8. [ 8 ] Cyclic quadrilateral ABCD has side lengths AB = 2, BC = 3, CD = 5, AD = 4. Find A B C D 2 sin A sin B (cot + cot + cot + cot ) . Your answer can be written in simplest form as 2 2 2 2 a/b . Find a + b . Solution: (This is somewhat similar to 2006 Geometry #8) Label side AB as a , BC as b , CD as c , AD as d . Label diagonal BD as f , and the circumradius as R . Then the area of triangle ABC is 1 1 ab sin B , and the area of triangle BCD is cd sin D . Since B and D are supplementary, 2 2 1 sin B = sin D . So the area of the quadrilateral, K , is ( ab + cd ) sin B . Similarly, if we split it 2 2 4 K 1 via the other diagonal, we get K = ( bc + ad ) sin A . So sin A sin B = 2 ( ab + cd )( bc + ad ) Note that since a + c = b + d , the quadrilateral is tangential (in other words, it contains an incircle). Draw the incircle with inradius r and center O : 5 Since AO , BO , CO , and DO are angle bisectors, we have: A B C D cot + cot + cot + cot 2 2 2 2 (( ) ( ) ( ) ( )) 1 A B B C C D D A = cot + cot + cot + cot + cot + cot + cot + cot 2 2 2 2 2 2 2 2 2 (( ) ( ) ( ) ( )) 1 w + ( a − w ) x + ( b − x ) y + ( c − y ) z + ( d − z ) = + + + 2 r r r r a + b + c + d = 2 r A B C D In addition, K = rs , where s is the semi perimeter. So cot + cot + cot + cot = 2 2 2 2 2 ( a + b + c + d ) . 4 K 2 4 4 4 K ( a + b + c + d ) ( a + b + c + d ) Multiplying what we have, the final result is = . 2 ( ab + cd )( bc + ad ) 16 K 4( ab + cd )( bc + ad ) Plugging in the original numbers, we have 4802 / 299. So a + b = 5101 . Alternate solution (by Elizabeth Yang): A B C D We can start by simplifying cot + cot + cot + cot using the half angle formula 2 2 2 2 x 1 + cos x cot = 2 sin x This gives 1 + cos A 1 + cos B 1 + cos C 1 + cos D

sin A sin B sin C sin D Since ABCD is cyclic, cos A = − cos C and cos B = − cos C , so we can cancel terms out. 2 2 + sin A sin B 6 Squaring and then multiplying by (sin A )(sin B ), we have ( ) sin A sin B 8 + 4 + sin B sin A ∼ Using Law of Sines on triangles ABD and ABC and the fact that ∠ BDA ∠ BCA , we get

the ratio AC BD

sin A sin B sin A AC

sin B BD Going back to the expression from the problem, we now have ( ) AC BD 8 + 4 + BD AC 22 We can sub in for BD (from Ptolemy’s), giving us AC ( ) 2 ( AC ) 22 8 + 4 + 2 22 ( AC ) Using Law of Cosines on triangles BCA and ACD to solve for AC and plugging in, we get 4802 . So a + b = 5101 . 299 7