Wednesday, October 24, 2012

Problem 818: Square, Triangle, Angles

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

Click the figure below to see the complete problem 818.

Online Geometry Problem 818: Square, Triangle, Angles

15 comments:

  1. http://www.wolframalpha.com/input/?i=tg%283a%29%3Dtg%28a%29%2B1

    alpha=0.30714520
    x=pi/2+alpha

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    Replies
    1. To Anonymous, problem 818: solution is not complete.

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  2. alpha=pi/12=15 G
    x=pi/2+alpha= 105 G

    I solve system of 3 equations

    c=tg(a)
    c+1=b*tg(3a)
    c+1=(b+1)*tg(2a)

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  3. In the triangle AFE
    AFE) = x = 180 - alpha -(90 - 2 * alpha) = 90 + alpha.
    Then (DCE) = (AFE) = 90 + alpha.

    In the triangle ADE :
    AD/sin(alpha) = DE/sin(2 alpha)
    Therefore:
    DE = 2 AD cos(alpha) ----- (1)

    In the triangle DCE :
    DC/sin(2 alpha) = DE/sin(90 + alpha)
    Therefore:
    DE = DC / ( 2 sen(alpha) ) ----- (2).

    DC = AD

    Therefore from (1) and (2) :
    2 cos(alpha) = 1/( 2 sen(alpha) ) ------ (3)

    If we simplify the (3):
    sin(2 alpha) = 1/2.
    Therefore: alpha = 15
    x = 90 + 15 = 105

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  4. Problem 818: Try to use elementary geometry (Euclid's Elements).

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    Replies
    1. Geometric solution by Michael Tsourakakis
      EC sectionAB=H
      (q) is Circumcircle of the scare ABCD and (q) section AE=K
      bisector of the angle EAD section (q) =M.
      angleDCK=angleKAD=2α .So angleDCM=angleDKM=angleCDK=angleKDM=α.Therefore, angleCBK=α
      But angleKBD=angleKAD=2α .So angleDBC=45=2α+α=3α. So,α=15
      angleCAK=α=angleCEA.So, angleHKA=2α =30
      From the triangle HCA : x=180-30-45=1050

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    2. http://1.bp.blogspot.com/-H0dtaRbb9bY/UMTFm5yQRyI/AAAAAAAAAAM/5Dco1bojssk/s1600/Untitled.jpg

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  5. Prof Radu Ion,Sc.Gim.Bozioru,BuzauDecember 21, 2012 at 3:36 AM

    Daca notam cu P intersectia dreptelor CF si AD vom obtine:2α=m(m( in triunghiul AFP dreptunghic in A, m(
    x=m(<AFE)=180-(90-α)=90+α

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  6. To: Michael Tsourakakis---In your solution you have implied that the ARC CKMD is split into 3 equal parts. This is not true based on your reasoning.

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    1. correct solution
      (q) is Circumcircle of the square ABCD and (q) section AE=K
      BD ,is mediator of AC and V is mediator of CE. BD section (v)=M.Then , CM is mediator of AE ,so CM is perpendicular of the AE.Beacause AC is ,diameter the circle (q) ,CK ,is perpendicular of AE and the points C,K,M , are collinear . The triangle ACE ,is isosceles.
      So, angle CAE=α and angleCAD=3α=45 . Τhen,α=15
      x=180-45-2α =135-30=105 degrees
      http://img191.imageshack.us/img191/5597/geogebra3.png
      (Michael Tsourakakis )

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    2. correction:
      The triangle CAE is isosceles And not "the triangle ΜΑΕ is isosceles"
      (Michael Tsourakakis

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  7. Anonymous

    If we say the cut of BD with CF = P and
    the cut AE with CD =R you have accept without verification that angle CPR= 2 alpha or equivalent that angle APR =90 degree .
    If you can not verify one of these two saying than the solution is wrong.

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    Replies
    1. you're right.Thanks for the correction.I'll look again
      HAPPY NEW YEAR
      Michael Tsourakakis

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  8. http://img839.imageshack.us/img839/3518/problem818.png

    Draw circumcircle of triangle CDE
    Note that angleDOG= 2 α and AE cut arc CD at midpoint G
    Since angle (GAD)= angle (DOG)= 2 α and AD//GO => ADOG is a parallelogram
    And OD=AD=CD => triangle DOC is a equilateral
    So 2 α= 30
    In triangle AFE , we have x=180- α-(90-2 α)= 90+ α= 105 degrees

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  9. http://www.mathematica.gr/forum/viewtopic.php?f=22&t=32386&p=149846

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