Let A and B be different points on a circle with center O. With only a straightedge and a compass, can you construct a straight line through O meeting the segment AB at C, C strictly between A and B, and meeting the circle at D, so that C is between O and D and the segments AC and AD have the same length?
From triangle CAD: <CAD + <ADC + < ACD = 180
From isosceles triangle ODA : <ODA = <OAD
Combining these with the fact that
<ADC = <ODA and <OAC = <OAB
gives
<ODA = <OAD = <OAC + <CAD = <OAC + (180  <ADC  <ACD)
= <OAB + (180  <ODA  <ACD)
or
<OAB = 2 <ODA + <ACD  180
Therefore,
AD = AC <=> <ACD = <ADC = <ODA
<=> <OAB = 3 (<ODA  60)
or
AD = AC <=> <OAB/3 = <ODA  60
Therefore, our construction is possible if and only if
<OAB can be trisected with a straightedge and compass.

Posted by Bractals
on 20050720 16:23:16 