Saturday, November 28, 2009

Problem 395: Square, 15 Degree, Equilateral triangle

Proposed Problem
Click the figure below to see the complete problem 395 about Square, 15 Degree, Equilateral triangle, Congruence.

 Problem 395: Square, 15 Degree, Equilateral triangle.
See also:
Complete Problem 395

Level: High School, SAT Prep, College geometry

12 comments:

  1. http://www.wolframalpha.com/input/?i=%281-1%2F2*tg%2815%29%29%5E2%2B1%2F4

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  2. http://geometri-problemleri.blogspot.com/2009/11/problem-52-ve-cozumu.html

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  3. in triangle ABE ang(ABE)=75;BE=AB/(2cos15)
    with the law of cosine
    AE²=AB²+BE²-2AB.BE cos75=
    AB²(1+(1/(2cos15)²-cos75/cos15)= AB² then
    cos(75)=cos(45+30)=(sqr6-sqr2)/4
    cos(15)=cos(45-30)=(sqr6+sqr2)/4
    same result for CE
    CE=AE=AD
    ADE is equilateral

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  4. build on each side of the square (from the inside) isosceles triangle with 15 degrees..from there its easy..

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  5. Consider the equilateral triangle BCF at the exterior of the square ABCD say of side a. The quandrilateral FEDC and FEAB are rhombuses, therefore

    ED = FC = a and EA = FB = a

    (This holds because ang(FEC) = 75 = ang(ECF), so FE = FC = a and FE parallel to CD = a)

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  6. See
    http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=472125

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  7. See the drawing : Drawing

    - ABCD is a square
    - Define J as the rotation of E by 90° from the center of the square
    - E becomes J => BE=CE=CJ=DJ and ΔECJ is equilateral
    - Define F the intersection of DJ et EC
    - EDA=60°
    - By symmetry, <EAD=60°

    Therefore ΔEAD is equilateral

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  8. See diagram here.

    - E can be uniquely defined as the intersection of the perpendicular bisector to BC and the ray from B forming a 15° angle with BC.
    - Make F the intersection of circular arcs BD with center A and AC with center D.
    - AD = AF = DF ⇒ ∆ADF is equilateral ⇒ F is on the perpendicular bisector to BC (same as that to AD) and ∠BAF = 30°.
    - But on circular arc BD, ∠BAF = 30° ⇒ ∠CBF = 15° since CB is tangent in B to arc BD.

    - So F = E and ∆AED is equilateral QED

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  9. Let length of square = a & mid-pt of BC & AD be F & G respectively
    tan15=EF/(a/2)
    EF=atan15/2
    EG=a(1-tan15/2)
    tan<EAD=(1-tan15/2)/(1/2)
    tan<EAD=2-tan15
    tan<EAD=2-(2-sqrt3)
    tan<EAD=sqrt3
    <EAD=60

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  10. Construct Equilateral Triangle BCF, F outside the square.

    Triangles CEF, BEF and ABE are all equilateral SAS (CF = BF = BA, CE = BE, included angle is 75 for all 3 triangles)

    So Triangles BFE and EFC are isosceles 30-75-75) and so AEB is also isosceles and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

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    Replies
    1. I should have said ,"Triangles CEF, BEF and ABE are all all congruent......." not equilateral

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