Tricky Triangle
Geometry
Level
5
A triangle ABC has positive integer sides . Find the minimum length of the perimeter of ABC.
Conditions:
The answer is 77.
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1 solution
From sine law a/sinA=b/sinB=c/sinC=2R.Because of A=2B,sinA=2sinBcosB.After replacing we get a/(2sinBcosB)=b/sinB,so a=2bcosB, so cosB=a/2b ,it's rational number. A+B+C=180, 2B+B+C=180, 3B=180-C<90 (because C>90),so B<30,so cosB>cos30=√3/2~0.866.Let cosB=7/8=0.875(very close rational number),so a=2b cosB=7b/4. For b=4 we get a=7. sinb=√(1-cosb cosb)=√15/8 .now we can calculate sinA=2sinBcosB=7√15/32,so cosA=2cosB cosB-1=17/32 Now we can calclulate 2R=a/sinA=7/(7√15/32)=32/√15 .Now From 2R=c/sinC we get c=2RsinC=2Rsin(180-A-B)=2Rsin(A+B)=2R(sinAcosB+sinBcosA)=(32/√15) (66√15/256)=33/4=8.25. So we find that a=7,b=4,c=8.25 satisfy conditions and now we just multiply all sides by 4. OUR solution is a=28,b=16,c=33 and result is 28+16+33=77. Minimum length of the perimeter of ABC is 77.