if you generalize the problem to n consecutive squares, the first and last of which touching the sides of the triangle, you get a side length of a=(6n-4*sqrt(3))/((n+1)*(3n²-4))

I'm liking these take-a-breather problems lately. The deep analysis stuff makes my eyes vibrate.

5:18 that 8 is gonna give me nightmares

Me before: math.. relaxing? Haha good joke m8 Me after: dang this is actually relaxing

I felt really proud of myself solving this just by looking at the thumbnail and then watching the video getting slightly excited each time I got a step right

6:08

engineer's approach: draw a straight line perpendicular to the base of the triangle, and then with a compass, measure how many a's fit into that line, set an approximation and boom! result.

I showed this problem to my kid and asked her, what is "a"? And she answered, "a" is the first letter of the Alphabet.

2 things. 1)It's nice when you change things up and do problems like this. So keep doing it. 2)I don't know if you already do. But, a "this is a good place to stop" merch-line wouldn't be a shabby idea. So maybe look into it (If you already haven't)

The decimal expansion of the answer starts off nicely, according to my calculator: 0.120345617...

Since the base side has length 1, we just have to represent the base as a function of a and equate the result to 1. (a/tan60) + a + 2a + 3a + (3a/tan60) = 1. Finish up.

your channel is really amazing... I just discovered my fascination with maths and your videos are giving me really comprehensible and easily explained maths problems... keep 'em coming bro !

I would have used just similar triangels. The ratio of the hight to half of the equilateral triagles lenght (to get a rectangular triangle) is sqrt(3)/2:1/2=sqrt(3). TIs have to be equal to the left, small triagle, hence sqrt(3)= a/x => x=a/sqrt(3). The same works with the right triangle, therefore sqrt(3)= 3a:(1-x-6a)= 3a:(1-a/sqrt(3)-6a). Solving for a lead to a=sqrt(3)/(4+6sqrt(3)), which is the same as your solution.

An alternative method would be to do this graphically. [1] Thinking of the bottom left corner of the triangle as (0, 0), we use y=mx+c to graph the left hand edge of the triangle in order to find the distance between the bottom left corner and the first box within the triangle. Substituting the coordinates of the top corner of the triangle (1/2, sqrt3) into y=mc+c, we get the equation y=x(sqrt(3)). As we know that the coordinates (d, a) [with "d" being the distance between the bottom left corner and the box with lengths a] touch y=x(sqrt(3)), we can substitute (d, a) into the equation to find that d(sqrt3)=a. [2] Now that we know what d is equal to, we can now find out what a is equal to by mapping the right edge of the triangle as an equation. As we know that the triangle is an equilateral, we know that the equation of the right hand edge will be a mirror of the left hand edge. Therefore m=-x(sqrt(3))+c, where "m" is just the y of the equation for the right hand edge. Using this function, we can substitute the coordinates of the top corner of the triangle (1/2, sqrt3) to find C and thus the equation m=-x(sqrt3)+sqrt3 (with C being sqrt3). [3] Using this new equation, we can then substitute where the right hand edge touches the third box. We know that the x coordinate of the third box will be 3a+2a+a+d. We calculated that d is a/sqrt3, so we can then substitute that and then get an x coordinate of 6a(sqrt3)+a/sqrt3. We know that the y coordinate will be 3a, as that is the height of the box where it touches the edge of the triangle. Using the coordinates we just figured out, (6a(sqrt3)+a, 3a), and the equation of the right hand side, m=-x(sqrt3)+sqrt3, we can then calculate the value of a. [4] Substituting the coordinates gives us: 3a=-(6a(sqrt3)+a/sqrt3)(sqrt3)+sqrt3, which will then simplify to a=(sqrt3)/(4+6(sqrt3)). Once you rationalise the surds, you will find that a will be equal to 0.1203, which can also be written as (9-2(sqrt(3))/46. If you made it this far, thank you for reading and I hope that I was able to write this in an intelligible way :D

Nice to be watching nice Math work on a Saturday evening....what else better to do....😁 thanks!!!!!

this was actually a nice problem to solve! thanks

Finally, a problem I can do by my own. :333

Extremely easy going by the standards this channel has set!!!

I love the geometry problems, you talk through them especially well

Easiest problem so far. Yes it was relaxing too.

It would be relaxing if my brain wasn’t screaming at “eh-quilateral” 😂

The bits on the end are similar to half an equilateral triangle, so the base is 1/sqrt(3) times the height. That means the base of the big triangle is (6+4/sqrt(3))a = 1 so a = 1/(6+4/sqrt(3)).

I paused this at the beginning and solved it myself took some different (but similar) routes (intercept theorem and law of sines) but at the end came to the exact same result, very relaxing indeed, thanks

As an engineer, I'm pulling my hair out everytime he simplifies after 3:44 😂 how many times can you rewrite the same thing?? A mathematician will show you!!

I am mostly impressed by how te erased the 6 at 4:04

every time you post a geometry problem like this one I go to Autocad to replicate it. I find quite fascinating to see how everything matches

The jump cut after “18^2?” night be the funniest thing I’ve seen today

後半の計算めっちゃ丁寧で優しいな

I'm confused, what principle allows you to change the order of subtraction?

Alright let me just really quickly summarize this problem: Use trig to calculate the length between the edges of the squares to the corners of the triangles, divide that by the amount of a, you get a.

Excellent video. Well done.

Rescale A=1 and find side L=1+2+3+(1+3)/root3. Dividing all lengths by L we get back required a=1/L

Professor grow your problem solving series. Please make a video on the extremal principle.

5:17 Did he just right an 8 by drawing two circles on top of each other. THEY AREN’T EVEN TOUCHING.

Respect from one mathematics teacher to another teacher ❤️

Nice bro! Simply elegant.

Thumbnail: "What is a?" me: "a wHAT?!"

Viewers: See's math equation Me: SHHHAAAAAAARRKK

I know most people here watched this video to make them better math students or something like that. But, as an engineer, I see this as an exercise to solve real-world problems, and intuitively look for similar shapes and relations to use in design, maintenance, programming and so on; thus, an answer such as (9-2*SQRT(3))/46 is not very practical, and I would never have stoped there. Now, .120 or 3/25 is close enough. Heck, in woodworking, in a 1" equilateral triangle, x could be 1/8" give it or take. I Still gave it a thumbs up.

Thanks, man! Entertaining video

a is a small green box, you can clearly see it on the picture

Better use X in terms of A (instead of A in terms of X). 4X + 6 ( sqr(3) X ) = 1 ( 4 + 6 sqr(3) ) X = 1 X = 1 / ( 4 + 6 sqr(3) ) A = sqr(3) X = sqr(3) / ( 4 + 6 sqr(3) ) A = 3 / ( 4 sqr(3) + 18 ) A = ( 9 - 2 sqr(3) ) / 46

I really gotta work on my geometry, all these Geometry problems on KZfaq are so interesting!

This is easily, one of the easiest questions on your channel but I still love your videos. ,👍👍👍

suggestion for a future problem: can this be generalized to n number of squares?

Rearranging the fraction at the end is at first confusing since you move all the signs. It looks wrong and would’ve been nice if you explained how you did it.

Hi Michael, I have an interesting problem. Inside a unit square (side length is 1), is there a point whose distance to all 4 vertices are rational ?

1=a+2a+3a+x+3x (x being as described in the video & 3x by similar triangles) x= a/tan(60)=a/sqroot(3) 1=6a+4a/sqroot(3) a=1/(6+4/sqroot(3)) 0.120 to 3 decimal places simples!

Now i understand what gura's "a" means

Very interesting and fun problem. :) I have a new question for you which I couldn't solve. It was on my latest Newtonian mechanics exam and it was listed as a question with a short answer i.e. it was considered easy. I hope you can help me because I have no idea how to solve it. Here goes: Which of the following differential equations systems have periodic solutions: a) x' = y y' = 3x^2 - y - y^5 b) x' = y y' = x^2 + y^2 + 1 c) x' = y y' = (x^2 +1)y - x^5

Really nice problem, perfect for me, easy to follow and understand.

Nice puzzle! I solved it just like you, but not so elegant. I didn't use sqrt (3) and I used a calculator to completely decimal it.

question: what is a gawr gura: "i happen to be an expert on this subject"

6:08 Wasn't the denominator supposed to be negative?

Wow! This is actually relaxing

So amazing.

'a' is smol shark

3:43 - That's actually a good place to stop.

I love geometry so much. Thank you for interesting video.

Can you help me with this? Any positive integer n can be written as (a 1) + (b 2) + (c 3) uniquely where 0

Not gonna lie I was expecting Gawr Gura's A

Wait why didn’t he make the 276 negative in the second to last step. Or am I just trippin

For generalized n number of squares a=(6n-4sqrt(3))/(3n^3+n^2-4) I dont think it can be simplified further

I like that the method you used to take 48 from 324 was the same way that I do my calculations mentally.

TBH: The nicest description of the number is the exercise to solve it.

Just as I thaugt I whould like trigonometry in high school...

Simple yet fun!

Amazing teacher

He sounds exactly like benson from regular show

A question similar to this appeared in the latest entrance exam of Nada Junior High School in Japan.So, I've already known how to solve this problem.

I wonder how I can get your shirts you showed in the end of this video if I live in Taiwan. Please tell me.

This was indeed vey relaxing

Me: "Yoda, what do think?" Yoda:"The math is strong in this one."

"a" is the first letter of English Alphabet.

good question, thank you very much 👌

quality video. short and sweet!!!!!!!!!!!! worth a sub

This one is primary school level compared with your other Olympiad problems.

So easy for Russian 9'th class

I'm in med school and I really don't know why am i here. But loving those videos. Hugs from Brazil.

watched only like 5 videos on this channel and this is 1st puzzle I knew how to solve straight away xD

Nice problem, overly complicated solution. tan 30 is about 0.58, so multiply 0.58 by 4 to find the missing a's on the base of the triangle. Add this to 6 and take the reciprocal. 1/(4tan30+6)

this should've be relaxing but i don't understand a thing (;´༎ຶٹ༎ຶ`)

Could you make the same problem more generical imputing a n number of squares. What would be the answer?

Gonna do this with my Class next week!

Nice job, although the problem is a bit too easy... Don‘t get me wrong but I came up with the same idea within 10 seconds of thinking and the rest would have been easy simplification. I‘m sure you are a bit inspired of the Chanel „mind your decisions“ and I mean you are showing us your face and it is a bit more funny, yes, but the problems are a bit too easy. However, Cool videos and I hope that you will reach quickly 100k and when this is done I‘m certain that you will reach more milestones. Have a nice day Best Regards

I just found another solution, a = a

Is there some way to get to see the answer? Can the logo that covers it be removed?

I think it is more aesthetically pleasing to consider what is the value of x, when n is the number of squares. In that case, x = 1/(1+ n)(1+ nα) Where α = tan(pi/3). This makes it easier to generalize to any value of n or any isosceles triangle, as this depends on both n and the tan of the angles. Great problem and video.

I missed the point that, at which point this problem was supposed to be relaxing?..

What is a? Baby don't hurt me Sorry I didn't actually watch the video

Similar triangles allows you to conclude that a = x*sqrt(3) without trig. Also, I think doing the other substitution (so you solve 4x + 6x*sqrt(3) = 1) is a little bit neater and gives a slightly simpler answer with less work: a = sqrt(3) / (4+6*sqrt(3))

A = 1/(6 +4tan(30)) Simplest form i could get it to. Doing a little trig to get the bottom distance of the left and right triangles, (left being tan(30) and right 3tan(30)) add them together with 6 then divide 1 by that total. I’d say it’s the quickest route but the fact that there’s more than one way to a correct answer is the beauty of math.

not relaxing when he mispronounces equilateral constantly.

Yay! I got this one!

Very helpful sir

Fun problem and good explanation. One nit: the graphic in the thumbnail image isn’t right. The sides of the 3a box are 2x those of the 2a box.

I solved by using 30 60 90 to find the lengths of x (a/√3) and 3x (a√3), then adding the whole side together.

I fucking love this channel. I wish they used colored chalk in my maths classes at school. I'm not even particularly involved with math on a daily basis, besides calculating which brand of pizza offers more area for the money.

I have a question: given that 4x+6a=1 doesn't it mean that x=a since for example 4×0.1+6×0.1=1 ? I have very little knowledge of math and even the math in this video is hard for me to follow, considering this please help me understand if my assumption above is wrong and if yes then why is it wrong. On the other hand if it is right then why is all the math manipulation shown in this video actually necessary? Thank you.

pretty proud tha i worked this out :)