英文解析

8. Let ABCD be a quadrilateral with an inscribed circle ω that has center I . If IA = 5 , IB = 7 , IC =
   4 , ID = 9, find the value of .AB
   Proposed by: Sam Korsky cD
   Answer:35
   The I -altitudes of triangles AIB and CID are both equal to the radius of ω , hence have equal length.36
   [ AIB ]
   Therefore = . Also note that [ AIB ] = IA · IB · sin AIB and [ CID ] = IC · ID · sin CID ,AB
   [ CID ] CDbut 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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