Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 883.
Wednesday, June 5, 2013
Problem 883: Five Tangential or Circumscribed Quadrilaterals, Circle, Tangent, Common Tangent
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ReplyDeleteLet b=BR=BZ , a=AY=AX, d=DV=DU and c=CS=CT
Let b’=B’M’=B’G’ , a’=A’E’=A’N’, d’=D’F’=D’Q’, c’=C’H’=C’P’ ( see attached sketch )
A’B’C’D’ is a tangential quadrilateral => B’C’+A’D’=A’B’+C’D’
Add a’+b’+c’+d’ to both sides of above equation and simplify it . we get G’H’+E’F’=M’N’+P’Q’ (1)
P’Q’ and TU are 2 external tangents to 2 circles so P’Q’=TU
Similarly E’F’=XV , M’N’=ZY and G’H’=RS
Replace it in (1) .Equation (1) become RS+XV=ZY+TU
Add a+b+c+d to both sides of above equation and simplify it . we get BC+AD=BA+CD
So ABCD is a tangential quadrilateral and AD=BA+CD-BC=2
guys, we lost imageshack.us. We need someone to bring Peter's solutions back.
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