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

Labels:
bicentric quadrilateral,
distance,
incenter,
inradius

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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