Absolutely love your channel Andy. The concise focus on the solution without a lot of tangents (pun intended), and the How Exciting tag-line gets me every time. Keep up the good work!
Во-первых, он явно знает про эту формулу. Не зря же он решил домножить именно на это число всю дробь. Просто у него в разборах задач он всегда всё делает на максимально доступном уровне, по этому он раскрывал скобки по честному. Во-вторых, у вас опечатка в формуле. P.S. Прошу прощения. Опечатка в формуле появляется только при переводе комментария на русский.
@@zOni413Understood I thought if he told people about that then people would become more efficient in doing problems like that rather than breaking it up which will take up some more time. Generally in entrance tests, people can't afford to lose time in places like that
Hey, Andy I like how you use the 90-60-30 triangle rule but it would be more interesting if you used the sin or cos theta so that one also gets to know how this 30-60-90 came into in the first place.
Just bisect an equilateral triangle with side length 2, and you get two 30-60-90 triangles. Take one of said triangles, and notice that the hypotenuse is 2 and the short side is 1. Use pythagorean theorem to get your 1-sqrt3-2 identity
@Randumbvideosss well if you look at a 30 60 90 triangle and mirror it over the longer cathetus (adjacent) you get a 60 60 60 thus equilateral triangle, thus you know that the shorter cathetus of the original triangle is 1/2 of the hypothenuse and that the longer cathetus of the original triangle is actually the height of the equilateral mirrored one. Using Pythagoras you get hypothenuse c squared equals c/2 squared plus height squared, from which height is (c x sqrt3) /2
3:09 "Now I want to get x by itself" and ask it to think long and hard about how its life is going, and if it really thinks it's going anywhere in the long-run. Gotta have a plan, x.
I got to a different equation for the answer: Instead of rationalising the fractions I left them as X/√3 and 3X/√3. Then the base of the triangle is 1=4X/√3 + 6X Then you can isolate the X to get the final answer: X=1/(4/√3 + 6) Which is a different equation, but still gets you that same 0.1203...
For the left triangles i divide x by tan(60) (which is sqrt3) so that's x/sqrt3. For the right triangle I divide 3x by sqrt3 so that's 3x/sqrt3. Everything inbetween is x+2x+3x=6x. x/sqrt3+3x/sqrt3=4x/sqrt3 (4/sqrt3)x+6x=1 x(4/sqrt3+6)=1 x=1/(4/sqrt3+6)
112k subs I knew you will grow rapidly Even you knew that because your videos are very simple and quick Last time I saw you had 37k subs some 15 days ago
Could possibly save a step by noticing the two triangles and similar, so the sides are in a ratio of 1:3… then rather than all the adding up you would know that the total unknown length is 4/route3
I’d like to know more about how you’re approaching this channel, your content production tools and process, and how it fits into the whole AndyMath empire. But I get that it’s none of my business.
I started with the same triangle on the left (sqrt(3)), but I managed to make it slightly more complicated afterwards. In the end, I got the exact same answer, with the same rationalisation, though.
Слишком лёгкая задача. Вспоминаем определение тангенса угла, вспоминаем чему равен тангенс 60° - всё, считай готово. Осталась только чистая арифметика. (Ну а про сумму углов треугольника и про равенство сторон квадрата мне даже и говорить неловко...)
I wonder if this picture would work with any other quadratically growing number of boxes, or would the edges of some boxes be sticking out of the triangle?
Hello there! Thank you for such an amazing video. Even with my poor english language knowledges i understood everything. Nice problem, nice solution. Thx for your work
I would have gone for the obvious that "??" would be 3 times "?" since they are similar triangles. "?" has the height as 1 and "??" has the height as 3. So ?? would be 3?.
Downrated because I dislike problems where you have to remember factoids to solve them. It would be better to derive what the relationships between the legs of the triangle are.
He was excited for real this time
yuuka pfp, how exciting
Absolutely love your channel Andy. The concise focus on the solution without a lot of tangents (pun intended), and the How Exciting tag-line gets me every time. Keep up the good work!
At 3:36, for the denominator you could just use a²-b²=(a+b)(a-b) so that would instantly give you 4²-36×3
Во-первых, он явно знает про эту формулу. Не зря же он решил домножить именно на это число всю дробь. Просто у него в разборах задач он всегда всё делает на максимально доступном уровне, по этому он раскрывал скобки по честному.
Во-вторых, у вас опечатка в формуле.
P.S. Прошу прощения. Опечатка в формуле появляется только при переводе комментария на русский.
i had that in school today thats funny
@@zOni413Understood I thought if he told people about that then people would become more efficient in doing problems like that rather than breaking it up which will take up some more time. Generally in entrance tests, people can't afford to lose time in places like that
Hey, Andy I like how you use the 90-60-30 triangle rule but it would be more interesting if you used the sin or cos theta so that one also gets to know how this 30-60-90 came into in the first place.
Just bisect an equilateral triangle with side length 2, and you get two 30-60-90 triangles. Take one of said triangles, and notice that the hypotenuse is 2 and the short side is 1. Use pythagorean theorem to get your 1-sqrt3-2 identity
tan theta
@@normalbattlecat8088 exactly. The a*sqrt(3) doesn't derive from trigonometry but from the Pythagorean theorem.
@@z000ey huh how? Would you please elaborate?
@Randumbvideosss well if you look at a 30 60 90 triangle and mirror it over the longer cathetus (adjacent) you get a 60 60 60 thus equilateral triangle, thus you know that the shorter cathetus of the original triangle is 1/2 of the hypothenuse and that the longer cathetus of the original triangle is actually the height of the equilateral mirrored one. Using Pythagoras you get hypothenuse c squared equals c/2 squared plus height squared, from which height is (c x sqrt3) /2
What software do you use for your videos?
I’m also curious about this
Whatever he uses I’m sure it requires a lot of design and planning. What he does is great.
its possible to use powerpoint to replicate his animations
3:09 "Now I want to get x by itself" and ask it to think long and hard about how its life is going, and if it really thinks it's going anywhere in the long-run.
Gotta have a plan, x.
I never want to get my x by itself. I want another responsible adult around just in case things go south.
cant belive i saw this the minute it was uploaded lol. keep making content love the math problems! cheers!!
He's going to quit because of this comment
ok your not funny
Love ur contet plz bring amc or any other Olympiad previous year questions
These questions are very fun to solve ngl
What software are you using on screen when you make the vids?
... I too would like to know what software is being used please.
PowerPoint can do this easily.
@@DandoPorsaco-ho1zs i see. It looked like something he was working with “live” rather than a preset presentation.
I got to a different equation for the answer:
Instead of rationalising the fractions I left them as X/√3 and 3X/√3.
Then the base of the triangle is 1=4X/√3 + 6X
Then you can isolate the X to get the final answer:
X=1/(4/√3 + 6)
Which is a different equation, but still gets you that same 0.1203...
For the left triangles i divide x by tan(60) (which is sqrt3) so that's x/sqrt3.
For the right triangle I divide 3x by sqrt3 so that's 3x/sqrt3.
Everything inbetween is x+2x+3x=6x.
x/sqrt3+3x/sqrt3=4x/sqrt3
(4/sqrt3)x+6x=1
x(4/sqrt3+6)=1
x=1/(4/sqrt3+6)
112k subs
I knew you will grow rapidly
Even you knew that because your videos are very simple and quick
Last time I saw you had 37k subs some 15 days ago
Could possibly save a step by noticing the two triangles and similar, so the sides are in a ratio of 1:3… then rather than all the adding up you would know that the total unknown length is 4/route3
plus 6x too because it also contains the length of 3 squares, 6x + 4x/root3 = 1
I’d like to know more about how you’re approaching this channel, your content production tools and process, and how it fits into the whole AndyMath empire. But I get that it’s none of my business.
*DO YOU MOONLIGHT FOR THE CIA? BECAUSE YOU'RE VERY NOSY!*
He was really excited for discovering something. 😅
I started with the same triangle on the left (sqrt(3)), but I managed to make it slightly more complicated afterwards. In the end, I got the exact same answer, with the same rationalisation, though.
As i did just used tan 60 i e rt 3 to find that small length and got ans very easy
Ngl these remind me of i solved olympiads, pretty fun
He looks like he actually had fun doing math.
Слишком лёгкая задача.
Вспоминаем определение тангенса угла, вспоминаем чему равен тангенс 60° - всё, считай готово. Осталась только чистая арифметика.
(Ну а про сумму углов треугольника и про равенство сторон квадрата мне даже и говорить неловко...)
I wonder if this picture would work with any other quadratically growing number of boxes, or would the edges of some boxes be sticking out of the triangle?
You're very good. Good explanation. Thanks
if it wasn't for the 0 and 1, the solution would be a perfect 0.1234567
3:35 you can just use the difference of two squares
Hello there! Thank you for such an amazing video. Even with my poor english language knowledges i understood everything. Nice problem, nice solution. Thx for your work
How exciting
How exciting 🔥🔥🔥
I love these!
brilliant
How radical
What is the program used on videos?
Did you find out?
@@altogeorge2406 sadly, no.
How exiting!
How exciting!
cool way simplifying of trigonometry.
When you discovered how to zoom in, you missed the perfect opportunity to say, "How exciting!"™😀
#TIL Square root of number can also be called as radical number..
Good question. Thank you buddy your content is amazing
Does Andy still check his Patreon? I am hopelessly fascinated with how he does this stuff and I have to know what goes on in his head.
You could have done 20 things wrong in that video, and i would have had no idea.
I feel like there are much easier and faster ways for this
Cool!👍👍
Good content
Nice 😮
where have you been?
I would have gone for the obvious that "??" would be 3 times "?" since they are similar triangles. "?" has the height as 1 and "??" has the height as 3. So ?? would be 3?.
That's basically already what you do whenever you use one of the standard triangles, honestly.
❤
Yeah, i don't know what is going on.
If that is true, then x = 0.1146
Downrated because I dislike problems where you have to remember factoids to solve them. It would be better to derive what the relationships between the legs of the triangle are.
You only got half the answer, as I always say to my students. You literally forgot units.😁
But there is no unit here
@@lucastoutcourt7991 So the answer is x = (9 - 2√3)/46 _units_ ≈ 0.1203 _units_
Annoying editing. Channel blocked.
Maths is too hard. Channel blocked.
There, I fixed that for you.
Actually, x is 1/10.
How exciting!