BMAD Posted March 7, 2016 Report Share Posted March 7, 2016 (edited) 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 March 7, 2016 by BMAD 1 Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 10, 2016 Report Share Posted March 10, 2016 On 3/9/2016 at 11:11 PM, BMAD said: I have it in a much smaller amount but not as small as our new friend above. My answer is... Reveal hidden contents 12 even spaced and parallel lines Expand I think your answer is right. Mine is overkill. Are you sure that Reveal hidden contents 11 won't do? Quote Link to comment Share on other sites More sharing options...
0 SamsonBrainPower Posted March 8, 2016 Report Share Posted March 8, 2016 3. On 3/8/2016 at 12:42 AM, SamsonBrainPower said: 3. Expand 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 Quote Link to comment Share on other sites More sharing options...
0 SamsonBrainPower Posted March 8, 2016 Report Share Posted March 8, 2016 On 3/8/2016 at 12:42 AM, SamsonBrainPower said: 3. 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 Quote Link to comment Share on other sites More sharing options...
0 BMAD Posted March 8, 2016 Author Report Share Posted March 8, 2016 On 3/8/2016 at 1:40 AM, 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 Expand care to have your father explain? I have a different solution. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted March 9, 2016 Report Share Posted March 9, 2016 Looks like ... Reveal hidden contents 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. Quote Link to comment Share on other sites More sharing options...
0 BMAD Posted March 9, 2016 Author Report Share Posted March 9, 2016 (edited) On 3/9/2016 at 10:42 PM, bonanova said: Looks like ... Reveal hidden contents 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. Expand I have it in a much smaller amount but not as small as our new friend above. My answer is... Reveal hidden contents 12 even spaced and parallel lines Edited March 9, 2016 by BMAD Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted March 10, 2016 Report Share Posted March 10, 2016 I think of 5..if 1 unit circle is measured on its diamater Reveal hidden contents Quote Link to comment Share on other sites More sharing options...
0 BMAD Posted March 10, 2016 Author Report Share Posted March 10, 2016 On 3/10/2016 at 12:22 PM, bonanova said: I think your answer is right. Mine is overkill. Are you sure that Reveal hidden contents 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. Expand Quote Link to comment Share on other sites More sharing options...
0 CaptainEd Posted March 11, 2016 Report Share Posted March 11, 2016 Perhaps Bonanova's point is that Reveal hidden contents 20/sqrt(3) gives the number of parallel stripes, 11+, but it takes one less line (11) to separate 12 stripes. Quote Link to comment Share on other sites More sharing options...
0 BMAD Posted March 16, 2016 Author Report Share Posted March 16, 2016 On 3/11/2016 at 6:10 PM, CaptainEd said: Perhaps Bonanova's point is that Reveal hidden contents 20/sqrt(3) gives the number of parallel stripes, 11+, but it takes one less line (11) to separate 12 stripes. Expand No... Reveal hidden contents 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 3/10/2016 at 3:16 PM, TimeSpaceLightForce said: I think of 5..if 1 unit circle is measured on its diamater Reveal hidden contents Expand with the height of the triangle at .173, this would not work, even in this context. Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted March 17, 2016 Report Share Posted March 17, 2016 (edited) did the CAD again to prove bonanova's 11 Reveal hidden contents Where can the equilateral triangle be without touching the 11 blue lines.. unless inside means not fully inside. Edited March 17, 2016 by TimeSpaceLightForce Quote Link to comment Share on other sites More sharing options...
0 Brainy Binary Posted March 17, 2016 Report Share Posted March 17, 2016 Where are the lines and triangles intersects ?? if height of triangle = distance between 2 lines Quote Link to comment Share on other sites More sharing options...
0 TimeSpaceLightForce Posted March 18, 2016 Report Share Posted March 18, 2016 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. Quote Link to comment Share on other sites More sharing options...
0 BMAD Posted March 18, 2016 Author Report Share Posted March 18, 2016 On 3/18/2016 at 4:49 PM, 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. Expand 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. Quote Link to comment Share on other sites More sharing options...
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BMAD
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?
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