Surprisingly this is one of the Golden Ratio triangles with n² = Φ = (√5 + 1)/2 as the square of two of the six roots. Proof: n⁶ - n⁴ - n² = 0 or n⁴ - n² - 1 = 0, which has the solution n² = (√5 + 1)/2
@@aidan2453 In the simplest way, dividing a section into two parts so that the ratio of the length of the longer one to the shorter one is the same as the ratio of the entire section to the longer part. But the get the actual mathematical terms you would need to google it on your own
@@aidan2453 Here is a riddle. Think of a positive number Φ. Subtract 1 from that number Φ-1 and invert it 1/(Φ-1) and you get back Φ=1/(Φ-1). What is that positive number? Rearrange this equation to Φ^2-Φ-1=0 and solve for Φ=(1+√5)/2 and there you have the famous golden ratio that whole books have been written about. Here is another riddle. Think of a positive number φ. Add 1 to that number φ+1 and invert it 1/(φ+1) and you get back φ=1/(φ+1). What is that positive number? Rearrange this equation to φ^2+φ-1=0 and solve for φ=(-1+√5)/2 and there you have the famous inverse golden ratio, φ=1/Φ=Φ-1. Just a bit of food for thought. sin18°=φ/2=1/(2Φ) and sin54°=Φ/2=1/(2φ) sin²18°+cos²54°=[(Φ-1)^2+Φ^2]/4=(2Φ^2-2Φ+1)/4=(2Φ+2-2Φ+1)/4=3/4
@@aidan2453it sneaks into so many things. You can generate it with the Fibonaccie sequence, which comes up in nature for example in human body proportions or how many rows are on a pineapple. A standard sized sheet of paper is based on the golden ratio put 2 sheets together side by side and rotate it 90⁰ and you have the same shape of twice the area. There is more interesting properties you can find by expanding powers of the Golden ration, but some of the simplest forms are φ2=φ+1 or 1/φ=φ-1 or (φ+1)(φ-1)=φ (this is all the equation rearranged, there's many more interning ones you can find at higher powers). Solve the quadratic x²-x-1=0 to get roots of φ and φ Look up Fibonacci spiral to see how the sequence generates the golden ratio.
Divide each side by n. This is allowed as it is congruent to the given triangle. So the side lengths in the new triangle are now 1, n and n². Now by the Pythagoras theorem (n²)² = n² + 1². So n⁴ - n² - 1 = 0 Let y = n², so y² - y -1 = 0. So, y = (1 ± √5)/2 = n², But n² > 0, so, n²= (1 + √5)/2 so, n = √((1 + √5)/2)
or formulate the pythagoras of the original triangle ( n^2+(n^2)^2=(n^2)^2 and then divide both sides by n^2 to obtain your 1+n=n^2 pythagoras relation of the 1/n-scaled down similar (not congruent!) triangle
Before watching: Pythagorean theorem tells us a^2+b^2=c^2. Then we have n^2 + n^4=n^6, since (a^m)^n = a^(mn). We group all the terms on one side and get n^6-n^4-n^2 = 0 Presuming n is not 0 (because if so then we're literally measuring nothing) we can divide by n^2 to get n^4-n^2-1= 0. Substituting in X = n^2, we write this as x^2-x-1= 0. We then use the quadratic formula to arrive at x= (1 +- sqrt 5)/2, where +- means '+ or -' because I can't use altcode on my phone. But this is X. Since x = n^2, we square root to find n. We can then disregard the 1- sqrt5 solution, as that number is negative and would yield a complex n. Instead we get n = +- sqrt ((1+sqrt5)/2). If we presume our length must be positive and real, we then have only n = sqrt((1+sqrt5)/2) Of the remaining 5 roots to n^6-n^4‐n^2=0; two are 0 (recall we accounted for that before dividing by n^2 earlier), two are complex (+- sqrt ((1-sqrt5)/2), and one is negative (-sqrt ((1+sqrt5)/2)). Thus, the solution listed above, sqrt ((1-sqrt 5)/2) is the only solution that gives us a positive real length.
In the initial "rejected" result, the issue is NOT that it would be negative length, when you evaluate it, as it is "n^2 = ...", you would have to square root the value to find 'n', so it is that you would have to take the square root of a negative number, making it imaginary, you would have a length with an imaginary number, so again, it isn't that you would have a negative length, it is that you would have an imaginary length.
Since n^2 is also a length in the triangle. you can reject it in this case, on the ground that the n^2 is negative and therefor one of the side would have to be negative.
Not to mention that the side n^2 is longer than side n but the diagram has it smaller. This made me question how n^2 could possibly be smaller than n. Which makes n^3 even smaller.
To answer the question, n^2 is smaller than n if n is in range 0 < n < 1. And as you pointed out, n^3 would be even smaller. With that and the fact, that n^3 must be greater than n, since it is the hypotenuse of the triangle we learn two things. First, n must be greater than 1, second, don't count on drawings.
Or you can use the property of the trigonometric function. let n²=k. sin²ø+cos²ø=1, so k/k³+k²/k³=1, 1/k²+1/k=1; 1+k=k²; k²-k-1=0 (k>0) you made the quadratic equation. so k equals to (1±sqrt(5))/2. n>0, so k=(1+sqrt(5))/2. Finally you can square root the value of k. (because k=n².) Thank you.
@@tharunsankar4926 Using the one of the properties of trigonometric functions is to derive quadratic equations, and this process makes the answers more accessible. Also, I wanted to show that solving this problem is not limited to one way. Why are you mad?
Second rejection 8:11 is not correct. As the answer can be positive also. I know this possibility was not neglected because the same value for n had appeared in the previous step. But mathematically it was not expressed rightly, I think.
By visual inspection of the thumbnail (as of writing) there is no positive real solution (not sure about complex solutions) as it mandates that n^3 > n > n^2
By Pythagoras theorem n²+n⁴=n⁶ n²+n²n²=n⁶ n²(1+n²)=n⁶ 1+n²=n⁴ n⁴-n²+1=0 Let n²=x, so we get x²-x+1=0 By quadratic formula x=(1±√5)/2 Meaning n²=(1±√5)/2 n=√phi
@@larswilms8275 Of course. But this level of diagram didn’t exaggerate it is misleading. Of course n greater than 1 then n^2 is greater than n. But as a potentially tricky problem this diagram is not just ‘not to scale’.
Assuming it's valid to say the square root of zero is zero (I can't remember), then N can be 0. I doubt it can be any other exact value. If the result isn't exact, then you get an F. LOL
@@DenWanDerEr в школе учат вычислять отдельно, тк проще для восприятия. Лично меня приучили так и воспринимать в одно выражение мне сложно. Но тут уже как кому удобнее
@@TheMapofMathematics We must be genious... But it seems you are not ! That's such a big mistake, could you suggest a good resolution for this equation ? Thanks !
Show me your right angled triangle with equal sides of length 1 unit? The hypotenuse of a right angled isosceles triangle with side 1 is the square root of 2. (a^2 + b^2 = c ^2 --> 1^2 + 1^2 = 2 --> c = sqrt(2))
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
Surprisingly this is one of the Golden Ratio triangles with n² = Φ = (√5 + 1)/2 as the square of two of the six roots.
Proof: n⁶ - n⁴ - n² = 0 or n⁴ - n² - 1 = 0, which has the solution n² = (√5 + 1)/2
What is the golden ratio? If you have time to explain.
@@aidan2453
In the simplest way, dividing a section into two parts so that the ratio of the length of the longer one to the shorter one is the same as the ratio of the entire section to the longer part. But the get the actual mathematical terms you would need to google it on your own
@@zipprime6956 thank you!!
@@aidan2453 Here is a riddle. Think of a positive number Φ. Subtract 1 from that number Φ-1 and invert it 1/(Φ-1) and you get back Φ=1/(Φ-1). What is that positive number? Rearrange this equation to Φ^2-Φ-1=0 and solve for Φ=(1+√5)/2 and there you have the famous golden ratio that whole books have been written about.
Here is another riddle. Think of a positive number φ. Add 1 to that number φ+1 and invert it 1/(φ+1) and you get back φ=1/(φ+1). What is that positive number? Rearrange this equation to φ^2+φ-1=0 and solve for φ=(-1+√5)/2 and there you have the famous inverse golden ratio, φ=1/Φ=Φ-1.
Just a bit of food for thought. sin18°=φ/2=1/(2Φ) and sin54°=Φ/2=1/(2φ)
sin²18°+cos²54°=[(Φ-1)^2+Φ^2]/4=(2Φ^2-2Φ+1)/4=(2Φ+2-2Φ+1)/4=3/4
@@aidan2453it sneaks into so many things. You can generate it with the Fibonaccie sequence, which comes up in nature for example in human body proportions or how many rows are on a pineapple. A standard sized sheet of paper is based on the golden ratio put 2 sheets together side by side and rotate it 90⁰ and you have the same shape of twice the area.
There is more interesting properties you can find by expanding powers of the Golden ration, but some of the simplest forms are φ2=φ+1 or
1/φ=φ-1 or
(φ+1)(φ-1)=φ
(this is all the equation rearranged, there's many more interning ones you can find at higher powers). Solve the quadratic x²-x-1=0 to get roots of
φ and φ
Look up Fibonacci spiral to see how the sequence generates the golden ratio.
Divide each side by n. This is allowed as it is congruent to the given triangle.
So the side lengths in the new triangle are now 1, n and n².
Now by the Pythagoras theorem
(n²)² = n² + 1².
So n⁴ - n² - 1 = 0
Let y = n², so y² - y -1 = 0.
So, y = (1 ± √5)/2 = n²,
But n² > 0, so, n²= (1 + √5)/2
so, n = √((1 + √5)/2)
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so the triangle is Φ:Φ²:Φ³?
@@penguincute3564 actually the square roots
of each of the three Phi powers
or formulate the pythagoras of the original triangle ( n^2+(n^2)^2=(n^2)^2 and then divide both sides by n^2 to obtain your 1+n=n^2 pythagoras relation of the 1/n-scaled down similar (not congruent!) triangle
It took me a while to realize that 'scared' really meant 'squared' 😂
Maths are scaring.
This doesn’t appear to be an olympiad problem. This is just a class IX problem
Olimpiyat seviyeye değil sana katılıyorum dostum@@ramanma9915
@@pierreduchesne0001 no
Before watching:
Pythagorean theorem tells us a^2+b^2=c^2. Then we have n^2 + n^4=n^6, since (a^m)^n = a^(mn). We group all the terms on one side and get n^6-n^4-n^2 = 0
Presuming n is not 0 (because if so then we're literally measuring nothing) we can divide by n^2 to get n^4-n^2-1= 0.
Substituting in X = n^2, we write this as x^2-x-1= 0. We then use the quadratic formula to arrive at x= (1 +- sqrt 5)/2, where +- means '+ or -' because I can't use altcode on my phone.
But this is X. Since x = n^2, we square root to find n. We can then disregard the 1- sqrt5 solution, as that number is negative and would yield a complex n. Instead we get n = +- sqrt ((1+sqrt5)/2). If we presume our length must be positive and real, we then have only n = sqrt((1+sqrt5)/2)
Of the remaining 5 roots to n^6-n^4‐n^2=0; two are 0 (recall we accounted for that before dividing by n^2 earlier), two are complex (+- sqrt ((1-sqrt5)/2), and one is negative (-sqrt ((1+sqrt5)/2)). Thus, the solution listed above, sqrt ((1-sqrt 5)/2) is the only solution that gives us a positive real length.
In the initial "rejected" result, the issue is NOT that it would be negative length, when you evaluate it, as it is "n^2 = ...", you would have to square root the value to find 'n', so it is that you would have to take the square root of a negative number, making it imaginary, you would have a length with an imaginary number, so again, it isn't that you would have a negative length, it is that you would have an imaginary length.
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Since n^2 is also a length in the triangle. you can reject it in this case, on the ground that the n^2 is negative and therefor one of the side would have to be negative.
Very good.
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Good job!
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Tres bien
Beautiful question!
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Not to mention that the side n^2 is longer than side n but the diagram has it smaller. This made me question how n^2 could possibly be smaller than n. Which makes n^3 even smaller.
To answer the question, n^2 is smaller than n if n is in range 0 < n < 1. And as you pointed out, n^3 would be even smaller. With that and the fact, that n^3 must be greater than n, since it is the hypotenuse of the triangle we learn two things. First, n must be greater than 1, second, don't count on drawings.
Reminder: Diagrams and drawings are not to scale.😁
Or you can use the property of the trigonometric function.
let n²=k. sin²ø+cos²ø=1, so k/k³+k²/k³=1,
1/k²+1/k=1; 1+k=k²; k²-k-1=0 (k>0)
you made the quadratic equation.
so k equals to (1±sqrt(5))/2.
n>0, so k=(1+sqrt(5))/2. Finally you can square root the value of k. (because k=n².) Thank you.
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Wtf use the Pythagorean theorem, same thing as what ever you wrote, what’s wrong with you????
@@tharunsankar4926
Using the one of the properties of trigonometric functions is to derive quadratic equations, and this process makes the answers more accessible. Also, I wanted to show that solving this problem is not limited to one way. Why are you mad?
@_Heing_ yeah I agree that the guy you're replying to shouldn't be so mad.
divide both sides by n and complete square
shrink each of the sides by n times. we get a right triangle of 1, n, n^2, thus n^4=n^2+1, solve this we can get the answer.
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Second rejection 8:11 is not correct. As the answer can be positive also. I know this possibility was not neglected because the same value for n had appeared in the previous step. But mathematically it was not expressed rightly, I think.
Thanks
Truly a golden traingle
1.272….. would have been nice for you to plug the math back in and confirm the original equation in camera once you found the solution. Great video
n=1.26 about n⁴ - n² - 1 = 0 fourth degree equation attributable to second degree
n has result decimal,could it be in practical the side length of a polygon triangle.
By visual inspection of the thumbnail (as of writing) there is no positive real solution (not sure about complex solutions) as it mandates that n^3 > n > n^2
Корень из золотого сечения, устно за прд минуты.
🎉
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help please.
if ab+bc+ca=1
prove:
sqrt( a + (1/a) ) + sqrt( b + (1/b) ) + sqrt( c + (1/c) ) >= 2( sqrt(a) +sqrt(b) + sqrt(c) )
By Pythagoras theorem
n²+n⁴=n⁶
n²+n²n²=n⁶
n²(1+n²)=n⁶
1+n²=n⁴
n⁴-n²+1=0
Let n²=x, so we get
x²-x+1=0
By quadratic formula
x=(1±√5)/2
Meaning
n²=(1±√5)/2
n=√phi
Interesting geometry! The answer works out to be the square root of the Golden Ratio
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The drawing isn't correct. It looks like side n^2 is smaller that n and side n^3 is larger than n and n^2
that's also what i saw. n^3>n^2 thus, n>1, thus, n^2>n
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Yeah. That made me initially consider n close to but less than 1.
drawings and diagrams are not to scale.
@@larswilms8275
Of course. But this level of diagram didn’t exaggerate it is misleading.
Of course n greater than 1 then n^2 is greater than n. But as a potentially tricky problem this diagram is not just ‘not to scale’.
Задача для восьмикласника....
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Don’t be scare homie, Nate Diaz
n=0 is a legitimate solution. Not an interesting one, but legitimate.
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Nope. n is the length of this triangle’s side. So n > 0.
That makes y (n^2) also > 0.
So we have only one valid root - 1,272
@@timotim8722no one said there can't be a triangle with sides equal to 0 tho
@@farqilion8747 A dot?
@@timotim8722 technically yes.
Just like circle is a regular polyhedron with infinite amount of sides, why not?
Gotta love how side length n is shown as bigger than n2...
Bu videoları çok seviyorum
Зачётно. Класс
Hmmmm....
if n is not n∈Z can Langth be complex number?
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Raíz cuadrada de fi
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N²+n⁴-n⁶=0
N²=t
T+t²-t³=0
T(1+t-t²)=0
T=0 equivalant to n²=0 thus n = 0 or n=0 first 2 solutions
Or
1+t-t²=0
T= (1±√5)/2
N=±√(1±√5)/2
4 more solutions
Довольно простой пример для 11 класса я считаю
Why cant be n=1?????
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@@TheMapofMathematics thank you ❤️❤️ love you dude
Isn't that the square root of the golden ratio phi?
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n=1
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Why just say rejected, its called an extraneous solution. A non real solution.
It’s rejected by the assumption that n, it’s square, represents a length, which is necessary a real solution
But thats not how you do maths or come to conclusions via mathemtical solutions. Its wrong terminology ontop of that lol. @@S8EdgyVA
Прикольненько) ❤
n=1/2(1± root 5)
Now, I'm scared!
n=sqrt( (1+sqrt(5))/2)
Assuming it's valid to say the square root of zero is zero (I can't remember), then N can be 0. I doubt it can be any other exact value. If the result isn't exact, then you get an F. LOL
Squfre root of phi.
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Is question me good tha kya
This was very easy
Usually "n" is for an integer
So this problem has no interest
Примерно 1,27. Если точно, то V((1+V5)/2)
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Можно заметить, что иностранцы не вычисляют дискриминант отдельно.
@@ivan_577можно же в 1 действие всё сделать. Зачем отдельно его вычислять?
@@DenWanDerEr в школе учат вычислять отдельно, тк проще для восприятия. Лично меня приучили так и воспринимать в одно выражение мне сложно. Но тут уже как кому удобнее
Patagoen tiuram=Pythagoras theorem
Scared=Squared
Qadretic fun=Quadratic function
Ecvayan=Equation
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@@TheMapofMathematics Oh thank you but why did you say that
I'm doing this solution in my head so why can't the solution be imaginary?
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square root of the golden ratio
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@@TheMapofMathematics Thank you 😘
n=( ( 1+5^0,5)/2)^0,5
why n is scared ? n scared of who ?
Doesn't look like an Olympiad peoblem. It's way straightforward for olympiad level problems.
n=1.272
Its a pity the answer has to be obscured by three random images.
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9:18 the problem is very interesting but the presentation is very dirty: bad handwriting, filthy board ....;
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1.272
Bro solving with sketches and taking 9:34 minutes
Meanwhile JEE adv aspirants solving that in mind laughing in the corner😂
Ответ3
c est faux des le debut , au lieux de n puissance 4 vous devais ecrire n a la puissance 3 dan l eq du second degres
Birim çember de bu islem yay uzunlugu hipotenus 1 + 1 =2 birimlik yeyi gorur bu ac dik acib90 derece
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Why is n so scared?
because o peed q.
This is not a math olympiad problem at all
Moi je vois une erreur on devrait normalement avoir n²(n³-n²-1)=0 et non n²(n⁴-n²-1)=0
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Tout à fait d'accord ! Y'a une énorme erreur dans son calcul... 😂 😂 😂
@@nlbdp oui et cela a faussé tout ce qu'il ou elle a fait
@@TheMapofMathematics We must be genious... But it seems you are not ! That's such a big mistake, could you suggest a good resolution for this equation ? Thanks !
n=0 within 2 seconds 😂
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Pythagore
Hey, when you factorized n^6 you said it is n^2 times n^4 but this is n^8
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You do not multiply powers, powers are added 2^2(2^4)=4×16=64 not 2^8=128.
Exactly. Any answer he gets after this mistake will be wrong
Lmao are you roasting him@@TheMapofMathematics
exponents are added when multiplying terms, and multiplied when a term is raised to another exponent
Never mind WHAT n is - let's address your spelling. The word is MATHS - it has an s at the end. Please GET IT RIGHT.
Yes I Noted my mistake.
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In USA, it is Math
Lots of substitutions and rejections which does not look mathematically correct.
Just swop n=1, and the answer will be 1
👍
Show me your right angled triangle with equal sides of length 1 unit?
The hypotenuse of a right angled isosceles triangle with side 1 is the square root of 2. (a^2 + b^2 = c ^2 --> 1^2 + 1^2 = 2 --> c = sqrt(2))
@@larswilms8275 there you go. See you solved it with 1 🥸
I like how he says "scared" instead of "squared".
Wrong!
Oui ce n'est pas logique
Amaçsız anlamsız saçma sapan işlem dolu.olimpiyat seviyesinde bir soru değil
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bro you re so slow
silly
Primitiv
🤮💩👎
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