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The Distance Between the Circumcenter and Any Point in the Plane of the Triangle

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The Distance Between the Circumcenter and Any Point in the Plane of the Triangle THE DISTANCE BETWEEN THE CIRCUMCENTER AND ANY POINT IN THE PLANE OF THE TRIANGLE Prof. Dasari Naga Vijay Krishna Department of Mathematics, Narayana Educational Institutions Machilipatnam, Bangalore, INDIA [email protected] Abstract:In this article we give a metric relation which gives the distance between circumcenter to any point in the plane of the triangle. Keywords: Circumcenter, Stewart’s theorem, circumradius, Euler’s Inequality. 1.Introduction It is well-known that the perpendicular bisectors of the sides of a triangle meet at a single point, which is the center of the circumcircle [8],[9]. In this article we first give a metric relation which finds the distance between the circumcenter and any point in the plane of the triangle which allows us to prove some theorems related to circumcenter and in conclusion of the article we will prove the famous ‘Euler’s Inequality’. 2.Notations: Let ABC be a triangle. We denote its side-lengths by a=BC, b=AC, c=AB, a b c and angles by A, Band C its semi perimeter by s , its area by , 2 Its classical centers are the circumcenter S, the incenter I, the centroid G, and the orthocenter O . The nine-point center N is the midpoint of SO and the center of the nine-point circle, which passes through the side-midpoints and the feet of the three altitudes. The Euler Line Theorem states that G lies on SO with OG : GS = 2 : 1 and ON : NG : GS = 3 : 1 : 2. We write I1, I2, I3 for the excenters opposite A, B, C, respectively, these are points where one internal angle bisector meets two external angle bisectors. Like I, the points I1, I2, I3 are equidistant from the lines AB, BC, and CA, and thus are centers of three circles each tangent to the three lines. These are the excircles. The classical radii are the circumradius R (= SA = SB = SC), the inradius r, and the exradii r1, r2, r3. Let D, E and F are the foot of the cevians drawn through S from the vertices A,B and C to the opposite sides BC,CA and AB respectively. 139 The following formulae are well known abc (a) rsrsa1 rsb 2 rsc 3 ssasbsc 4R abc 2 (b) sin2A sin2 B sin2 C 4sin ABC sin sin 2R3 R 2 (c) sin2A sin2 B sin2 C 4cos ABC cos sin (d) sin2A sin2 B sin2 C 4cos ABC sin cos (e) sin2A sin2 B sin2 C 4sin ABC cos cos r (f) cosA cos B cos C 1 R r (g) cosB cos C cos A 1 1 R r (h) cosA cos C cos B 2 1 R r (i) cosA cos B cos C 3 1 R 22r 2 2 r 2 2 r (j) AIrsa,BI rsb and CI rsc A B C sin sin sin 2 2 2 2 2r 22 r 2 2 r (k) AI r s1,B I r s c 1 and C I r s b 1 1 1A 1 1 B 1 1 C sin cos cos 2 2 2 A sbs( c ) (l) Tan 2s ( s a ) 3.BASIC LEMMA’S Lemma -1 If S is the circumcenter of triangle ABC (acute or right), the cevians through S from the vertices A, B and C intersects opposite sides BC,CA and AB at the points D,E and F respectively then BD sin2C CE sin2A AF sin2B (1.1) = , = and = DC sin2B EA sin2C FB sin2A 140 Proof: Since S is circumcenter, we have ASB 2 C and ASC 2 B It implies BAD 90 C and CAD 90 B So by applying the sine rule for the triangles ABD and ACD, BD sinCcosC sin 2C We can show DCsinBcosB sin 2 B Which proves the required results. The above results are also true even if the triangle is obtuse, It is clear that the ratios are internal if the triangle is acute or right and they are external if the triangle is obtuse. 141 Lemma -2 If S is the circumcenter, then Rsin2A Rsin2B Rsin2C (2.1) SD = , SE = and SF = sin2B + sin2C sin2A+ sin2C sin2A+ sin2B 2Δ 2Δ 2Δ (2.2) AD = , BE = and CF = R sin2B+ sin2C R sin2A+ sin2C R sin2A+ sin2B Proof: BDsin 2 C By lemma-1(1.1) we have DCsin 2 B a sin 2C a sin 2B So BD and CD sin 2C sin 2 B sin 2C sin 2 B And Now for the triangle ADC, the line BSE acts as transversal AE CB DS So by Menelaus theorem we have 1 EC BD SA By replacing the all known relations and by little algebra we can arrive at the required conclusions of (2.1) And by using (b), and the fact AD = AS+SD, with little algebra we can prove the conclusions of (2.2) . 4.MAIN RESULT THEOREM – 1 If S is the circumcenter of an acute or right triangle and M be any point in the plane of triangle, then 2 2R 2 2 2 SM = sin2A.AM + sin2B.BM + sin2C.CM - 2Δ 2Δ 142 Proof: Let M be any point in the plane of triangle, Now from triangle BMC, DM is cevian, Hence by Stewart’s theorem 2 2 2 BDCM.. CDBM We have DM BDDC. BC BC 2 2sin2C 2 sin2 B 2 aBC sin2sin2 DM CM BM 2 sin2CB sin2 sin2 CB sin2 sin 2C sin 2 B 143 Now from triangle ADM, SM is cevian, Hence again by Stewart’s theorem 2 2 2 AS.. DM SD AM We have SM AS. SD AD AD 2 2 2Rsin2 C sin2 B sin2C 2 sin2 B 2 aBC sin2 sin2 SM CM BM 2 2 sin2CB sin2 sin2 CB sin2 sin2C sin2 B 2 2 RAsin 22 RA sin 2 AM 2 sin 2C sin 2 B Further simplification gives 2 2R 2 2 2 SMsin2 A .AM sin2 B .BM sin2 C .CM 2 2 Hence proved. The Theorem -1is also true even if the triangle is obtuse. 5.APPLICATIONS (I). Due to EULER the following are well known (IA). SI2 R 2 2 Rr 2 2 (IB). SI1 R 2 Rr 1 Proof: By theorem-1, 2 2R 2 2 2 We have SMsin2 A .AM sin2 B .BM sin2 C .CM 2 2 Now let us fix M as I, 2 2R 2 2 2 So SIsin2 A .AI sin2 B .BI sin2 C .CI 2 2 144 2 2 2 R r Using (j), SIsin 2 A . 2 2 A 2 sin 2 Using (a), (b), (f) and (k) and by brute force computation We can prove the required conclusion (IA) In the similar manner if we fix M as I1, and by computation we can arrive at the conclusion (IB). (II). If X is any point on the circumcircle of triangle ABC, then sin2A. AX2 + sin2B.BX 2 + sin2C.CX 2 = 4Δ Proof: By Theorem-1, 2 2R 2 2 2 We have SMsin2 A .AM sin2 B .BM sin2 C .CM 2 2 let us fix M as X, 2 2R 2 2 2 SXsin2 A .AX sin2 B .BX sin2 C .CX 2 2 and X is a point on the circumcircle of triangle so SX = R 2 2R 2 2 2 Hence Rsin2 A .AX sin2 B .BX sin2 C .CX 2 2 Further simplification gives the desired result. (III). Circumcenter lies on the line join of Weill’s point and incenter. Proof: We have by Theorem -1, 2 2R 2 2 2 SMsin2 A .AM sin2 B .BM sin2 C .CM 2 2 145 Let us fix M as Weill’s point (We), then 2 2R 2 2 2 SWesin2.A A W e sin2.B B W e sin2.C C W e 2 2 2 2 2 And by replacing the values of AWe , BWe and CWe and by some computation 2 2r 2 We can prove that SWe 1 SI 3R 2 2r 2 And we have IWe SI 3R Now it is clear that We I IS W e S Hence circumcenter lies on the line join of Weill’s point and incenter. Another way of proving the above statement is available in [3]. In the similar argument whatever we adopted for proving the above statement by using Theorem-1, we can also prove that the circumcenter lies on the line join of different triangle centers Such as The Circumcenter lies on the Line through the Centroid and the Nine-Point Center The Circumcenter lies on the Line through the Centroid and the Exeter Point. The Circumcenter lies on the Line through the Centroid and the Schiffler Point. The Circumcenter lies on the Line through the Centroid and the Gibert Point. The Circumcenter lies on the Line through the Centroid and the Skordev Point. The Circumcenter lies on the Line through the Orthocenter and the de Longchamps Point. The Circumcenter lies on the Line through the Orthocenter and the Schiffler Point. The Circumcenter lies on the Line through the Orthocenter and the Skordev Point. The Circumcenter lies on the Line through the Nine-Point Center and the Orthocenter. The Circumcenter lies on the Line through the Centroid and the Far-Out Point The Circumcenter lies on the Line through the Mittenpunkt and the Orthocenter of the Extouch Triangle ……. Etc. For Further details, see [7] and [9]. 146 (IV) EULER’S INEQUALITY If R is the Circumradius and r is the Inradius of a non-degenerate triangle then due to EULER we have an inequality referred as “Euler’s Inequality” which states that R 2 r , and the equality holds when the triangle is Equilateral.
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