Results 1 to 15 of 15

Math Help - square incribed in cricle

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    6

    square inscribed in circle

    I have a problem here that I haven't been able to solve.
    You're given a circle with radius r and a square is inscribed on it.
    The center of the circle is at the origin and the edges of the square do not lie on coordinate axis. Given co-ordinates (x1,y1), tell whether the co-ordinates lie on any side of the square.
    I hope my question is clear.
    Thanks in advance.
    Last edited by padfoot; February 6th 2009 at 05:58 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    I see you are a newbie very warm Welcome to the Forum


    Hint: There can be infinite squares in a circle
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member star_tenshi's Avatar
    Joined
    Jan 2009
    Posts
    32
    Quote Originally Posted by padfoot View Post
    I have a problem here that I haven't been able to solve.
    You're given a circle with radius r and a square is inscribed on it.
    The center of the circle is at the origin and the edges of the square do not lie on coordinate axis. Given co-ordinates (x1,y1), tell whether the co-ordinates lie on any side of the square.
    I hope my question is clear.
    Thanks in advance.
    This is my interpretation of your question:

    since we know that it is a square inscribed in a circle centered at the origin, then \theta = 45 degrees = \frac{\pi}{4}. Thus x & y can be calculated:
     x = r \cdot cos\theta = r \cdot cos\frac{\pi}{4}

     y = r \cdot sin\theta = r \cdot sin\frac{\pi}{4}
     \Rightarrow x = y

    So any coordinate with either the x or y coordinates equation to r \cdot cos\frac{\pi}{4} or r \cdot sin\frac{\pi}{4} would lie on the edges of the square.

    Hope it helps!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member star_tenshi's Avatar
    Joined
    Jan 2009
    Posts
    32
    Please excuse my previous post. I looked at ADARSH's post and realized my mistake. There are infinitely many squares inscribed in a circle. I was just looking at a particular case.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,868
    Thanks
    746
    Hello, padfoot!

    I don't know what you're asking, but I'll take a guess.


    You're given a circle with radius r and a square is inscribed on it.
    The center of the circle is at the origin.
    Given co-ordinates (x1,y1) of one vertex, locate the other vertices.
    Code:
                    Y
                    |
                  * * *
              *     |     * P
            * - - - + - - - o(x1,y1)
           *|       |     * |*
            |       |  r*   |
          * |       | * θ   | *
      - - * | - - - + - - - | * - - X
          * |      O|       | *
            |       |       |
           *|       |       | *
            * - - - + - - - *
              *     |     *
                  * * *
                    |

    The coordinates of the one vertex is: . P(x_1,y_1)

    \text{Let }r\, = \,\text{distance }OP.
    \text{Let }\theta \,=\,\angle POX,\,\text{ where }\tan\theta \,=\,\frac{y_1}{x_1}


    Then P has coordinates: . \left(r\cos\theta,\:r\sin\theta\right)


    The four vertices are: . \bigg[r\cos\!\left(\theta + \frac{\pi}{2}n\right),\: r\sin\!\left(\theta + \frac{\pi}{2}n\right)\bigg]\;\text{ for }n = 0,1,2,3

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Feb 2009
    Posts
    6
    Thanks for the replies. I tried to use it to solve the problem but unfortunately it didn't work out.
    I'll try to tell the question here.

    There is a circular island of radius r (centered at origin). The owner wants to build a square tower inscribed on it. Given two points E1 (x1,y1) and E2 (x2,y2) for entrance and exit, is it possible to build a tower such that the two points (E1 and E2) lie on the tower's foundation ?

    Hopefully, it made more sense this time around.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    From your reply it seems either you are hesitating to give complete question or question is wrong

    If its first case - Dont worry we are not gonna spoon feed you

    If its second - Crush the question with your foot, "padfoot"
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Feb 2009
    Posts
    6
    I just rephrased the question. I removed the unnecessary parts. If you guys are still having trouble understanding it, I'll post the original one.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    The square will have diameter 2r
    So
    \sqrt{2r^2}= \text{side of square}

    Now ther can be infinite such squares(after rotating)
    Locus of all such squares is a circle of radius equal to half the side of square ie; \frac{r}{\sqrt{2}}
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Feb 2009
    Posts
    6
    I'm not sure how to apply that.

    I'll give an example.
    r = 2
    (x1,y1) = 1,1
    (x2,y2) = 1,-1

    The answer is Yes.

    You get this by taking the vertices of the square as A(2,0), B(0,2), C(-2,0), D(0,-2). The points (1,1) and (1,-1) lie on lines AB and AD respectively.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by padfoot View Post
    I'm not sure how to apply that.

    I'll give an example.
    r = 2
    (x1,y1) = 1,1
    (x2,y2) = 1,-1

    The answer is Yes.

    You get this by taking the vertices of the square as A(2,0), B(0,2), C(-2,0), D(0,-2). The points (1,1) and (1,-1) lie on lines AB and AD respectively.
    I've attached a sketch.

    You easily can determine the area where the points E_1 and E_2 must be located so that the condition is satisfied.
    Attached Thumbnails Attached Thumbnails square incribed in cricle-quad_einbeschrieben.png  
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Feb 2009
    Posts
    6
    I'm sorry but I still didn't get it. How do you get the area where E1 and E2 must be located ?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    -Since you didn't specify the real values of points to be checked we took it as (x,y)

    - Since the square could be rotated we rotated it and hence(as in earboth's figure) every point of that area got included

    -the mid point of a square inscribed in a circle is the point nearest to it

    -Joining the midpoints of a particular side of a rotating square square we get a cicle of radius \frac{r}{\sqrt{2}}

    - All points in the area between these two circles can be called E1 and E2
    ie;  r^2 \ge x^2 +y^2 \ge \frac{r^2}{2}

    Every (x,y) satisfying above is your answer (except those for which either is 0)
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Newbie
    Joined
    Feb 2009
    Posts
    6
    Thanks a lot. It makes sense now. I have one doubt though.
    Can't the two points E1 and E2 satisfy that condition but be part of two different squares ?
    For example, is it possible to get a point E1 that lies on the line of a square S1 and E2 lie on a square S2 but not on S1 ?
    I hope I'm not sounding like an idiot...
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    -See every point that satisfies that condition can be called E1 or E2

    -This does NOT mean that every point that satisfies it are part of same square they may not be so

    -the thing that satisfies that condition is an area (of which every point ) is your answer
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. midpoint of a cricle.
    Posted in the Geometry Forum
    Replies: 4
    Last Post: September 3rd 2011, 07:59 AM
  2. circle incribed in a circular sector
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: August 24th 2011, 08:29 PM
  3. graphing a cricle
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 3rd 2009, 03:50 PM
  4. How do you tell a cricle by the formula?
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: April 26th 2009, 04:38 AM
  5. [SOLVED] Incribed Rectangle and Triangle
    Posted in the Geometry Forum
    Replies: 4
    Last Post: March 28th 2009, 05:12 AM

Search Tags


/mathhelpforum @mathhelpforum