Proposed Problem

Click the figure below to see the complete problem 423 about Triangle, 30, 40 degree, Angle, Congruence.

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Level: High School, SAT Prep, College geometry

## Wednesday, February 3, 2010

### Problem 423: Triangle, 30, 40 degree, Angle, Congruence

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ReplyDeleteit's definitely a great solution, but it's far-fetched i would've done what you just did.

ReplyDeletei'm wondering if you have another solution.

can you give me any mail so that we can talk?

a more simpler solution added to blog, thank you for your question. it made me think again.

ReplyDeletehere is my mail:

geometri-problemleri@hotmail.com

i am some sick, maybe we can not talk for now. but you can send mail to me

is the answer 85 degrees?

ReplyDeleteTo anonymous:

ReplyDeleteThe answer is not 85 degrees.

in tr(ABD), AD/BD=sin(x-30)/sin30

ReplyDeletein tr(BDC), BC/BD=AD/BD=sinx/sin40

thus, sinx=2sin40sin(x-30)

or

cotx=((3^0.5)sin40-1)/sin40

=(2(3^0.5)sin40cos40-2cos40)/2sin40cos40

=((3^0.5)sin80-2sin(80-30))/sin80

=cot80

therefore, x=80

Hmm, this one was tricky, I feel like there has to be a simpler solution.

ReplyDeleteLet E be on the extension of AC such AB=BE, and let F be on the opposite side of BE as D such that triangle BCF is congruent to triangle ADB, with angle FBC=30. Let AF and BE meet at G. Then triangles ABE, ABF, and FBE are isoceles, from which we find angle GBC= angle GAC=10, so ABGC is cyclic, so angle GCB= angle GAB=20. We also find angle GFE=60=angle GCD, so FGCE is cyclic. Thus angle GCF= angle GEC=80. Since triangle BCF is congruent to triangle ADB, we find x=80.

Firstly, you must now a very importan triangle: 30-40-110, wich has a crucial property, "Let tr ABC with <A=30 and <B=40, if we extend AB beyond B to X such that BC=BX, then CX=AB"

ReplyDeletethe prove for this in spanish si here:

http://www.youtube.com/watch?v=2_Kq1_QU-xQ

Now for the problem, lets make the same construction (extend AC to X such that CX=BC), then AC=DX=BX, so <BDX=<DBX=80 because <X=20.

AEC , is equilateral triangle. Because ,angle CAB=30, AB is mediator of EC.So , EB=BC and angleBCE=angleBEC=20

ReplyDeleteCZ=//BD (DBCZ , is parallelogram).Therefore,BC=DZ=DA . Because angleCDZ=40,then,angleDAZ=angleDZA=20

So, the triangles, EBC,ADZ , are equal.Therefore,AZ=EC. But EC=AC.So,AZ=AC .

Therefore,angle AZC=angleACZ=x=80 degrees

see the image: http://img842.imageshack.us/img842/3751/p423triangleangle3040.gif

Extend ADC to E such that AB = AE. Draw CM perpendicular to AB to meet BE at M. Let the circle with B as centre and AB = BE as centre cut AM extended at N.

ReplyDeleteSince B is the centre of circumcircle of Tr. AEN, < ANE = 1/2 < ABE = 60 implying that CENM is cyclic with < CME = 30 = < BAC so ACMB is cyclic from which we deduce that < BCM = 20 = < BAN = < BNA = < NBM.

Now we see that Tr. ABD & CBN are congruent SAS so < ABD = < BNC = 50 and so x = 80.

Sumith Peiris

Moratuwa

Sri Lanka

Should read AB = BE as radius and not as centre as line 4 states. The error is regretted.

ReplyDeleteSimpler geometric proof

ReplyDeleteLet E be the circumcentre of Tr. ABC. Then < BEC = 2 < BAC = 60 so Tr. BCE is equilateral. Hence BC = AD = AE and since < CAE = 20 < AED = 80. But < AEB = 80 so B,D,E are collinear implying that x = 80 (180 - 60 - 40 in Tr. DBC)

Sumith Peiris

Moratuwa

Sri Lanka

Problem 423

ReplyDeleteForming the equilateral triangle AEC (E is above the point Β) then triangleABC=triangleABE

And BE=BC,<BEC=<BCE=20.Draw EK perpendicular to AC (K is on AC) and point F on EK

such that AF=FC=BC=AD=BE. So < ADF=80=<FBC then FBCD is cyclic.Therefore <BDC=<BFC=80 (<FCB=20 ).

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE