Tuesday, May 13, 2014

Geometry Problem 1014. Square, Diagonal, 45 Degrees, Angle, Equal Areas, Triangle, Quadrilateral

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

Click the figure below to enlarge the figure of problem 1014.

Online Geometry Problem 1014. Square, Diagonal, 45 Degrees, Angle, Equal Areas, Triangle, Quadrilateral.

Posted by Antonio Gutierrez at 8:07 PM
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3 comments:

  1. Peter TranMay 13, 2014 at 9:23 PM

    Since ∠ (EAH)= ∠ (EBH)=45 => ABEH is cocyclic
    And ∠ (EHA)= 90 => AEH is a right isosceles triangle
    AE/AH=sqrt(2)
    Similarly we also have AF/AG= sqrt(2)
    Triangle AGH similar to triangle(AFE) with ratio of similarity= sqrt(2)
    And area(AEF)=2. Area(AGH)
    So S1=S2

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  2. Jacob HA (EMK2000)May 13, 2014 at 9:55 PM

    It suffice to prove that (AE×AF)/(AG×AH) = 2.

    Using trigonometry,
    Let ∠BAE=α and ∠DAF=β, then α+β=45°.

    (AE×AF)/(AG×AH)
    = AE/AG × AF/AH
    = (AB+BE)/AB × (AD+DF)/AD --- Angle bisector theorem
    = (1 + BE/AB) × (1 + DF/AD)
    = (1 + tanα) × (1 + tanβ)
    = 1 + tanα tanβ + tanα + tanβ
    = 1 + tanα tanβ + tan(α+β) [1 − tanα tanβ] --- Compound angle formula
    = 1 + tanα tanβ + 1 − tanα tanβ
    = 2

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  3. Sumith PeirisApril 9, 2016 at 5:05 AM

    AGFD is cyclic GF is perpendicular to AG. Let AG = GF = a and let GE = b

    ABEH is cyclic so AH is perpendicular to EH and AH = EH = (a+b)/sqrt2

    S1 = S(AGF) X AH/AF = a^2/2 X (a+b)/sqrt2 / (sqrt2 a)= a(a+b)/4...(1)

    S1+S2 = 1/2 a(a+b)...(2)

    From (1) and (2) S1 + S2 = 2S1 and hence S1 = S2

    Sumith Peiris
    Moratuwa
    Sri Lanka

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