Proposed Problem
See complete Problem 256 at:
gogeometry.com/problem/p256_equilateral_triangle_circumcircle_distance_vertex.htm
Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Tuesday, February 24, 2009
Problem 256: Equilateral Triangle, Circumcircle, Point, Vertices, Distances
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Use Ptolemy's therem by which in this case we've:
ReplyDeleteAD*BC = CA*BD + AB*CD or d*BC = e*CA + f*AB which leads us to d = e + f since AB = BC = CA
Ajit: ajitathle@gmail.com
according to ptolemy's theorem, AB.CD + BD.AC = BC.AD. but triangle ABC is equilateral. so AB = BC = CA so we have AD = BD + CD that is d = e + f
ReplyDeleteQ. E. D.
Obviously ∠BDA = ∠ADC = 60°.
ReplyDeleteRotate 60° anti-clockwise about D.
Let B→B'. Then ΔBDB' is equilateral, and DB' = e.
Rotate 60° anti-clockwise about B.
Then B'→D, A→C, so B'A = DC = f.
As a result, d = DA = DB' + B'A = e + f.
Problem 256 (solution 1)
ReplyDeleteFollowing on the DB to point B take the point E such that DE=AD (<EDA=<BDA=<BCA=60).Then triangleADE is equilateral .But triangleAEb=triangleADC
(AE=AD,AB=AC,<BAE=60-<BAD=<DAC ).Therefore AD=DE=BD+BE=BD+DC.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
Problem 256 (solution 2)
ReplyDeleteFollowing on the BD to point D take the point F such that DF=DC (<BDA=<CDA=<CDF=60).Then triangleCDF is equilateral .But triangleBCF=triangleADC
(BC=AC,CF=DC,<ACD=60+<BCD=<BCF ).Therefore AD=BF=BD+DF=BD+DC.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
Problem 256 (solution 3)
ReplyDeleteOn the side AB ( not to extend ) get the point K such that BD=BK.Then triangle BKD is equilateral(<BDK=60=
<BKD ),so triangleABK=triangleCBD (AB=BC,BK=BD,<ABK=60-KBC=<CBD), so AK=DC. Therefore AD=AK+KD=DC+DB.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
Extend CD to F such that BF//AD.
ReplyDeleteThen Tr. BDF is equilateral and
Tr.s ABD ≡ CBF.
Hence AD = CF = CD + DF so d = e + f
Sumith Peiris
Moratuwa
Sri Lanka
See the drawing
ReplyDeleteCAD intercepts arc CD such as CBD => ∠CAD= ∠CBD
Define D’ n AD such as AD’C is congruent to BDC
=> AD’=f and ∠ACD’= ∠BCD
∠ACD= ∠ACB+ ∠BCD=Π/3+ ∠BCD
And ∠ACD= ∠ACD’+ ∠D’CD=∠BCD+ ∠D’CD => ∠D’CD = Π/3
CDA intercepts arc CA such as ABC => ∠CDA= ∠ABC=Π/3
∠D’CD = Π/3 and ∠CDA= ∠CDD’= Π/3 => ΔD’CD is equilateral
=>D’D=CD=e
d=AD=AD’+D’D therefore d=e+f
Draw AE such that AE=e and forms a regular trapezium ADBE
ReplyDeleteBDC and AEB are congruent => EB=f
Drop Perpendiculars from B and E on AD and denote them as B' and E'
=>B'E'=f
Since m(BCA)=60=> m(BDA)=60=>Tr.BDB' is a 30-60-90 triangle
=>DB'=E'A=e/2
AD=d=DB'+B'E'+E'A=e+f
See diagram here.
ReplyDelete∆ABE ~ ∆CDE ~ ∆ADB and ∆AEC ~ ∆BED
Therefore a/AE = f/CE = d/a and a/AE = e/BE
So a/AE = f/CE = e/BE = (f+e)/(BE+EC) = (f+e)/a = d/a ⇒ d = e+f
Here is my solution...
ReplyDeletehttps://www.youtube.com/watch?v=tB1FLsVUoio
See my sketch of P257
ReplyDeleteLet length of the triangle be x
ReplyDelete<ADC=<ACB=60 & <ADB=<ACB=60
By cosine law, x^2=d^2+f^2-2dfcos60 & x^2=d^2+e^2-2decos60
so, d^2+f^2-2dfcos60=d^2+e^2-2decos60
f^2-df=e^2-de
de-df=e^2-f^2
d(e-f)=(e+f)(e-f)
d=e+f