An Excursion through Elementary Mathematics, Volume II by Antonio Caminha Muniz Neto

Author:Antonio Caminha Muniz Neto
Language: eng
Format: epub
Publisher: Springer International Publishing, Cham

As we stressed before, our final example makes use of the property of angle invariance (more precisely, of orthogonality invariance) of inversions.

Example 9.9 (IMO—Adapted)

Let ABC be an acute triangle and D be an interior point, with and . Prove that the tangents to the circumcircles of triangles ACD and BCD at a common point are perpendicular to each other.

Proof

We let the figure below represent the described situation, and let Γ be the circumcircle of ACD and Σ be that of BCD.

To say that the tangents to Γ and Σ at D are perpendicular is the same as to say that Γ and Σ are orthogonal. In turn, Theorem 9.8 and the subsequent discussion assure that this is so if and only if an inversion of center C transforms these circles into two perpendicular straightlines. As we shall now see, this gives a clue to the solution.

Indeed, upon the inversion of center C and ratio , let A′ be the inverse of A and B′ be that of B. Since D is its own inverse, Proposition 9.4 guarantees that is the inverse of Γ and is that of Σ. Thus,

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