Two tangents from an external point P are drawn to a circle and intersect it at A and B. A third tangent meets the circle at T, and the tangents PA and PB at points Q and R, respectively. Find the perimeter of Triangle PQR.

Question

anonymous · Answer

The problem involves Power of a Point theorem.

anonymous · Answer

Here is the diagram that I made for the question. This should make things easier to understand. Also, this problem requires the use of the Power of a Point theorem, so keep that in mind.

anonymous · Answer

@satellite73

anonymous · Answer

@campbell_st 

@Luis_Rivera 

@phi

campbell_st · Answer

let PQ = a and AQ = b

well you need to know that tangents from an external point are equal 

so you know 

AQ = QT and RB = RT  and PA = PB 

therefore PQ = RQ

so the triangle is isosceles 

then the perimeter is 

P = 2a + 2b

thats about the best I can do off the top of my head.

anonymous · Answer

I did everything you did, the only thing that worried me was that the Perimeter should be some numbers. Now of course that doesn't have to be the case, but I noted everything you did so I am just trying to make sure that the Perimeter in fact is not a number which you also seem to have concluded. If that's the case, then I guess things should be fine.

campbell_st · Answer

oopss... here is the better solution 

since QT = AQ and RT = RB

then the perimeter is  P = PQ + QT + PR + TR 

or 

P = PQ + QA + PR + TB  = PA + PB

since tangents from an enternal point are equal then P = 2PA

that makes more sense... 

hope you can follow it.

campbell_st · Answer

oops should read 

P = PQ + QA + PR + RB = PA + PB

sorry for the typo

campbell_st · Answer

then P = 2PA

anonymous · Answer

Yeah that makes sense. I now realise the only thing that threw me off is the fact that the question is supposed to involve Power of a Point theorem and it doesn't really have much of an application to this question. That's why I kept trying hard to make some sense out of it but turns that the question really doesn't have much to do with it lol. Anyway ty.

campbell_st · Answer

I'm not familiar with power of a point... I know tangents from an external point are equal... and the logic  works... 

glad to help

anonymous · Answer

Yes it does indeed. The whole power of a point theorem threw me off and made me think that it was what I needed to solve this question when it actually was not =,=