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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 735 details.
B,D,B1,E lie on the circle on BB1 as diameter.By sine rule DE = BB1.sin(B)in ∆BDE.2∆ = b. BB1DE:2∆ = sin(B):b = 1:2RDE = ∆:R = 2R.sin(A).sin(B).sin(C)In ∆AB1C1, by sine rule B1C1:sin(A)=AB1:sin(C)=c.cos(A):sin(C)=2R.cos(A)∴a1=B1C1=2R.cos(A).sin(A)=R.sin(2A)Similar expressions for b1, c1.∴ a1+b1+c1=R[(sin(2A)+sin(2B)+sin(2C)]=R.4.sin(A).sin(B).sin(C)= 2.DE
Let be P and Q on ray B1E and B1D such that A1P=A1B1 and C1Q=C1B1. Note that D and E are midpoints of QB1 and B1P because triangles QC1B1 and PC1B1 are isosceles and C1D with A1E are its respectively altitude.Then 2DE=QP, but QP=QC1 + C1A1 + A1P=a1+b1+c1, so DE=(a1+b1c1)/2.Greetings.
Greetings to Eder Contreras Ordenes.Your proof is nice.Let X,Y be the midpoints of B₁C₁,B₁A₁ respExtend YX to meet BA at D'YXD'∥A₁C₁implies ∠C₁D'X = ∠BC₁A₁ = CAlso YXD'∥A₁C₁ and B₁D'∥CC₁ imply ∠B₁D'X =∠CC₁A =90°-CFollows (by adding the angles)∠B₁D'C₁ = 90°, B₁D'⊥ AB and so D’ coincides with DSo ∆B₁C₁D is right angled at D and DX=XB₁=XC₁=B₁C₁/2=a₁/2Similarly EY=A₁B₁/2=c₁/2Already XY=A₁C₁/2=b₁/2Hence DE=DX+XY+YE=(a₁+b₁+c₁)/2
Here the video of my solution (in spanish):http://www.youtube.com/watch?v=rbZq7dsmHnMGreetings Go-solvers!
Let DE intersect CB and AB at X and Y respectively.Let H denote the orthocentre of ∆ABCA,B₁,H,C₁ are concyclic.So ∠AC₁B₁ = ∠AHB₁AA₁∥B₁ESo ∠AHB₁ = ∠HB₁E = ∠BB₁EB,D,B₁,E are concyclicSo ∠BB₁E = ∠BDEIt follows ∠AC₁B₁ = ∠BDESame as ∠DC₁X = ∠C₁DXTherefore DX = C₁X.Similarly DX = B₁XSo X is the midpoint of (the hypotenuse) B₁C₁ (of the right ∆B₁DC₁)DX = (1/2)B₁C₁Similarly Y is the midpoint of A₁B₁; EY = A₁Y = B₁Y =(1/2)A₁B₁ Also XY = (1/2)A₁C₁ (X, Y being the midpoints of B₁C₁,B₁A₁ resp.)Hence DE = DX + XY + YE = (1/2)(B₁C₁ + A₁B₁ + A₁C₁)= (a₁+b₁+c₁)/2.