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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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 A much easier way to solve this problem | Comment 8 of 12 |

This is one of those questions where, at first glance, it seems that there is not enough information given to us to solve the problem.  We know that the triangle is a right triangle, but we aren't given any information regarding the other two angles, those at P and R.

In this case, we have no choice but to assume that the value of k must be independent of the values of those angles.  That is, regardless of what angle we have at point P, k will be invariant.

With this in mind, let's go to extremes!  Allow the angle at P to get smaller and smaller, until, in the limiting case, the hypotenuse PR becomes coincident with side PQ.  (Drawing a quick sketch will help you to visualize this.)  At this point, we can let PQ = 3, so PR = 3, and QA = 2, QB = 1.  (Check this!)

So k = (QA^2+QB^2)/PR^2

=(2^2 + 1^2)/3^2

=5/9

Of course, the value of k agrees with those arrived at in earlier comments, but I believe that this is a far simpler and clearer solution method.  (Having said that, I thought that Richard's argument involving PR being placed on the real axis diameter of a unit circle in the complex plane was quite clever!)

Thanks for the problem Chris.

-John

 Posted by John Reid on 2006-02-04 22:47:12

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