Drawing the square root of 3

[BMAD]

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

[ShadowAngel7]

This would be so easy if it was "length 2 yields sqrt(3)"

[Rob_Gandy]

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.

[phil1882]

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.

[k-man]

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).

[bonanova]

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.

[bhramarraj]

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.

[bonanova]

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.

[bhramarraj]

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?

[bhramarraj]

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.