Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 909.
Wednesday, August 7, 2013
Problem 909: Bicentric Quadrilateral, Incircle, Circumcircle, Circumscribed, Inscribed, Tangent, Incenter, Inradius, Distance
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∠IHA = ∠CGI = 90
ReplyDelete∠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
A+C=180
ReplyDeletesin(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.
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
ReplyDelete