if we find one side we find them all. anyway the circle is 5 meters in its radius so is you put another line the same length vertical against the square you can pretty much tell that the square is only about 2/5 the height of 5meters meaning that the square is 2 by 2. 2x2=4 so the answer is 4 squared I hope this helped

[@mums_poop_sock]

​@@Kingsidtheslothnope it's 5m squared. The square fits between the two circles with perfect symmetry because the circles are the same size right? So the circle touches the square at the 'half way point' between centre and the bottom which means it's half the radius so 2.5m height

[@Kingsidthesloth]

@@mums_poop_sock I like your theory but if you watch the whole vid the answer is four squared he just did it a different way than I did

[@silvo4880]

Solved it by realising square was actually rectangle with length twice of width. Had (5-A)^2 + (5-2A)^2 = 5^2. Realised you ended up with 3, 4, 5 triangle and then it was pretty simple to solve. Nice video, subbed. 👍🏻

[@44yash60]

woah i didnt think of that.. nice

[@lucassoriano839]

Me ha gustado bastante el problema. Yo lo resolví así: primero sistema de ecuaciones con cos(a)=1-x/10 Cos(90-a)=1-x/5 Elevas al cuadrado ambas ecuaciones y las sumas, y obtienes que 1=(1-x/10)^2 +(1-x/5)^2. Y de aquí llegas a la misma ecuación de segundo grado y listo...

[@handledav]

2 Circles 1 Square

[@dot-ln7jc]

lmao

[@theonewhocaredandasked9126]

2 gi-

[@manolodantonio]

👯🍵

[@stoicmadi9704]

this channel is pure gold. God bless you brother ...

[@js_2911]

You can also solve it quite quickly by using the equation of a circle (x-5)^2+(y-5)^2. Here you are cutting the square in half so you know at the point where the square touches the circle y=2x. Plug that in and solve for x, use the smaller value of x, x=1 so the whole side length is 2, 2^2 is 4.

[@mikkl99]

You can make it much easier without calculations: One side of the triangle = 5, one side of the triangle = 5-x, one side of the triangle we can call 5-1/2x (instead of y, because the other half occupies the area next to the other circle). As a conclusion c^2 (=25) = (c-1/2x)^2 + (c-x)^2 And anyone that has calculated the most basic numbers of this formula sees that 5^2= 4^2 + 3^2 . That shows us x= 5-3 = 2 . I m not a math pro but I thought maybe it can make things more visible. Keep going my friend you make a lot of people stimulate their brain. Much love from romania and sorry for my bad English! ❤

[@Quickdrawmcgraw406]

A right triangle with a hypotenuse=5 does not always have sides of 4 and 3, however. So you still have to do most of the calculations

[@adam_ibrr]

for the longest time i believed i could not do maths at all, but when i found this video i actually felt good knowing that i understood the equation and worked it out. giving me hope fr

[@10054]

Video title is genius.

[@outterspace7240]

And cursed at the same time.

[@flavrt]

Always enjoy your thoughtful content, Andy. How about this? Distance from the circle center to the center of square bottom side is 5√2. Then a little isosceles triangle inside the square has hypotenuse = 5√2 - 5 = √2. So the sides of this little triangle = 1. And since they are half the side of the square, the square side = 2x1 = 2 and area = 2x2.

[@dhruvanaidu9738]

5sqrt2 -5 does not equal sqrt2

[@requiem1k87]

Bro 5 root 2 - 5 is not root 2

[@nayanbora7781]

if we see these taking angles into account...then in the rt angled triangle hypotenuse is 5 so one angle would be 53 degree and other one will be 37 so the (5-x) is 3 and y is 4 just by looking at it so the x =2 henxe x²= 4 Saves time from all those eequations ....this method will take just a few seconds

[@stathislourantos1194]

If θ is the angle defined by the horizontal radius of the left circle and its common point with the square, then sinθ=(5-x)/5 and cosθ=(5-x/2)/5. Then squaring both sides of the equations and adding them together and because sin^2(θ)+cos^2(θ)=1, solving for x we get x=2.

[@Fasciseus]

Easy method to calculate in your head: recognize 0:49 a right triangle with a hypotenuse of 5 units is a 3-4-5 triangle with the green side equal to 4 units and the (short) blue side equal to 3 units, and so radius of 5 units less the 3 units gives x (red line) of 2 units, the length of the square; 2 squared is 4 units^2.

[@harshkumbhar3015]

How did you prove that it is a 3-4-5 right triangle?

[@ESPHMacD]

​@harshkumbhar3015 You don't have to, it's a pythagorean triple already proven by the pythagorean theorem to always be true. If the hypotenuse is 5 in a right triangle, then the sides are always 3 and 4, because when you square them 9+16=25. So by having the hypotenuse being 5 you know immediately 5-x will be either 3 or 4. If it's visually accurate then I would assume the longer side to be 4, so the 5-x=3 so x=2. Then for comfortability you can check your answer with his bit at the end of 2y+x=5+5 which is 2y+x=10, and if y=4 because of that pythagorean triple then you get x=2 same as having 25-16=(5-x)^2=3^2, so 5-x=3, so again x=2. It's one of those geometry constants my teacher had us memorize since it will save you time on tests. Simplest and most common being the (3,4,5), (5,12,13), (7,24,25)... but there are others too. Not all right triangles are pythagorean triples, but the ones that are make life so much easier since they will always be the same whole number every time. If you see one side then you know the other two sides.

[@harshkumbhar3015]

​@@ESPHMacDI get your explanation. I was trying to not depend on the visual queues, but I guess it helps. It is just that I was assuming that triangle to not necessarily have pythagorean triples as sides just because the hypotenuse is of 5 units. Thank you.

[@ESPHMacD]

@harshkumbhar3015 Valid. Its at the very least nice to remember because then you can use it as a way to check your answer afterwards. I just like to start there because I'm a bit lazy lol. If I can figure it out without all that polynomial math and substitutions then I'll at least give it a try.

[@benl2140]

You solution was quite a bit easier and more intuitive than mine, but I'll still present it here: We can imagine the shapes in a xy-plane, with y=0 being the base of the square (and bottoms of the circles), and x=0 being the midpoint of the square (and where the circles meet). I'll use s as the side length of the square (since x is already being used). Then the top right corner of the square is at (s/2,s). The center of the right circle is at (5,5), and it has a radius of 5, so it's described by the equation (x-5)^2 + (y-5)^2 = 5^2. The point where this circle touches the square is the top right corner, which we already know is at x=s/2,y=s. Substituting those into the circle's equation, we get (s/2-5)^2 + (s-5)^5 = 25. This simplifies to (5/4)s^2 - 15s - 25 = 0. Multiplying by 4, we get 5s^2 - 60s - 100 = 0, and after that, my solution is the same as yours.

[@pulledpork9778]

Awesome video, the method was very interesting yet easy to understand! I think you could have divided both sides by (10-x) at 2:21 for further coolness B)

[@prototypesoup1685]

A good rule is "do not divide using the variable", just in case you divide by zero on accident. In this case, since the polynomial does have 10 as solution for x, then by dividing by "10 - x", you'd be dividing by zero

[@LumosTerris]

I solved it a bit differently! I saw that the two circles were touching tangentially. If you draw that tangent line, you'll see that since the box is a square, that tangent line must divide the box into two equal slices. I decided to make the width of each half equal to x, so the area of the box was 4x^2. Most of the rest of the setup was the same, except y was set as 5-x. Solving for that was much easier :DD

[@UCHTT]

I saw that line too but decided to guess as I know nothing about maths 😂

[@flames9310]

I saw the figure thought the same thing! But is there a generalized proof or property, or a way of proving this (that the square is divided into equal parts)? I kinda fell short on that.

[@Yt_sebi]

there’s a little shortcut here at 0:54 , if you remember that 3,4, and 5 is a pythagorean triple; meaning that the shortest side of the triangle has to be 3, and 5 - x = 3 means that x = 2, giving us the answer without all the extra stuff

[@phucwang3882]

this reminded me with our 8grades math. so memorable

[@chochmah]

You could simply use pytagoras on the triangle in view: (5-x)^2 + (5-x/2)^2 = 5^2

[@MrSwordfish1988]

Actually not valid since the diagonal of the square would not go through the center of the circle (if you would make the diagonal longer)

[@lucdrouin2625]

Sir, you made an old mathematics professor cry. I'm so thankful you exist and that you are passionate about what you do! Will you one day consider making videos about the people in the list below? I think you will consider some you have not heard of. Alan Mathison Turing (1912-1954) Amalie Emmy Noether (1882-1935) Archimedes of Syracuse (c. 287 BC - c. 212 BC) Aryabhata (476-550) Bernhard Riemann (1826-1866) Bhāskara II (also known as Bhaskaracharya) (1114-1185) Blaise Pascal (1623-1662) Brahmagupta (598-668) Carl Friedrich Gauss (1777-1855) Euclid of Alexandria (fl. 300 BC) Georg Ferdinand Ludwig Philipp Cantor (1845-1918) Ghiyath al-Din Abu'l-Fath Umar ibn Ibrahim Al-Nisaburi al-Khayyami (Omar Khayyam) (1048-1131) Gottfried Wilhelm Leibniz (1646-1716) Isaac Newton (1643-1727) Jacob Bernoulli (1655-1705) Leonhard Euler (1707-1783) Leonardo of Pisa (Fibonacci) (c. 1170-c. 1250) Marie-Sophie Germain (1776-1831) Muhammad ibn Musa al-Khwarizmi (c. 780-c. 850) Pythagoras of Samos (c. 570-495 BC) René Descartes (1596-1650) Sir Andrew John Wiles (born 1953) Abu Ali al-Hasan ibn al-Haytham (Alhazen) (c. 965-c. 1040) Abu Rayhan Muhammad ibn Ahmad al-Biruni (973-1048)

[@Abrogator91]

I had a different solution. For this, im going to use S for the side length of the square. The reason will be clear in a moment. Take the center of the left circle to be the point (0,0). With this origin specified, i can state that the equation for the left circle is x²+y²=25 I can see that the corner of the square touches this circle at the following coordinates: x=5-s/2 y=-5+s Plugging these in, you get (5-s/2)²+(-5+s)²=25 Multiply out thise squares 5² + s²/4 - 5s + 5² + s² - 10s =25 Combining like terms 5s²/4 - 15s + 25 = 0 Multiply everything by 4, divide all by 5 S² - 12s + 20 = 0 Which is of course the same quadratic you had

[@cassandra_classic]

How exciting indeed. I never liked math, yet this was still quite an interesting video. Goes to show just how important algebra is.

[@Юлиус-п6ц]

I wish my math teacher was you. Marvelous explanation! Appreciate that!

[@veryseriousperson_]

The title of the video reminded me of some things...

[@SiddhantYadav]

A simpler way is to drop a perpendicular line from the intersection down to the baseline through the square. This line would be the same as the radius so 5m. This point at the baseline would be the centre of the baseline of the square and would form a triangle with the centre of the circle and would be going through the corner of the square. The length of this side can be calculated by the Pythagoras theorem. It would be approx 7m. Now the part of this line within the square ie from the angle of the square to the middle of the baseline side of the square would be 7 minus the radius which would be 2m. A line from the angle of the square to the centre of the side on the baseline would be the same length as the side of the square. Because of you were to consider a triangle within the square with its tip at the middle of the baseline of the square and one side formed by the top line of the square it would be an equilateral triangle since it would always be 60 degrees to reach the midpoint of the side of a square to the angle of the square. Since we know the length of one side of the equilateral triangle is 2m. So we know the side of the triangle that makes the top line of the square is also 2m. So area of 4 sq m I liked the more mathy explanation too.. I guess this was more geometric and intuitive to me. 😊

[@darainsyed7515]

You cant assume the line intersects at the centre of the square. It works for this scenario but if the example was not to scale then your assumption would lead you to the wrong answer

[@SiddhantYadav]

​@@darainsyed7515if the square is between two equal circles and two angles of the square touch the circles then wouldn't a tangent from the intersection always fall at the midpoint of the square?

[@Eddie42023]

I was building rectangles and triangles, realized a 5(sqrt[2]) -5 was the diagonal of the square. Then I thought: 'are you going to tell me it's 2^2?' I have a BA in physics, so I have that kind of intuition. Your solution was very well done.

[@barryomahony4983]

I got the same when I paused, but just used the quadratic formula to get the roots. I also threw away the mirror image and solved for a rectangle of width x/2 and height x. Keeps this 65 year old brain in shape. 😉

[@jama211]

Nice!

[@MCAzertox]

I had the exact same approach, nice Great job by the way

[@Diam0nd_King]

Is it just me or does his room look like the room from fnaf 4?

[@dheeraj711]

I was thinking the same too😂

[@youraverageslime]

genuinely thought rhis was gonna include some sort of formula that i had never heard of in my life, but i was quite surprised when you used simple algebra to solve this

[@Omerfurkan40]

This question used to be a simple question in the general university exam prepared by the ÖSYM institution in Turkey, almost like a snack.

[@polat6Alemdar7778]

neler çekioz biz be kanka ya

[@alexp7274]

Pythagorean would have given the length of 3 right away at 0:54. Going the 5-x route complicated things unnecessarily imo.

[@wilsonwang8653]

How do you know it’s the 3/4/5 without the restrictions from the circles?

[@alexp7274]

@@wilsonwang8653 the definition of the problem dictates it. Look again.

[@kitsune0689]

what Pythagorean?? youre trying to solve a 2 variable problem. You need to know x to know y or y to know x... 0:57 or you need to change y to a term with x to transform the problem into a single variable one, which he does

[@alexp7274]

@@kitsune0689 Again, overcomplicating things. You do you though.

[@kitsune0689]

@@alexp7274 solve it with your method real quick then

[@Hieenr-]

This guy went straight to the point. I like it!!

[@Мопс_001]

I went the other way and decided to do it like this: I take X as an angle between the two radii in the triangle in the video (vertical radius and the one that connects with the vertex of the square). In fact, I get the "y" as sin(x)*5 and 5 - 5*cos(x) is the vertical side of the square The I write it down like this 10 - 2*5*sin(x) = 5 - 5*cos(x) 2 * (1 - sin(x)) = 1 - cos(x) I won't write the whole thing but in the end x ≈ 53.13 degrees so 5 - 5*(cos(53.13)) = 2

[@LockenJohny101]

I acctually tried the same first lol. Then I saw 2cos(x)-1-sin(x)=0 and though... nah screw that and found the aproach with the pythagorean very similar to the video. Its much cleaner. But anyway, how did you solve the equation 2cos(x)-1-sin(x)=0?

[@Мопс_001]

@@LockenJohny101 well, that's a matter of choices I guess. My way seems to me much cleaner without tons and tons of substitutions, quadratic equations and square roots. Just solve for the angle without all this. Tbh, I haven't solved these for a long time, so I just used photomath (to prove atleast that this is the right way) which suggested universal trigonometric substitution. You put 2t/(1+t^2) instead of sin(x) And (1-t^2)/(1+t^2)instead of cos(x) I believe there is much simpler solution to it though.

[@KGTiberius]

This was my approach, too. Well, 1st thought. Then geometric approach appeared faster, because the radii were equal. 1st thought is better for unequal values of radius.

[@rbarr775]

Yeah, I went directly for a trig solution. To obtain a value for the angle, I used an HP41 Solve routine. Kinda cheating, but once I have the algebraic solution, I consider it conquered.

[@OmnipotentJC]

I did the same thing and used a half angle identity. Starting where you left off: 2 - 2sinx = 1 - cosx -> 1 + cosx = 2sinx -> 1/2 = sinx / (1 + cosx) -> 1/2 = tan(x/2) -> arctan(1/2) = x/2 -> 2arctan(1/2) = x.

[@rexmundi8154]

I’m a machinist and we use trig a lot, it’s pretty much the basis of all shop math. A practical way to solve this in the shop quickly would be to draw a third circle with its center on the top line of the square and tangent to the point the 2 original circles touch. That radius gives you everything you need to know to solve with normal trig. For some really fun math, you should check out the formulas section of the Machinery’s Handbook. It’s the Bible for manufacturing

[@DanAMU-e8p]

Good, however, this is a trick question which you can solve visually very quickly, you have a square 5 by 5 and one leg bisects the small red square so 1 to the left one to the right is 2,also just draw 5 horizontal lines and it will math the height of of the red square, its risky but in a time exam the less time you spend in an easy problem the more time you can use on the very hard ones that requires alot of computation 😊

[@DanAMU-e8p]

photos.app.goo.gl/PPoohZ4GYyq2DEJ58

[@JohnDoe-qr4xu]

I would have done this a different way. As it's a square, all sides are equal. The square is the middle of the 2 circles vertically so when you draw a line down it splits the square into 2. As you can use pythagoras to figure out the distance to the cube and also you know the radius, simply subtract the 2 and you get 1, therefore double that is 2 the length of one side of the square. Easy.

[@JohnDoe-qr4xu]

Seriously though, you've over complicated this 😂. My way is 100x quicker

[@JohnDoe-qr4xu]

The other way is to simply take 40 percent of the diameter from ine Circle. This works in every case where the circles are equal.

[@davidtudios]

Can you explain better?

[@JohnDoe-qr4xu]

​@davidtudios the diameter of the circles is 10m. 40 percent of 10 is 4. The area of the cube is 4 :). This will work always in this kind of puzzle. Try it for other diameters. It's always 40 percent of the diameter.

[@davidtudios]

@@JohnDoe-qr4xu Do you know why it always work?

[@Reaver-Altar]

Honestly, I guessed 4 immediately since the image appeared to be drawn to scale. The square is short of being half of the line, which would be 2.5 so the side of the square is 2 meters. I remember in high school our teachers would either draw things not to scale intentionally or they'd tell us to show our work so we didn't just guess correctly.

[@placidqualm]

Exactly how i got my solution. If it wasn’t to scale i’d be screwed

[@SunshineCasy]

Same thing, but I didn't even think about the chance it might not be to scale. 😅

[@pofdsjoijsodfjsoidf]

Congratulations, the numerical value of the result is useless.

[@hammy_and_britney]

The power of assumption 😎

[@ltu42]

I was surprised the result is such a round number.

[@jessvarey2597]

I’m a designer and have a good eye. I estimated the side of the square was 2/5ths the length of the 5m line, making the area 4m^2. I have no idea how you did what you did, but I think this is why my maths teachers hated me. I got the right answers, but couldn’t show my work. 😂 fascinating video.

[@phillipglennon4299]

What I did wrong was assuming that the angle from the horizontal plane of the box to the radius was 45 degrees Using that I found the sine of 45 times 5 (or square root of 2 times 5), double that, subtract 10 by that, then second power that getting 8.58 meters squared

[@ThoseInterestingStories]

I did the exact same thing lol I looked in Desmos that would’ve been true if we were looking at a circle inside a square but it wouldn’t have been true in this problem

[@The_NSeven]

I also did that.. except a bit more complicated haha

[@dhirajsaxena1971]

As a person that doesn't understand math and sines, yeh same lol.

[@krithikkumar9887]

When u half the image and plot the right side in the first quadrant u will find that y=2x will intersect the circle not y =x hence not 45 degrees. Since when u half the square height remains the same but only the length halves. So tanx is 2 and u will get x = 63 degrees.

[@AnasHart]

Yeah, I realised my mistake when I was solving the quadratic lol

[@kakaomago715]

2x+y=10 (1°); Trace uma diagonal no retângulo formado; perceba que se vc traçar uma paralela entre o ponto de tangência das circunferências, temod um quadrado e o seu lado passa pelo quadrado menor. A diagonal do retângulo passa pegando o vértice de cima do quadrado menor. Temos um caso de semelhança com os dois triângulos formados. O primeiro com os catetos x e y; o maior com os catetos 10 e 5; x/10=y/5; x=2y (2°). Pegamos a equação (1°) e substituímos!!!! 2x+y=10; 2(2y)+y=10; 5y=10; y=2. y²=4!!!!

[@norsharn2883]

No Ingles? 😂

[@kakaomago715]

@@norsharn2883 🤣🤣🤣 My english isn't good. One day I'll be able!!!

[@drunkirk3472]

I tried another approach, got the same ans. Made hypotenuse from top corner of square to center. Got the right angle triangle as (5-x) height, (5-(x/2)) as base and 5 radius as hypotenuse. Using hypotenuse formula (5)² = (5-(x/2))² + (5-x)², gives the same values of x as 10 and 2

[@blue_birb]

You could also think of the circles as a function. Let's say the "ground" line is the x axis and the origin is the first circle's intersection point with the ground line. Cut off the top half of the circles so we have an actual function, defined from -5 to 15 Now we just need to analyze the function a bit, let's call it f, and we need to find where f(x) = 2(5-x) And the are will be f(x)^2

[@MulloyDIY]

Turns out the first triangle he drew was a perfect 3-4-5 triangle. My basic math intuition says nearly anything else would have turned into a disaster of decimal points. Great solution.

[@jollyjoker6340]

It's a right triangle with hypothenuse 5 and legs 5-x and 5-x/2. I think x=2 and 3-4-5 is the only solution here. He could have started with 5-x/2 instead of y. 25 = (5-x)^2 + (5-x/2)^2 = 25-10x+x^2 + 25-5x+x^2/4 = 50 - 15x -+5/4(x^2) 5 = 10 - 3x + (x^2)/4 0 = 20 - 12x + x^2 (x-10)(x-2) = 0

[@EricLedman]

Hey, I don't do videos, but you can update yours with the 5 second solution, and here it is: Draw a horizontal line from the center of one circle to the other. We can call this the horizontal for reference. It's 10 long (2 x radius). You know that the line from the center of the circle to the top corner of the square is also the radius of the circle, or 5. Make that the hypotenuse of a right triangle with a vertical leg up to "the horizontal". If you make a right triangle with a hypotenuse of 5, you know the legs are 3 and 4. The longer leg is on the horizontal, so the part of that radius over the square is 5-4 = 1. Same on the other side, so X=2 and area =4.

[@needamuffin]

I saw that a single circle defined the length of a side of the square vertically and half the length of a side of the square horizontally, so instead of using an arbitrary y variable, I set it equal to 5 - x (x in this case being half of your x). I then followed the same process you did and got x = 5, 1 (since mine were half what yours were).

[@yashveersingh1858]

2 circles 1 square ☠️☠️☠️☠️☠️

[@haydenhuss8758]

You just blew my mind, man. I freakin’ love math!

[@letothemon2746]

I'm thinking: is y necessary? If we assumed that the square is smack in the middle of the two circles, we could say that y is really just 5 - (x/2).

[@nol2521]

y is not necessary

[@coder436]

He uses the equivalent equation 2y + x = 10 at around 1:42

[@janesh8788]

"2 girls 1 cup" 💀

[@slowlearner46]

I was having a lot of fun solving simple equations to find the needed armor rating in a video game, and I want to thank you for reminding me just how much I hate geometry and that I should abandon math as a hobby

[@Dagonius.]

So as a rule, if two Circles are the same size: x = r(2/5) That's easy to remember and could come in handy someday.

[@WifeWantsAWizard]

Yes, I use this formula all the time when I'm playing mini-golf.

[@Dagonius.]

@@WifeWantsAWizard I'm a machinist, so it is quite useful actually...

[@majorfails5512]

If you realize the right triangle you used in the beginning is a 60/30/90 triangle, then its clear the side lengths are 3, 4, and 5. So x = 5-3, which is 2. Much faster. Edit: My bad, it's not a 30 60 90 triangle, but the idea of recognizing it as a 3 4 5 side length triangle still stands.

[@abdu4729]

but how do you know that the triagle is 60/30/90?

[@chrollo1867]

​@@abdu4729it's actually 37/53/90 triangle and it's a special triangle where the hypotenuse is 5 and the other sides are 4 and 3

[@Stirdix]

It's not 30/60/90; 3-4-5 corresponds to a 36.9 degree angle. But you're correct that you can get if quickly if you recognize it as a 3-4-5 triangle, which is an easy guess (given that the problem-maker chose the radius as 5) that is straightforward to confirm [using 5-4=2*(5-3) to confirm that the middle object is indeed a square].