Math prove marathon

Fast review:

In this note I want to start a prove chain. This means if you want to continue this chain you should prove something, but you must use the result of the previous approach. For example if the first person proved that ab=2Aa\cdot b=2A in a right triangle and you want to prove that the area of a square is a2a^2, then this is a good approach:

If we cut the square into two right triangles: As=2At;ab=2At;a=bAs=2aa2=a2A_s=2\cdot A_t;a\cdot b=2A_t;a=b\Rightarrow A_s=2\cfrac{a\cdot a}{2}=a^2

This can be just a fun note. If you are boring or you want to rest for a few minute, then this is a chance to prove something what you want(and you can). You can prove basic, easy or hard things.

Rules:

  • If you want to share something, what isn't continue this chain, then use the reply button under one of the approaches
  • Don't share messages to your friends
  • Don't prove something again in this note
  • If your approach is wrong, then don't delete your comment. Mention somebody to help you or solve alone
  • To avoid two approaches at the same time you should reply for the previous approach(use the given text) before you continue this chain
  • Don't prove evil things what others can't continue

Temporary rule(s)\color{#eeaa00}\text{Temporary rule(s)}

  • The first ten approaches should not create a chain\color{#eeaa00}\text{The first ten approaches should not create a chain}
  • You can share not original, but not famous approches\color{#eeaa00}\text{You can share not original, but not famous approches}

Current theme(s)\color{#D61F06}\text{Current theme(s)}

  • Number theory\color{#D61F06}\text{Number theory}
  • Algebra\color{#D61F06}\text{Algebra}

Event(s)\color{#3D99F6}\text{Event(s)}

  • There has been no events yet(after 10 active members will be the first event)\color{#3D99F6}\text{There has been no events yet(after 10 active members will be the first event)}

I will change the colored sections sometimes. If you don't want to skip them, then you can subscribe above the comments or if you write an approach, then you will get notifications too. When I change the temporary rule(s) I will notificate you.

You can also write proofs that do not belong to the given theme(s)\color{#D61F06}\text{You can also write proofs that do not belong to the given theme(s)}

You can:

  • Use things that have already been proven before you(you should name which one is that)
  • Mention people to join
  • Mention somebody to help you if your approach is wrong
  • Mention somebody to help you continue this chain if you wrote an approach, but you are bad in LaTeX\LaTeX
  • You can prove evil things, but then you must continue the chain and prove something easier to help others
  • If this is possible you can split up your approach into two approaches

How to:

Mention somebody:

@<name>
Example:
@Páll Márton

Write an approach:

  1. Reply for the last approach: I want to continue this chain. If somebody already working on this you should wait.
  2. Write your approach. Use this form below:
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# ID: <serial number of this approach>
## <write a short review about your approach or write the formula>
## <serial number of used formulas>
<write here your approach>

> <P.S. or notes>

When you write an approach don't use the reply button. Just write as a comment.

Use pictures:

The code of your picture will be: ![<text under the picture>](<link to the picture>){: .<position>}

  1. Profile->+Note(in the bottom)->Upload an image
  2. Copy the code of the picture
  3. Paste/Insert your code
  • If your picture is uploaded to the internet, then you should just paste the link.
  • The [] section can be empty
  • The <position> can be center,left,right.

Use LaTeX\LaTeX:

Without spaces:

  • \ ( <LaTeX code> \ ) - left aligned
  • \ [ <LaTeX code> \ ] - center aligned

If you are bad in LaTeX\LaTeX, then you can use online editors or web pages to learn.

Use other formatting:

Here you can see how to use links, use bold and italistic styles, use quotes etc.

The code of the first comment is(without spaces between \ and ( or ) ):

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# ID: 1
## Graphically prove that \ (a^2-b^2=(a-b)(a+b)\ )
## I won't use other formulas from this note
Assume that \ (a>b\ ). In this picture below \ (T_?\ ) means territory/area:
![](https://ds055uzetaobb.cloudfront.net/uploads/ZmvJbP3yh4-0.PNG){: .center}
In this picture  \ (T_1+T_3+T_4=a^2\ ) and \ (T_4=b^2\implies T_1+T_3=a^2-b^2\ ). We know \ (T_1+T_2=(a+b)(a-b)\ ) from the side length of the rectangle. And we know \ (T_2=T_3\ ), because both of them is equal to \ (b(a-b)\ ). Therefore \ (T_1+T_3=T_1+T_2=(a+b)(a-b)\ ). So \ (a^2-b^2=(a-b)(a+b)\ ).

> P.S.: If \ (a=b\ ) then \ (a^2-b^2=(a-b)(a+b)=0\ ). And if \ (a<b\ ) then, then we can multiply by -1. After that we get this: \ (b^2-a^2=(b-a)(a+b)\ ). If we swap a and b in the picture we get the same result.

Good luck!

Active members:

  • If you post an approach, I will give you 5 points, except if your solution is wrong, not complete or confusing.
  • If you post an approach about the current theme, then you can get 3 extra points.
  • If you write something interesting about the approach (for example ID-3 by David Vreken), then you can get 1 extra points.
NamePosted approachesPoints
David Vreken16
Páll Márton5
Vinayak Srivastava15

If we reach the 10 active members, then I will mark the 3 most active people. And I will mark the winner of the last event too.

Note by Páll Márton (no activity)
4 months, 2 weeks ago

No vote yet
1 vote

  Easy Math Editor

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ID: 1

Graphically prove that a2b2=(ab)(a+b)a^2-b^2=(a-b)(a+b)

I won't use other formulas from this note

Assume that a>ba>b. In this picture below T?T_? means territory/area: In this picture T1+T3+T4=a2T_1+T_3+T_4=a^2 and T4=b2    T1+T3=a2b2T_4=b^2\implies T_1+T_3=a^2-b^2. We know T1+T2=(a+b)(ab)T_1+T_2=(a+b)(a-b) from the side length of the rectangle. And we know T2=T3T_2=T_3, because both of them is equal to b(ab)b(a-b). Therefore T1+T3=T1+T2=(a+b)(ab)T_1+T_3=T_1+T_2=(a+b)(a-b). So a2b2=(ab)(a+b)a^2-b^2=(a-b)(a+b).

P.S.: If a=ba=b then a2b2=(ab)(a+b)=0a^2-b^2=(a-b)(a+b)=0. And if a<ba<b then, then we can multiply by -1. After that we get this: b2a2=(ba)(a+b)b^2-a^2=(b-a)(a+b). If we swap a and b in the picture we get the same result.

Páll Márton (no activity) - 4 months, 2 weeks ago

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@Vinayak Srivastava What about this note?

Páll Márton (no activity) - 4 months, 2 weeks ago

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Good, I will try to find something good! Nice note!

Vinayak Srivastava - 4 months, 2 weeks ago

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@Vinayak Srivastava Thank you!

Páll Márton (no activity) - 4 months, 2 weeks ago

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@Páll Márton (no activity) I have proof of something else, can I write it in a separate comment?

Vinayak Srivastava - 4 months, 1 week ago

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@Vinayak Srivastava Yeah. I will see that :)

Páll Márton (no activity) - 4 months, 1 week ago

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@Páll Márton (no activity) Done!

Vinayak Srivastava - 4 months, 1 week ago

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ID:2

Proof of sin2θ+cos2θ=1{\sin}^2{\theta}+{\cos}^2{\theta} =1

I won't use other formulas from this note


We are usually taught that the proof of the identity sin2θ+cos2θ=1{\sin}^2{\theta}+{\cos}^2{\theta} =1 is this:

In any right triangle, O2+A2=H2 (Where H is the hypotenuse, O is opposite, and A is adjacent.)O^2+A^2=H^2 \text{ (Where H is the hypotenuse, O is opposite, and A is adjacent.)}     O2H2+A2H2=1\implies \dfrac{O^2}{H^2}+ \dfrac{A^2}{H^2} =1     sin2θ+cos2θ=1\implies {\sin}^2{\theta}+{\cos}^2{\theta} =1

However this proof is not complete! What about an angle θ>90{\theta} >90^{\circ}? It won't form any right triangle!


Here is where our unit circle helps:

In our unit circle, for any angle 360θ0,360^{\circ}\geq {\theta}\geq 0^{\circ}, we have the particular point on its circumference representing cosθ,sinθ\cos{\theta},\sin{\theta}, as its co-ordinates, I have shown in my diagram:

But we know that any point on a unit circle satisfies the equation x2+y2=1x^2+y^2=1

Hence, for any angle 360θ0,360^{\circ}\geq {\theta}\geq 0^{\circ}, we get sin2θ+cos2θ=1\Huge{{\sin}^2{\theta}+{\cos}^2{\theta} =1 \blacksquare}

Vinayak Srivastava - 4 months, 1 week ago

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Or: (sin(a))2+(cos(a))2=a2+b2c2=1(sin(a))^2+(cos(a))^2=\cfrac{a^2+b^2}{c^2}=1, where a2+b2=c2a^2+b^2=c^2 :)

Páll Márton (no activity) - 4 months, 1 week ago

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This is the common notation there

Páll Márton (no activity) - 4 months, 1 week ago

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@Páll Márton (no activity) Yes, but I wanted to show that sin=o/h and cos=a/h.

Vinayak Srivastava - 4 months, 1 week ago

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@Páll Márton (no activity) Please invite more people, it will be fun to prove many things!

Vinayak Srivastava - 4 months, 1 week ago

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@Vinayak Srivastava Can you help me in it?

Páll Márton (no activity) - 4 months, 1 week ago

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@Páll Márton (no activity) I have some work of school. Sorry! But, mention few people at a time, it will work.

Vinayak Srivastava - 4 months, 1 week ago

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@Vinayak Srivastava Yeah. I have to read 10+ books, but I hate reading :) I will try mention people

Páll Márton (no activity) - 4 months, 1 week ago

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@Páll Márton (no activity) Same with me! :) But I have to write also! :(

Vinayak Srivastava - 4 months, 1 week ago

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@Vinayak Srivastava Write a book? :)

Páll Márton (no activity) - 4 months, 1 week ago

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@Páll Márton (no activity) No! Write answers(at least 20 in one chapter)!

Vinayak Srivastava - 4 months, 1 week ago

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@Vinayak Srivastava I should write a journal about all of them

Páll Márton (no activity) - 4 months, 1 week ago

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@Páll Márton (no activity) LOL 18 books :) I won't read them. I want to learn english :)

Páll Márton (no activity) - 4 months, 1 week ago

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Now I can't post any approaches with pictures. Can you post an approach about sin(90α)=cos(α)\sin(90^{\circ}-\alpha)=\cos(\alpha)? Same thing with cos and tangent, and with 180. I think you can make amazing pictures with unit circles :) I will use them in my approach, but I think we can't skip them.

Páll Márton (no activity) - 4 months, 1 week ago

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Actually, I don't know this formula! I'll search it, and if I understand, I will try.

Vinayak Srivastava - 4 months, 1 week ago

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ID: 3

James A. Garfield's Proof of the Pythagorean Theorem (reproduced from here)

I won't use other formulas from this note

Start with a right triangle with legs aa and bb and hypotenuse cc. Extend leg aa by bb units and construct a duplicate right triangle along this extension. Upper leg aa is parallel to the original leg bb since, in a plane, if a line is perpendicular to each of two lines, then the two lines are parallel. Draw segment XYXY to close the figure. The resulting quadrilateral is a trapezoid with bases aa and bb and altitude a+ba+b.

The trapezoid is composed of two congruent right triangles and right XYZ\triangle XYZ. XYZ\triangle XYZ is isosceles since two of its sides have length cc. XZY\angle XZY is a right angle since 1+XZY+2=180°\angle 1 + \angle XZY + \angle 2 = 180° and 1+2=90°\angle 1 + \angle 2 = 90°. (The acute angles of a right triangle are complementary.) Therefore, the area of the trapezoid equals the sum of the areas of the three right triangles of which it is composed, two of which are congruent, so that:

AVWXY=AZVY+AXWZ+AXYZ12(a+b)(a+b)=12ab+12ab+12c2(a+b)(a+b)=ab+ab+c2a2+2ab+b2=2ab+c2a2+b2=c2\begin{aligned} A_{VWXY} &= A_{\triangle ZVY} + A_{\triangle XWZ} + A_{\triangle XYZ} \\ \frac{1}{2}(a + b)(a + b) &= \frac{1}{2}ab + \frac{1}{2}ab + \frac{1}{2}c^2 \\ (a + b)(a + b) &= ab + ab + c^2 \\ a^2 + 2ab + b^2 &= 2ab + c^2 \\ a^2 + b^2 &= c^2 \end{aligned}

P.S. Written in 1876, James A. Garfield was the only U.S. president to contribute a proof to a mathematical theorem.

David Vreken - 4 months, 1 week ago

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Yeah. In America the presidents are smarter than in Europe. Ferenc Gyurcsány said 11 is the biggest 2-digit prime number :) But the basic approach should be c2=a2+b22abcos(c);cos(90)=0c^2=a^2+b^2-2ab cos(c);cos(90^{\circ})=0

Páll Márton (no activity) - 4 months, 1 week ago

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Typically the approach to proving the law of cosines is to use the Pythagorean Theorem, so if you use the law of cosines to prove the Pythagorean Theorem you would have circular reasoning. (Now this almost becomes a philosophical conversation, because it all depends on which premises you want to assume are true.)

David Vreken - 4 months, 1 week ago

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By the way, if you put a slash in front of cos in latex, it won't be italicized (like this: "\cos")

David Vreken - 4 months, 1 week ago

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Wait, Hungary is in Europe! Ugh..............I need to work on my Geography, I thought Hungary was in China or something. I hate my Geo teacher, I think she's a dracanae, trying to kill me.

Percy Jackson - 4 months ago

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@Percy Jackson LOL I learned the location of all the countries. And also the states of the United States But I don't remember too much

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@Páll Márton (no activity) Wow, that must be hard, it probably is with a dracanae as your teacher.................

Percy Jackson - 4 months ago

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@Percy Jackson No, I went to a competition. But I failed on evil geometry. I proved twice this is impossible to solve the problems.

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@Páll Márton (no activity) Most of my teacher's(Gym coach is not, coz he likes how much I can swim and my stamina) are monsters, that's why I'm not good at any subject, and that isn't an excuse for my horrible math and science skills(all right it is an excuse, I don't like studying in english letters, they get jumbled up coz of my dyslexia)

Percy Jackson - 4 months ago

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@Percy Jackson Hmmm. My teacher who "teached" me sports was released from sports :) BTW he teached me Ukranian language three years long, but he don't have degree to teach them

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@Páll Márton (no activity) Released from sports??? @Páll Márton

Percy Jackson - 4 months ago

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@Percy Jackson I can't translate it :( When he was young he couldn't play sports.

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Good proof Sir! I remember that I once saw a proof in which the triangle is copied 4 times to form a square.

Vinayak Srivastava - 4 months, 1 week ago

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Yeah. And there are this formula: the length of the two sides of the parallelogram on the square is equal to the sum of the diagonals on the square.

Páll Márton (no activity) - 4 months, 1 week ago

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@Páll Márton (no activity) I haven't heard of anything like this. Can you give a link, or explain yourself?

Vinayak Srivastava - 4 months, 1 week ago

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@Páll Márton, @David Vreken, @Vinayak Srivastava - Garfield(the mingy, but hilarious cat) was a president of the United States, am I missing something here? And how do you know that Garfield's first name is James?

Percy Jackson - 4 months ago

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ID: 4

Prove that 2a2+2b2=e2+f22a^2+2b^2=e^2+f^2, where aa and bb is the two sides of the parallelogram and ee and ff is the diagonals of it

I will use this formula from there: ID-3

From ID-3:

  • In ΔBFC\Delta BFC: x2+m2=b2x^2+m^2=b^2
  • In ΔAFC\Delta AFC: (a+x)2+m2=f2(a+x)^2+m^2=f^2
  • In ΔEBD\Delta EBD: (ax)2+m2=e2(a-x)^2+m^2=e^2

The sum of the two last statement: e2+f2=2a2+2x2+2m2e^2+f^2=2a^2+\blue{2x^2+2m^2}. From the first statement: e2+f2=2a2+2b2e^2+f^2=2a^2+2b^2

You can use this formula too: c2=a2+b22abcos(γ)c^2=a^2+b^2-2ab\cdot cos(\gamma)

Páll Márton (no activity) - 4 months, 1 week ago

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Where is point FF?

Vinayak Srivastava - 4 months, 1 week ago

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Ohhh. Wait a minute :)

Páll Márton (no activity) - 4 months, 1 week ago

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@Páll Márton (no activity) See now. Thank you!

Páll Márton (no activity) - 4 months, 1 week ago

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Nice proof! I had not heard of this theorem, but seems interesting!

Vinayak Srivastava - 4 months, 1 week ago

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Yeah. I hate it. My math teacher always use this. Once she allowed me to prove one formula on her lesson, so I started to write a long chain(hence the idea) and I proved this formula

Páll Márton (no activity) - 4 months, 1 week ago

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I will post the next approach :)

Páll Márton (no activity) - 4 months, 1 week ago

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ID: 5

Prove that: a2=b2+c22bccos(α)a^2=b^2+c^2-2bc\cdot cos(\alpha)

I will use these formulas from there: ID-2, ID-3

Assume that A>B>C\angle A>\angle B>\angle C There are three cases:

1.case A=90\angle A=90^{\circ}:

From ID-3, this is true, because cos(90)=0cos(90^{\circ})=0

2.case A<90\angle A<90^{\circ}:

Swap A and B\huge\text{Swap A and B} AD=AB  cos(A)AD=AB\cdot\;cos(\angle A) and BD=AB  sin(A)BD=AB\cdot\;sin(\angle A)
From the 1. case: BC2=BD2+DC2=BD2+(ACAD)2=(ABsin(A))2+AC2+(ABcos(A))22ACABcos(A)=AB2(cos2(A)+sin2(A))+AC22ACABcos(A)\begin{aligned} BC^2&=BD^2+DC^2\\ &=BD^2+(AC-AD)^2\\ &=(AB\cdot sin(\angle A))^2+AC^2+(AB\cdot cos(\angle A))^2-2AC\cdot AB\cdot cos(\angle A)\\ &=AB^2(cos^2(\angle A)+sin^2(\angle A))+AC^2-2AC\cdot AB\cdot cos(\angle A) \end{aligned} From ID-2: BC2=AB2+AC22ACABcos(A)BC^2=AB^2+AC^2-2AC\cdot AB\cdot cos(\angle A)

3.case A>90\angle A>90^{\circ}:

Same way: AD=AB  cos(BAD)AD=AB\cdot\;cos(\angle BAD) and BD=AB  sin(BAD)BD=AB\cdot\;sin(\angle BAD)
From the 1. case: BC2=BD2+DC2=BD2+(AC+AD)2=(ABsin(BAD))2+AC2+(ABcos(BAD))2+2ACABcos(BAD)=AB2(cos2(BAD)+sin2(BAD))+AC2+2ACABcos(BAD)\begin{aligned} BC^2&=BD^2+DC^2\\ &=BD^2+(AC+AD)^2\\ &=(AB\cdot sin(\angle BAD))^2+AC^2+(AB\cdot cos(\angle BAD))^2+2AC\cdot AB\cdot cos(\angle BAD)\\ &=AB^2(cos^2(\angle BAD)+sin^2(\angle BAD))+AC^2+2AC\cdot AB\cdot cos(\angle BAD) \end{aligned} From ID-2: BC2=AB2+AC2+2ACABcos(BAD)BC^2=AB^2+AC^2+2AC\cdot AB\cdot cos(\angle BAD), but cos(180α)=cos(α)cos(180^{\circ}-\alpha)=-cos(\alpha). So BC2=AB2+AC22ACABcos(A)BC^2=AB^2+AC^2-2AC\cdot AB\cdot cos(\angle A)

Páll Márton (no activity) - 4 months, 1 week ago

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I will go to my grandmother, so I won't post approaches in the next couple of days with pictures. Now I will upload an image, to my next approach :) And I will insert it.

Páll Márton (no activity) - 4 months, 1 week ago

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And see the actual theme :)

Páll Márton (no activity) - 4 months, 1 week ago

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@Páll Márton (no activity)

I will prove these things:

  • sin(α2)=±1cos(α)2sin(\cfrac{\alpha}{2})=\pm\sqrt{\cfrac{1-cos(\alpha)}{2}}
  • same thing with cos, tan, cot
  • 1tan2(α)=cos(2α)cos2(α)1-tan^2(\alpha)=\cfrac{cos(2\alpha)}{cos^2(\alpha)}
  • same thing with cot

Sorry if there are typos!

Páll Márton (no activity) - 4 months, 1 week ago

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@Páll Márton (no activity) And cos(α)=cos2(α/2)sin2(α/2)cos(\alpha)=cos^2(\alpha /2)-sin^2(\alpha /2)

Páll Márton (no activity) - 4 months, 1 week ago

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For case 2 I don't think your letters match with your picture (specifically BC2=BD2+DC2BC^2 = BD^2 + DC^2)

David Vreken - 4 months, 1 week ago

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Sorry. Swap A and B :) I will edit it, when I can

Páll Márton (no activity) - 4 months, 1 week ago

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@Páll Márton (no activity) that should do it!

David Vreken - 4 months, 1 week ago

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ID:6

Prove these formulas:

  • sin(α±360k)=sin(α)\sin(\alpha\pm360^{\circ}\cdot k)=\sin(\alpha) and cos(α±360k)=cos(α)\cos(\alpha\pm360^{\circ}\cdot k)=\cos(\alpha), where kk is an integer
  • sin(180α)=sin(α)\sin(180^{\circ}-\alpha)=\sin(\alpha) and cos(180α)=cos(α)\cos(180^{\circ}-\alpha)=-\cos(\alpha)
  • sin(90α)=cos(α)\sin(90^{\circ}-\alpha)=\cos(\alpha) and cos(90α)=sin(α)\cos(90^{\circ}-\alpha)=\sin(\alpha)
  • tan(α)=sin(α)cos(α)\tan(\alpha)=\cfrac{\sin(\alpha)}{\cos(\alpha)}

I won't use other formulas from there

In this picture we can see the sin(x)\sin(x) and cos(x)\cos(x) functions. We should calculate in radian. This means:

2π=3602\pi=360^{\circ}, π=180\pi=180^{\circ}, π2=90\cfrac{\pi}{2}=90^{\circ} We can see if we shift the functions by 2π2\pi, then we get the same functions. Same way if we shift these functions by 4π,6π,8π,10π4\pi,6\pi,8\pi,10\pi, then we get the same functions.


From the unit circle:

sin(180α)=sin(α)\sin(180^{\circ}-\alpha)=\sin(\alpha) and cos(180α)=cos(α)\cos(180^{\circ}-\alpha)=-\cos(\alpha)

If α>180\alpha>180^{\circ}, then you should mirror the image, and you will get the same result.


From interpretation of sin,cos\sin,\cos: sin(α)=ac,cos(β)=ac,β=90α    cos(90α)=sin(α)\sin(\alpha)=\cfrac{a}{c},\cos(\beta)=\cfrac{a}{c},\beta=90^{\circ}-\alpha\implies\cos(90^{\circ}-\alpha)=\sin(\alpha). Same way sin(90α)=cos(α)\sin(90^{\circ}-\alpha)=\cos(\alpha) .


From similarity: sin(x)cos(x)=tan(x)1\cfrac{\sin(x)}{\cos(x)}=\cfrac{\tan(x)}{1}

Páll Márton (no activity) - 4 months, 1 week ago

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ID:7

Prove these formulas:

  • sin(α2)=1cos(α)2\sin(\cfrac{\alpha}{2})=\sqrt{\cfrac{1-\cos(\alpha)}{2}}
  • cos(α)=1+cos(α2)2\cos(\alpha)=\sqrt{\cfrac{1+ \cos(\cfrac{\alpha}{2})}{2}}
  • tan(α2)=1sin(α)1+cos(α)\tan(\cfrac{\alpha}{2})=\sqrt{\cfrac{1-\sin(\alpha)}{1+\cos(\alpha)}}
  • cot(α2)=1+cos(α)1sin(α)\cot(\cfrac{\alpha}{2})=\sqrt{\cfrac{1+\cos(\alpha)}{1-\sin(\alpha)}}
  • cos(2α)=cos2(α)sin2(α)\cos(2\alpha)=\cos^2(\alpha)-\sin^2(\alpha)
  • 1tan2(α)=cos(2α)cos2(α)1-\tan^2(\alpha)=\cfrac{\cos(2\alpha)}{\cos^2(\alpha)}
  • 1cot2α=cos(2α)sin2(α)1-\cot^2{\alpha}=-\cfrac{\cos(2\alpha)}{\sin^2(\alpha)}

I will use these formulas: ID-5, ID-6

a) In this picture AC=BC=x and BD=z From ID-5: AD2=x2+(xz)22x(xz)cos(α)AD2=z2+AB22zABcos(β)AB2=x2+x22xxcos(α)=x2(22cos(α))\begin{aligned} AD^2&=x^2+(x-z)^2-2x(x-z)\cdot\cos(\alpha)\\ AD^2&=z^2+AB^2-2z\cdot AB\cdot\cos(\beta)\\ AB^2&=x^2+x^2-2x\cdot x \cdot \cos(\alpha)=x^2(2-2\cos(\alpha)) \end{aligned} The left side of the first and the second equations are the same, and we can substitute the third equation to the second, so we can get this long equation: x2+(xz)22x(xz)cos(α)=x2+x22x2cos(α)+z22zx22cos(α)cos(β)x^2+(x-z)^2-2x(x-z)\cos(\alpha)=x^2+x^2-2x^2\cdot\cos(\alpha)+z^2-2zx\sqrt{2-2\cos(\alpha)}\cdot\cos(\beta) x2+x2+z22xz+2x2cos(α)+2xzcos(α)=x2+x22x2cos(α)+z22zx22cos(α)cos(β)\blue{x^2+x^2+z^2}-2xz+\blue{2x^2\cdot\cos(\alpha)}+2xz\cdot\cos(\alpha)=\blue{x^2+x^2-2x^2\cdot\cos(\alpha)+z^2}-2zx\sqrt{2-2\cdot\cos(\alpha)}\cdot\cos(\beta) We can simplify that. We can substract from both sides the blue section and after that we can divide both sides by 2xz-2xz: 1cos(α)=22cos(α)cos(β)1-\cos(\alpha)=\sqrt{2-2\cdot\cos(\alpha)}\cdot\cos(\beta) cos(β)=1cos(α)22cos(α)=1cos(α)211cos(α)=1cos(α)2=1cos(α)2\begin{aligned} \cos(\beta)&=\cfrac{1-\cos(\alpha)}{\sqrt{2-2\cos(\alpha)}}\\ &=\cfrac{1-\cos(\alpha)}{\sqrt{2}\sqrt{1-1\cos(\alpha)}}\\ &=\cfrac{\sqrt{1-\cos(\alpha)}}{\sqrt{2}}\\ &= \sqrt{\cfrac{1-\cos(\alpha)}{2}} \end{aligned} We know these things:

  • β=180α2=90α2\beta=\cfrac{180^{\circ}-\alpha}{2}=90^{\circ}-\cfrac{\alpha}{2}
  • sin(90α)=cos(α)\sin(90^{\circ}-\alpha)=\cos(\alpha) sin(α2)=1cos(α)2\sin(\cfrac{\alpha}{2})=\sqrt{\cfrac{1-\cos(\alpha)}{2}}

b) From ID-2: sin2(α)+cos2(α)=1\sin^2(\alpha)+\cos^2(\alpha)=1 From ID-7a: 1cos(α2)2+cos2(α)=11cos(α2)+2cos2(α)=2cos2(α)=1+cos(α2)2cos(α)=1+cos(α2)2\begin{aligned} \cfrac{1-\cos(\cfrac{\alpha}{2})}{2}+cos^2(\alpha)&=1\\ 1- \cos(\cfrac{\alpha}{2})+2\cos^2(\alpha)&=2\\ \cos^2(\alpha)&=\cfrac{1+ \cos(\cfrac{\alpha}{2})}{2}\\ \cos(\alpha)&=\sqrt{\cfrac{1+ \cos(\cfrac{\alpha}{2})}{2}} \end{aligned}


c) From ID-7a and ID-7b: tan(α2)=sin(α2)cos(α2)=1cos(α)21+cos(α)2=1sin(α)1+cos(α)\begin{aligned} \tan(\cfrac{\alpha}{2})&=\cfrac{\sin(\cfrac{\alpha}{2})}{\cos(\cfrac{\alpha}{2})}\\ &=\cfrac{\sqrt{\cfrac{1-\cos(\alpha)}{\cancel{2}}}}{\sqrt{\cfrac{1+\cos(\alpha)}{\cancel{2}}}}\\ &=\sqrt{\cfrac{1-\sin(\alpha)}{1+\cos(\alpha)}} \end{aligned}


d) 1tan(α)=cot(α)\cfrac{1}{\tan(\alpha)}=\cot(\alpha). So from ID-7c this is true.


e) From ID-7b: cos(2α)=2cos2(α)1=2cos2(α)(sin2(α)+cos2(α))=cos2(α)sin2(α)\begin{aligned} \cos(2\alpha)&=2\cos^2(\alpha)-\blue{1}\\ &=2\cos^2(\alpha)-(\sin^2(\alpha)+\cos^2(\alpha))\\ &=\cos^2(\alpha)-\sin^2(\alpha) \end{aligned}


f) 1tan2(α)=1sin2(α)cos2(α)=11cos(2α)2cos2(α)=cos2(α)1cos(2α)2cos2(α)=12cos2(α)cos(2α)2cos2(α)=24cos2(α)2cos2(α)=2cos2(α)1cos2(α)=cos(2α)cos2(α)\begin{aligned} 1-\tan^2(\alpha)&=1-\cfrac{\sin^2(\alpha)}{\cos^2(\alpha)}\\ &=1- \cfrac{\cfrac{1-\cos(2\alpha)}{2}}{\cos^2(\alpha)}\\ &=\cfrac{ \cos^2(\alpha)-\cfrac{1-\cos(2\alpha)}{2}}{\cos^2(\alpha)}\\ &=-\cfrac{1-2\cos^2(\alpha)-\cos(2\alpha)}{2\cos^2(\alpha)}\\ &=-\cfrac{2-4\cos^2(\alpha)}{2\cos^2(\alpha)}\\ &=\cfrac{2\cos^2(\alpha)-1}{\cos^2(\alpha)}\\ &=\cfrac{\cos(2\alpha)}{\cos^2(\alpha)} \end{aligned}


g) 1cot2α=sin2(α)1+cos(2α)2sin2(α)=1cos(2α)21+cos(2α)2sin2(α)=cos(2α)cos(2α)2sin2(α)=cos(2α)sin2(α)\begin{aligned} 1-\cot^2{\alpha}&=\cfrac{\sin^2(\alpha)-\cfrac{1+\cos(2\alpha)}{2}}{\sin^2(\alpha)}\\ &=\cfrac{ \cfrac{1-\cos(2\alpha)}{2}- \cfrac{1+\cos(2\alpha)}{2}}{\sin^2(\alpha)}\\ &=\cfrac{ -\cos(2\alpha)-\cos(2\alpha)}{2\sin^2(\alpha)}\\ &=-\cfrac{\cos(2\alpha)}{\sin^2(\alpha)} \end{aligned}

Páll Márton (no activity) - 4 months, 1 week ago

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I'm working on this yet :) I should prove that, when α>180\alpha>180^{\circ}.

Páll Márton (no activity) - 4 months, 1 week ago

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ID:8

Proof that 2\sqrt{2} is irrational

I will not use any formulas from other proofs

This is a proof by contradiction, so let's assume that 2\sqrt{2} is rational.

If so, then it can be expressed in an equation as pq\dfrac{p}{q} where pp and qq are co-primes - 2=pq2=p2q22p2=q2\begin{aligned} \sqrt{2} &= \dfrac{p}{q} \\ 2 &= \dfrac{p^{2}}{q^{2}} \\ 2p^{2} &= q^{2} \end{aligned}

By this simplification, we know that q2q^{2} even, because it is 22 multiplied by some other number. So qq must also be an even number. That means qq can be represented as (2m)(2m).

Back to the equation, if we use the new value of qq and simplify further - 2p2=q22p2=(2m)22p2=4m2p2=2m2\begin{aligned} 2p^{2} &= q^{2} \\ 2p^{2} &= (2m)^{2} \\ 2p^{2} &= 4m^{2} \\ p^{2} &= 2m^{2} \end{aligned}

As pp can also be represented like this, pp must also be even.

This creates a contradiction, as we had earlier said that pp and qq were co-primes, but now we say they are both even, which means they have a common divisor of 22. Since this creates a contradiction, our original assumption must have been wrong...

Q.E.D - 2 is irrational!\text{Q.E.D - } \sqrt{2} \text{ is irrational!}

This was originally proved by the Greek Mathematician and Father of Geometry, Euclid of Alexandria


@Páll Márton (no activity) - Thoughts? Also, do I have to wait till ID:10 to start chains?

Percy Jackson - 4 weeks, 1 day ago

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ID:9

Graphical Proof of Circle Area formula - πr2\pi r^{2}

I will not use any formulas from other proofs

Let's cut a circle into 88 equal sectors and arrange them as shown in the image below -

Image credits - https://medium.com Image credits - https://medium.com

As you can see, it forms a figure that looks very close to a parallelogram that is going towards becoming a rectangle. The next two images show what happens when we take even smaller arcs.

Image credits - https://medium.com Image credits - https://medium.com Image credits - https://medium.com Image credits - https://medium.com

In the last image, you see an figure that is almost a rectangle. TO get a perfect rectangle, we would need infinitesimally small sectors, but we can evaluate the area of the circle using this figure, as we know the formula for the area of a rectangle - A=bhA = bh

The height is easy to find, as it is just the radius. The base, if you look closely at the next image, is actually half of the circumference or half of 2πr2 \pi r or πr\pi r

Image credits - https://medium.com Image credits - https://medium.com

Thus, the formula for the area of a circle πr×r=πr2\pi r \times r = \boxed{\pi r^{2}}

Percy Jackson - 4 weeks ago

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Excellent animations!

David Vreken - 4 weeks ago

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@David Vreken Quite true. How do they even do that?

Bithiah Koshy - 4 weeks ago

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@Bithiah Koshy Its a mystery, but I'm sure our buddy Mahdi Raza could do much better.

Percy Jackson - 4 weeks ago

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@Percy Jackson mmmmmmmmmmmmm!!!!!

Bithiah Koshy - 4 weeks ago

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@Bithiah Koshy @Percy Jackson1-2 more mins untill daily challenge

NSCS 747 - 4 weeks ago

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@Nscs 747 Yess!!!!!

Percy Jackson - 4 weeks ago

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@Nscs 747 @NSCS 747 yea for Australian timing right?

Bithiah Koshy - 4 weeks ago

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@Bithiah Koshy um yea

NSCS 747 - 4 weeks ago

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@Nscs 747 K that's good! I have recently learned Aussie slang such as dead horse! Even though I am a migrant Australia who has stayed in the land of Oz for 7 years!!!!!

Bithiah Koshy - 4 weeks ago

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@Bithiah Koshy i live in NSW sydney so its 10:02 right now

NSCS 747 - 4 weeks ago

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@Nscs 747 @NSCS 747 I live in Victoria. Why do so many people live in NSW?!!

Bithiah Koshy - 4 weeks ago

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@Bithiah Koshy i was born here so i just like it here i havnt been to any other state or city except for newcastle and canberra

NSCS 747 - 4 weeks ago

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@Nscs 747 Ah I see I have only stayed in Victoria since I cam to Australia.

Bithiah Koshy - 4 weeks ago

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@Bithiah Koshy how easy is icas for you?

NSCS 747 - 4 weeks ago

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@Nscs 747 Like moderately easy. Got High distinction for MATH ONLY< ONLY!!!!

Bithiah Koshy - 4 weeks ago

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@Bithiah Koshy for me i always get credit lol and for some reason i always get 81% for maths like always even two part exams part a and part b i get 81%

NSCS 747 - 4 weeks ago

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@Nscs 747 oh wait cus the mean is 81% frik im dumb

NSCS 747 - 4 weeks ago

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Yes! I found them on medium.com, so I image credited each of them under the pics.

Percy Jackson - 4 weeks ago

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ID:10

Algebraic Proof of following Trig Identities

  • a) sin(90β)=cos(β)\sin(90- \beta) = \cos(\beta)

  • b) cos(90β)=sin(β)\cos(90-\beta) = \sin(\beta)

I will not use any formulas from other proofs

a) and b) can be proven using the sine and cosine angle subtraction formulas

  • cos(αβ)=cos(α)cos(β)+sin(α)sin(β)\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)

  • sin(αβ)=sin(α)cos(β)cos(α)sin(β)\sin(\alpha - \beta) = \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta)


a) sin(90β)=cos(β)\sin(90- \beta) = \cos(\beta)

Using the above formula, substituting α\alpha with 9090,

sin(90β)=sin(90)cos(β)cos(90)sin(β)sin(90β)=1×cos(β)0×sin(β)sin(90β)=cos(β)\begin{aligned} \sin(90- \beta) &= \sin(90)\cos(\beta) - \cos(90)\sin(\beta) \\ \sin(90- \beta) &= 1 \times\cos(\beta) - 0 \times\sin(\beta) \\ \color{#3D99F6}\sin(90- \beta) \color{#3D99F6} &= \color{#3D99F6}\cos(\beta) \end{aligned}


b) Using the above formula again, substituting α\alpha with 9090,

cos(90β)=cos(90)cos(β)+sin(90)sin(β)cos(90β)=0×cos(β)+1×sin(β)cos(90β)=sin(β)\begin{aligned} \cos(90- \beta) &= \cos(90)\cos(\beta) + \sin(90)\sin(\beta) \\ \cos(90- \beta) &= 0 \times \cos(\beta) + 1 \times \sin(\beta) \\ \color{#3D99F6}\cos(90- \beta) \color{#3D99F6} &= \color{#3D99F6}\sin(\beta)\end{aligned}


@Páll Márton (no activity) - We have reached ID:10!!! Can we start the chain proofs now?

Percy Jackson - 4 weeks ago

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ID:11

Proof that 2cos1n+2sin1n=π2\cos^{-1} n+2 \sin^{-1} n=\pi in radians

Not using other identities

Construct an isosceles triangle ABC\triangle ABC.
In isoscelesABC\triangle ABC, BCBC is the base. Construct the median AMAM where MM is the midpoint of BCBC.

Since ABC\triangle ABC is isosceles, AMB\angle AMB is a right angle.
AMB\Rightarrow \triangle AMB is a right angled triangle.
This part uses basic definitions of sin & cos.
Call the length of the hypotenuse of RtAMBRt\triangle AMB kk. Call the length of median AMAM ll.
BAM=cos1lk.\Rightarrow \angle BAM=\cos^{-1} \frac{l}{k}.
BAC=2BAM=2cos1lk.\Rightarrow \angle BAC=2\angle BAM=2\cos^{-1} \frac{l}{k}.
Since AMB\triangle AMB is right angled, and ABC\triangle ABC is isosceles,
sinB=sinC=lk.\sin \angle B=\sin \angle C=\frac{l}{k}.
B=C=sin1lk.\Rightarrow \angle B=\angle C=\sin^{-1} \frac{l}{k}.
Letlk=n.\frac{l}{k}=n..
Finally, since A,B,C\angle A,\angle B,\angle C are the interior angles of a triangle, i.e. they add up to π(rad)\pi (rad), 2cos1n+2sin1n=π2\cos^{-1} n+2 \sin^{-1} n=\pi in radians.

Jeff Giff - 3 weeks, 6 days ago

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I think this is a new identity that has never been discovered ago?

Jeff Giff - 3 weeks, 6 days ago

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Maybe, but if so, you're a mathematician officially.

Percy Jackson - 3 weeks, 6 days ago

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@Percy Jackson OMG :P

Jeff Giff - 3 weeks, 6 days ago

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@Jeff Giff lol, yes OMG

Percy Jackson - 3 weeks, 6 days ago

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@Percy Jackson Thx for deleting the comments :)

Jeff Giff - 3 weeks, 6 days ago

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@Jeff Giff No prob, sorry bout the notifs

Percy Jackson - 3 weeks, 6 days ago

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Notifications View all (97 not shown) ...

NSCS 747 - 3 weeks, 6 days ago

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see the solution

Jeff Giff - 3 months, 2 weeks ago

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You can add one more rule: use buttons to decrease length. :)

Jeff Giff - 3 months, 2 weeks ago

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[[show-hide-button|H]]
[[start-hidden]]
my proof
[[end-hidden]]

Jeff Giff - 3 months, 2 weeks ago

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Whup!

Jeff Giff - 3 months, 2 weeks ago

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@Jeff Giff - Buttons don't work in chat, only in notes and wikis, not even problems :)

Percy Jackson - 3 months, 2 weeks ago

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You can use buttons only in notes :)

Páll Márton (no activity) - 3 months, 2 weeks ago

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