Sunday, May 18, 2008

Elearn Geometry Problem 3



See complete Problem 3
Triangle, Angles, Median, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

26 comments:

  1. Solution to this posted at http://www.osinfofrom.us/prob003.html

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  2. In your solution:
    Why "ABC appears to be a right triangle"?
    Why triangles ABD and DBC are congruent?

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  3. For lack of better ideas, here's a solution by analytical geometry: Let D be the origin and A:(-1,0) & B;(1,0) and let angles BAC & BCA be 3m & m resply. Also let tan(m)=t. Now BD:y=-x while BC:y=(1-x)t. Thus, B is (t/(t-1),-t/(t-1)). Since co-ordinates of A & B are known we've Slope BA = t/(1-2t) = tan(3m) = (3t-t^3)/(1-3t^2)
    In other words, (1-2t)(3-t^2)= (1-3t^2) which yields three solutions of which the only admissible solutions is t=tan(m)=(2)^(1/2) - 1
    In other words, m=angle BCA = 22.5 deg.
    Ajit:ajitathle@gmail.com

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  4. Attempted second explanation at http://www.osinfofrom.us/prob003.html as Antonio pointed out a flaw in the first attempt to prove the value of x.

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  5. Are Joe and Ajit synonymous?

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  6. They're the same person, not synonymous though!
    Ajit

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  7. If you want to use trigonometry then:
    In Tr. ABC -- AC/sin(3x+x)= AB/sin(x)
    or AC/AB =sin(4x)/sin(x)
    In Tr. ABD -- AD/sin(3x+45)= AB/sin(45)
    or AD/AB =AC/2AB =sin(3x+45)/sin(45)
    or AC/AB=2sin(3x+45)/sin(45)
    or sin(4x)/sin(x)=2sin(3x+45)/sin(45)
    And this yields,x=0.3926988845 radians
    = 22.5 degrees as the only admissible solution.

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  8. http://geometri-problemleri.blogspot.com/2009/05/problem-12-ve-cozumu.html

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  9. Chose point E on BC so that CAE = x. BEA is exterior angle of AEC so that BEA = 2x and AB = BE. Let F is midpoint of AE clearly triangles AFB and EFB are congruent. If H and G are midpoint of AB and BE respectly then AHF, BHF, BGF and EGF are congruent so that HAF = HBF = GBF = GEF = 2x.
    So finally x + x + 2x + 2x + 2x = 8x = 180 deg  x = 22.5 deg.
    wisnab36@gmail.com

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  10. Why are AHF, BHF, BGF and EGF congruent?

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  11. Thank’s Mr. Antonio for your comment. I will try to justify that.
    Triangle AFB is right at F and HF is median sothat HF = AH = HB. By the same reason we can find that GF = BG = GE.
    AFB and EGF are equilateral triangles so that anggle AFH = anggle EFG = 2x.
    From these facts, HF // BE and AB // BG so that BGFH is a rhombus which has properties HG = BF ……………………………. (1)
    In ABE triangle, because H and G mid points of AB and BE respectly then HG // AE and HG = ½ AE = AF ………………… (2)
    From (1) and (2) we get BF = AF = FE ……………………….. (3)
    From (3) anggle ABF = anggle EBF = 2x.
    We had showed that HB = HF = BG = GF
    From these we get angle HFB = angle BFG so that AHF, BHF, BGF and EGF congruent

    From (3) we can find another solution.
    Because F is midpoint of AE and AF = FE = BF then ABE = 90 deg.
    So that 2x + x + x = 4x = 90 deg  x = 22.5 deg
    Thank’s
    wisnab36@gmail.com

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  12. angle ABD=135-3x & DBC=45-x
    so angle ABD=3z & DBC=z
    area of ABD= area of DBC there fore
    1/2AB.BD.sin3z=1/2BC.BD.sinz
    so sin3z/sinz=BC/AB
    due to sine theorem in ABC sin3x/sinx=BC/AB
    there fore sin3x/sinx=sin3z/sinz
    therefore 3-4sin^2(x)=3-4sin^2(z)
    x and z are oblique so x=z
    in triangle ABD 3x+3x+45=135 so x=22.5
    please tell me how to send another picture solutions?(henrik_ir@yahoo.com)

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  13. Correção: de Valdir Marques Faria

    A resolução de Anonymou não é consistente quando justifica a congruência dos triângulos. O ângulo de 45° é determinante na resolução do problema. Assim, teremos:
    No triângulo BAD, triângulo ABD = 135°– x e, no triângulo BDC, triângulo DBC = 45°+x. Aplicando a lei dos senos nos triângulos ABD e BDC, teremos:
    sen(3x)/sen(135°-3x) = sen(x)/sen(45°-x) ...
    sen(3x).sen(45°-x) = sen(135°-3x).senx ...
    sen(3x).(sen45°.cosx – cos45°.senx) = senx(sen135°.cos(3x) – sen(3x).cos135°) ...
    sen(3x).(cosx – senx) = senx.(cos3x + sen3x) ...
    sen(3x).cosx – cos(3x).senx = 2.sen(3x).senx ...
    sen(2x) = cos(2x) – cos(4x) ...
    sen(2x) = cos(2x) + quadr.(sen(2x)) – quadr(cos(2x)) ...
    sen(2x) – cos(2x) = (sen(2x) + cos(2x)).(sen(2x) – cos(2x))
    Se sen(2x) for diferente de cos(2x) teremos:
    sen(2x) + cos(2x) = 1 , e assim x = 0 ou x = 45° o que não convém.
    Se sen(2x) = cos(2x), teremos 2x = 45 e x = 22,5°.

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  14. I'm brazilian , and my english i'snt good,but i saw a beautifull solution for it, really fast.However i'll explain in portuguese, please translate it.
    Pronlogando o lado AB, traçamos uma reta partindo de C, quer corte perpendicularmente o prolongamento do lado AB, fprmando o ponto E.Se traçarmos o segmento ED, este será a emediana relativa a hipotenusa do triângulo retângulo ACE, portanto terá a mesma medida de AD.Tendo o ângulo C como 90-3x, aparecerá o triângulo isósceles CDE cujo ângulo EDC também será 90-3x.
    Portanto, 90-3x+90-3x+135=180, logo = -6x=-135, portanto x=22,5

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  15. The solution of mr. Anonymous (wisnab36@gmail.com) is nog correct!
    He says:
    "From these facts, HF // BE and AB // BG so that BGFH is a rhombus which has properties HG = BF ."
    That's NOT true!
    Yes, it's a rhombus, but NOT HG = BF!!!
    It's also cleat that his proof can't be correct, because he doesn't use the fact that angle ADB = 45°. IF his proof was solid, then it would be true voor EVERY triangle with angles x and 3x, that x = 22,5°.

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  16. Build quadrilateral ABEC recordable => [DB - bisector [EB - bisector x = 22.5

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  17. http://www.youtube.com/watch?v=kBh_rdTqAvc

    Greetings

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  18. Video solution (Spanish version) by Eder Contreras and Cristian Baeza at Geometry problem 3. Thanks Eder.

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  19. http://henrik-geometry.webs.com/3.htm

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  20. Select Point E at BC so that EAC=x
    So, BEA=2x
    therefore AB=BE (as BEA=BAE=2x)
    Since EAC=ECA=x
    While Drawing a perpendicular to AC from E it meets AC at D
    Now Draw a perpendicular From B to AC,
    Say it is BG
    Since Tri.BGD is 90-45-45
    BG=GD
    Now Draw a parallel to DG through E, so it meets BG at F
    so FE=GD
    ie:BG=EF
    BEF=ECD=x (Corresponding angles, EF//AC)
    so Tri BAG and Tri.BEF are congruent
    therefore ABG=x
    GBD=45, DBC=45-x
    So, ABC=x+45+45-x
    ABC=90 Deg.
    Now draw a semicircle having its center D and radius DA.
    it passes through A,B,C as AD=DC and ABC=90Deg
    SInce the angle subtended at center =2.angle subtended at circumference
    BDA=2.BCA
    2.BCA=45
    2x=45Deg,
    so, x=22.5 Deg.

    Very nice Question, Thanks Mr.Antonio Gutierrez.
    .................

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  21. Problem 003 revisited.
    Draw through B a parallel to AC and trough D a perpendicular do AC. These two lines meet at E, and DE meets BC at G.
    If BF is the altitude of ABC, then BEDF is a square.
    We have AG = CG, then ang(GAC) = x and ang(BGA) = ang(BAG) = 2x.
    So ABG is isosceles with AB = BG and triangles ABF and GBE are congruent.
    Since ang(BAF) = 3x and
    ang(BAF) = ang(BGE) = ang (CGD) = 90 – x,
    then 3x = 90 – x, or x = 22,5.

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  22. Drop BN ⊥ AC
    cot x - cot 3x = (NC - NA)/BN = 2ND/BN = 2
    sin 3x cos x - cos 3x sin x = 2 sin 3x sin x
    sin 2x = 2 sin 3x sin x = cos 2x - cos 4x
    cos 4x = cos 2x - sin 2x
    cos²2x - sin²2x = cos 2x - sin 2x
    (cos 2x - sin 2x) (cos 2x + sin 2x) = (cos 2x - sin 2x)
    cos 2x + sin 2x = 1 or cos 2x - sin 2x = 0

    If sin 2x = 1 - cos 2x then
    2 sin x cos x = 2 sin²x
    cos x = sin x, x = 45° not admissible

    Hence cos 2x - sin 2x = 0
    tan 2x = 1,
    2x = 45°,
    x = 22.5°

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  23. Mi solución con geo clasica :p (en español) http://i50.tinypic.com/2jfalbp.png

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  24. Let ED be the perp bisector of AC, E being on BC. Perp.s from A and E to BD are equal hence Tr ABD is congruent to Tr EBD and the result follows.

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  25. I have a bad English. See Whether you could understand my solution.
    Mark a point E on the side BC such that angle BAE = angle BEA = 2x
    so that BE = BA
    Since Tri AEC is isosceles , the perpendicular drawn from E to BC meet the side BC at D.
    This implies angle BDE = 45 deg.
    Now we shall extend the line DE to F such that DF = AD
    it's obvious that Tri BAD congruent to Tri BDF (S.A.S)
    This implies BF = BA = BE and angle BFD = 3x
    In Tri BEF, angle BEF = angle BFE = 3x
    angle BEF = angle DEC = 3x (vertically opposite angles)
    So, in Tri DEC 3x + x = 90 degree
    4x = 90 degree
    therefore x = 22.5 degree

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  26. https://www.youtube.com/watch?v=7FyrBy5tERU

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