Saturday, February 21, 2009

Problem 17: Right triangles and Angles

Proposed Problem
Problem 17: Right triangles and Angles.

See complete Problem 17 at:
gogeometry.com/problem/problem017.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Posted by Antonio Gutierrez at 9:50 PM
Labels: ,

16 comments:

  1. AjitApril 1, 2009 at 12:42 AM

    From Tr. ABC, AB/AC = sin(2θ)
    AB = AC sin(2θ)=2ACsin(θ)cos(θ)
    = 2AC * DE/DC * DC/AC from Tr. CED & ADC
    Thus, AB = 2DE
    Ajit: ajitathle@gmail.com

    ReplyDelete
  2. AnonymousMay 11, 2009 at 7:00 PM

    This comment has been removed by a blog administrator.

    ReplyDelete
  3. AnonymousJuly 2, 2010 at 2:58 AM

    http://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdf

    ReplyDelete
  4. Nilton LapaJanuary 2, 2012 at 6:14 AM

    Nilton Lapa
    Take F on de extension of BC, so that CF=AC. Tr. ACF is isosceles, with angle(CAF)=angle(CFA)=(theta). Let BC be the median of ACF. Draw GH perpendicular to CF. DAC and GCF are congruent so GH=DE. As F is middle point of CF, we have GH=AB/2 and AB=2.DE.

    ReplyDelete
  5. PravinJanuary 2, 2012 at 7:54 PM

    AB/DE = AB/AC. AC/CD . CD/DE
    = sin 2θ. sec θ. cosec θ = 2

    ReplyDelete
  6. Nilton LapaJanuary 6, 2012 at 11:54 AM

    Where I said "Let BC be the median of ACF", please change to "Let CG be ..."
    Where I said "As F is middle point of CF", please change to "As G is middle poit of AF".
    Sorry about that.
    Nilton Lapa

    ReplyDelete
  7. UnknownApril 10, 2012 at 4:18 AM

    Solution is uploaded to the following link

    https://docs.google.com/open?id=0B6XXCq92fLJJU2s3bDJBRDJBX2c

    ReplyDelete
  8. AnonymousJanuary 10, 2015 at 2:00 PM

    Take F - reflection of A in CD abd G projection of F onto AC. Easily FG=2DE and triangles CFG and CAB are congruent, hence FG=AB, done.

    Best regards,
    Stan Fulger

    ReplyDelete
  9. Sumith PeirisJuly 21, 2015 at 1:44 AM

    Let AC, DB meet at F and let AG drawn parallel to BC meet BD at G.

    ABCD is con cyclic and Tr.s ADF, BCF & AFG are all isoceles.

    Now Tr.s DEF & BAG being similar AB/DE = AG/EF = 2. AG/AF = 2 since AG=AF

    Hence AB = 2.DE

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  10. Loukas SidiropoulosApril 19, 2016 at 2:27 PM

    https://www.facebook.com/photo.php?fbid=10206193945516253&set=a.10205987640598759.1073741831.1492805539&type=3&theater

    ReplyDelete
  11. Stan FulgerApril 22, 2016 at 10:42 PM

    Another idea: take F reflection of D in E; AFBCD is cyclic and we need DF=AB. Since CF is angle bisector of <ACB, there exists the equality BF=AF; but as constructed AF=AD, thus DAFB is an isosceles trapezoid, having equal diagonals, DF=AB, done.

    ReplyDelete
  12. Sumith PeirisDecember 14, 2017 at 10:30 AM

    Solution 2

    Following on from my previous solution

    Let AE = EF = p and let CF = BC = q
    AB^2 = (2p+q)^2 - q^2 = 4p(p+q) = 4DE^2 and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  13. SequeirosJune 13, 2020 at 2:04 PM

    Sea P el reflejo de D respecto a la recta AC, es fácil ver que los puntos A,P,B,C,D pertenecen a una misma circunferencia de diámetro aC.
    Por ser P el reflejo de D, el ángulo PCA es igual al ángulo ACD igual a 2•teta por lo que al arco PB mide 4•teta, y por letra sabemos que el arco AB es 4•teta y de la misma circunferencia por lo que AB=PD y como P es el reflejo de D se sabe que PD=2•DE por lo que AB=2•DE

    ReplyDelete
  14. Sailendra TDecember 7, 2020 at 4:55 PM

    ABCD is cyclic.Let D' be the reflection of D along AC. ADD' is isosceles & D' lies on the same circle.
    CD' is now angle bisector of m(ACB). Observe that B is the reflection of A along CD'. Hence AB=DD' and AB=2DE

    ReplyDelete
  15. MarcoFebruary 17, 2021 at 12:26 AM

    [For easy typing, I use x instead of theta]
    Let AC=a
    Consider triangle ABC
    AB=asin2x, AC=acos2x

    Consider triangle ACD
    CD=acosx

    Consider triangle CDE
    sinx=DE/acosx
    DE=asinxcosx
    =(a/2)*(2sinxcosx)
    =(a/2)*(sin2x)
    =AB/2

    ReplyDelete
  16. rv.littlemanDecember 17, 2021 at 6:47 AM

    See the drawing

    ∠ABC=∠ADC = π/2 => ABCD are concyclic with center O
    => OA=OD=OC=OB

    Define D’ such as ΔCAD is congruent to ΔCAD’
    ∠CDA=∠CD’A = π/2 =>D’ is on the circle O
    => ∠ACD= ∠ACD’= α => ∠BCD’= α
    =>ΔECD is congruent to ΔECD’ (SAS) => ED=ED’

    ∠ACD= ∠ACD’= ∠BCD’= α => ∠AOD= ∠AOD’= ∠BOD’= 2α
    ∠DOD’= ∠AOB= 4α =>ΔDOD’ is congruent to ΔAOB (SAS)
    => DD’=AB
    DD’=2DE

    Therefore AB=2 DE

    ReplyDelete

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