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 Tough Isosceles (Posted on 2005-07-20)
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?

 See The Solution Submitted by McWorter No Rating

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 Conditional Solution | Comment 2 of 6 |
` `
`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 2005-07-20 16:23:16

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