Friday, November 4, 2011

Problem 685: Isosceles Triangle, 100-40-40 Degrees, Angles, Congruence

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 685.

Online Geometry Problem 685: Isosceles Triangle, 100-40-40 Degrees, Angles, Congruence

20 comments:

  1. Awaiting a synthetic Proof:
    Let AB = BC = a and let AC = b
    Given AC = BC + AD
    b = a + AD, AD = b - a.
    Note BD = AB + AD = BC + AD = AC = b
    Apply Sine Rule in triangle ABC:
    a/sin 40 = b/sin 100
    = b/sin 80 = b/2sin40cos40
    So b = 2acos40
    Apply Sine Rule in triangle BDC:
    a/sinx = b/sin(80-x)
    So b sinx = a(sin80cosx-cos80sinx)
    2acos40sinx = a(sin80cosx-cos80sinx)
    tanx = sin80/(2cos40+cos80)

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  2. Problem 685 - To Pravin: and your answer is?

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  3. The answer is 30°, but I can't prove it yet, not that easy at first glance, haha!

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  4. x = 30 deg ?
    Unable to derive it from from above expression for tanx

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  5. cirumcenter of Tri ADC

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  6. AB=a
    AC=2acos40
    Let point E on AC, and CE=AB
    AD=AE ---> ang ADE= ang AED=20
    AE=AD=a(2cos40-1)=a(4cos^2(20)-3)
    DE=2AEcos20=2a(4cos^3(20)-3cos(20))=2a(cos60)=a
    ---> tri EDC=isosceles ---> ang DEC=160 ---> ang EDC=10
    ang ADC=ang ADE + ang EDC = 30

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  7. Locate point F on AC such that AD=AF,join B & D to F.We've /_ADF=/_AFD=20 and /_BFC=/_FBC=70. Now, DF/sin(40)=AF/sin(20) or DF=2AFcos(20)-----(1). Further, FC/BF=sin(70)/sin(40) and BF/AF=sin(40)/sin(30);hence FC=AFsin(40)*sin(70)/[sin30)sin40)]=2AFsin(70)=2AFcos(20)-----(2). By (1) & (2), DF=FC but /_DFC=160 hence FDC=10 and thus /_ADC = 20 + 10 = 30

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  8. Hello Pravin,

    sin80=cos10

    2cos40+cos80=2cos40+2(cos40)^2-1
    2(2(cos20)^2-1)+2((2(cos20)^2-1)^2)-1
    =4(cos20)^2-2+8(cos20)^4-8(cos20)^2+2-1
    =8(cos20)^4-4(cos20)^2-1
    =2cos20(4(cos20)^3-3cos20+cos20)-1
    =2cos20(cos60+cos20)-1
    =cos20(1+2cos20)-1
    =(2cos20-1)(cos20+1)
    =(4(cos10)^2-3)(2(cos10)^2)
    =(4(cos10)^3-3cos10)(2cos10)
    =2cos30cos10

    sin80/(2cos40+cos80)
    =cos10/(2cos30cos10)=1/(2cos30)=1/root3

    tanx=1/root3 ---> x=30

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  9. There's yet another way of determining /_ADC which is to locate F on AC with AF=AD as b4. Now let /_FDC=z. Using Ceva's Theorem in trigonometric form: [sin(20)/sin(z)][sin(20-z)/sin(40)][sin(70)/sin(30)]= 1 or sin(20-z)=sin(z) on simplification which gives 20 - z = z or z = 10 hence /_ADC = 20 + z = 20 + 10 = 20

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  10. Let E be a point on AC so that AE=AD, hence EC=BC. Connect DE, BE and CD.
    Costruct the altitude CH of the isosceles triangle BCE.
    Also the triangle ADE is isosceles.
    Because angle A = angle C = 40, then the angle ECH = 20 and the angle EBC=70.
    The angle ABE=30 and the angle ADE=AED=20.
    In right triangle ECH we have EH = ½BE = EC sin20.
    Apply Sine Rule in triangle BDE:
    DE/sin30 = BE/sin20 →
    DE=½BE/sin20 = EC sin20/sin20 = EC.
    Hence DE=EC, then also the triangle DEC is isosceles,
    angle EDC=10 and angle ADC=x=20+10=30.

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  11. http://www.imagengratis.org/images/p685.png

    MIGUE.

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  12. Κατασκευάζω ισόπλευρο τρίγωνο CDZ στο ημιεπίπεδο
    AC,B και παίρνω πάνω στη πλευρά ΑΒ ευθύγραμμο τμήμα ΑΕ=ΑD. Παρατηρώ ότι,CBZ=ABC ,EBZ ισόπλευρο,
    EDZ,ΕDB ισοσκελή και εύκολα φαίνεται ότι η γωνία
    ADB=30 μοίρες

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  13. Extend AB and let X and Y be a points on line AB so that x=30°.

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  14. To Eder: Problem 685. Please send again your solution as I have received only one line.

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  15. I am Ankur Guria, (brother of Ankan Guria) class 8 from India with a purely geometrical solution. My email i.d. is ankanguria.ju@gmail.com


    SOLUTION: Cut BC=CE on AC.
    So, left AD=AE.
    So, in triangle ADE, angles ADE and AED are equal to 20deg. Bisector of Angle BCA is drawn, meeting AB at F. Therefore triangles FBC and FEC are congruent. So, angle EFC = angle BFC = 60deg. Triangles DFE and EFC are congruent. Thus DE=EC. Angle DEC=160deg. So, in triangle DEC, angles EDC and ECD are equal to 10deg. Thus, angle ADC = angle ADE + angle EDC = 20 + 10 = 30deg= x

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  16. Solution 1
    Forming equilateral triangle BCE internally the triangle ABC, then triangle DBE=triangle ABC (DB=AC ,BE=AB ,<DBE=100-60=40=<BAC). Therefore DE=BC=AB=BE so E is circumcenter BCD so
    <BDC=<BEC/2=60/2=30.

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  17. Forming equilateral triangle BDE (interior triangle BCD). Then triangle ABC=triangle BCE (BE=AC, AB=BC,<CAB=40=<EBC).So BC=CE, but DB=DE then triangle DBC=triangle DEC.
    Therefore <BDC=60/2=30.

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  18. Solution 3
    Forming equilateral triangle BDE (interior triangle BCD). Then triangle ABC=triangle BCE (BE=AC, AB=BC,<CAB=40=<EBC).So BC=CE, but DB=DE then triangle DBC=triangle DEC.
    Therefore <BDC=60/2=30.

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