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Geometry Level 2

How many sides does a polygon have if it has 2015 diagonals?


The answer is 65.

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Lovelykelly Oracion by Lovelykelly Oracion , 23, Philippines

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4 solutions

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Sujoy Roy
Dec 16, 2014

In number of sides of the polygon is n n , then

n C 2 − n = 2015 ^{n}C_{2}-n=2015

or, n ( n − 3 ) = 65 ∗ 62 n(n-3)=65*62

or, n = 65 n=\boxed{65}

18 Helpful 0 Interesting 0 Brilliant 0 Confused

For those who don't know, n C r ^nC_r or ( n r ) \dbinom{n}{r} refers to the number of ways r r items can be selected from n n items at once (irrespective of the arrangement in which they are selected). This is known as Combination .

For an n n -sided polygon with d d diagonals, we have ( n 2 ) − n = d \dbinom{n}{2} - n = d , because, by definition, a diagonal of a polygon is the line segment other than the sides of the polygon made by joining any 2 2 points of the polygon.

P.S - Sorry for bad grammar! :P

Prasun Biswas - 6 years, 5 months ago

Instead of checking via brute force, note that 4030 ≈ 63.482 \sqrt{4030} \approx 63.482 is slightly below n n . Since 4030 is a multiple of 10, the only possibility is n = 65 n = 65 .

Jake Lai - 6 years, 5 months ago

Another way is to use the quadratic formula and neglect the negative root of the equation n 2 − 3 n − 4030 = 0 n^2-3n-4030=0 giving us the value of n = 65 n=\boxed{65} .

Prasun Biswas - 6 years, 5 months ago

I use this way to solve :)

christoffer santos - 6 years, 5 months ago

You're right

Lovelykelly Oracion - 6 years, 5 months ago
William Isoroku
Dec 17, 2014

Use the formula to find the number of diagonals: n ( n − 3 ) 2 \frac { n(n-3) }{ 2 } where n n is the number of the sides. Given the diagonals, just solve for n n , which is 65 \boxed{65}

6 Helpful 0 Interesting 0 Brilliant 0 Confused
Vaibhav Kandwal
Dec 17, 2014

The number of diagonals in a polygon = n ( n − 3 ) / 2 n(n-3)/2

Equating, 2015 = n ( n − 3 ) / 2 2015=n(n-3)/2 ,

We get n = 65 \boxed{n=65}

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Minyoung Kim
Dec 21, 2014

To solve it intuitively, first try count every lines possible. To make a line, there should be at least two points. If its n-polygon, than one point can have (n-1) lines. As there are 'n' points, The number of total line is n(n-1). However, as it is duplicated, it should be divided by two. To briefly say, n-poligon has n(n-1)/2 -n diagonals, counting out its circumferential segments.

if solve 2015=n(n-1)/2-n, than n=65 or -62. However, as 'n' should be positive number, the answer is 65.

Sorry for poor grammar :-)

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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