Area of Pedal Triangles - Ivan Borsenco - MR | Triangle | Sine

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  On the area of a pedal triangleIvan Borsenco Geometry has been always the area of mathematics that attracted problemsolvers with its exactness and intriguing results. The article presents one of such beautiful results - the Euler’s Theorem for the pedal triangle and itsapplications. We start with the proof of this theorem and then we discussOlympiad problems. Theorem 1.  Let  C  ( O,R ) be the circumcircle of the triangle  ABC  .Consider a point  M   in the plane of the triangle. Denote by  A 1 ,B 1 ,C  1  theprojections of   M   on triangle’s sides. The following relation holds S  A 1 B 1 C  1 S  ABC  =  | R 2 − OM  2 | 4 R 2  . Proof:  First of all note that quadrilaterals AB 1 MC  1 ,  BC  1 MA 1 ,  CA 1 MB 1 are cyclic. Applying the extended Law of Cosines in triangle  AB 1 C  1  we get B 1 C  1  =  AM   sin α . Analogously  A 1 C  1  =  BM   sin β   and  B 1 C  1  =  CM   sin γ  . Itfollows that B 1 C  1 BC   =  AM  2 R , A 1 C  1 AC   =  BM  2 R , A 1 B 1 BC   =  CM  2 R . Suppose  AM,BM,CM   intersect the circumcircle  C  ( O,R ) at points  X,Y,Z  . Mathematical Reflections  2  (2007)  1  Angle chasing yields ∠ A 1 B 1 C  1  = ∠ A 1 B 1 M  + ∠ MB 1 C  1  = ∠ A 1 CM  + ∠ MAC  1  = ∠ ZYB + ∠ BYX   = ∠ ZYX. Similarly,  ∠ B 1 C  1 A 1  =  ∠ YZX   and  ∠ B 1 A 1 C  1  =  ∠ YXZ  . Thus triangles A 1 B 1 C  1  and  XYZ   are similar and A 1 B 1 XY   =  R A 1 B 1 C  1 R . Because triangles  MAB  and  MYX   are also similar we have XY AB  =  MX MB. Combining the above results we obtain S  A 1 B 1 C  1 S  ABC  =  RR A 1 B 1 C  1 · A 1 B 1 · B 1 C  1 · A 1 C  1 AB · BC  · AC   =  MX MB · MA 2 R  · MB 2 R  =  MA · MX  4 R 2  =  | R 2 − OM  2 | 4 R 2  . As we can see, the proof does not depend on the position of   M   (inside oroutside the circle). Corollary 1.  If   M   lies on the circle, the projections of   M   onto triangle’ssides are collinear. This fact is known as Simson’s Theorem.One more theorem we want to present without proof is the famous La-grange Theorem. Theorem 2.  Let  M   be a point in the plane of triangle  ABC   with barycen-tric coordinates ( u,v,w ). For any point  P   in the plane of   ABC   the followingrelation holds u · PA 2 + v · PB 2 + w · PC  2 = ( u + v  + w ) PM  2 +  vwa 2 + uwb 2 + uvc 2 u + v  + w . The proof can be found after applying Stewart’s Theorem a few times. Thecorollary of this theorem in which we are interested is the case when  P   coin-cides with the circumcenter  O . We get R 2 − OM  2 =  vwa 2 + uwb 2 + uvc 2 ( u + v  + w ) 2  . From this fact and the Euler’s Theorem for pedal triangle we obtain the nextresult: Mathematical Reflections  2  (2007)  2  Theorem 3.  Let  M   be a point in the plane of triangle  ABC   withbarycentric coordinates ( u,v,w ). Denote by  A 1 ,B 1 ,C  1  the projections of   M  onto triangle’s sides. Then S  A 1 B 1 C  1 S  ABC  =  vwa 2 + uwb 2 + uvc 2 4 R 2 ( u + v  + w ) 2  . From the above results we can see that Theorem 1 and Theorem 3 give usinsight on the area of pedal triangles. The Euler’s Theorem for the area of a pedal triangle is a useful tool in solving geometry problems. The first oneapplication we present is about Brocard points.Let us introduce the definition of a Brocard point. In triangle  ABC   thecircle that passes through  A  and is tangent to  BC   at  B , circle that passesthrough  B  and is tangent to  AC   at  C   and the circle that passes through  C  and is tangent to  AB  at  A  are concurrent. The point of their concurrencyis called Brocard point. In general, we have two Brocard points, the secondone being obtained by reversing clockwise or counterclockwise the tangency of circles. Problem 1.  Let Ω 1  and Ω 2  be the two Brocard points of triangle  ABC  .Prove that  O Ω 1  =  O Ω 2 ,  where  O  is the circumcenter of   ABC  . Solution:  From the definition of the Brocard points we see that ∠ Ω 1 AB  = Ω 1 BC   = Ω 1 CA  =  w 1 ,  and similarly  ∠ Ω 2 BA  = Ω 2 AC   =Ω 2 CB  =  w 2 . Let us prove that  w 1  =  w 2 .Observe that  S  B Ω 1 C  S  ABC  =  B Ω 1 · BC   sin w 1 AB · BC   sin β   = sin 2 w 1 sin 2 β   and, analogously, S  C  Ω 1 A S  ABC  = sin 2 w 1 sin 2 γ  , S  A Ω 1 B S  ABC  = sin 2 w 1 sin 2 α . Mathematical Reflections  2  (2007)  3  Summing up the areas we obtain1 =  S  B Ω 1 C   + S  C  Ω 1 A  + S  A Ω 1 B S  ABC  = sin w 21 sin 2 α  + sin w 21 sin 2 β   + sin w 21 sin 2 γ   or1sin w 21 = 1sin 2 α  + 1sin 2 β   + 1sin 2 γ . The same expression we get summing the areas for Ω 2 :1sin w 22 = 1sin 2 α  + 1sin 2 β   + 1sin 2 γ  From two equalities we conclude that  w 1  =  w 2  =  w .The idea for the solution comes from Euler’ Theorem for pedal triangles.Observe that Ω 1  and Ω 2  always lie inside triangle  ABC  , because every circlelies in that half of the plane that contains triangle  ABC  . It follows that thepoint of their intersection Ω 1  or Ω 2  lies inside the triangle. In order to provethat  O Ω 1  =  O Ω 2 ,  we prove that their pedal areas are equal. If so, then wehave  R 2 − O Ω 21  =  R 2 − O Ω 22  and therefore  O Ω 1  =  O Ω 2 . Denote by  A 1 ,B 1 ,C  1  the projections from Ω 1  onto  BC,AC,AB , respec-tively. Then using the extended Law of Sines we get  A 1 C  1  =  B Ω 1  sin b ,because  B Ω 1  is diameter in the cyclic quadrilateral  BA 1 Ω 1 C  1 . Also from theLaw of Sines in triangle  AB Ω 1  we have B Ω 1 sin w  =  c sin b. It follows that A 1 C  1  =  B Ω 1  sin b  =  c sin w. Similarly we obtain  B 1 C  1  =  b sin w  and  A 1 B 1  =  a sin w . It is not difficultto see that triangle  A 1 B 1 C  1  is similar to  ABC   with ratio of similarity sin w .From the fact  w 1  =  w 2  =  w  we conclude that pedal triangles of Ω 1  and Ω 2 have the same area. Thus  O Ω 1  =  O Ω 2 ,  and we are done. Remark:  The intersection of symmedians in the triangle is denoted by K   and is called Lemoine point. One can prove that  K  Ω 1  =  K  Ω 2 .  Moreover O, Ω 1 ,K, Ω 2  lie on a circle with diameter  OK   called Brocard circle.We continue with an interesting approach to another classical problem. Mathematical Reflections  2  (2007)  4
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