BMAD 63 Report post Posted May 28, 2013 (edited) Given a compass, a straightedge, and a line segment AB of length 3, how could you draw a line segment with a precise length of the square root of 3? Edited May 28, 2013 by BMAD Quote Share this post Link to post Share on other sites
0 ShadowAngel7 0 Report post Posted May 28, 2013 This would be so easy if it was "length 2 yields sqrt(3)" Quote Share this post Link to post Share on other sites
0 Rob_Gandy 2 Report post Posted May 28, 2013 First trisect the segment AB giving you segments of length 1. From that segment make a 1-1-sqrt(2) right triangle. Then using the sqrt(2) segment make a 1-sqrt(2)-sqrt(3) right triangle. Quote Share this post Link to post Share on other sites
0 phil1882 13 Report post Posted May 29, 2013 trisect the line of length 3 to get a line of length 1. add 1 ot the length to get a line of length 4. contruct a circle with a center at the midpoint. contruct a line perpendicular to the line at 3. the length of the perpendicular line is sqrt 3. Quote Share this post Link to post Share on other sites
0 k-man 26 Report post Posted May 29, 2013 1. Set the compass to the length of AB 2. Draw 2 circles centered at A and B with the radius of 3 finding points of their intersection C and D. 3. Draw a circle centered at C with the radius of 3 finding the point of intersection with B circle and call it E. 4. Draw line segments CD and AE. The intersection of these two line segments is O. O is the center of the equilateral triangle ABC. AO=BO=CO=sqrt(3). Quote Share this post Link to post Share on other sites
0 bonanova 83 Report post Posted May 30, 2013 I won't spoiler this answer cuz it's offered in jest. Start instead with a line AB of length 1 and construct a line of length sqrt(1). Then scale up by factor of three. Quote Share this post Link to post Share on other sites
0 bhramarraj 4 Report post Posted May 30, 2013 It seems to me as easy one, so I doubt following solution is correct. Draw a line at 30degree to AB at A, and make a perpendicular line at B. Say both lines meet at point C.The length BC = sqrt3. Quote Share this post Link to post Share on other sites
0 bonanova 83 Report post Posted May 30, 2013 It seems to me as easy one, so I doubt following solution is correct.Draw a line at 30degree to AB at A, and make a perpendicular line at B. Say both lines meet at point C.The length BC = sqrt3. Step 1 would be the problem. Quote Share this post Link to post Share on other sites
0 bhramarraj 4 Report post Posted May 30, 2013 It seems to me as easy one, so I doubt following solution is correct.Draw a line at 30degree to AB at A, and make a perpendicular line at B. Say both lines meet at point C.The length BC = sqrt3. Step 1 would be the problem. If compass is given, there should be no problem, an equilateral triange has 60 degree angles. OR I am missing some thing? Quote Share this post Link to post Share on other sites
0 bhramarraj 4 Report post Posted May 30, 2013 It seems to me as easy one, so I doubt following solution is correct.Draw a line at 30degree to AB at A, and make a perpendicular line at B. Say both lines meet at point C.The length BC = sqrt3. Step 1 would be the problem. If compass is given, there should be no problem, an equilateral triange has 60 degree angles. OR I am missing some thing? Make a perpendicular line AD of length 3 at A and make an equilateral trangle ADE (in the first quadrant). Angle BAE will be 30 degree. Make a perpendicular to BA at point B, and extend AE to meet this perpendicular at C. The length BC = sqrt3. Quote Share this post Link to post Share on other sites
Given a compass, a straightedge, and a line segment AB of length 3, how could you draw a line segment with a precise length of the square root of 3?
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