Friday, July 13, 2018

Geometry Problem 1362: Equilateral Triangle Inscribed in a Circle, Lunula, Ratio of Areas, Tangent, Sketch, iPad Apps

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

Geometry Problem 1362: Equilateral Triangle Inscribed in a Circle, Lunula, Ratio of Areas, Tangent, Sketch, iPad Apps.

4 comments:

  1. Let be P the point Aq meet the circle Q.Join F to Q, F to P
    The triangle PFQ eqilateral => Q midpoint of PQ
    => r = 2/3 (R)

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  2. Let the side of the equilateral triangle ABC be 3 units and AE=AF=x => FC=3-x
    AEF is equilateral (since AE=AF) and AEQF and ABDC are similar cyclic quadrilaterals.
    Also AOQD are collinear
    Since m(EQF)=120=>m(EDF)=60 --------(1)
    Consider the triangle ADF, since m(FAD)=m(FDA)=30
    => m(DFC)=60 -------(2)
    and DF=AF=AE=EF and EDF is an equilateral triangle --------(3)
    Since m(AFQ)=m(ACD)=90, the triangle DFC is a 30-60-90 triangle
    => DF/FC=2
    => x/(3-x)=2
    => x=2
    So the ratios of the radius of the circles O and Q are 3/2
    Areas = 9/4
    =>Ar.of Blue lunula/Ar. of Yellow region=5/4

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  3. Let circle Q be of radius r and circle O be of radius R.

    Since QA & OA both bisect <BAC it follows that A,O,Q,D are collinear points.

    Since QAF is a 30-60-90 Tr.,
    AQ = 2QF.
    So R +(R-r) = 2r
    Hence R/r = 3/2

    So A(Circle O) / A(Circle Q) = 9/4

    Therefore S/S1 = 5/4

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  4. Q and O are on AD; AE = AF;
    QE = QF; thus ∠ EOA = 90°.
    ∠ EBO = 30° = ∠OAE. Thus ∠ AEO = 60°,
    but ∠ AEQ = 90°, thus ∠OEQ = 30°.
    BE = .5 * OB * sec(30°) = OE
    QE = OE * sec(30°)
    =.5 * OB * sec(30°) * sec(30°)
    = OB * 2 / 3. Thus S / S_1 = 5 / 4.

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