## Thursday, March 26, 2009

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

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

1. how this is solved?

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

3. what is this problem called?

4. solution?

5. 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.

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

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.

6. 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

7. Solution 3 problem 274
The cycle (A, AB) intersects the BC, EC and AB in points K, L, M respectively.Τhen <CAK=
MK=MA .Then the point K is the center of the circumscribed triangle AMD ,so <MDA=

8. 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

9. Here is my solution...

10. Draw F on BC with ang(CAF)=20 ; Draw G on AB with AG=AC; draw AF,FG
In 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.

11. This is my solution to this beautiful problem:
https://mega.nz/#!1xIElAoT!xYKR1wZeoySshIxhfv3Eu__zNCBS6yFSwY1kzCymHM8

Pedro Miranda