The 3-square problem (classic geometry question)
ฝัง
- เผยแพร่เมื่อ 12 เม.ย. 2020
- I will show you both the geometry way and the complex numbers way to compute arctan(1/2)+arctan(1/3). Enjoy!
Outro song by Motormusic: • Math Bros - Gotta Get ...
💪 Join our channel membership (for as low as $0.99 per month) to unlock several perks, including special badges and emojis during the livestreams: / @blackpenredpen
🏬 Shop math t-shirt & hoodies: teespring.com/stores/blackpen...
10% off with the code "TEESPRINGWELCOME10"
😎 IG: / blackpenredpen
☀️ Twitter: / blackpenredpen
Equipment:
👉 Expo Markers (black, red, blue): amzn.to/2T3ijqW
👉 The whiteboard: amzn.to/2R38KX7
👉 Ultimate Integrals On Your Wall: teespring.com/calc-2-integral...
Try the problem at 8:00
x=13 right?
@@thapakaji8579 yes
@@thapakaji8579 Yes I got same
x=5/3.
Given the problem that we have to solve for x in tan(1/2)+tan(1/4)+tan(1/x)=pi/4
Because arctan(1/2)+arctan(1/3)=pi/4,
arctan(1/4)+arctan(1/x)=arctan(1/3).
This implies:
arctan(1/x)=arctan(1/3)-arctan(1/4)
1/x = tan(arctan(1/3)-arctan(1/4)).
Given that
tan(a-b) = (tan(a)-tan(b))/(1+tan(a)tan(b)),
we get 1/x = 1/13.
Therefore:
x=13
Q.E.D.
@@thapakaji8579 Yes!
BUT WHERE'S YOUR PROOF?
A solution to arctan(1/2) + arctan(1/4) + arctan(1/x) = pi/4:
1. Convert arctan(1/2) + arctan(1/4) + arctan(1/x) into (2 + i)(4 + i)(x + i)
2. Multiply the (2 + i)(4 + i)(x + i) to get 7x + 7i + 6xi - 6
3. Group the real and imaginary parts to get (7x - 6) + (7 + 6x)i
4. Since we want the argument of the complex number to be pi/4, this would have to mean that the a and b of the complex number a + bi would have to be the same.
5. Set the a and b of (7x - 6) + (7 + 6x)i equal to each other to get 7x - 6 = 7 + 6x
6. Solve for x to get x = 13 as the answer.
Yep--that is exactly how I did it too, since doing it through geometry alone would be a huge mess!
Smart
You can't quarantine math
You can't quarantine blackpenredpen
math = blackpenredpen
@@budtastic1224 math = blackpenredpen + 3blue1brown
'Complex numbers aren't complex,they're just different.'
~A guy sitting in the library
The concept of complex relates to the fact that Complex numbers supposed to represent an additional dimension to real numbers. Hence, the number line is now a number plain.
Solution to question at 8:00 :
atan(1/2) + atan(1/4) = atan((1/2+1/4)/(1-(1/2*1/4))
(by atan(x) +- atan(y) = atan((x+-y)/(1-+(xy)) which is derived from the regular angle sum formula for tangent)
= atan((3/4)/(7/8)) = atan(6/7)
Then we have atan(6/7) + atan(1/x) = pi/4
Using the sum of atan formula again, and applying the regular tangent to both sides, we get
(6/7+1/x)/(1-6/7x) = 1 => 6/7 + 1/x = 1 - 6/7x => 1/x + 6/7x = 1/7 => 13/7x = 1/7 => x=13
Mystery Biscuits yes that’s correct
He mentions to use complex numbers but the tangent sum formula is derived from the complex number multiplication so it’s the same thing.
We can also use trig formula tan[x+y] = [tanx + tan y]/[1 - tanxtany] and letting x = arctan(1/2) and y = arctan(1/3) to solve for x +y
Works for two angles perfectly!
But what if you have tan(x+y+z)?
Then you're forced to use complex or apply the identity twice. Just my thought... you had a good observation either way!
Perfect!!!
Nah! 👍🏻
Claire Li --I appreciate your approach
@@budtastic1224 Thks
In general,
arctan(x) + arctan(y) =
arctan[(x+y)/(1-xy)]
Please proof this
@@sueyibaslanli3519 it directly follows from tangent of sum. You just need to prove boundaries
@sheyyub aslanli let expression equal to thetha and take tangent both side and use formula Tan(x+y)
@@sueyibaslanli3519 you could proof this with the formula for tan(x+y), but its much cooler to do it with complex numbers like in the video:
arctan(x) is the argument of the complex number 1+ix and arctan(y) is the argument of the complex number 1+iy.
If you multiply two complex numbers their arguments will add, so you get:
arctan(x)+arctan(y) = arg((1+ix)*(1+iy)) = arg(1-xy + i*(x+y)) = arctan((x+y)/(1-xy))
QED :)
I believe you have to check intervals where x and y belongs to. It may results in some missing additive coefficient to correct the formula
Solution to the problem on 8:00
Rearranging gives
arctan(1/2) + arctan(1/4) = π/4 - arctan(1/x)
Taking tan on both hands, we get
(1/2 + 1/4)/(1 - 1/8) = (1 - 1/x)/(1 + 1/x)
A simple algebra manipulation leads to x = 13
He is showing alternate methods to do the problem inverse trigo is known by all..
This is equivalent to the three square geometry problem from Numberphile. Nice!
Oh yea I just checked! I didn’t know they have a video on this already. But again, this is a classic.
@@blackpenredpen I just remembered, a solution with complex numbers is in this Zach Star video: th-cam.com/video/8tiQl6T-inE/w-d-xo.html
@@blackpenredpen It is not surprising that a person like you should want to cover this classic problem. It is MATH FOR FUN, after all!
solution for 8:00,
as arctan(1/2)+arctan(1/4)+arctan(1/x)=π/4=arctan(1/2)+arctan(1/3)
so arctan(1/4)+arctan(1/x)=arctan(1/3)
tan the both side,
tan(arctan(1/4)+arctan(1/x))=1/3
Apply the tan(a+b)=(tana+tanb)/(1-tan(a)*tan(b)) formula,
(1/4+1/x)/(1-(1/4)(1/x))=1/3
solve the eqaution and get x=13
Great video blackpenredpen.
Another sol:
Special right triangle 37/2° = 18,5°. The tangent of this angle is: tan(37/2°) = 1/3. -------> arctan(1/3) = 37/2°. Another special right triangle 53/2° = 26,5°. The tangent is: tan(53/2°) = 1/2 -------> arctan(1/2) = 53/2°.
Now
arctan(1/2) + arctan (1/3) = (37/2)° + (53/2)° = (90/2)° = 45°
But 45° pi/4.
They complex versious of this question is just great, switch the real and imaginary parts and you get two problems for one! Haha
Thank you BPRP!!
Wow I didn't think I'd understand the complex numbers one more than the geometry one
usually mechanical is easier
8:02 me every 5 seconds when I'm recording a video
6:00 The collective chuckle when he almost forgets the i
I loved it. Any time maths is translated into visual things, I get both more confidence and deeper understanding.
You can also prove it algebraically using the tangent addition formula:
Let a = arctan(1/2) and b = arctan(1/3)
Then tan(a+b) = (tan(a)+tan(b))/(1-tan(a)tan(b)) = (1/2+1/3)/(1-1/2*1/3) = (5/6)/(5/6) = 1
So a+b = arctan(1) = pi/4
Thank you very much man! Best math videos
Really love that you're uploading regularly again!
Knowing the sum of two arctan's formula, it's actually easy to solve the 8:00 problem
You made it simple.. intuitive, thank for sharing ideas.
Use tan(a+b) = (tan(a)+tan(b))/(1-tan(a)tan(b)). Ans = arctan(1) = \pi/4.
Numberphile also made a video about this!
Well explained... I thought complex numbers would be hard to deal with, but it was quiet easier...Thank you Sir...
So much fun. Thank you Steve.
Thanks a lot! I forgot how to do these since I left school. Good to relearn what we have learned.
let arctan(1/2)=x and arctan(1/3)=y
we have to find x+y and let that equal to u and take tan of both sides
tan(x+y)=tanu
use tan sum formula (tanx+tany)/(1-tanxtany)=tanu
plug the values in and you get tanu=1 so u must be pi/4
That's how i solved it too!
@@szerednik.laszlo yeah it's good way to solve these kind of problems
You'll get the answer just by using the double angle formula for tangent
tan(A+B) = ( tanA + tanB )/( 1 - tanAtanB )
Consider A = arctan½ , B = arctan⅓
Funny but soo true.
He is noob is did question in 4 min but using formula I did it in 20 seconds.
Idiot he is taking the challenge to do the problem without the inverse trigo formula we all know the inverse formula I myself know it from class 11
@@chandrabitpal9151 idiot he never said to do without inv trigo now go run rat
I think I had the simplest solution ever.
ang(2+i) = arctan(1/2)
ang(4+i) = arctan(1/4)
ang(x+i) = arctan(1/x)
By adding, you multiply.
(ang(a+bi)=arctan(b/a))?
For ang(c) = pi/4, real(c) = img(c) / i (c is a complex number)
(2+i)(4+i)(x+i) = 7x+6xi+7i-6
(wow, six and sevens!)
For the angle to be pi/4, then
7x-6=6x+7
Do a simple algebra
7x-6x = 7+6
x = 13
there you have it, 13.
Wow right after Michael Penn's video about the same topic. What a coincidence 😂
Zsigmond Telek I know!!
Just like 3b1b’s newest video. “Probability being 0 doesn’t imply impossible”.
7:36 "And you're done!" - Nobody says it quite like you do. It's your style, inimitable, achieving cult status almost instantaneously.
That was neat 👌
There are formulas too converting .but your process with the complex number is really amazing .
Let a=arctan(1/2) and b=arctan(1/3) so tana=1/2 and tanb=1/3.
Then, tan(a+b)=(tana + tanb)/(1-tanatanb)=(1/2+1/3)/(1-1/2.1/3)=(5/6)/(5/6)=1,
so a+b=pi/4.
EDIT: +kpi
Let x = the expression. Apply tangent on both sides. Use the tan(a+b) formula. X= pi/4
Math is a beauty
Nice! Geometry way is most of the students learn that earlier tha complex number theory. Right order and nicely done!
Very Interesting Video
Simple trigonometry:
Let x=arctan(1/2)
=> tan(x) =1/2
Let y=arctan(1/3)
=>tan(y) =1/3
tan(x+y) =
{tan(x) +tan(y)}/{1-tan(x)*tan(y) }
={(1/2) +(1/3) }/{1-(1/2) *(1/3) }
={5/6}/{5/6}
=1
=>x+y=arctan(1) =π/4
or
arctan(x) +arctan(y) =π/4
solution to the problem at 8:00:
using bprp's method, one defines 3 complex numbers z1, z2 & z3 with z1 = 2+i ; z2= 4 +i & z3 = x+i. As multiplying complex numbers is equivalent to adding the angle of each number and scaling the length to the product, the argument of the product of z1, z2 and z3 should equal pi/4. Thus, arg( (2+i)(4+i)(x+i) ) = pi/4 or better arg( 7x - 6 + i(7+6x)). Finally, one must recognize that an angle of pi/4 requires an right triangle with adjacent and opposite sides equalling one another. Therefore, the equation 7x - 6 = 7 + 6x must be solved, leaving the result of x = 13.
Yes I have seen this one
Answer for 8:00 is x=13
Regarding the problem at 8:00 :
As shown in this video, subtracting arctan(1/2) on both sides yields
arctan(1/4) + arctan(1/x) = arctan(1/3)
arctan(1/x) = arctan(1/3) - arctan(1/4) .
I. e. arctan(1/x) will be the argument of the complex number (3+i)/(4+i) .
(3+i)/(4+i) = (3+i)(4-i)/[(4+i)(4-i)]
= (13+i)/17 .
So arg[(13+i)/17] = arctan(1/x) .
Since the factor 1/17 does only affect the absolute value r but not the argument θ, we can ignore it:
arg[(13+i)/17] = arg(13+i) , which analogically to the other arcustangens values equals arctan(1/13) .
Therefore, x=13 .
Very nice!!
Good Job
tan(arctan(1/2) + arctan(1/3)) = 1/2+1/3/1-1/2*1/3 = 1 (by tan(a+b) formula)
so arctan(1/2) + arctan(1/3) = arctan(1) = 45
hey blackpenredpen how are you doing?! hope you are safe and quarantined. i had a question for you, have you ever attempted the putnam examination?
Nice
I am Japanese. I am sorry I am not good at English. My solution may have arleady been written, but I wrote my solution here!!!!
arctan(1/2)=a arctan(1/3)=b とおく
tan(a+b)=(1/2+1/3)/1-(1/2)(1/3)=1
a+b=π/4+kπ (k=0,±1,±2...)
using the complex numbers method the solution is : x=13, correct ?
兄弟，你是真的在造福人类！！！
On algebra class we had to solve this using complex numbers
In fact we had to use multiplication of complex numbers
For additional practice, LetsSolveMathProblems says: arctan(1/3) + arctan(1/4) + arctan(1/5) + arctan(1/n) = pi/4.
Aww, the cute old problem.
Hi blackpenredpen the answer of question at 8:00 is x=13.
Nobody : Maths is one of the toughest subjects.
Blackpenredpen : Hold my pen.
Good 😆👍
Or you can simply equate the answer to x and solve tan x which is tan (A+B)=(tanA+tan(B))/(1-tanAtanB)...=1 ==>x=arctan1=pai/4
Mast bro
I watched first half of the video and i was like "man, why didn't you just multiply complex numbers" :D
I wish you began with: Alright, I'll show you, guys, two ways to do this *Riemann's hypothesis*
Oh rascal it's not that easy.
Oh man
Or we can add an substract arctan(1)+arctan(2)+arctan(3) and use the fact that
arctan(x)+arctan(1/x)=π/2
Could you solve combination of (n;i)?
That reminds me of this game I downloaded. It's called Pythagorea.
arctan(1/2) + arctan(1/4) + arctan(1/x) = pi/4
Arg(2+i) + Arg(4+i) + Arg(x+i) = pi/4, such that -pi/2 < Arg(z) < pi/2
Arg((2+i)(4+i)(x+i)) = pi/4, using property of Argument (similar to log properties)
Arg((8+6i-1)(x+i)) = pi/4, expansion
Arg((7+6i)(x+i)) = pi/4, simplification
Arg(7x+7i+6xi-6) = pi/4, expansion
Arg(7x-6 + (6x+7)i) = pi/4, grouping real and imaginary parts
arctan((6x+7)/(7x-6)) = pi/4, converting back into inverse tangent
(6x+7)/(7x-6) = tan(pi/4), taking the tangent of both sides since -pi/2 < pi/4 < pi/2
(6x+7)/(7x-6) = 1, evaluating exact value of tan(pi/4)
6x+7 = 7x-6, cross multiplying
x = 13, moving x to one side and constants to the other.
I've never seen the three square geometry problem adapted to trig and complex numbers. Very nice :) Also (semi spoiler alert ahead),
for the problem at the end, since it uses atan(1/2) and the result of π/4, would the observation that atan(1/4)+atan(1/x)=atan(1/3) be true? Then we can manipulate tan(A-B) = (tanA - tanB) / (1+tanA*tanB) to give a formula for combining added arctangents to a single arctan & get x that way?
EDIT: I solved by simply isolating the atan(1/x) term and taking tangent of both sides to get tan(π/4+atan(1/4)+atan(1/2)) and got the same magnitude, but opposite sign. Did I just do a stupid with arctans on the above method?
(2+i)*(4+i)*(x+i)= 7x-6 + i(6x+7); 7x-6 = 6x+7 x=13
The complex number trick was obvious but that geometry truck was cool ,,,hero😎😎
Impressive! But that ending... haha
3:54 "But we're adults now" xddd
the thumbnail challenged me to do it with complex numbers, so I did that without realising there was a geometric proof of it.
My way was much more complex than yours *badum tish*, because I didn't realise I was going to get a recognisable angle like 45 degrees
I used the complex exponential form of tan to write:
theta_a = (-i/2)*ln((4+3i)/5)
theta_b = (-i/2)*ln((3+4i)/5)
Add those together, simply with log rules, and remember that ln(i)=(pi/2)*i, and you get :
theta_a + theta_b = pi/4 = 45 degrees
13
(53/2)° + (37/2)° = 45°
Arctan[ (0.5+0.333333)/(1-0.5*0.3333333)]=
arctan (1) = π/4
I thought it was a numberphile video. They also have almost the same video where the imo gold medalist lady had solved the geometry problem the same way. But anyhow the complex no. Part was nice
Sir how you can do these type of problems ? I am from india
Answer for 8:00 is 13
This angle has to be what? 一瞬間的一句話 我感覺到了滿滿的台灣味 熟悉的感覺
Solv Int suq arctanx ?
Can you do sqrt(i+sqrt(i+sqrt(i+...)))
At 8:00 is the answer to the problem = 13 ?
3:55 I was a bit nervous then, I thought perhaps we weren't all adults anymore
Instead of using geometry you could do it let = x and then tan on both sides simplify then arctan both sides....I would say that complex number proof was rather simple than thought
IIT-JEE Aspirants: that was easy
Answer to 8:00 is 13
Could have been really easy by considering arctan(tan(arctan(1/2)+arctan(1/3))), and using the formula for tan(a+b), because here tan(arctan(1/2)+arctan(1/3)) is precisely equal to 1, and arctan(1) is pi/4.
What happened to your outro music?
I did Tan(x+y) and showed it is 1 so x+y must be pi/4
*Ruler: Am I a joke to you*
The answer of 8:00 is 13
arctan(1/2) + arctan(1/3) = arctan(1)
Bprp, why don't use the formula:
arctan(x) + arctan(y) = arctan(x+y/1-xy)
This would be the 3rd method
Btw for the problem, is the x = 13?
You can actually derive this formula with his second method
@@maxsch.6555 ok. That's another another perspective for me. What I know is we can derive that formula from regular tan (p+q) formula, by setting p = arctan a ; q = arctan b
i got 13 for the problem @8:00
But,
I simply applied the formula:
arctan(x)+arctan(y)=arctan(x+y/1-xy) and got the answer as π/4
Can't we just use the formula?
arctan(a)+arctan(b)=arctan(a+b/1-ab)
Putting a=1/2 , b=1/3
This gives arctan(1) = π/4
If you’re one of the people who memorised this formula for some reason, sure!
The answer is 13!!
What a coincidence! I just watched Michael Penn’s video, also released today about the same topic : th-cam.com/video/zLEJVW3SeAU/w-d-xo.html
With both of you, I can't forget anymore.
I know. We were both shocked too!
Help me to solve this pls,
a + b + c = 4
a^2 + b^2 + c^2 =66
a^3 + b^3 + c^3 = 280
Sir, what is your name?
tan^-1 A + tan^-1 B = tan^-1 {(A+B)/(1 - AB)}
Put A= 1/2 and B = 1/3, you will readily get π/4
Why do such complicated method???
The value of x is 13