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 Sum Segment Length (Posted on 2016-06-25)
PQR is a triangle with ∠QPR = 60o.

The point S lies inside the triangle in such a way that:
∠PSQ = ∠QSR = ∠RSP = 120o.

M is the midpoint of QR, and PM= x

Find SP+SQ+SR in terms of x.

 No Solution Yet Submitted by K Sengupta No Rating

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This is a tricky problem.

I try to deal with it in two ways:

a) looking at features of the triangles and sides on the drawn figure. It seems that things are clear but I wasn't able to obtain a clear relation between SP+SQ+SR and x.

b) As point S is the Fermat point, it is possible to use
the equilateral-triangle feature of the Fermat point to solve it. Here is better to place a coordinate system centred in P, with the axis X in PR. The goal is to get the intersection between the circle rounding the equi-triangle PRV and the line QV (this is quite a good way but with the inconvenience that using abstract quantities, the formulas get easily complex, and lead to err).

To have a clear orientation I decided to use the second way but defining more the problem. So I add PR=10 and QP=6. This was interesting, with nice numbers and lead to what I had suspected using the first way. I get that SP+SQ+SR=14 and x=7. So the Fermat sum is twice the median.

Then I reviewed from there what I got before, using the first way, and I understood why this is a tricky problem. I explain now why:

For the cosine rule you see that: PQ^2=SQ^2+SP^2+SQ*SP. (cos 120=-1/2) [1]
Similar formulas are easily deduce for PR and QR. In addition, the median formula stablish that:
PM^2=x^2= (2PQ^2+2PR^2-QR^2)/4. [2]

Applying [1 and similar] to [2] and putting simpler:
4x^2=(SP+SQ+SR)^2+SP^2-SQ*SR
So you don't really arrive to express SP+SQ+SR as an x function.

Where is the trick? The trick is that SP^2=SQ*SR, so that their difference is 0, but I didn't found this easy to prove.
Anyway, it's possible to show this at least (but probably there are other simplier ways to do the same, i.e: using the areas) recalling other formula for the median PM, so to have a check between both formulas. For ex: to apply the cosine rule to the median. As the angle involved here is also 120 degrees, it comes that:
PM^2=PQ^2+PR^2+PQ*PR [3]

From [3] and [2] PQ*PR=PQ^2+PR^2-QR^2. As you can express all this in terms of SP, SQ, SR take a good breath and compare both formulas. What do you get? SQ^2=SP*SR

Then PM^2=x^2=(SP+SQ+SR)^2 and so
PM=x=(SP+SQ+SR)/2

Edited on July 10, 2016, 9:52 am
 Posted by armando on 2016-07-10 06:43:19

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