Problem: P P A B C D . A B C D . A P βΎ A P β³ A P B β³ A P B a + b Ο β c d e e a + b Ο β c d β β a , b , c , d , e a , b , c , d , e gcd β‘ ( a , b , c , e ) = 1 , g cd( a , b , c , e ) = 1 , d d a + b + c + d + e ? a + b + c + d + e ?
Answer Choices:
A. 25 2 5 26 2 6 27 2 7 28 2 8 29 2 9
Solution:
Without loss of generality, let the side length of the square A B C D A B C D 2 2 2 2 A A B B G G A B A B A G = B G = 1 A G = B G = 1 F F C D C D E E F G F G
Since A E = B E = 2 A E = B E = 2 A G = 1 A G = 1
G E = A E 2 β A G 2 = 4 β 1 = 3 , β G A E = 6 0 β , β D A E = 3 0 β . G E = A E 2 β A G 2 β = 4 β 1 β = 3 β , β G A E = 6 0 β , β D A E = 3 0 β .
We can calculate the area of the blue region as
Area of the blue region = [ β³ G E B ] + [ sector A E B ] β [ β³ A E B ] . Area of the blue region = [ β³ G E B ] + [ sector A E B ] β [ β³ A E B ] .
We compute each part individually
[ β³ G E A ] = 1 2 β
A G β
G E = 1 2 β
1 β
3 = 3 2 , [ β³ G E B ] = 3 2 , [ β³ G E A ] = 2 1 β β
A G β
G E = 2 1 β β
1 β
3 β = 2 3 β β , [ β³ G E B ] = 2 3 β β ,
[ β³ A E B ] = = 1 2 β
2 3 = 3 , [ β³ A E B ] = = 2 1 β β
2 3 β = 3 β ,
[ sector A E B ] = 6 0 β 36 0 β β
Ο β
2 2 = 1 6 β
4 Ο = 2 Ο 3 . [ sector A E B ] = 3 6 0 β 6 0 β β β
Ο β
2 2 = 6 1 β β
4 Ο = 3 2 Ο β .
Hence
Area of the blue region = 3 2 + 2 Ο 3 β 3 = 2 Ο 3 β 3 2 . Area of the blue region = 2 3 β β + 3 2 Ο β β 3 β = 3 2 Ο β β 2 3 β β .
We can compute the area of the purple region as well:
Purple = [ A G F D ] β [ sector A D E ] β [ β³ A G E ] . Purple = [ A G F D ] β [ sector A D E ] β [ β³ A G E ] .
Here A G F D A G F D A G = 1 A G = 1 A D = 2 A D = 2 [ A G F D ] = 2 [ A G F D ] = 2
[ sector A D E ] = 3 0 β 36 0 β β
Ο β
2 2 = 1 12 β
4 Ο = Ο 3 , [ β³ A G E ] = 3 2 . [ sector A D E ] = 3 6 0 β 3 0 β β β
Ο β
2 2 = 1 2 1 β β
4 Ο = 3 Ο β , [ β³ A G E ] = 2 3 β β .
Thus
Purple = 2 β Ο 3 β 3 2 . Purple = 2 β 3 Ο β β 2 3 β β .
Total desired area = Pink + Purple = ( 2 Ο 3 β 3 2 ) + ( 2 β Ο 3 β 3 2 ) = 2 + Ο 3 β 3 . Total desired area = Pink + Purple = ( 3 2 Ο β β 2 3 β β ) + ( 2 β 3 Ο β β 2 3 β β ) = 2 + 3 Ο β β 3 β .
Since the square has area 4 4
P = 2 + Ο 3 β 3 4 = 1 2 + Ο 12 β 3 4 = 6 + Ο β 3 3 12 . P = 4 2 + 3 Ο β β 3 β β = 2 1 β + 1 2 Ο β β 4 3 β β = 1 2 6 + Ο β 3 3 β β .
Therefore, in the form a + b Ο β c d e e a + b Ο β c d β β
a = 6 , b = 1 , c = 3 , d = 3 , e = 12 , a = 6 , b = 1 , c = 3 , d = 3 , e = 1 2 ,
so
a + b + c + d + e = (A) 25 a + b + c + d + e = (A) 25 β
The problems on this page are the property of the MAA's American Mathematics Competitions
