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On the Area of a Pedal Triangle Ivan Borsenco Geometry Has Been

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On the Area of a Pedal Triangle Ivan Borsenco Geometry Has Been On the area of a pedal triangle Ivan Borsenco Geometry has been always the area of mathematics that attracted problem solvers with its exactness and intriguing results. The article presents one of such beautiful results - the Euler’s Theorem for the pedal triangle and its applications. We start with the proof of this theorem and then we discuss Olympiad 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 A1,B1,C1 the projections of M on triangle’s sides. The following relation holds 2 2 SA1B1C1 |R − OM | = 2 . SABC 4R Proof: First of all note that quadrilaterals AB1MC1, BC1MA1, CA1MB1 are cyclic. Applying the extended Law of Cosines in triangle AB1C1 we get B1C1 = AM sin α. Analogously A1C1 = BM sin β and B1C1 = CM sin γ. It follows that B C AM A C BM A B CM 1 1 = , 1 1 = , 1 1 = . BC 2R AC 2R BC 2R Suppose AM, BM, CM intersect the circumcircle C(O, R) at points X, Y, Z. Mathematical Reflections 2 (2007) 1 Angle chasing yields ∠A1B1C1 = ∠A1B1M+∠MB1C1 = ∠A1CM+∠MAC1 = ∠ZYB+∠BY X = ∠ZY X. Similarly, ∠B1C1A1 = ∠YZX and ∠B1A1C1 = ∠YXZ. Thus triangles A1B1C1 and XYZ are similar and A B R 1 1 = A1B1C1 . XY R Because triangles MAB and MYX are also similar we have XY MX = . AB MB Combining the above results we obtain 2 2 SA1B1C1 R A1B1 · B1C1 · A1C1 MX MA MB MA · MX |R − OM | = · = · · = 2 = 2 . SABC RA1B1C1 AB · BC · AC MB 2R 2R 4R 4R As we can see, the proof does not depend on the position of M (inside or outside the circle). Corollary 1. If M lies on the circle, the projections of M onto triangle’s sides 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 following relation holds vwa2 + uwb2 + uvc2 u · PA2 + v · PB2 + w · PC2 = (u + v + w)PM 2 + . u + v + w The proof can be found after applying Stewart’s Theorem a few times. The corollary of this theorem in which we are interested is the case when P coin- cides with the circumcenter O. We get vwa2 + uwb2 + uvc2 R2 − OM 2 = . (u + v + w)2 From this fact and the Euler’s Theorem for pedal triangle we obtain the next result: Mathematical Reflections 2 (2007) 2 Theorem 3. Let M be a point in the plane of triangle ABC with barycentric coordinates (u, v, w). Denote by A1,B1,C1 the projections of M onto triangle’s sides. Then 2 2 2 SA1B1C1 vwa + uwb + uvc = 2 2 . SABC 4R (u + v + w) From the above results we can see that Theorem 1 and Theorem 3 give us insight 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 one application we present is about Brocard points. Let us introduce the definition of a Brocard point. In triangle ABC the circle that passes through A and is tangent to BC at B, circle that passes through 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 concurrency is called Brocard point. In general, we have two Brocard points, the second one being obtained by reversing clockwise or counterclockwise the tangency of circles. Problem 1. Let Ω1 and Ω2 be the two Brocard points of triangle ABC. Prove that OΩ1 = OΩ2, where O is the circumcenter of ABC. Solution: From the definition of the Brocard points we see that ∠Ω1AB = Ω1BC = Ω1CA = w1, and similarly ∠Ω2BA = Ω2AC = Ω2CB = w2. Let us prove that w1 = w2. 2 SBΩ1C BΩ1 · BC sin w1 sin w1 Observe that = = 2 and, analogously, SABC AB · BC sin β sin β 2 2 SCΩ1A sin w1 SAΩ1B sin w1 = 2 , = 2 . SABC sin γ SABC sin α Mathematical Reflections 2 (2007) 3 Summing up the areas we obtain 2 2 2 SBΩ1C + SCΩ1A + SAΩ1B sin w1 sin w1 sin w1 1 = = 2 + 2 + 2 or SABC sin α sin β sin γ 1 1 1 1 2 = 2 + 2 + 2 . sin w1 sin α sin β sin γ The same expression we get summing the areas for Ω2: 1 1 1 1 2 = 2 + 2 + 2 sin w2 sin α sin β sin γ From two equalities we conclude that w1 = w2 = w. The idea for the solution comes from Euler’ Theorem for pedal triangles. Observe that Ω1 and Ω2 always lie inside triangle ABC, because every circle lies in that half of the plane that contains triangle ABC. It follows that the point of their intersection Ω1 or Ω2 lies inside the triangle. In order to prove that OΩ1 = OΩ2, we prove that their pedal areas are equal. If so, then we 2 2 2 2 have R − OΩ1 = R − OΩ2 and therefore OΩ1 = OΩ2. Denote by A1,B1,C1 the projections from Ω1 onto BC, AC, AB, respec- tively. Then using the extended Law of Sines we get A1C1 = BΩ1 sin b, because BΩ1 is diameter in the cyclic quadrilateral BA1Ω1C1. Also from the Law of Sines in triangle ABΩ1 we have BΩ c 1 = . sin w sin b It follows that A1C1 = BΩ1 sin b = c sin w. Similarly we obtain B1C1 = b sin w and A1B1 = a sin w. It is not difficult to see that triangle A1B1C1 is similar to ABC with ratio of similarity sin w. From the fact w1 = w2 = w we conclude that pedal triangles of Ω1 and Ω2 have the same area. Thus OΩ1 = OΩ2, 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 KΩ1 = KΩ2. Moreover O, Ω1, K, Ω2 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 Problem 2. Consider a triangle ABC and denote by O, I, H its circum- center, incenter, and orthocenter, respectively. Prove that OI ≤ OH. Solution: Many solvers will start to play with formulas for OI and OH. We choose another way, we consider the pedal areas for I and H. Clearly, the incenter is always inside the triangle ABC and if OH ≥ R, then we are done. Therefore assume that OH < R. It follows that in this case we have an acute-angled triangle. In order to prove that OI ≤ OH, we can use Euler’s Theorem for pedal triangles and prove that the area of pedal triangle I, SI is greater than the area of the pedal triangle of H, SH . This is because if 2 2 2 2 SI ≥ SH then R − OI ≥ R − OH and therefore OH ≥ OI. Let us find both areas. Denote by A1,B1,C1 the projections of H and A2,B2,C2 the projections of I onto BC, CA, AB, respectively. Recalling the fact that A1,B1,C1 lie on the Euler circle of radius R/2 we get A B · B C · A C S = 1 1 1 1 1 1 . A1B1C1 4 · R/2 Using the Law of Sines it is not difficult to see that B1C1 = a cos α, A1C1 = b cos β, A1B1 = c cos γ and therefore abc · cos α cos β cos γ S = = 2S cos α cos β cos γ. A1B1C1 2R To calculate the area of triangle A2B2C2, observe that 1 r2(a + b + c) S = S +S +S = r2(sin α+sin β+sin γ) = . A2B2C2 A2IB2 A2IC2 B2IC2 2 4R r α β γ Recalling the formula = sin sin sin we obtain 4R 2 2 2 r α β γ S = 2S · = 2S sin sin sin . A2B2C2 4R 2 2 2 It follows that it is enough to prove that α β γ sin sin sin ≥ cos α cos β cos γ. 2 2 2 Now we use the classical Jensen Inequality. Consider the concave function π f : (0, 2 ) → R, f(x) = ln cos x. Then a + b b + c a + c f( ) + f( ) + f( ) ≥ f(a) + f(b) + f(c). 2 2 2 α β γ Thus ln sin sin sin ≥ ln(cos α cos β cos γ), and the problem is solved. 2 2 2 The next two examples feature Olympiad problems where Euler’s Theorem for pedal triangle is probably the shortest way to solve them. Mathematical Reflections 2 (2007) 5 Problem 3. (Balkan Mathematical Olympiad) Let ABC be an acute triangle. Denote by A1,B1,C1 the projections of centroid G onto triangle’s sides. Prove that 2 S 1 ≤ A1B1C1 ≤ . 9 SABC 4 Solution: As we will see this problem requires a direct application of Euler’s Theorem for pedal triangles. Applying it we have to prove 2 R2 − OG2 1 ≤ ≤ . 9 4R2 4 The right-hand side immediately follows. To prove the left-hand side recall the well known formula 9OG2 = OH2 = R2(1 − 8 cos α cos β cos γ). Because triangle ABC is acute-angled we have cos α, cos β, cos γ ≥ 0, hence 8 cos α cos β cos γ ≥ 0, and thus 9OG2 ≤ R2 or OG2 ≤ R2/9.
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