1, a, b, and A are in geometric progression, as you can see from the stack of inscribed squares in the diagram. (imagine a square with side length A at the bottom)
@@GraemeMcRae The geometric progression makes for an elegant way to find a, since the height of the large triangle is the infinite sum, a² + a + 1 + 1/a + 1/a² + ... The value of this sum is therefore given by a²/(1 − 1/a) = a³/(a−1). Since the height of the triangle is also given by A√3/2 = a³√3/2, we have, a³/(a-1) = a³√3/2. The a³ cancels on both sides and solving for a gives a = 1 + 2/√3 = (3 + 2√3) / 3.
At about 5:30 you missed the beautiful fact that b =(1+2/rt3)^2, which = a^2. Then you missed that A = (1+2/rt3)^3 which = a^3. By insisting on "rationalising the denominator" and then using FOIL every time you missed some beautiful mathematics. As others have said the process of growing squares multiplies each square by the same factor (1+2/rt3). You could have written them all down in a lovely concise form, and then if you have to rationalise them (but why?) do this at the end! I am enjoying your videos a lot, keep them coming!
Good job ! If you construct a forth square under de base of the greatest triangle, it is easy to show that each square is the image of the previous by the same homothety whose ratio is a. So, their side lenghts form a geometric serie and then b=a² and A=a³(avoids to do three times the same reasoning).
Nice problem, touching on plane geometry concepts and properties of triangles too.... Interestingly, 1, a, b and A form a Geometric Progression (G.P.) with common ratio as [(2+√3)/√3] This also makes b = a² and A = a³, where a = [(2+√3)/√3] It can also be inferred from the fractal like appearance of the squares.... If we continue downward and have the next big equilateral triangle, the square inscribed will have side length of A, and the side of the equilateral triangle would be A*[(2+√3)/√3] Also, as b/2 is the circum-radius and r is the in-radius of equilateral triangle with side c, r = (b/2)*Cos(60) = b/4, and c = 2*(b/2)*Cos(30) = (b√3)/2
At 12:04 I didn't notice him calling the circle a triangle but when I looked at the little text saying '*circle' I genuinely laughed lol. : ) Thanks Professor Penn for such awesome vids!
I know lots of viewers find these geometry problems to be pretty simple but I, nevertheless, enjoy solving them. Even though I find myself making numerous dumb arithmetic errors along the way.
I know why he says those fractions can’t be reduced anymore, but my OCD really wants the left halves of those fractions reduced to integers when possible.
around 13:00 you could be a little more clever and know that the triangle with angles pi/6, pi/2, pi/3 has the relationship that the hypotenuse is twice the short leg. c and r come for free as r = b/4 and r^2 + c^2 = b^2 -> c = sqrt(3) b/2
hi Michael Penn..awesome ..learnt so much math concepts..just curious could we find a sugwara tablet where all values turn out to be integers...just curious..thnq
When he said purple triangle I thought about board games and how a circle becomes the shape of the space. Using all points equal distance from a point (x,y) as the definitely of a circle: Square spaces, all the squares at range 3 form a square, but they are all equal distance from the center so its a circle. Same with hexagon maps but then the circle has 6 sides. Wouldn't be that hard to define coordinates so a triangle is a circle.
These are always fun, but sometimes the method is odd. We know the result for a based upon the unit square, so b is just scaled by a, which we eventually did, and similarly for the final. Always curious why some obvious shortcuts are not utilized instead of algebra.
In my humble opinion, a good place to stop would be after you have told us what the actual answers are. a = 2.15 ; b = 4.64 ; A = 10.0037 ; c = 4.02 ; r = 1.16.
@michaelpennmath What are the height and width of your blackboard? I'm thinking of getting one for my apartement and yours seems to be just the right size
Hi, 0 "and so on and so forth", 11 "go ahead and ...", including 5 "let's go ahead and ...", inluding 1 "let's go ahead and do that", 7 "great", and the usual "and that's a good place to stop".
First squaring on both sides and then you get a sextic equation and polynomial equations >4 are insolvable by Abel's and Ruffini's theorem. The given equation has two positive roots and two negative roots by DESCARTES RULE OF SIGNS.Using Newton-Raphson interpolation method X=1.46104 and X = 1.89111 and four complex roots
Teacher : Those who pause the video please inform. Type down your channel and Category. Tomorrow is geometry test on livestream. Michael : Hi teacher! Good evening (turned off the video) Teacher : 🙄 who is Michael in my class?
Because x is on the adjacent side, cot (pi/3) = adj/opp = x/1. Its just one less step. One could have done tan (pi/3) = 1/x. It is the same calculation.
b = a²; A = a³; r = b/4; c = b√3/2. So much nicer when you see it written this way.
1, a, b, and A are in geometric progression, as you can see from the stack of inscribed squares in the diagram. (imagine a square with side length A at the bottom)
@@GraemeMcRae The geometric progression makes for an elegant way to find a, since the height of the large triangle is the infinite sum, a² + a + 1 + 1/a + 1/a² + ... The value of this sum is therefore given by a²/(1 − 1/a) = a³/(a−1). Since the height of the triangle is also given by A√3/2 = a³√3/2, we have, a³/(a-1) = a³√3/2. The a³ cancels on both sides and solving for a gives a = 1 + 2/√3 = (3 + 2√3) / 3.
@@ib9rt "elegant". But it was where I was going as well. Guess some of us prefer algebra to geometry.
Love these types of geometry videos. Not hard to follow, and relaxing to watch
14:49
Also, we reached ¾ of 100K subs
Good Place To Stop ok great
Going strong
I adore your use of board space. Super clean and organized, and so clear. Will use this spatial skill in my teaching.
In simple terms ( expressing everything in terms of 'a' )-
a=(2/√3+1), b=a^2, A=a^3, c=√3a^2/2, r=a^2/4
Very good question from traditional Japanese maths .
Now all values in terms of a:
a=a (lol)
b=a^2
A=a^3
c= a*(2+sqrt(3))/2
or: c=a^2 * sqrt(3)/2
r=(a^2)/4 , which is really interesting
At about 5:30 you missed the beautiful fact that b =(1+2/rt3)^2, which = a^2. Then you missed that A = (1+2/rt3)^3 which = a^3. By insisting on "rationalising the denominator" and then using FOIL every time you missed some beautiful mathematics. As others have said the process of growing squares multiplies each square by the same factor (1+2/rt3). You could have written them all down in a lovely concise form, and then if you have to rationalise them (but why?) do this at the end! I am enjoying your videos a lot, keep them coming!
Please do videos about elliptical integrals and how to solve elliptic integral problems
+1.
Good job !
If you construct a forth square under de base of the greatest triangle, it is easy to show that each square is the image of the previous by the same homothety whose ratio is a. So, their side lenghts form a geometric serie and then b=a² and A=a³(avoids to do three times the same reasoning).
Nice problem, touching on plane geometry concepts and properties of triangles too....
Interestingly,
1, a, b and A form a Geometric Progression (G.P.) with common ratio as [(2+√3)/√3]
This also makes b = a² and A = a³, where a = [(2+√3)/√3]
It can also be inferred from the fractal like appearance of the squares.... If we continue downward and have the next big equilateral triangle, the square inscribed will have side length of A, and the side of the equilateral triangle would be
A*[(2+√3)/√3]
Also, as b/2 is the circum-radius and r is the in-radius of equilateral triangle with side c,
r = (b/2)*Cos(60) = b/4,
and
c = 2*(b/2)*Cos(30) = (b√3)/2
My solve attempt went surprisingly well. Only got r wrong due to forgetting a square root. Thanks for the video, that was a fun problem!
At 12:04 I didn't notice him calling the circle a triangle but when I looked at the little text saying '*circle' I genuinely laughed lol. : )
Thanks Professor Penn for such awesome vids!
I know lots of viewers find these geometry problems to be pretty simple but I, nevertheless, enjoy solving them. Even though I find myself making numerous dumb arithmetic errors along the way.
I know why he says those fractions can’t be reduced anymore, but my OCD really wants the left halves of those fractions reduced to integers when possible.
Love your work!
It's much simpler if you observe the chain of similar triangles and just square/cube "a" rather than recalculating them all from scratch
Simply beautiful
around 13:00 you could be a little more clever and know that the triangle with angles pi/6, pi/2, pi/3 has the relationship that the hypotenuse is twice the short leg. c and r come for free as r = b/4 and r^2 + c^2 = b^2 -> c = sqrt(3) b/2
Good Video! The last part however could easily be drawn out with sin(pi/6) giving the relation r = (b/2)*sin(pi/6) = b/4.
Please do IMO 2020.
It might have been slightly easier to calculate r from b since sin(pi/6) = r/(b/2) -> 1/2 = 2r/b -> 2r = b/2 -> r = b/4.
For 'r'.. couldn't you have just used 'b/4' since 'r' is was the opposite side to the pi/6 angle... and is half the 'b/2' hypotenuse?
Yes, he could have.
Me : Saw the thumnail
*illuminati confirmed
hi Michael Penn..awesome ..learnt so much math concepts..just curious could we find a sugwara tablet where all values turn out to be integers...just curious..thnq
When he said purple triangle I thought about board games and how a circle becomes the shape of the space.
Using all points equal distance from a point (x,y) as the definitely of a circle:
Square spaces, all the squares at range 3 form a square, but they are all equal distance from the center so its a circle.
Same with hexagon maps but then the circle has 6 sides. Wouldn't be that hard to define coordinates so a triangle is a circle.
These are always fun, but sometimes the method is odd. We know the result for a based upon the unit square, so b is just scaled by a, which we eventually did, and similarly for the final. Always curious why some obvious shortcuts are not utilized instead of algebra.
In my humble opinion, a good place to stop would be after you have told us what the actual answers are. a = 2.15 ; b = 4.64 ; A = 10.0037 ; c = 4.02 ; r = 1.16.
@michaelpennmath What are the height and width of your blackboard? I'm thinking of getting one for my apartement and yours seems to be just the right size
damn i found all my value by using the quadratic formula and pythagorean theorem and it took me aproximatively half an hour to do it
The topmost square is obviouslt not to scale. :-) Which I think is good.
Sir, please record video on ancient Indian calculus and geometry, they seem to be very fascinating
So b = 4r, seems there should be some way to get r from b?
b=(2+root3)^2/3
CHINA TST please
If you had a series of squares, the length of a square would be the geometric mean of its neighboring squares' lengths.
They form a geometric series. Pretty cool and not obvious to me
Hi,
0 "and so on and so forth",
11 "go ahead and ...", including 5 "let's go ahead and ...", inluding 1 "let's go ahead and do that",
7 "great",
and the usual "and that's a good place to stop".
awesome 👍😊👍👏👏 👏
Is there ever a bad place to stop, and you stopped there?
Plz do this question.
√(5-x)=5-x³
Solve for x.
First squaring on both sides and then you get a sextic equation and polynomial equations >4 are insolvable by Abel's and Ruffini's theorem. The given equation has two positive roots and two negative roots by DESCARTES RULE OF SIGNS.Using Newton-Raphson interpolation method X=1.46104 and X = 1.89111 and four complex roots
I think blackpenredpen done a video on that
@@bebos3001 no he's not.
r=(2+root3)^2/12
Oda of root 3
Teacher : Those who pause the video please inform. Type down your channel and Category. Tomorrow is geometry test on livestream.
Michael : Hi teacher! Good evening (turned off the video)
Teacher : 🙄 who is Michael in my class?
Good
Great dear 👍
And if we had n squares and have to calculate A? :)
Now imagine not knowing trig and attempting this problem :D
The only trig used was sin π/6 = √3/3, which is easy enough to replace with isoceles triangle theorem, pythagorean theorem, and similar triangles.
Done
I think examiner ll just ask the value of r only.
So r is 1/4 of b. Coincidence?
2:41 why do u use cot? surely it should just be tan?
Because x is on the adjacent side, cot (pi/3) = adj/opp = x/1. Its just one less step. One could have done tan (pi/3) = 1/x. It is the same calculation.
Not first but quick
So how about a pin instead of heart for my speed
you have to try harder next time
Hi. Sorry, that was true.
10:29 why are you always doing this the hard way in your videos? Use the fact that the centre is the medicentre and so the height is 3b/4.
(Just for fun)
Critics be like:
That's illuminati symbol
No. of likes rn: 666
AAAAAAAAAhhhhhhh!
Nice vídeo as always, but risking sounding pedantic, pronouncing the su in Sugawara like the su in sugar made me cringe hard
Why do you keep saying "zee" instead of "z"?
US English vs British English
That really wasn’t difficult at all. Just a lot of work lol
FIRST!!!
First! NOTIFICATION SQUAD!
My name is also arnav
Congratulations 🎊
@@ashimdey1004 we can see a Arnav just had a birth
@@ashimdey1004 No, ONLY I AM ARNAV
This thread had enormous potential, please continue it
Sabh kuch bhi bolo, binod to ek hi hai
Where are the olympiad problems?? This one was only calculation, no fun at all.