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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?

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 solution | Comment 5 of 6 |

Angle ADC is a base angle of the desired isosceles triangle ADC.  It also coincides with angle ADO, which is a base angle of the isosceles triangle ADO, so triangles ADC and ADO are similar, with angle DAC = angle AOD.

But angle DAC is on the circumference of the circle and subtends arc DB, which therefore has twice the measure of angle DAC (i.e., twice the measure of angle AOD).  But the measure of angle AOD is the measure of arc AD.  Therefore arc AD is 1/3 the measure of ard AB, and angle AOD is 1/3 the measure of angle AOB.

Thus, if the construction were to work for arbitrary points A and B, it would trisect arbitrary angle AOB, which is impossible with compass and straightedge.

So the answer is No, you cannot construct the desired isosceles triangle with a compass and straightedge.

 Posted by Charlie on 2005-07-20 18:56:12

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