Problem: Ο 1 Ο 1 β Ο 2 Ο 2 β O 1 O 1 β O 2 O 2 β Ξ© Ξ© O 1 O 1 β O 2 O 2 β Ο 1 Ο 1 β B B C C Ο 2 Ο 2 β A A D D A B = 2 , O 1 O 2 = 15 A B = 2 , O 1 β O 2 β = 1 5 C D = 16 C D = 1 6 A B O 1 C D O 2 A B O 1 β C D O 2 β
Solution:
First observe that A O 2 = O 2 D A O 2 β = O 2 β D B O 1 = O 1 C B O 1 β = O 1 β C A β² A β² B β² B β² A A B B Ξ© Ξ© O 1 O 2 βΎ O 1 β O 2 β β A β² A β² B β² B β² Ξ© Ξ© A β² B β² = A B = 2 A β² B β² = A B = 2 A B O 1 O 2 A B O 1 β O 2 β A β² B β² O 2 O 1 A β² B β² O 2 β O 1 β A B O 1 C D O 2 A B O 1 β C D O 2 β B β² A β² O 1 C D O 2 B β² A β² O 1 β C D O 2 β A β² O 1 ^ = D O 2 ^ A β² O 1 β β = D O 2 β β O 1 C ^ = O 2 B β² ^ O 1 β C β = O 2 β B β² β B β² D = A β² C B β² D = A β² C B β² A β² C D B β² A β² C D
Because A β² O 1 = D O 2 A β² O 1 β = D O 2 β A β² O 1 D O 2 A β² O 1 β D O 2 β A β² D = O 1 O 2 = 15 A β² D = O 1 β O 2 β = 1 5 B β² C = 15 B β² C = 1 5 B β² A β² C D B β² A β² C D B β² D β
A β² C + 2 β
16 = 1 5 2 B β² D β
A β² C + 2 β
1 6 = 1 5 2 B β² D = A β² C = B β² D = A β² C = 193 1 9 3 β Ξ± = β B β² A β² D Ξ± = β B β² A β² D β³ B β² A β² D β³ B β² A β² D
cos β‘ Ξ± = 1 5 2 + 2 2 β ( 193 ) 2 2 β
2 β
15 = 3 5 cos Ξ± = 2 β
2 β
1 5 1 5 2 + 2 2 β ( 1 9 3 β ) 2 β = 5 3 β
and hence sin β‘ Ξ± = 4 5 sin Ξ± = 5 4 β R R A β² B β² A β² B β² A β² R βΎ β₯ D R βΎ A β² R β₯ D R β³ A β² R D β³ A β² R D cos β‘ ( β D A β² R ) = 3 5 cos ( β D A β² R ) = 5 3 β A β² R = 9 A β² R = 9 D R = 12 D R = 1 2 A β² B β² βΎ A β² B β² C D βΎ C D D R = 12 D R = 1 2 B β² A β² C D B β² A β² C D 1 2 β
12 β
( 2 + 16 ) = 108 2 1 β β
1 2 β
( 2 + 1 6 ) = 1 0 8
Now let O 1 C = O 2 B β² = r 1 O 1 β C = O 2 β B β² = r 1 β O 2 D = O 1 A β² = r 2 O 2 β D = O 1 β A β² = r 2 β Ο 1 Ο 1 β Ο 2 Ο 2 β r 1 + r 2 = 15 r 1 β + r 2 β = 1 5 β B β² O 2 D β B β² O 2 β D β B β² A β² D β B β² A β² D A β² B β² O 2 D A β² B β² O 2 β D β B β² O 2 D β B β² O 2 β D 18 0 β β Ξ± 1 8 0 β β Ξ± β³ B β² O 2 D β³ B β² O 2 β D
193 = r 1 2 + r 2 2 β 2 r 1 r 2 ( β 3 5 ) = ( r 1 2 + 2 r 1 r 2 + r 2 2 ) β 4 5 r 1 r 2 = ( r 1 + r 2 ) 2 β 4 5 r 1 r 2 = 225 β 4 5 r 1 r 2 . 1 9 3 β = r 1 2 β + r 2 2 β β 2 r 1 β r 2 β ( β 5 3 β ) = ( r 1 2 β + 2 r 1 β r 2 β + r 2 2 β ) β 5 4 β r 1 β r 2 β = ( r 1 β + r 2 β ) 2 β 5 4 β r 1 β r 2 β = 2 2 5 β 5 4 β r 1 β r 2 β . β
Thus r 1 r 2 = 40 r 1 β r 2 β = 4 0 β³ B β² O 2 D β³ B β² O 2 β D 1 2 r 1 r 2 sin β‘ Ξ± = 16 2 1 β r 1 β r 2 β sin Ξ± = 1 6 108 + 2 β
16 = 140 1 0 8 + 2 β
1 6 = 1 4 0 β
The problems on this page are the property of the MAA's American Mathematics Competitions
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