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

Labels:
arc,
circumcircle,
distance,
equilateral,
point,
triangle,
vertex

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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

ReplyDeletedrawingCAD 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+fDraw 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+fHere 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