In an equiangular pentagon, the sum of the squares of the side lengths equals 308 3 0 8 800 8 0 0 m n m n β m m n n n n m + n m + n
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Let
v β 1 = A B β , v β 2 = B C β , v β 3 = C D β , v β 4 = D E β , v β 5 = E A β . v 1 β = A B , v 2 β = B C , v 3 β = C D , v 4 β = D E , v 5 β = E A .
The diagonals correspond to the vectors v β 1 + v β 2 , v β 2 + v β 3 , v β 3 + v β 4 , v β 4 + v β 5 v 1 β + v 2 β , v 2 β + v 3 β , v 3 β + v 4 β , v 4 β + v 5 β v β 5 + v β 1 v 5 β + v 1 β
β i = 1 5 β£ v β i + v β i + 1 β£ 2 = β i = 1 5 ( β£ v i β β£ 2 + β£ v i + 1 β β£ 2 ) + 2 β
β i = 1 5 v β i β
v β i + 1 = 2 ( β i = 1 5 β£ v i β β£ 2 + β i = 1 5 v β i β
v β i + 1 ) = 800 i = 1 β 5 β β£ v i β + v i + 1 β β£ 2 β = i = 1 β 5 β ( β£ v i β β β£ 2 + β£ v i + 1 β β β£ 2 ) + 2 β
i = 1 β 5 β v i β β
v i + 1 β = 2 ( i = 1 β 5 β β£ v i β β β£ 2 + i = 1 β 5 β v i β β
v i + 1 β ) = 8 0 0 β
Here we have adopted the convention that v β k = v β k β 5 v k β = v k β 5 β k > 5 k > 5 β i = 1 5 β£ v i β β£ 2 = 308 β i = 1 5 β β£ v i β β β£ 2 = 3 0 8 β i = 1 5 v β i β
v β i + 1 = 92 β i = 1 5 β v i β β
v i + 1 β = 9 2
Now, note that since the pentagon is closed:
v β 1 + v β 2 + v β 3 + v β 4 + v β 5 = 0. v 1 β + v 2 β + v 3 β + v 4 β + v 5 β = 0 .
β£ v β 1 + v β 2 + v β 3 + v β 4 + v β 5 β£ 2 = β i = 1 5 β£ v i β β£ 2 + 2 β
β i = 1 5 v β i β
v β i + 1 + 2 β
β i = 1 5 v β i β
v β i + 1 = 308 + 2 β
92 + 2 β
β i = 1 5 v β i β
v β i + 2 = 0 β£ v 1 β + v 2 β + v 3 β + v 4 β + v 5 β β£ 2 β = i = 1 β 5 β β£ v i β β β£ 2 + 2 β
i = 1 β 5 β v i β β
v i + 1 β + 2 β
i = 1 β 5 β v i β β
v i + 1 β = 3 0 8 + 2 β
9 2 + 2 β
i = 1 β 5 β v i β β
v i + 2 β = 0 β
and so β i = 1 5 v β i β
v β i + 2 = β 246 β i = 1 5 β v i β β
v i + 2 β = β 2 4 6
Finally, the square of the perimeter is given by:
( β£ v β 1 β£ + β£ v β 2 β£ + β£ v β 3 β£ + β£ v β 4 β£ + β£ v β 5 β£ ) 2 = β i = 1 5 β£ v β i β£ 2 + 2 β i = 1 5 β£ v β i β£ β
β£ v β i + 1 β£ + 2 β i = 1 5 β£ v β i β£ β
β£ v β i + 2 β£ = β i = 1 5 β£ v β i β£ 2 + 2 cos β‘ 7 2 β β i = 1 5 v β i β
v β i + 1 + 2 cos β‘ 14 4 β β i = 1 5 v β i β
v β i + 2 = 308 + 2 ( 5 + 1 ) β
92 β 2 ( 5 β 1 ) β
( β 246 ) = 676 5 ( β£ v 1 β β£ + β£ v 2 β β£ + β£ v 3 β β£ + β£ v 4 β β£ + β£ v 5 β β£ ) 2 β = i = 1 β 5 β β£ v i β β£ 2 + 2 i = 1 β 5 β β£ v i β β£ β
β£ v i + 1 β β£ + 2 i = 1 β 5 β β£ v i β β£ β
β£ v i + 2 β β£ = i = 1 β 5 β β£ v i β β£ 2 + cos 7 2 β 2 β i = 1 β 5 β v i β β
v i + 1 β + cos 1 4 4 β 2 β i = 1 β 5 β v i β β
v i + 2 β = 3 0 8 + 2 ( 5 β + 1 ) β
9 2 β 2 ( 5 β β 1 ) β
( β 2 4 6 ) = 6 7 6 5 β β
so m + n = 676 + 5 = 681 m + n = 6 7 6 + 5 = 6 8 1 β
Let the side lengths of the pentagon be a , b , c , d , e a , b , c , d , e a 2 + b 2 + c 2 + d 2 + e 2 = 308 a 2 + b 2 + c 2 + d 2 + e 2 = 3 0 8 cos β‘ 10 8 β = 1 4 ( 1 β 5 ) cos 1 0 8 β = 4 1 β ( 1 β 5 β ) a a b b
diagonal 2 = a 2 + b 2 β 2 a b β
cos β‘ angle between a and b = a 2 + b 2 β 2 a b β
1 4 ( 1 β 5 ) = a 2 + b 2 + a b β
5 β 1 2 diagonal 2 = a 2 + b 2 β 2 a b β
cos angle between a and b = a 2 + b 2 β 2 a b β
4 1 β ( 1 β 5 β ) = a 2 + b 2 + a b β
2 5 β β 1 β
Thus adding these up, we get
β diagonal 2 = 2 ( a 2 + b 2 + c 2 + d 2 + e 2 ) + ( a b + b c + c d + d e + e a ) β
5 β 1 2 = 800 β diagonal 2 = 2 ( a 2 + b 2 + c 2 + d 2 + e 2 ) + ( a b + b c + c d + d e + e a ) β
2 5 β β 1 β = 8 0 0
Thus we get
a b + b c + c d + d e + e a = ( 800 β 2 β
308 ) β
( 5 β 1 2 ) β 1 = 184 β
1 + 5 2 = 92 ( 1 + 5 ) a b + b c + c d + d e + e a = ( 8 0 0 β 2 β
3 0 8 ) β
( 2 5 β β 1 β ) β 1 = 1 8 4 β
2 1 + 5 β β = 9 2 ( 1 + 5 β )
Now orient the pentagon so that the side a a ΞΈ = 7 2 β ΞΈ = 7 2 β x x
Side Length Angle ( ΞΈ ) Horizontal Component ( s cos β‘ ΞΈ ) Vertical Component ( s sin β‘ ΞΈ ) 1 a 0 β a 0 2 b 7 2 β b cos β‘ 7 2 β b sin β‘ 7 2 β 3 c 14 4 β c cos β‘ 14 4 β c sin β‘ 14 4 β 4 d 21 6 β d cos β‘ 21 6 β d sin β‘ 21 6 β 5 e 28 8 β e cos β‘ 28 8 β e sin β‘ 28 8 β Side 1 2 3 4 5 β Length a b c d e β Angle ( ΞΈ ) 0 β 7 2 β 1 4 4 β 2 1 6 β 2 8 8 β β Horizontal Component ( s cos ΞΈ ) a b cos 7 2 β c cos 1 4 4 β d cos 2 1 6 β e cos 2 8 8 β β Vertical Component ( s sin ΞΈ ) 0 b sin 7 2 β c sin 1 4 4 β d sin 2 1 6 β e sin 2 8 8 β β β
Note that cos β‘ ΞΈ = cos β‘ ( 36 0 β β ΞΈ ) cos ΞΈ = cos ( 3 6 0 β β ΞΈ )
a + ( b + e ) cos β‘ 7 2 β + ( c + e ) cos β‘ 14 4 β = 0 a + ( b + e ) cos 7 2 β + ( c + e ) cos 1 4 4 β = 0
0 + ( b + e ) sin β‘ 7 2 β + ( c + e ) sin β‘ 14 4 β = 0 0 + ( b + e ) sin 7 2 β + ( c + e ) sin 1 4 4 β = 0
If we square and add these, we get
β ( side 2 β
2 ΞΈ + side 2 β
2 ΞΈ ) + 2 β side 1 β
side 2 β
( cos β‘ ΞΈ 1 cos β‘ ΞΈ 2 + sin β‘ ΞΈ 1 sin β‘ ΞΈ 2 ) β ( side 2 β
cos 2 ΞΈ + side 2 β
sin 2 ΞΈ ) + 2 β side 1 β
side 2 β
( cos ΞΈ 1 β cos ΞΈ 2 β + sin ΞΈ 1 β sin ΞΈ 2 β )
Using the fact that cos β‘ ( ΞΈ 1 β ΞΈ 2 ) = cos β‘ ΞΈ 1 cos β‘ ΞΈ 2 + sin β‘ ΞΈ 1 sin β‘ ΞΈ 2 cos ( ΞΈ 1 β β ΞΈ 2 β ) = cos ΞΈ 1 β cos ΞΈ 2 β + sin ΞΈ 1 β sin ΞΈ 2 β 2 ΞΈ + 2 ΞΈ = 1 sin 2 ΞΈ + cos 2 ΞΈ = 1
0 = β side 2 + 2 β side 1 β
side 2 β
cos β‘ ( angle between side 1, 2 ) 0 = β side 2 + 2 β side 1 β
side 2 β
cos ( angle between side 1, 2 )
The angle between sides is 7 2 β 7 2 β 144 4 β 1 4 4 4 β
0 = ( a 2 + b 2 + c 2 + d 2 + e 2 ) + cos β‘ 7 2 β β
( a b + b c + c d + d e + e a ) + cos β‘ 14 4 β β
( a c + b d + c e + d a + e b ) 0 = ( a 2 + b 2 + c 2 + d 2 + e 2 ) + cos 7 2 β β
( a b + b c + c d + d e + e a ) + cos 1 4 4 β β
( a c + b d + c e + d a + e b )
Thus, we get that as cos β‘ 14 4 β = β cos β‘ 3 6 β = β 1 4 ( 5 + 1 ) cos 1 4 4 β = β cos 3 6 β = β 4 1 β ( 5 β + 1 )
a c + b d + c e + d a + e b = ( 308 + 5 β 1 2 β
( 92 + 92 5 ) ) β
( 5 + 1 2 ) β 1 = 246 5 β 246 a c + b d + c e + d a + e b = ( 3 0 8 + 2 5 β β 1 β β
( 9 2 + 9 2 5 β ) ) β
( 2 5 β + 1 β ) β 1 = 2 4 6 5 β β 2 4 6
Thus
( a + b + c + d + e ) 2 = 308 + 2 ( 92 + 92 5 ) + 2 ( 246 5 β 246 ) = 676 5 ( a + b + c + d + e ) 2 = 3 0 8 + 2 ( 9 2 + 9 2 5 β ) + 2 ( 2 4 6 5 β β 2 4 6 ) = 6 7 6 5 β
for an answer of so m + n = 676 + 5 = 681 m + n = 6 7 6 + 5 = 6 8 1 β
Suppose that the sides of the pentagon are a 1 , a 2 , β¦ , a 5 a 1 β , a 2 β , β¦ , a 5 β
b β a = a 1 c β b = a 2 β
Ο d β c = a 3 β
Ο 2 e β d = a 4 β
Ο 3 a β e = a 5 β
Ο 4 b β a = a 1 β c β b = a 2 β β
Ο d β c = a 3 β β
Ο 2 e β d = a 4 β β
Ο 3 a β e = a 5 β β
Ο 4
where Ο = e 2 Ο i / 5 Ο = e 2 Ο i / 5
a 1 + a 2 β
Ο + a 3 β
Ο 2 + a 4 β
Ο 3 + a 5 β
Ο 4 = 0 a 1 β + a 2 β β
Ο + a 3 β β
Ο 2 + a 4 β β
Ο 3 + a 5 β β
Ο 4 = 0
Note that:
β i = 1 5 β£ a i β
Ο i β 1 β£ 2 = β i = 1 5 a i 2 = 308 i = 1 β 5 β β£ a i β β
Ο i β 1 β£ 2 = i = 1 β 5 β a i 2 β = 3 0 8
and
β i = 1 5 β£ a i β
Ο i β 1 + a i + 1 β
Ο i β£ 2 = β i = 1 5 β£ a i + a i + 1 β
Ο β£ 2 = β i = 1 5 ( a i + a i + 1 β
Ο ) β
( a i + a i + 1 β
1 Ο ) = 2 β i = 1 5 a i 2 + ( Ο + 1 Ο ) β i = 1 5 a i a i + 1 = 2 β
308 + ( 2 cos β‘ 7 2 β ) β
β i = 1 5 a i a i + 1 = 800 i = 1 β 5 β β£ a i β β
Ο i β 1 + a i + 1 β β
Ο i β£ 2 β = i = 1 β 5 β β£ a i β + a i + 1 β β
Ο β£ 2 = i = 1 β 5 β ( a i β + a i + 1 β β
Ο ) β
( a i β + a i + 1 β β
Ο 1 β ) = 2 i = 1 β 5 β a i 2 β + ( Ο + Ο 1 β ) i = 1 β 5 β a i β a i + 1 β = 2 β
3 0 8 + ( 2 cos 7 2 β ) β
i = 1 β 5 β a i β a i + 1 β = 8 0 0 β
and so:
β i = 1 5 a i a i + 1 = 92 cos β‘ 7 2 β = 92 ( 5 + 1 ) i = 1 β 5 β a i β a i + 1 β = cos 7 2 β 9 2 β = 9 2 ( 5 β + 1 )
Now, note that:
a 1 + a 2 β
Ο + a 3 β
Ο 2 + a 4 β
Ο 3 + a 5 β
Ο 4 = 0 a 1 β + a 2 β β
Ο + a 3 β β
Ο 2 + a 4 β β
Ο 3 + a 5 β β
Ο 4 = 0
and so:
( a 1 + a 2 β
Ο + a 3 β
Ο 2 + a 4 β
Ο 3 + a 5 β
Ο 4 ) β
( a 1 + a 2 β
1 Ο + a 3 β
1 Ο 2 + a 4 β
1 Ο 3 + a 5 β
1 Ο 4 ) = β i = 1 5 a i 2 + 2 ( Ο + 1 Ο ) β i = 1 5 a i a i + 1 + 2 ( Ο 2 + 1 Ο 2 ) β i = 1 5 a i a i + 2 = 308 + 2 β
92 + 2 β
cos β‘ 14 4 β β i = 1 5 a i a i + 2 = 0 = β ( a 1 β + a 2 β β
Ο + a 3 β β
Ο 2 + a 4 β β
Ο 3 + a 5 β β
Ο 4 ) β
( a 1 β + a 2 β β
Ο 1 β + a 3 β β
Ο 2 1 β + a 4 β β
Ο 3 1 β + a 5 β β
Ο 4 1 β ) i = 1 β 5 β a i 2 β + 2 ( Ο + Ο 1 β ) i = 1 β 5 β a i β a i + 1 β + 2 ( Ο 2 + Ο 2 1 β ) i = 1 β 5 β a i β a i + 2 β = 3 0 8 + 2 β
9 2 + 2 β
cos 1 4 4 β i = 1 β 5 β a i β a i + 2 β = 0 β
so β i = 1 5 a i a i + 2 = β 492 2 β
cos β‘ 14 4 β = β 246 ( 5 β 1 ) i = 1 β 5 β a i β a i + 2 β = β 2 β
cos 1 4 4 β 4 9 2 β = β 2 4 6 ( 5 β β 1 )
( a 1 + a 2 + a 3 + a 4 + a 5 ) 2 = β i = 1 5 a i 2 + 2 β i = 1 5 a i a i + 1 + 2 β i = 1 5 a i a i + 2 = 308 + 2 β
( 92 ( 5 + 1 ) ) + 2 β
( β 246 ( 5 β 1 ) ) = 308 + 184 5 + 184 β 492 + 492 5 = 676 5 ( a 1 β + a 2 β + a 3 β + a 4 β + a 5 β ) 2 β = i = 1 β 5 β a i 2 β + 2 i = 1 β 5 β a i β a i + 1 β + 2 i = 1 β 5 β a i β a i + 2 β = 3 0 8 + 2 β
( 9 2 ( 5 β + 1 ) ) + 2 β
( β 2 4 6 ( 5 β β 1 ) ) = 3 0 8 + 1 8 4 5 β + 1 8 4 β 4 9 2 + 4 9 2 5 β = 6 7 6 5 β β
so m + n = 676 + 5 = 681 m + n = 6 7 6 + 5 = 681 β
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