About a Strengthened Version of the Erdos …

1 Forum GeometricorumVolume 17 (2017) 197 GEOMISSN 1534-1178 About a Strengthened Version ofthe Erd os-Mordell InequalityDan S tefan Marinescu and Mihai MoneaAbstract. In this paper, we use barycentric coordinates to prove the strength-ened Version of the Erd os-Mordell inequality, proposed by Dao, Ngyuen andPham in [3].One of the most beautiful results in geometry is represented by the Erd os-Mordell ([4]) inequality that for any pointPinside a triangleABC,PA+PB+PC 2d(P, AB) + 2d(P, BC) + 2d(P, CA),whered(P, AB)denotes the distance from the pointPto the lineAB. There area number of references on this result; see, for example, [1, 5]. Recently, Dao,Nguyen and Pham [3] improved the Erd os-Mordell inequality by replacing thelengthsPA,PB,PCby the distances fromPto the tangents to the circumcircleatA,B, aim of this paper is to prove a further Strengthened Version of the theoremof Dao-Nguyen-Pham.

2 We use barycentric coordinates to obtain new inequalities(Corollaries 4, 5), and the inequality of Dao-Nguyen-Pham in Corollary 6. Finally,we complete with an interesting application (Corollary 7).In this paper,X [Y, Z]means thatX,Y,Zare collinear, andXis an interioror a boundary point of the segmentY start with the following ,B,Cbe points on a line andB [A, C]andk:=ABACbe theratio of directed lengths. Thend(B, ) = (1 k)d(A, ) +kd(C, ). byU,V,Wthe orthogonal projections of the pointsA,B,Cre-spectively onto the line . LetT [C, W]such thatAT CWandAT BV={S}(see Figure 1).ThenAUV SandSV WTare rectangles andAU=SV=TW. On the otherside, ASB ATC. ThenBSCT=ABAC=k; soBS=k CT. Furthermore,(1 k)d(A, ) +kd(C, ) = (1 k)AU+kCW= (1 k)SV+kSV+kCT=SV+BS=BV=d(B, ). Publication Date: June 6, 2017. Communicating Editor: Paul S.

3 Marinescu and M. MoneaABCSTUVW Figure 1We recall that for any pointPinside or the sides of triangleABC, there arex, y, z [0,1]withx+y+z= 1such thatx PA+y PB+z PC= numbers are unique and are called thebarycentric coordinates ofPwithreference totriangleABC. Moreover, we havex=[PBC][ABC], y=[PCA][ABC], z=[PAB][ABC],where[XY Z]denotes the (oriented) area of triangleXY a triangle with vertices on the same side of a line , andPa point inside or on the sides of the triangle. Ifx,y,zare the barycentriccoordinates ofPwith reference toABC, thend(P, ) =xd(A, ) +yd(B, ) +zd(C, ).ABCDP Figure BC={D}so thatx=[PBC][ABC]=PDAD. From Lemma 1,d(P, ) = (1 x)d(D, ) +xd(A, ).(1) About a Strengthened Version of the Erd os-Mordell inequality199On the other hand,y=[PCA][ABC]andz=[PAB][ABC], so thatyz=[PCA][PAB]=CDBD, andCDCB=yy+z.

4 From Lemma 1,d(D, ) =(1 yy+z)d(C, ) +yy+zd(B, ).Sincex+y+z= 1, this is equivalent to(1 x)d(D, ) = (y+z)d(D, ) =zd(C, ) +yd(B, ).Together with (1), this givesd(P, ) =xd(A, ) +yd(B, ) +yd(B, ) +zd(C, ). Consider triangleABCwithA [B, C],B [A, C], andC [A, B]. Let , , R, andPbe a point in the plane of the triangle. We investigate theinequality: 2d(P, BC) + 2d(P, AC) + 2d(P, AB) 2 d(P, B C ) + 2 d(P, A C ) + 2 d(P, A B ).(2)Proposition following assertions are equivalent:(a)For any pointPinside or on the sides of triangleA B C , the inequality(2)holds.(b)For any pointP {A , B , C }, the inequality(2)holds, , for , , R, 2d(A , AC) + 2d(A , AB) 2 d(A , B C ), 2d(B , BC) + 2d(B , AB) 2 d(B , A C ), 2d(C , BC) + 2d(C , AC) 2 d(C , A B ).Proof.(a) (b): clear.(b) (a). Letx,y,zbe the barycentric coordinates of the pointPwith referenceto triangleA B C.

5 By Lemma 2, we haved(P, BC) =xd(A , BC) +yd(B , BC) +zd(C , BC)=yd(B , BC) +zd(C , BC),and analogous results for the linesCA,ABreplacingBC. Then 2d(P, BC) + 2d(P, AC) + 2d(P, AB)= 2(yd(B , BC) +zd(C , BC)) + 2(xd(A , AC) +zd(C , AC))+ 2(xd(A , AB) +yd(B , AB))=x( 2d(A , AC) + 2d(A , AB)) +y( 2d(B , BC) + 2d(B , AB))+z( 2d(C , BC) + 2d(C , AC)) x 2 d(A , B C ) +y 2 d(B , A C ) +z 2 d(C , A B ).(3)200D. S. Marinescu and M. MoneaSinced(P, B C ) =xd(A , B C ) +yd(B , B C ) +zd(C , B C ) =xd(A , B C ),and similarlyd(P, C A ) =yd(B , A C ),d(P, A B ) =zd(C , A B ), the lastterm of (3) is equal to2 d(P, B C ) + 2 d(P, A C ) + 2 d(P, A B ).This completes the proof of (b) (a). Corollary the incircle of triangleABCtouch the sidesBC,CA,ABatA ,B ,C respectively. The inequality(2)holds for any pointPinside or on the sidesof triangleA B C.

6 ABCA C B C A B Figure using Proposition 3, it enough to prove the inequality (3) only forP {A , B , C }. We supposeP=A . Denote byA ,B ,C the orthogonalprojections of the pointA onto the linesB C ,AC,ABrespectively. Letrbe theradius of the incircle of the triangleABC. ThenA C =A C sinC C A =A C sinA B C = 2rsin2A B C .Similarly,A B = 2rsin2A C B . Now we have2 A A = A C sinA C B + A B sinA B C = 2 rsinA B C sinA C B + 2 rsinA B C sinA C B = 4 rsinA B C sinA C B 2 2rsin2A B C + 2 2rsin2A C B = 2A C + 2A B .Also, 2d(A , AB) + 2d(A , AC) 2 d(A , B C ),and the proof is complete. About a Strengthened Version of the Erd os-Mordell inequality201 Corollary the incircle of triangleABCtouch the sidesBC,CA,ABatA ,B ,C respectively. For any pointPinside or on the sides of triangleA B C ,d(P, BC)+d(P, AC)+d(P, AB) 2d(P, B C )+2d(P, A C )+2d(P, A B ).

7 Apply Corollary 4 for = = = 1. Now, the inequality of Dao-Nguyen-Pham ([3]) is an easy consequence of theprevious 6(Dao-Nguyen-Pham [3]).LetABCbe a triangle inscribed in a circle(O), andPbe a point inside the triangle, with orthogonal projectionsD,E,FontoBC,CA,ABrespectively , andH,K,Lonto the tangents to(O)atA,B,Crespectively. ThenPH+PK+PL 2(PD+PE+PF). conclusion follows by using Corollary 5 for the triangle determined byall three tangents, and the fact that the circle(O)is the incircle of this triangle. In fact, Corollary 4 and a similar reasoning lead us to the weighted Version ofthe previous inequality(see [3, Theorem 4]). Now, we conclude our paper with thefollowing application, motivated by a recent problem posed in American Mathe-matical Monthly ([2]).Corollary a triangle inscribed into a circle(O), andPbe a pointinside the triangle, with orthogonal projectionsD,E,Fonto the tangents to(O)atA,B,Crespectively.

8 ThenPDa2+PEb2+PFc2 1R,whereRis the circumradius of circumcircle(O)is the incircle of the triangle bounded by the threetangents at the vertices. Applying Corollary 4 with =1a, =1b, =1c, we havePDa2+PEb2+PFc2 2d(P, BC)bc+2d(P, AC)ac+2d(P, AB)ab=2abc(a d(P, BC) +b d(P, AC) +c d(P, AB))=2abc(2[PBC] + 2[PCA] + 2[PAB])=4[ABC]abc=1R,and the proof is complete. 202D. S. Marinescu and M. MoneaReferences[1] C. Alsina and R. B. Nelsen, A visual proof of the Erd os-Mordell inequality,Forum Geom., 7(2007) 99 102.[2] N. Anghel and M. Dinc a, Problem 11491,Amer. Math. Monthly, 117 (2010) 278; solution, ibid.,119 (2012) 250.[3] T. O. Dao, T. D. Nguyen, and N. M. Pham, A Strengthened versionof the Erd os-Mordell inequal-ity,Forum Geom., 16 (2016) 317 321.[4] P. Erd os, L. J. Mordell and D. F. Barrow, Problem 3740,Amer. Math. Monthly, 42 (1935), 396;solutions, ibid.

9 , 44 (1937) 252 254.[5] H. J. Lee, Another proof of the Erd os-Mordell theorem,Forum Geom., 1 (2001) 7 S tefan Marinescu: National College Iancu de Hunedoara of Hunedoara, RomaniaE-mail Monea: University Politehnica of Bucharest and National College Decebal of Deva, Ro-maniaE-mail