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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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 No Subject | Comment 9 of 12 |

PA = AB = BR = 1/3 PR

We will draw a perpendicular from Q on PR and name the point C where it meets PR.

Now

QA^2 = CA^2 + QC^2

QB^2 = CB^2 + QC^2

QA^2 + QB^2 = CA^2 + 2*QC^2 + CB^2

QC^2 = QP^2 - PC^2

QC^2 = QR^2 - CR^2

QA^2 + QB^2 = QP^2 - PC^2 + QR^2 - CR^2 + CA^2 + CB^2

Since

QP^2 + QR^2 = PR^2

QA^2 + QB^2 = PR^2 - (PA - CA)^2 - (CA + AR)^2 + CA^2 + (CA + AB)^2

QA^2 + QB^2 = PR^2 - PA^2 - CA^2 + 2*PA*CA + - CA^2 - AR^2  - 2*CA*AR + CA^2 + CA^2 + AB^2 + 2*CA*AB

QA^2 + QB^2 = PR^2 - (1/9 PR^2 + 4/9 PR^2 - 1/9 PR^2) + 2*( CA (PA+ AB - AR)

Since

PA + AB = PB = CR

Therefore

QA^2 + QB^2 = PR^2 - 4/9 PR^2

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

Therefore,

k=5/9

 Posted by akash on 2006-02-06 11:41:06

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