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Question: Let O be the centre of a regular pentagon ABCDE and $\overline{OA} = \overline{a}$. then $\overline{...

Let O be the centre of a regular pentagon ABCDE and OA‾=a‾\overline{OA} = \overline{a}. then AB‾+2BC‾+3CD‾+4DE‾+5EA‾\overline{AB} + 2\overline{BC} + 3\overline{CD} + 4\overline{DE} + 5\overline{EA} equals:

A

0

B

4a‾4\overline{a}

C

5a‾5\overline{a}

D

6a‾6\overline{a}

Answer

5a‾5\overline{a}

Explanation

Solution

Let the position vectors of the vertices A, B, C, D, E with respect to the center O be a‾,b‾,c‾,d‾,e‾\overline{a}, \overline{b}, \overline{c}, \overline{d}, \overline{e} respectively. So, OA‾=a‾\overline{OA} = \overline{a}.

We can express each vector in the given sum in terms of these position vectors:

AB‾=OB‾−OA‾=b‾−a‾\overline{AB} = \overline{OB} - \overline{OA} = \overline{b} - \overline{a}

BC‾=OC‾−OB‾=c‾−b‾\overline{BC} = \overline{OC} - \overline{OB} = \overline{c} - \overline{b}

CD‾=OD‾−OC‾=d‾−c‾\overline{CD} = \overline{OD} - \overline{OC} = \overline{d} - \overline{c}

DE‾=OE‾−OD‾=e‾−d‾\overline{DE} = \overline{OE} - \overline{OD} = \overline{e} - \overline{d}

EA‾=OA‾−OE‾=a‾−e‾\overline{EA} = \overline{OA} - \overline{OE} = \overline{a} - \overline{e}

Substitute these into the expression:

S=(b‾−a‾)+2(c‾−b‾)+3(d‾−c‾)+4(e‾−d‾)+5(a‾−e‾)S = (\overline{b} - \overline{a}) + 2(\overline{c} - \overline{b}) + 3(\overline{d} - \overline{c}) + 4(\overline{e} - \overline{d}) + 5(\overline{a} - \overline{e})

Expand and group terms by position vectors:

S=b‾−a‾+2c‾−2b‾+3d‾−3c‾+4e‾−4d‾+5a‾−5e‾S = \overline{b} - \overline{a} + 2\overline{c} - 2\overline{b} + 3\overline{d} - 3\overline{c} + 4\overline{e} - 4\overline{d} + 5\overline{a} - 5\overline{e}

S=(−a‾+5a‾)+(b‾−2b‾)+(2c‾−3c‾)+(3d‾−4d‾)+(4e‾−5e‾)S = (-\overline{a} + 5\overline{a}) + (\overline{b} - 2\overline{b}) + (2\overline{c} - 3\overline{c}) + (3\overline{d} - 4\overline{d}) + (4\overline{e} - 5\overline{e})

S=4a‾−b‾−c‾−d‾−e‾S = 4\overline{a} - \overline{b} - \overline{c} - \overline{d} - \overline{e}

S=4a‾−(b‾+c‾+d‾+e‾)S = 4\overline{a} - (\overline{b} + \overline{c} + \overline{d} + \overline{e})

For a regular pentagon with its center at the origin, the sum of the position vectors from the center to its vertices is the zero vector:

OA‾+OB‾+OC‾+OD‾+OE‾=0‾\overline{OA} + \overline{OB} + \overline{OC} + \overline{OD} + \overline{OE} = \overline{0}

a‾+b‾+c‾+d‾+e‾=0‾\overline{a} + \overline{b} + \overline{c} + \overline{d} + \overline{e} = \overline{0}

From this, we can write:

b‾+c‾+d‾+e‾=−a‾\overline{b} + \overline{c} + \overline{d} + \overline{e} = -\overline{a}

Substitute this back into the expression for S:

S=4a‾−(−a‾)S = 4\overline{a} - (-\overline{a})

S=4a‾+a‾S = 4\overline{a} + \overline{a}

S=5a‾S = 5\overline{a}