Monday, May 19, 2008

Elearn Geometry Problem 44



See complete Problem 44
Angles, Triangle. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Posted by Antonio Gutierrez at 9:14 PM
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10 comments:

  1. AjitMarch 28, 2009 at 5:29 AM

    Tr. ABC gives us, AC/AB = sin(x)/sin(2α)
    Tr. ACD gives us, AC/CD = sin(3α)/sin(2α)
    But AB=CD, so sin(x)/sin(2α)= sin(3α)/sin(2α)
    Thus, sin(x)=sin(3α)or x=3α. Now in Tr. ABC,
    3α+4α+2α =180 or α =20 deg. and x=3α=60 deg.
    Ajit:ajitathle@gmail.com

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  2. AnonymousJune 2, 2009 at 11:52 AM

    This comment has been removed by a blog administrator.

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  3. AnonymousJune 2, 2009 at 2:46 PM

    This comment has been removed by a blog administrator.

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  4. UnknownApril 27, 2012 at 11:46 PM

    The solution is uploaded to the following link:
    https://docs.google.com/open?id=0B6XXCq92fLJJWHh3X19VYWFYNjQ

    Sir, My i send my solution in tamil language as a copy, so that my students can easily understand the way.
    With regards,
    PradheepKannah.

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  5. Sumith PeirisOctober 13, 2015 at 5:29 AM

    Let AD meet BC at E. Drop perpendiculars DH and DJ from D to AC and AB respectively. Drop another perpendicular BFG from B to AF to meet the bisector of < BAD at F and AD at G

    Since D is the incentre of Tr. ABC < ABD = x/2 = 90-3@. Since < ABG = AGB = 90-@, < DBG = 2@ and so BD = BG.

    Now Tr. s CDH, ABF & AFG all being congruent ASA, BG = BD = 2r where r is the inradius. But JD also = r so in right Tr. BDJ, BD = 2DJ hence this Tr. is 30-60-90.

    So x/2 = 30 and x = 60

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  6. Sumith PeirisFebruary 4, 2019 at 4:18 AM

    Solution 2

    Draw DE parallel to AC, E on BC.
    ACED is an isosceles trapezoid.
    < AEB = 3@ = x so 3@ + 2@ + @ = 180,
    @ = 20 and so x = 3@ = 60

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  7. Sumith PeirisFebruary 8, 2019 at 8:04 AM

    Or simply since ACED is an isosceles trapezoid with equal diagonals, Tr. ABE is equilateral

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  8. GregJuly 8, 2019 at 9:44 AM

    Together with my friend rv.littleman we came up with this :

    Draw AE bisector of Ang(CAD). Ang(AEB) = Ang(EAC) + Ang(ECA) = 3α = Ang(EAB). So ABE is isoscele in B and AB = EB.
    ADC and AEC have same base AC and same base angle pairs α and 3α. They are similar and therefore CD = AE which is equal to AB (given).
    So AB = EB = AE, ABE is equilateral and x = 60°.

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