Proposed Problem
Try to use elementary geometry (Euclid's Elements.)
See complete Problem 274 at:
gogeometry.com/problem/p274_isosceles_triangle_80_80_20_angles.htm
Level: High School, SAT Prep, College geometry
Thursday, March 26, 2009
Problem 274: Isosceles Triangle, 80-80-20, Angles
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how this is solved?
ReplyDeleteThis is "World's Hardest Easy Geometry Problem".
ReplyDeletewhat is this problem called?
ReplyDeleteSee more 80-20-80 problems.
ReplyDeletesolution?
ReplyDelete
ReplyDeleteExterior design of the triangle ABC ,equilateral triangle BEL .Join the L to C intersecting AB at K and AD to F .Then triangle BEC=triangle LEC (LE=BE, CE=CE, <CEB=150=<CEL=360-60-150 ).So LC=BC=AB, <ECK=10, <AKC=40 and triangle KFD is equilateral (triangle AKC=
Triangle ADC, AKDC is isosceles trapezoid),or KF=KD=FD. But <AFC=60=2.30=2.<AEC, so AF=
FC=AC=FE and <KEF=<KAF=20.Therefore <KFE=20 so KE=KF=KD=FD.Then <DEF=<DKF/2=
60/2=30. But <FEC=<FCE=10 ,so <DEC=x=30-10=20.
HI, here you stated twice, "KE=KF=KD=FD.Then <DEF=<DKF/2", "<AFC=60=2.30=2.<AEC, so AF=FC=AC=FE", whys is that? I mean the double angle then AC = FE, vice verse. Thanks
Deleteexodus.bo@gmail.com
Here you stated twice, "<AFC=60=2.30=2.<AEC, so AF=FC=AC=FE " and "KE=KF=KD=FD.Then <DEF=<DKF/2". Why is that? I mean the double angle, then AC = FE and vice verse. Thanks.
DeleteLet point K is symmetric of C with respect to AB then KB=CB=AB and triangle KEC is
ReplyDeleteEquilateral or KE=KC=EC. Forming an equilateral triangle LBD (L is the left of the E).
Triangle LBK= triangle ABD (KB=AB, LB=BD, <LBK=20=<ABD). Then KL=AD=BD=LB=LD
Therefore the point M is the center of the circumscribed triangle KBD, then <DKB=
(<DLB)/2=60/2=30 and <CKD=70-30=40. Then <KDC=180-70-40=70=<KCD . Therefore
KD=KC=KE and the point K is the center of the circumscribed triangle CDE.Then
<CED=x=(<CKD)/2=40/2=20.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORRYDALLOS GREECE
Solution 3 problem 274
ReplyDeleteThe cycle (A, AB) intersects the BC, EC and AB in points K, L, M respectively.Τhen <CAK=
<KAL=<LAD=<DAM=20 ,and triangle AKM is equilateral,but <KAD=<KDA=40, so DK=AK=
MK=MA .Then the point K is the center of the circumscribed triangle AMD ,so <MDA=
(<MKA)/2=60/2=30. But triangle AMD= triangle ALD (MA=AK, AD=AD,<MAD=20=<LAD)
Then <ADL=<ADM=30 ,so <AEL=<ADL=30. Therefore <LED=LAD=20.
Solution 4 problem 274
ReplyDeleteFollowing on ΑD get the point M such that AB=AF then know that AC=BE,but triangle ABC=
Triangle ABF so AC=BE=BF. The cycle (B, BF) intersect the AF in K,then <KBF=20 and
Triangle BEK is equilateral or BE=BK=EK=DK (<KDB=<KBD) so the point K is the center of the circumscribed triangle EBD ,then <EDB=(<EKB)/2=60/2=30. Therefore <DEC=
<EDB-<ECD=30-10=20.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORRYDALLOS GREECE
Here is my solution...
ReplyDeletehttps://www.youtube.com/watch?v=I46JZ90vHRY
Draw F on BC with ang(CAF)=20 ; Draw G on AB with AG=AC; draw AF,FG
ReplyDeleteIn triangle AFC is ang(AFC)=ang(ACF)=80 => AC=AF; ang(GAF)=60 => triangle AFG is equilateral.=> ang(ADC=180-6--80=40 ; ang(FAD)=60-20=40 => AF=DF => DF=GF
ang(GFD)=180-60-80=40 => ang(FDG)=ang(DGF)=70
draw CG: In triangle AGC is ang(AGC)=ang(ACG)=50 (AC=AG and ang(CAG)=80
draw line from C perpendicular on AB. Drwa line DG . These two lines intersect in H.
HC intersects AB in M. Draw EH. Now is:
ang(HGM)=180-60-70=50; in triangle HCG is GM angular bisector of ang(HGC) => HG=GC and HM=CM
ang(AEC)=180-80-70=30 ; E on GM => ang(HEM)=ang(CEM)=180-80-70=30=> ang(HEC)=60 triangle HCE is equilateral =>ang(CHE)=60
In triangle AMC is ang(MCA)=10 ; in triangle HCD is ang(HCD)=70; ang(CDH)=70 =>ang(CHD)=40 => ang(DHE)=20; HC=EC=EH and HC=HD => HD=HE => ang(HED)=80 => x=80-60=20.
This is my solution to this beautiful problem:
ReplyDeletehttps://mega.nz/#!1xIElAoT!xYKR1wZeoySshIxhfv3Eu__zNCBS6yFSwY1kzCymHM8
Pedro Miranda