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geerky42 Group Title

Six circles are tangent to each other and an equilateral triangle is inscribed around them as shown. What percent of the area of triangle is NOT shaded?

  • one year ago
  • one year ago

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  1. geerky42 Group Title
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    Bad drawing, sorry. Hopefully you know what I'm trying to draw...|dw:1352593402510:dw|

    • one year ago
  2. Dido525 Group Title
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    Do you know any other information?

    • one year ago
  3. geerky42 Group Title
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    No, this is all given information I have.

    • one year ago
  4. Dido525 Group Title
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    Well it's equilateral so all of it is the same...

    • one year ago
  5. geerky42 Group Title
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    I guess so.

    • one year ago
  6. Dido525 Group Title
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    The area of an equilateral triangle is : \[\frac{ s^2\sqrt{3} }{ 4 }\]

    • one year ago
  7. geerky42 Group Title
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    And circles are tangent to each other, so they all should be the same.

    • one year ago
  8. Dido525 Group Title
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    Mhmm... Let me see. You post very intrsting questions.

    • one year ago
  9. geerky42 Group Title
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    And how can I applying to it?

    • one year ago
  10. Dido525 Group Title
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    I am thinking...

    • one year ago
  11. geerky42 Group Title
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    Sorry for interruption, but I need your help... @Hero @tcarroll010 @AccessDenied @AriPotta

    • one year ago
  12. geerky42 Group Title
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    Any ideas, hints, tips?

    • one year ago
  13. tcarroll010 Group Title
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    Are all those circles of the same radius?

    • one year ago
  14. jon.stromer.galley Group Title
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    if you know the area of the triangle and the area of the circle(s) what can you say about the area not covered by the circles?

    • one year ago
  15. geerky42 Group Title
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    Well, we only know that they are tangent to each other and in the shown image on my paper, they appear that they also tangent to the sides of triangle too, so I guess yeah. This is literally the only given information I have...

    • one year ago
  16. geerky42 Group Title
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    Go on... @jon.stromer.galley

    • one year ago
  17. Dido525 Group Title
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    Well if they are tengential to each other I assume they have the same radius.

    • one year ago
  18. geerky42 Group Title
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    I suppose.

    • one year ago
  19. geerky42 Group Title
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    So where should we start?

    • one year ago
  20. tcarroll010 Group Title
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    I am exploring a relationship between the triangle and a second triangle formed by connecting the centers of the circles.

    • one year ago
  21. geerky42 Group Title
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    Why not the relationship between the area of circle to the area of triangle? I think this is good start, but I'm not sure.

    • one year ago
  22. geerky42 Group Title
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    I don't know, lol.

    • one year ago
  23. Dido525 Group Title
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    I am thinking about saying that: |dw:1352594259221:dw|

    • one year ago
  24. Dido525 Group Title
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    I don't think so though...

    • one year ago
  25. geerky42 Group Title
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    I doubt it. |dw:1352594316081:dw|

    • one year ago
  26. Dido525 Group Title
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    Yeah I know. Was just wondering...

    • one year ago
  27. AccessDenied Group Title
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    A better picture perhaps: http://puu.sh/1oLWD

    • one year ago
  28. geerky42 Group Title
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    I think @tcarroll010 has a good point, perhaps we should determine the relationship between the triangle and a second triangle formed by connecting the centers of the circles.

    • one year ago
  29. geerky42 Group Title
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    Much better. @AccessDenied

    • one year ago
  30. Dido525 Group Title
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    LOT better.

    • one year ago
  31. AccessDenied Group Title
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    Oh wow, this is sort of interesting: I draw in all of the tangents of the circles to the exterior triangle and an interior triangle: http://puu.sh/1oM08

    • one year ago
  32. Dido525 Group Title
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    Where are you getting these questions from? O_o .

    • one year ago
  33. Dido525 Group Title
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    Hmm!

    • one year ago
  34. Dido525 Group Title
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    Similar triangles perhaps?

    • one year ago
  35. jon.stromer.galley Group Title
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    Ok, so all you know for sure is that the triangle is equilateral and that the circles have a radius r. That is enough to express the combined circle area, but to answer the question you need to know how big the triangle is (how long the sides are) in terms of R. Let's look at the lower left corner.

    • one year ago
  36. jon.stromer.galley Group Title
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    |dw:1352594743279:dw|

    • one year ago
  37. Dido525 Group Title
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    We don't know that...

    • one year ago
  38. AccessDenied Group Title
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    Hint: you have an equilateral triangle for the exterior, so you know some angles. ;)

    • one year ago
  39. tcarroll010 Group Title
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    I've got what might be a good idea. You use trig. And 30-60-90 right triangles in the corners. I was independently working a similar diagram to Accessd.

    • one year ago
  40. jon.stromer.galley Group Title
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    let's draw it again...|dw:1352594832325:dw|

    • one year ago
  41. Dido525 Group Title
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    Ohh yeah!!!

    • one year ago
  42. Dido525 Group Title
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    Isn't that also one of those special triangles?

    • one year ago
  43. jon.stromer.galley Group Title
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    At this point if you have been shown you have enough data to figure out the base of the triangle. The length of a side of the over all triangle will be then 3r + 2( length of the base of the triangle we just drew)

    • one year ago
  44. tcarroll010 Group Title
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    It looks like a whole bunch of us about simultaneously hit on the right idea! Fun trig at this point.

    • one year ago
  45. geerky42 Group Title
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    I have sense I'm getting closer to solution, yet I have no idea where to start. I have to find the relationship between the radius and length of triangle side, right?

    • one year ago
  46. Dido525 Group Title
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    |dw:1352594992142:dw|

    • one year ago
  47. Dido525 Group Title
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    Now we know the radius of all those circles are 1!

    • one year ago
  48. AccessDenied Group Title
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    not necessarily. The 30-60-90 rule only expresses a ratio. :P

    • one year ago
  49. geerky42 Group Title
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    And from this, we can find the ratio of the total area of six circle to the area of triangle, right?

    • one year ago
  50. Dido525 Group Title
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    Can we not assume? O_o .

    • one year ago
  51. jon.stromer.galley Group Title
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    the radius is still "1" r but now the base can be expressed in terms of r as well

    • one year ago
  52. Dido525 Group Title
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    Not sure is the side length would just be 2r...

    • one year ago
  53. tcarroll010 Group Title
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    To be pure, you would use "r" for radius, but you are free to call radius "1" for your purposes, so, yes you can make the assumption, with that proviso. And then it's simple to get your triangle side lengths. Nice little problem!

    • one year ago
  54. Dido525 Group Title
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    Well looking at @AccessDenied Drawing... : Since the radius is 1 we can say the side length's of the smaller, inside triangle are 3, 3 and 2 going clockwise.

    • one year ago
  55. Dido525 Group Title
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    4 4 and 4 sorry.

    • one year ago
  56. Dido525 Group Title
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    |dw:1352595331990:dw|

    • one year ago
  57. Dido525 Group Title
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    I am assuming the height would be 3...

    • one year ago
  58. Dido525 Group Title
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    |dw:1352595405601:dw|

    • one year ago
  59. Dido525 Group Title
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    Now I can't get a ratio....

    • one year ago
  60. tcarroll010 Group Title
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    With a circle radius of "r", being the "short leg", then the "long leg" will be r times sqrt(3). So a triangle side will be 2r(sqrt(3)) + 2r.

    • one year ago
  61. Dido525 Group Title
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    So it would just be 2root(3)+2 .

    • one year ago
  62. tcarroll010 Group Title
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    If you are going with your assumption or requirement that r is "1", then, yes.

    • one year ago
  63. tcarroll010 Group Title
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    So, you should be able to get the altitude using Pythagorean and then area of triangle, and areas of circles. You're just about done!

    • one year ago
  64. jon.stromer.galley Group Title
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    I calculate that the length of a side of the master triangle is 6.4641r Anyone else? tanks to: adjacent = r / tan(30)

    • one year ago
  65. jon.stromer.galley Group Title
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    or alternatively by expressing the length of the line as: 2r * root(3) + 3r that plugs in nicely to the formula for the area of an el triangle: (length of side * root(3)) / 4 so that tells us the area is...

    • one year ago
  66. AccessDenied Group Title
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    I missed the answer placed by the original poster, but my work in ms paint is here: http://puu.sh/1oNgt

    • one year ago
  67. geerky42 Group Title
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    I got the same answer. Thank you, everyone, for your times. I appreciate it.

    • one year ago
  68. AccessDenied Group Title
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    You're welcome! :)

    • one year ago
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