Wednesday, August 7, 2013

Problem 909: Bicentric Quadrilateral, Incircle, Circumcircle, Circumscribed, Inscribed, Tangent, Incenter, Inradius, Distance

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

Click the figure below to see the complete problem 909.

Online Geometry Problem 909: Bicentric Quadrilateral, Incircle, Circumcircle, Circumscribed, Inscribed, Tangent, Incenter, Inradius, Distance.

Posted by Antonio Gutierrez at 3:57 PM
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3 comments:

  1. nakaAugust 7, 2013 at 4:24 PM

    ∠IHA = ∠CGI = 90
    ∠IAH = 1/2∠BAD = 1/2(180-∠BCD) = 90-1/2∠BCD = 90-∠ICG = ∠CIG
    So, ∆IAH ~ ∆ICG
    (a/c) = r/(sqrt(c^2 - r^2)) = (sqrt(a^2 - r^2))/r
    => a^4/c^4 = (a^2 - r^2)/(c^2 - r^2)
    => r^2 = (a^4*c^2 - c^4*a^2)/(a^4 - c^4)
    => 1/r^2 = (a^4 - c^4)/[a^2*c^2*(a^2 - c^2)]
    => 1/r^2 = (a^2+c^2) / a^2*c^2

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  2. Jacob HA (EMK2000)August 7, 2013 at 5:59 PM

    A+C=180
    sin(A/2) = cos(C/2)

    1/a^2 + 1/c^2
    = sin^2(A/2)/r^2 + sin^2(C/2)/r^2
    = [cos^2(C/2) + sin^2(C/2)]/r^2
    = 1/r^2

    Similarly for 1/b^2+1/d^2=1/r^2.

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  3. Ivan BazarovJanuary 10, 2014 at 8:02 PM

    a/r=c/sqrt(c^2-r^2)=csc(A/2), solving for r^2 gives r^2=(c^2a^2)/(c^2+a^2). The same reasoning gives r^2=(b^2d^2)/(b^2+d^2). The result follows immediately

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