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Finding a triangle in a circle


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Suppose there is an equilateral triangle (side length 1/5 unit) inside a unit circle.  Drawing straight lines through the circle, what is the least amount of lines that it would take to ensure we intersected (found) the triangle?

Edited by BMAD
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13 hours ago, BMAD said:

I have it in a much smaller amount but not as small as our new friend above. 

 

My answer is...

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12 even spaced and parallel lines

 

I think your answer is right. Mine is overkill. Are you sure that

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11 won't do?

 

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16 hours ago, SamsonBrainPower said:

Also I didn't solve it, my dad did. He doesn't have an account on this website, so I give all the credit to jjuddjjudd@hotmail.com

 

care to have your father explain?  I have a different solution.

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Looks like ...

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33 lines

Diameter of the circle iss 2.
Height of triangle is (1/5) (sqrt(3)/2) = sqrt(3)/10.

The circle will accommodate 20/sqrt(3) = 11.54 parallel lines.
To create triangular areas we do thise in directions parallel to
the three sides of the triangle.

33 lines will then divide the unit circle into triangles 0.2 units on a side.
now decrease the spacing of the lines by an infinitessimal amount
so that the triangle will not fit, undetected, in any of those spaces.

tri-circle.jpg.f108cf02da7e890aee4e776a1

 

 

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45 minutes ago, bonanova said:

Looks like ...

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33 lines

Diameter of the circle iss 2.
Height of triangle is (1/5) (sqrt(3)/2) = sqrt(3)/10.

The circle will accommodate 20/sqrt(3) = 11.54 parallel lines.
To create triangular areas we do thise in directions parallel to
the three sides of the triangle.

33 lines will then divide the unit circle into triangles 0.2 units on a side.
now decrease the spacing of the lines by an infinitessimal amount
so that the triangle will not fit, undetected, in any of those spaces.

tri-circle.jpg.f108cf02da7e890aee4e776a1

 

 

I have it in a much smaller amount but not as small as our new friend above. 

 

My answer is...

Spoiler

12 even spaced and parallel lines

 

Edited by BMAD
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5 hours ago, bonanova said:

I think your answer is right. Mine is overkill. Are you sure that

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11 won't do?

I don't believe it will as the diameter/height of the triangle is 11.63 meaning that 11 lines will not be enough to cover it.  Unless using parallel lines is not the way to go.

 

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On 3/10/2016 at 1:10 PM, CaptainEd said:

Perhaps Bonanova's point is that

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20/sqrt(3) gives the number of parallel stripes, 11+, but it takes one less line (11) to separate 12 stripes.

 

 

No...

Spoiler

11 parallel lines in a unit circle would put the distance between each line at .182 with the height of the triangle being .173, the triangle could actually fit between the lines and be missed.  However with one more, we will ensure that we hit the target

On 3/9/2016 at 10:16 AM, TimeSpaceLightForce said:

I think of 5..if 1 unit circle is measured on its diamater 

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ct.jpg.367a539e0bd091e5f83feb03edcc5bf1.

 

with the height of the triangle at .173, this would not work, even in this context.

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35 minutes ago, TimeSpaceLightForce said:

They overlapped on one side (blue)while intersects on the other 2 sides (red) near the vertex. ..  the triangle height =0.7132 while the line distances =0.713 only .. can't fit between. 

 

 

 

I disagree with your numbers but I concede your point.  I was doing 2/11 to see that eleven lines wouldn't work when I should have been doing 2/12 to see that dividing the space into 12 areas would.

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