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## Sunday, October 8, 2017

### Geometry Problem 1349: Three Squares, 90 Degrees, Perpendicular Lines, Diagonal. Math Infographic

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diagonal,
geometry problem,
perpendicular,
three squares

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Problem 1349

ReplyDeleteLet < AHB = < HBG = < AGB = θ.

AC2 = 2.CE2 = CE.CG, hence < CAE = < AGB = θ.

Since < CBP = < CAP =θ,

ABCP is concyclic and so

< CPE = < ABC = 90

Sumith Peiris

Moratuwa

Sri Lanka

Interestingly also AE = 5.CP

DeleteProblem 1349

ReplyDeleteIn the extension of AB to B above point K such that AB=BK=>triangleKAD=triangleAFR=E.So

KD=AE and KD perpendicular to AE.If the AE intersects the KD in Q and the BC in M. Then

BM=MC and EQ/QA=EM/AD=3/2 =>EQ/EA=3/5 (1).Is EP/PA=BE/AH=2/3 =>EP/EA=2/5(2).

From (1) and (2)=>EP/EQ=2/3=>EP/PQ=2/1=EC/CM=>CP//MQ=>CP is perpendicular in AE.

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

https://photos.app.goo.gl/vydebtzBccTnATBf1

ReplyDeleteLet BH meet EF at K

Let AB=BC=CD=AD= 1

So EK= 2/3 and AE= sqrt(5)

Triangle ABP similar to EKP ( case AA)

PE/PA= EK/AB= 2/3

So PE=2/sqrt(5) and PA=3/sqrt(5)

And PE/EB= 1/sqrt(5) and CE/AE= 1/sqrt(5)

Triangle ECP similar to EBA ( case SAS)

So ∠ (CPE)= ∠ (EBA)= 90 degrees

Prove by similar triangles: △CPE similar to △ABE

ReplyDeleteLet CE=1. Then BE=2, AH=3

△BPE similar to △HPA (AAA) so AP:PE=3:2

AE=SQRT(5)

PE=2(SQRT(5)) / 5

Since <AEB=<CEP (common angle)

AE/BE= SQRT(5)/2

CE/PE= 1/ 2(SQRT(5)) / 5 = 5 / 2(SQRT(5)) = SQRT(5) / 2

Hence △CPE similar to △ABE ( ratio of 2 sides, inc.<)

Since <ABC=90 degrees, <CPE is therefore 90 degrees. (proved)

Problem !349 is more amenable for analytic proof:

ReplyDeleteLet A = (0,0), D = (1,0), B = (0,1)

Then C = (1,1), E = (2,1), F = (2,0), H = (3,0)

Equations: BH: x + 3y = 3, AE: x - 2y = 0,

their intersection P = (6/5,3/5)

Slopes: of AE = 1/2, of CP = -2

Hence their product being -1, they are perpendicular.

Great work! you guys really know what you are doing! WOW just WOW!

ReplyDelete