A geometry problem by arko roychoudhury

Geometry Level 3

A rectangle with positive integer side lengths in cm \text{cm} has area A A cm 2 \text{cm}^{2} and perimeter P P cm \text{cm} . Which of the following numbers cannot equal A + P A+P ?

104 102 100 108 106

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1 solution

Let the positive integer side lengths of the rectangle be x , y x,y . Then A = x y A = xy and P = 2 x + 2 y P = 2x + 2y , and so

A + P = x y + 2 x + 2 y = ( x + 2 ) ( y + 2 ) 4 A + P + 4 = ( x + 2 ) ( y + 2 ) A + P = xy + 2x + 2y = (x + 2)(y + 2) - 4 \Longrightarrow A + P + 4 = (x + 2)(y + 2) .

So since x , y > 0 x,y \gt 0 , we must be able to represent A + P + 4 A + P + 4 as a product m × n m \times n with integers m , n > 2 m,n \gt 2 . Now as examples for the given options we have that

108 + 4 = 112 = 8 × 14 , 104 + 4 = 108 = 9 × 12 , 100 + 4 = 104 = 8 × 13 108 + 4 = 112 = 8 \times 14, 104 + 4 = 108 = 9 \times 12, 100 + 4 = 104 = 8 \times 13 and 106 + 4 = 10 × 11 106 + 4 = 10 \times 11 , and so it is possible for A + P A + P to be any of these options.

However, the only possibility for 102 + 4 = 106 102 + 4 = 106 is 2 × 53 2 \times 53 , as both 2 2 and 53 53 are prime, which does not satisfy the requirement that m , n > 2 m,n \gt 2 . Thus A + P A + P cannot equal 102 \boxed{102} .

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