Hardest Math Problem there is! Please help!

[TheRealFern]

I don’t know how to solve this I’ve been working on it for days now, can someone help me further. Please show work!

Given points O(0,0) and A(7,1) , segment OA is rotated 60∘ in a counterclockwise direction around point O such that O is fixed and A moves to A′ . A′ lies in the first quadrant. In simplest terms the x -coordinate of A′ is a−b√c . What is a+b+c?

 

[pka]

I don’t know how to solve this I’ve been working on it for days now, can someone help me further. Please show work! (We don't do work for you!)
Given points O(0,0) and A(7,1) , segment OA is rotated 60∘ in a counterclockwise direction around point O such that O is fixed and A moves to A′ . A′ lies in the first quadrant. In simplest terms the x -coordinate of A′ is a−b√c . What is a+b+c?

You must use a rotation matrix: \(\displaystyle A' = \left[ {\begin{array}{*{20}{c}} {\cos ({{60}^0})}&{ - \sin ({{60}^0})} \\ {\sin ({{60}^0})}&{\cos ({{60}^0})} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1 \\ 7 \end{array}} \right]\)

 

[TheRealFern]

It’s not a homework question rather than a math Competition question I don’t know how to solve I’ve never herd of a rotation matrix

 

[Deleted member 4993]

I don’t know how to solve this I’ve been working on it for days now, can someone help me further. Please show work!

Given points O(0,0) and A(7,1) , segment OA is rotated 60∘ in a counterclockwise direction around point O such that O is fixed and A moves to A′ . A′ lies in the first quadrant. In simplest terms the x -coordinate of A′ is a−b√c . What is a+b+c?

Please share your work/thoughts about this assignment.

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[pka]

It’s not a homework question rather than a math Competition question I don’t know how to solve I’ve never herd of a rotation matrix

Well in that case, you are seemingly not prepared to participate in that contest.

 

[TheRealFern]

I know and I asked 5 of my schools most intelligent teachers and they couldn’t solve it

 

[TheRealFern]

That’s why I want to find the solution and show them. We spend hours on it

 

[Deleted member 4993]

I know and I asked 5 of my schools most intelligent teachers and they couldn’t solve it

I think your teacher/s know how to solve it - they just don't want to solve "competition" problem!

After the competition is over, I think they will show you the process.

 

[TheRealFern]

This was from a practice completion and we looked at the solution and it was 12 but we don’t know how to get there

 

[pka]

I know and I asked 5 of my schools most intelligent teachers and they couldn’t solve it
That’s why I want to find the solution and show them. We spend hours on it

Show those teachers the solution I gave you. See if they can handle it.
Then show them this link.
The exact answer is \(\displaystyle \left(\dfrac{7-\sqrt3}{100},\dfrac{1+7\sqrt3}{100}\right)\)

 

[LCKurtz]

Not sure where those [MATH]100[/MATH]'s came from, but I think the answer to the specific question is [MATH]7[/MATH].

 

[Dr.Peterson]

It’s not a homework question rather than a math Competition question I don’t know how to solve I’ve never herd of a rotation matrix

If you want actual help, you'll need to tell us what you do know that might be relevant. How much trigonometry have you learned? How about analytic geometry? There are several ways this could be done, so whatever your level of knowledge, there may be a way.

Of course, if you just want an answer, you've got it; and if you want to know what subject to study in order to solve it most easily, you have an idea where to go now (try searching for "rotation matrix", and take links to more general topics until you reach one you're ready for). This is one of the ways I started getting good at math -- not internet searches, which didn't exist yet, but discovering interesting problems and looking for books so I could learn what I wasn't yet learning in school.

 

[infinity1010]

It’s not a homework question rather than a math Competition question I don’t know how to solve I’ve never herd of a rotation matrix

A rotation matrix is what the name implies. If you look at the (x, y) Euclidian plane, you can do a matrix but without rotating the x axis its just a rotation of 0, which keeps things they way they are. However, once you rotate a vector from the origin with respect to the x-axis, you have to consider the angle of rotation because the angle is no longer 0. Before the rotation it was simply the x-axis where the cos(0) = 1 and the sine(0) = 0. Therefore the matrix would just be an identity matrix. Once you rotate it though, the angle is no longer 0 that forms an identity matrix, its now the cosine and sine of that new angel which puts the vector on completely different points.