A geometry problem by Rahil Sehgal

Geometry Level 2

In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 17. What is the greatest possible perimeter of the triangle?


The answer is 49.

Rahil Sehgal by Rahil Sehgal , 20, India

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1 solution

Viki Zeta
Mar 10, 2017

Let x, 3x, and 17 be lengths os sides of Triangle Perimeter, P = 3 x + x + 17 = 4 x + 17 Now, using properties of triangle Sum of any 2 sides is greater than the third side x + 3 x > 17 4 x > 17 x > 4.25 > 4 ∴ x > 4 Difference of any 2 sides is lesser than the third side 3 x − x < 17 2 x < 17 x < 8.5 < 9 x < 9 ∴ 4 < x < 9 ⟹ x = 5 , 6 , 7 , 8 Largest perimeter is possible when x = 8 ∴ P = ( 4 × 8 ) + 17 = 49 \text{Let x, 3x, and 17 be lengths os sides of Triangle} \\ \text{Perimeter, P } = 3x + x + 17 = 4x + 17 \\ \text{Now, using properties of triangle} \\ \text{Sum of any 2 sides is greater than the third side} \\ x + 3x > 17 \\ 4x > 17 \\ x > 4.25 > 4 \\ \therefore x > 4 \\ \text{Difference of any 2 sides is lesser than the third side} \\ 3x - x < 17 \\ 2x < 17 \\ x < 8.5 < 9 \\ x < 9 \\ \therefore 4 < x < 9 \\ \implies x = {5, 6, 7, 8} \\ \text{Largest perimeter is possible when x = 8} \\ \boxed{\therefore P = (4 \times 8) + 17 = 49}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice Solution, but you could have just used 17 + x > 3 x 17+x > 3x ,it would have yield max value(integral) as x = 8 x=8

Achal Jain - 4 years, 3 months ago

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