Harvey Littleman



From Gogeometry Using Area of a triangle Tangents to a circle Inscribed circle of a triangle b=ACAN=AM, AF=AE, MF=NE,CP=CO, CF=CG, PF=OG,BE=BGr=DF=DE=DGx=JM=JN=LO=LPMP=2x S=[ABC]=r(AB+BC+AC)/2 => S=r(AB+CB+b)/2 AB=AE+EB=AF+EBCB=CG+GB=CF+GBAB+CB=AF+CF+EB+GB=b+2BE=> (1) S=r(b+BE) But S=[AMJN]+[MFDENJ]+[PFDGOLP]+[EBGD]S=xAM+(r+x)MF+xCP+(r+x)PF + rEB, knowing that [EBGD]=(rEB+rBG)/2=rEBS=x(AM+MF+CP+PF)+r(MF+PF) + rEB(2) S=xb+2xr + rEB (1) and (2) => S=r(b+BE)=xb+2xr+ rEBbr=xb+ 2xr Therefore x=br/(b+2r)

Gogeometry Problem 418


From Gogeometry Using Inscribed angles in a circle Equilateral triangle Concyclic points ∠BAD= π/2 => B, A and D are concyclic, circle with center F middle of BD=> FB=FA=FD and ∠BAF=15°In the same way, GC=GA=GE and ∠CAG=15°=>∠FAG=90-15-15=60°ΔFAG is isosceles in A (AF=AG) and ∠FAG=60° => ΔFAG is equilateral Therefore x=FG=BD/2=CE/2=1

Gogeometry Problem 406




Definition : In euclidean geometry, a kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other. Property : Every kite is orthodiagonal, meaning that its two diagonals are at right angles to each other

Kite




Using Additional points congruent triangles Area of a triangle [OQL]=OL . KL/2=bc/2[OPQ]+[LKQ]=PQ . KL/2+QK.KL/2=(PQ+QK).KL/2=PK.KL=bc/2=>[OQL]=[OPQ]+[LKQ]=bc/2[OPKL]= [OQL]+[OPQ]+[LKQ]=2bc/2=bc In the same way [LMHI]= [IJEF]= [FGNO]= [OPKL]= bc[QRST]= [OLIF] + [OQL] + [LRI] + [ISF] + [FTO][QRST]= c^2+2bc=c(c+2b) Therefore [QRST]= ac (Situation A) But, this is not true if b is too big, then […]

HP008


From Gogeometry Using Additional points Inscribed angles in a circle central and inscribed angle congruent triangles Concyclic points ∠ABC=∠ADC = π/2 => ABCD are concyclic with center O=> OA=OD=OC=OB Define D’ such as ΔCAD is congruent to ΔCAD’∠CDA=∠CD’A = π/2 =>D’ is on the circle O=> ∠ACD= ∠ACD’= α => […]

Gogeometry Problem 017




From Gogeometry Using Additional points Pythagoras congruent triangles Similar triangles Equilateral triangle Let b=AD=DC and d=BCΔACB and ΔBCD are similar since ∠BAC=∠DBC=x and ∠BCD=∠BCAThen AC/BC=CB/CD=AB/BD => 2b/d=d/b => d=b√2 Define C’ such as DC’ ⊥ AC and DC’=b∠BDC=∠ADC – π/4= π – π/4 = 3 π/4∠BDC’=∠BDA + ∠ADC’ = π/4 […]

Gogeometry Problem 001