Given a line segment AB, describe how to construct a segment whose length is [math]\frac{\sqrt{3}-1}{\sqrt{2}-1}[/math] times the length of AB, using a compass and unmarked straightedge.[br][br]1. Create a line segment with unit length 1[br][br][br]2. Make another line segment twice the unit length perpendicular to the original length[br][br][br]3. Connect to make a right triangle[br][br][br]4. Create an altitude from the hypotenuse to the right angle, label it x[br][br][br]5. Solving for x using similar triangles gives us √2[br][br][br]6. Repeat steps 1 – 5 with segments of length 1 and 3[br][br][br]7. Solving for x gives us a segment with length √3[br][br][br]8. Move new segments from steps 5 and 7, taking off a length of 1 from each.[br][br][br]9. Connect segments from step 8 at one point O forming an arbitrary acute angle[br][br][br]10.  Segment √2 –1 will be OP and segment √3 – 1 will be OQ[br][br][br]11.  Extend the segment √2 –1 to have a length of 1,call it OR[br][br][br]12.  Connect P and Q, make a parallel line to PQ going through R[br][br][br]13.  Extend the segment √3 –1 to intersect with parallel line going through R, call point of intersection S[br][br][br]14.  Connect point O with S, this is line segment x[br][br][br]15.  Now you can create a ratio with [math]\frac{OS}{OR}=\frac{OQ}{OP}[/math][br][br][br]