Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1220.
http://s33.postimg.org/bhlq9q7an/pro_1220.pngExtend C1E to G such that C1E=EGTriangle ABG is equilateral and ∠ (ABG)= ∠ (CBA1)= 60 degreesSo ∠ (ABC)= ∠ (GBA1)= 60- ∠(CBG)Triangle ABC congruent to GBA1 ( case SAS)So GA1=AC and ED= ½.AC=1/2.GA1= EFSimilarly we also have FD=1/2. C1H= ½ ACSo EFD is equilateral
Let K, M, L is within the sides AC1, AC, CA1 respectively .It is known that the triangles KMF, MFL is equilateral. Then the triangles KFE ,AKM and FDL,MCL are equal.(KFKM,KE=KA,<FKE=60-<MKE=<AKM and FL=ML,DL=CL, <FLD=60-<MLD=<MLC ).Therefore EF=AM=MC=FD.Then triangle KFE= triangle MFD.So EF=FD and <EFD=60. Therefore triangle EFD is equilateral.MANOLOUDIS APOSTOLIS 4th HIGH SCHOOL OF KORYDALLOS -PIRAEUS-GREECE
We use complex numbers.As triangle AB(C1) is equilateral: c1 – a = (b – a)[e^i(pi)/3], hence c1 = a + (b – a)[e^i(pi)/3].Similarly, a1 = b + (c – b)*[e^i(pi)/3]Midpoint of A1C1 is found by (a1 + c1)/2 = [e^i(pi)/3]*(c – b + b – a)/2 + (a + c)/2 = [e^i(pi)/3]*(c – a)/2 + (a + b)/2As f is midpoint of AB, and since we say that P is midpoint of AC, we can say: (f – e)/(p – a) = e^i(pi)/3Length of AP = Length of ED because PE is parallel to CD by the fact that P is midpoint of AC, E is midpoint of AB. Thus, d – e and p – a have equal magnitudes and equal directions. This is paramount to saying d – e = p – a, as complex numbers have vector properties.Conclusion: (f – e)/(d – e) = e^i(pi)/3This is equivalent to saying that triangle FED is an isosceles triangle with 60 degrees = angle FED. That is equivalent to saying that FED is an equilateral triangle.
Ugh what an embarrassing typo. When I said "This is paramount to saying d – e = p – a", I meant to say "This is equivalent to saying d – e = p – a". I typed that up when I was half-awake...