Question

question is based from book: Euclidean Plane and its Relatives, A Minimalist Introduction by Anton Petrunin

(18) (Modified Exercise 9.20, Fermat-Torricelli point) Let AABC be nondegenerate. Let P.Q. and R be such points that triangles BPC, CQA, and ARB are regular and 4BPC = LCQA= ARB = 5 (a) Prove that the circumcircles of triangles BPC, CQA, and ARB intersect in one point T, this point is called a Torricelli point of AABC (there is another Torricelli point when all angles given above are -3). (b) Prove that the lines (AP), (BQ), and (CR) all contain the point T. Hint: Intersect two of these circles and look for angles at the point of intersection. You can also look on Wikipedia for a nice picture that can help in your proof.

Answer 1

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Part(a)

Let $\Delta$ ABC is given triangle . we draw equilateral triangles $\Delta$ ARB , $\Delta$ BPC and $\Delta$ CQA on the sides AB , BC and CA of $\Delta$ ABC.

Let us draw circumcircles of $\Delta$ ARB and $\Delta$ BPC . Let it intersect at T .

Quadrilateral ARBT and Quadrilateral BPCT are cyclic .

Hence $\angle$ ATB = $\angle$ BTC = 120 because $\angle$ ARB = $\angle$ BPC = 60 .

Hence $\angle$ ATC = 120 and quadrilateral AQCT is cyclic because $\angle$ AQC = 60 .

Hence the Toricelli point is intersection of circumcircles of $\Delta$ ARB , $\Delta$ BPC and $\Delta$ CQA .

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To prove AP , BQ and CR contains T means AP , BQ and CR are straight lines and T is common intersection point.

It is already explained that $\angle$ ATC = 120 . We see from the figure $\angle$ PBC and $\angle$ PTC are angle subtended by chord PC in a circle . since $\angle$ PBC = 60 , then $\angle$ PTC = 60 . Hence $\angle$ ATC + $\angle$ PTC = 180 and AP is a straight line.

Similarly we can prove BQ and CR are straightlines . Hence T is common intersection point of AP, BQ and CR