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 Trisected Triangle (Posted on 2006-02-03)
A right triangle PQR, has its hypotenuse, PR, trisected at points A and B. Two lines, QA and QB are then drawn and k is such that QA^2 + QB^2 = (PR^2) * k. Find the value of k.

 No Solution Yet Submitted by Chris, PhD Rating: 4.5000 (4 votes)

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 Simplify by Going Complex | Comment 3 of 12 |

k has the value 5/9 independent of the shape of the right triangle.

A simple demonstration results when the triangle is placed in the complex plane, inscribed in the unit circle with its hypotenuse PR the diameter along the real axis. Q is then exp(it) for some angle of t radians in the interval (0,pi), A is at 1/3, and B is at -1/3.  Thus QA^2=[exp(it)-1/3][exp(-it)-1/3] and QB^2=[exp(it)+1/3][exp(-it)+1/3] so that QA^2+QB^2=2 2/9=20/9. Since PR^2=4, we then have k=5/9.

Edited on February 3, 2006, 5:47 pm
 Posted by Richard on 2006-02-03 17:45:59

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