8. Let ABCD be a quadrilateral with an inscribed circle ω that has center I . If IA = 5 , IB = 7 , IC = AB 4 , ID = 9, find the value of . CD Proposed by: Sam Korsky 35 Answer: 36 The I -altitudes of triangles AIB and CID are both equal to the radius of ω , hence have equal length. [ AIB ] AB Therefore = . Also note that [ AIB ] = IA · IB · sin AIB and [ CID ] = IC · ID · sin CID , [ CID ] CD but since lines IA, IB, IC, ID bisect angles ∠ DAB, ∠ ABC, ∠ BCD, ∠ CDA respectively we have that ◦ ◦ ◦ ∠ AIB + ∠ CID = (180 − ∠ IAB − ∠ IBA ) + (180 − ∠ ICD − ∠ IDC ) = 180 . So, sin AIB = sin CID . [ AIB ] IA · IB Therefore = . Hence [ CID ] IC · ID AB IA · IB 35 = = . CD IC · ID 36 ( )

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