已知在△ABC的三边a>b>c且a+c=2b,A-C=π|2,求a：b：c

1.a+c=2b
=>sinA+sinC=2sinB
=>2sin(A+C)/2cos(A-C)/2=4sin(A+C)/2cos(A+C)/2
又因0cos(π-B)/2=√2/4
sinB/2=√2/4
cosB/2=√[1-sin²(B/2)]=√14/4
sinB=2(√2/4)(√14/4)=√7/4
sinA*sinC=[cos(A-C)-cos(A+C)]/2
=[2cos²(A-C)/2-1-2cos²(A+C)/2+1]/2
=cos²(A-C)/2-cos²(A+C)/2=(√2/2)²-(√2/4)²=3/8
则sinA,sinC是x²-√7x/2+3/8=0的解(又sinA>sinC)
sinA=[√7/2+√(7/4-3/2)]/2,sinC=[√7/2-√(7/4-3/2)]/2
=>sinA=(√7+1)/4,sinC=(√7-1)/4
sinA:sinB:sinC=(√7+1):√7:(√7-1)
2.(a+b+c)(a—b+c)=3ac
=>(a+c)²-b²=3ac
=>a²+c²-b²=ac
cosB=(a²+c²-b²)/2ac=1/2
B=π/3
tanA+tanC=3+√3
=>(sinAcosC+sinCcosA)/cosAcosC=3+√3
=>sin(A+C)/cosAcosC=3+√3(A+C=π-B=2π/3)
=>cosAcosC=√3/2/(3+√3)=(√3-1)/4
又2cosAcosC=cos(A+B)+cos(A-C)=-1/2+cos(A-C)
则cos(A-C)=(√3-1)/2+1/2=√3/2
因-π