THE HADWIGER FINSLER INEQUALITY REVERSE IN AN ACUTE TRIANGLE sorin radulescu, marian dinca, marius dragan, eduard puschin

THE HADWIGER-FINSLER INEQUALITY REVERSE IN AN ACUTE TRIANGLE Prof.dr Sorin Rădulescu,Prof Marian Dincă,Prof.Marius Drăgan,Prof Eduard Puschin In the paper1 the authors proposed the following inequality: In any acute triangle ABC are true the following inequality: a + b + c ≤ F + 2− a − b + b − c + c − a 1 3−2 Abstract: In this paper for begining we shall prove the inequality 1 for the isosceles acute triangle and then for any acute triangle Theorem 1.In any isoscel acute triangle ABC with the sides of lengths a, b, c and area F and b = c are true the inequality 1 Proof :The inequality 1 may be written as : a +b +c −4 F ≤ 2− 2 a−b +b−c +c−a 3−2 Using the notation b = c = x, because F = inequality 2 may be written as a +b +c −4 F a−b +b−c +c−a ; 4x sin = +2x −2 x sinA A −x 4−4 cosA⋅ 12 + 2 2xsin x sinA and a = 2xsin A2 , the right side ratio of 4sin = A +2−2 sinA = 2 2sin A2 −1 21−cosA+2−2 sinA 2sin A2 −1 3 4−2 cosA+ sinA = = A 2 2sin A2 −1 π cos A4 − 12 A cos π + 12 = 1+cos = 1+cos ⋅sinA = A A 2 2sin A2 −1 − π6 1+cos = + π6 1+cos We shall consider the function f : 0, A A π  π sin A2 − 12 − π6 + π6 8sin 1−cos A− π3 −1 +1 = → R, fA = A = sin A2 −sin π6 cos A2 − π6 −cos 1+cos 2sin A2 sin π6 32sin − π6 A 2 + π6 A = + π6 32sin A A π − 12 cos π − 12 cos A A +1 +1 1+cos A2 + π6  5π < π2 , or 12 = 3π + 2π = in on equivalent form As A ≤ π2 , we have A2 + π6 ≤ π4 + 12 12 A π π π A π cos + ≥cos + also sin ≤sin it follow that fA ≤ f π2  and we have the inequality of the statement Theorem For any triangle ABC of semiperimeter s with incircle CI, r and circumcicle CO, R there exist two isosceles triangles A B C and A B C of semiperimeter s and s with incircle CI, r and circumcicle CO, R So as inequality exist : s2 ≤ s ≤ s1 4 Proof The straight line OI cat the circumcircle in it A and A The tangents to the incircle from a point A , cat the circumcicle in B and C The straight line B C is tangent to the incircle ,acording to Poncelet’s closure theorem The triangle A B C is isosceles Similarly the tangents to the incircle from a point A cat the circumcircle in B and C The straght line B C is tangent to the incircle ,acording to Poncelet’s closure theorem it follows that A B C is isoscel As a = B C = 2rR+r+d , b = c = A C = A B = R+r+dR+d , where 2 2 R+d −r R+d −r d =∣ OI ∣= R − 2Rr and a = B C = 2rR+r−d , b2 = c2 = A2C2 = A2B2 = 2 R−d −r s1 = a + b + c  = R+r−dR−d R−d −r and as 2R + 10Rr − r + 2R − 2r R − 2Rr π − 12 π + 12 s = 12 a + b + c  = 2R + 10Rr − r − 2R − 2r R − 2Rr acording with the W.J.Blundon inequality it follows that the inequality 4 is true Proof of the inequality (1) If the point O exterior of the incircle C(I,r),rezult∣ OI∣= d ≥ r , and the tangent to the incircle from a point O,cat the circumcircle in B and C and the tangent to the incircle C(I,r) from a points B and C cat the circumcircle C(O,R) in A , acording the Poncelet’s closure theorem.The triangle A B C it is right angled ∢B A C = π2 , Let s the semiperimeter of the acute angled triangle ABC and s the semiperimeter of the triangle A B C We shall prove that s ≥ s 5 Let the triangle A B C tangent to the incircle C(I,r) in the points ; D.E.F ;where D ∈ B C ; E ∈ C A and F ∈ B A we have A F = A E = x = rctg π4 = r, B F = B D = y = rctg B23 , C D = C E = z = rctg C23 , As B C = B D + C D = y + z = 2R ,we shall obtain :s =x + y + z = 2R + r ,it rezults that the inequality 5 is equivalent to : R ∑ sinA ≥ r + 2R ,or ∑ sinA ≥ Rr + = ∑ cosA + 1, or ∑ sinA −cosA ≥ ,6 lets ciclic π−α ,B ciclic ciclic ciclic = π−β , C = π−γ , as A, B, C ∈ 0, π2  2 we have that α, β, γ ∈ 0, π ,and α + β + γ = π The inequality 6 is equivalented to ∑ cos α2 −sin α2  ≥ 1,or ∑ cos α2 −tg π4 sin α2  ≥ 1, or A= ciclic ciclic ∑ cos α2 + ciclic π  ≥cos π4 7, let α + π = u, β + π = v and γ + π = w and using well-know identity: cosu +cosv +cosw +cosu + v + w = ∏ cos u+v  it follows that: ciclic ∑ cos α2 + ciclic π  +cos ∑  α2 + or ∑ cos α2 + ciclic π  ciclic −cos π4 π  = ∏ cos α+β + ciclic = ∏ cos π2 − ciclic γ π  = ∏ cos π−γ + ciclic π   = ∏ sin γ4 ≥ we shall obtain the ciclic inequality 7 Using well-known identity: ab + bc + ca = s + r + 4Rr and a + b + c = 2s − 2r − 8Rr and F = sr, let 2− = λ 3−2 The inequality 1 is equivalented to sr + 2λ − 12s − 2r − 8Rr − 2λs + r + 4Rr ≥ or λ − 1s + sr + 1 − 3λr + 4 − 12λRr = Es, R, r ≥ 8 As λ > 1, we shall obtain : Es, R, r ≥ Es , R, r 9 In order to prove inequality 8 it will be sufficient to prove inequality Es , R, r ≥ ,what reprezent the inequality 1 for right angled triangle A3B3C3 bc 2 We have F = , and a = b + c We shall obtain the following inequality: b c bc + 2λ − 12b + c  − 2λ b + c b + c + bc ≥ 0, let b+c = x and b+c =y obtain xy + 2λ − 121 − 2xy − 2λ − 2xy + xy ≥ or xy + − 5λ + 2λ − ≥ λ − 2xy ,let xy = t As x + y = 1, we shall obtain : t ∈ 0; 14  ,let − 2xy = θ ,θ ∈  , 1, it will rezult that :xy = 1−θ 1−θ  2 ,we shall obtain: + − 5λ + 2λ − − λ ⋅ θ ≥ or 5λ − − 2θ − 2λθ − λ + ≥ , the equation 5λ − − 2θ − 2λθ − λ + = has a root θ = it follows that θ = −λ+  5λ− −2 and θ θ = −λ+ 5λ− −2 and θ < and 5λ − − 2θ − 2λθ − λ + = 5λ − − 2θ − θ θ − θ  ≥ For ∣ OI ∣= d ≤ r, it shall rezult that any triangle is acute, let ∢B A C = α and ∢B A C = β r r sin α2 = R+d < =sin π4 , it follows that α2 < π4 and α < π2 As sin β2 = R−d ≤ or 2 r ≤ R − d , or d ≤ R − r , or d ≤ R − r  or R − 2Rr ≤ R − 2 Rr + 2r or  − 1R ≤ r or R ≤  + 1r ,this inequality is true,because d ≤ r imply R − 2Rr ≤ r , or R − r ≤ 2r who imply R ≤  + 1r we shall obtain: β sin ≤ =sin π4 , it follows that β ≤ π2 Because Es, R, r ≥ Es , R, r and Es , R, r ≥ because is an inequality for acute triangle and isosceles triangle We have inequality of the statement References 1 Cezar Lupu,Constatin Mateescu,Vlad Matei,Mihai Opincariu:Refinements of the Finsler-Hadwiger reverse inequality,Gazeta Matematica seria A,nr 1-2(2010)p.49-53 2 P.G.Popescu,I.V.Maftei,J.L.Diaz-Barrero,M.Dincă:Inegalitati Matematice;Modele Inovatoare, Editura Didactica şi Pedagogica,2007 3 Marian Dincă,J.L.Diaz-Barrero: A new proof of an Inequality of Oppenheim, vixra org:1008.0013 4 Marian Dincă,Mihaly Bencze:New generalisation Finsler-Hadwiger inequality:Octogon Mathematical Magazine,2002 ... to the incircle from a point O,cat the circumcircle in B and C and the tangent to the incircle C(I,r) from a points B and C cat the circumcircle C(O,R) in A , acording the Poncelet’s closure theorem .The. .. acording with the W.J.Blundon inequality it follows that the inequality 4 is true Proof of the inequality (1) If the point O exterior of the incircle C(I,r),rezult∣ OI∣= d ≥ r , and the tangent... theorem .The triangle A B C it is right angled ∢B A C = π2 , Let s the semiperimeter of the acute angled triangle ABC and s the semiperimeter of the triangle A B C We shall prove that s ≥ s 5 Let the