Thursday, March 26, 2009

Problem 274: Isosceles Triangle, 80-80-20, Angles

Proposed Problem
Try to use elementary geometry (Euclid's Elements.)
Problem 274: Isosceles Triangle, 80-80-20, Angles.

See complete Problem 274 at:
gogeometry.com/problem/p274_isosceles_triangle_80_80_20_angles.htm

Level: High School, SAT Prep, College geometry

11 comments:

  1. how this is solved?

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  2. This is "World's Hardest Easy Geometry Problem".

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  3. Exterior 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.

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    Replies
    1. 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
      exodus.bo@gmail.com

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    2. 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.

      Delete
  4. Let point K is symmetric of C with respect to AB then KB=CB=AB and triangle KEC is
    Equilateral 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

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  5. Solution 3 problem 274
    The 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.

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  6. Solution 4 problem 274
    Following 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

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