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Trisected Triangle (Posted on 2006-02-03) Difficulty: 3 of 5
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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No Subject | Comment 10 of 12 |

DRAW THE TRIANGLE FIRST and observe AP=AB=BR plus

PQ^2 + QR^2 = PR^2

in triangle PQB , QA is median so

2(QA^2 + AP^2) = PQ^2 + QB ^2 (Apollonius theorem)

Similarly in triangle AQR

2(QB^2 + AB^2)=QA^2 + QR^2 (Apollonius theorem)

ADD em nd get the result

QB^2 + QA^2 = 5/9 PR^2


  Posted by phi on 2006-03-09 05:22:58
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