New Interpolation Inequalities to Euler’s R 2

1 Forum GeometricorumVolume 17 (2017) 149 GEOMISSN 1534-1178 New Interpolation Inequalities to euler sR 2rDorin Andrica and Dan S tefan MarinescuAbstract. The purpose of this paper is to obtain some Interpolation inequalitiesto the well-known euler s inequalityR 2rin terms of new geometric ele-ments given by the radiiRA,RB,RCof the tangent circles at the vertices to thecircumcircle of a triangle and to the opposite sides. The main results are givenin Theorems IntroductionAt the first 2015 Romanian IMO Team Selection Test the first author of thispaper has proposed the following problem: LetRAbe the radius of the tangentcircle atAto the circumcircle of triangleABCand to the sideBC.

2 Similarly,define the radiiRBandRC. The following inequality holds1RA+1RB+1RC 2r,whereris the inradius of this short paper we discuss some proofs to the above inequality and we com-plete it to the left hand-side in order to get a new Interpolation for the well-knownEuler s inequalityR 2r, whereRis the circumradius of triangleABC. Also, wegive other Interpolation Inequalities to the euler s inequality in terms of the radiiRA,RB,RC. For other Interpolation and improvements Inequalities to the euler sinequality we refer to the excellent monograph [2].2. Some auxiliary resultsAs usual, we denote bya,b,cthe lengths of the sides opposite to the verticesA,B,C, respectively, and byK[ABC]the area of triangleABC.

3 We need thefollowing helpful triangleABCdenote byha,hb,hcthe lengths of the altitudes fromthe verticesA,B,C, respectively. The relation1r=1ha+1hb+ use the formula for the area of a triangle. Publication Date: May 21, 2017. Communicating Editor: Paul Andrica and D. S. MarinescuLemma the radius of the interior (exterior) tangent circle atAto thecircum- circle of triangleABCand to the side BC, thenrRA=ascos2B C2,wheresdenotes the semiperimeter of , then clearly we haveRA=h2. Using the relationsK[ABC] =sr=aha2, the conclusion , let us suppose thatB > C. ConsiderTthe intersection point ofthe common tangent line atAto the two circles with the lineBC(see Figure 1).

4 In triangleT AB, we have T=B Cand from the Law of Sines we obtaincsin(B C)=T AsinB= T A=bc2 Rsin(B C).BCAOTOAA A IFigure 1 BecausetanB C2=RATA, it follows thatRA=bc2 Rsin(B C) sinB C2cosB C2=bc2 Rcos2B ,rRA=rbccot 4 Rcos2B C2=ar4 RKcos2B c2=ascos2B C2,whereK=K[ABC], and the proof is complete. New Interpolation Inequalities to euler sR 2r151 Proof Abe the circle tangent atAto the circumcircle of triangleABCandtangent atTto the lineBC. Assume thatR= 1, and consider the inversion of poleAand unit power. In what follows,X will denote the image of the pointX6=Aby this this inversion, the lineBCis transformed into a circleAB C centeredat some point.

5 The circleABCis transformed into the lineB C , and Aistransformed into a line throughT and parallel toB C .LetDbe the orthogonal projection ofAon the lineBC. ThenAD=1AD=1ha,wherehais the length of the altitude from the vertexAin the triangleABC, and T = A= , letA1be the antipode ofAin circle A, soA 1is the orthogonal projectionofAon line , andAA 1=1AA 1= , letOdenote the circumcenter of the triangleABCand notice the anglesOAD, AA 1are both congruent to the absolute value of the difference of theinternal angles of triangleABCatBandC, to obtaincos(B C) =AA 1 T A=12RA 12ha12RA=haRA 1 =2 KaRA 1,whereK=K[ABC]and the desired formula follows after standard transforma-tions.

6 Lemma every triangleABCthe following inequality holdscos2B C2 have equality if and only if2a=b+ havecosB C2= cosB2cosC2+ sinB2sinC2= cosB2cosC2 sinB2sinC2+ 2 sinB2sinC2= cosB+C2+ 2 sinB2sinC2= sinA2+ 2 ,cosB C2 2 2 sinA2sinB2sinC2= 2 2 r4R= 2rR,and the conclusion follows. The equality holds if and only ifsinA2= 2 sinB2sinC2,that isa2= 4(s a)2, hence2a=b+c. 152D. Andrica and D. S. MarinescuProof the incenter of triangleABC, and considerA the intersectionpoint of the rayAIwith the circumcircle of triangleABC. We haveA A2= (A I+AI)2 4A I AI= 8Rr,where the last equality is obtained from the power ofIwith respect to the cir-cumcircle of triangleABC.

7 Clearly, the equality holds if and only ifA I= = 2 Rsin(B+A2)= 2 RcosB C2,hence the desired inequality follows. As we already mentioned, the equality holdsif and only ifA I=AI, that isAA = 2IA = 2BA , socosB C2= 2 obtainsinA2= 2 sinB2sinC2,therefore2a=b+c. 3. The main resultsThe first Interpolation result is directly connected to the original problem men-tioned in the introduction and it is contained in the following the above notations the following Inequalities hold4R 1RA+1RB+1RC 2r.(1)We have equality if and only if the triangleABCis Lemma 2 we haverRA as, with equality if and only ifB= ,rRB bswith equality whenC=A, andrRC cswith equality whenA=B.

8 Summing up these Inequalities it follows the right hand-side inequality,with equality if and only ifA=B=C, that is the triangle is Lemma 3 and Lemma 2 we haverRA as 2rR, with equality if and only if2a=b+c, and two analogous Inequalities for the radiiRBandRC. Summing upthese Inequalities we obtain1RA+1RB+1RC 2R a+b+cs=4R,and we are done. Remark.(1) It is possible to give a direct geometric argument for the right hand-side inequality in (1). ConsiderOAto be the center of the tangent circle atAto thecircumcircle of triangleABCand to the sideBC, andA the tangency point ofthis circle with the lineBC(see Figure 1).

9 Using the triangle inequality in triangleAOAA we haveha AA leAOA+OAA = 2RA, hence we obtain12RA 1ha,and other two similar Inequalities forRBandRC. Summing up these inequalitiesthe conclusion follows from Lemma Interpolation Inequalities to euler sR 2r153 Theorem the above notations the following Inequalities hold2rR K[ABC]3 abcRARBRC 1.(2)We have equality if and only if the triangleABCis the Inequalities obtained from Lemma 2, we obtainr3 RARBRC abcs3,henceK[ABC]3 abcRARBRC the other hand, multiplying the Inequalities obtained from Lemma 2 and usingLemma 3, it follows thatabcs3 8r3R3 is2rR K[ABC]3 abcRARBRC,and we complete the left hand-side of (2).

10 Clearly, the equality holds if and only ifthe triangleABCis equilateral. From the relationrRA=as cos2B C2proved in Lemma 2, we obtainRA=Kacos2B the second proof of Lemma 3 we have shown thatAA = 2 Rsin(B+A2)=2 RcosB C2, hencecosB C2=AA 2R. It is clear that the pointA is the feet of thebisector of the angleAof triangleABC. Denote by athe length of bisector ofangleAof triangleABC, the length of the segment[AA ]. TrianglesAA OAandAA Oare similar, therefore we obtainRAR= aAA = 2a a AA .From the Law of Sines in triangleACA , it follows that asinC=bsin(C+A2).But, clearly we havesin(C+A2)= sin(B+A2)= cosB C2,hence a AA = 2 RbsinC.