See complete Problem 141
Triangle, Incircle, Tangent and parallel to side, Perimeter. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Thursday, July 24, 2008
Elearn Geometry Problem 141
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A hexagon EFGHMD with the inscribed circle O tangent te sides at I,J,K,L,N and T:
ReplyDelete1.We have:
AB=AM+MN+ND+DB
AC=AH+HK+KG+GC
BC=BE+EI+IF+FC
MH=ML+LH
FG=FJ+JG
DE=DT+TE
&IF=FJ,GJ=GK,KH=HL,LM=MN,ND=DT,TE=EI.
p=AB+AC+BC
p=AM+MN+ND+DB+AH+HK+KG+GC+BE+EI+IF+FC
p=AM+LM+DT+DB+AH+HL+GJ+GC+BE+TE+FJ+FC
p=AM+LM+HL+AH+DT+TE+BE+DB+GJ+FJ+FC+GC
p=(AM+MH+AH)+(DE+BE+DB)+(FG+FC+GC)
p=p1+p2+p3.
2.we have:
EI=HL & FI=ML => EF=MH
DT=GK & ET=HK => DE=GH
DN=GJ & MN=FJ => DM=FG
yass_ghaz_57@hotmail.fr
Solution to problem 141.
ReplyDeleteLet P, Q, R, S, T, U be the tangent points of the incircle with segments HM, EF, DE, FG, DM respectively.
1) As it was proved in problem 140, (P1)/2 = AU = AS, (P2)/2 = BU = BQ, and (P3)/2 = CQ = CS. Then (P1+P2+P3)/2 = AU+BU+CQ and (P1+P2+P3)/2 = AU+BQ+CS. Adding these two equalities we get
P1 + P2 + P3 = AU + BU + CQ + AS + BQ + CS = AB + BC + AC = P.
2) For triangles OPH and OQE we have ang(OPH) = ang(OQE) = 90 and ang(PHO) = ang(QEO), since HM and BC are parallel. Furthermore OP = OQ = r (inradius). Then triangles OPH and OQE are congruent, thus PH = QE. Similarly we can prove that PM = QF so EF = HM. Also similarly, DM = FG and DE = GH.