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# HOT Competition 28 Aug/8AM: If a convex polygon has 88 more diagonals

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Re: HOT Competition 28 Aug/8AM: If a convex polygon has 88 more diagonals [#permalink]
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Quote:
If a convex polygon has 88 more diagonals than sides, how many vertices does it have?

=> let x be the number of sides.
=> Then , number of diagonals will be 88+x
=> Sum of the sides and diagonal = 88+x+x
=> 88+2x

Joining any 2 vertices will either result in side or diagonal, and it should be greater than 88 and should be even,

=> 12c2 is (12*11)/2 = 66, whcih itself if lower than 88
Try, 16, 16c2 = (16*15)/2 = 120; this is one possible solution.
Try, 18, 18c2, = (18*17)/2 = 153; 153-88 = 65, x = 32.5, x has to be integer.

Therefore, only possible solution is 66. Option D. 16
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Re: HOT Competition 28 Aug/8AM: If a convex polygon has 88 more diagonals [#permalink]
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given 88 diagonals than sides
so
n(n-3)/3=n+88
by solving n^2-5n-176=0
n=16,-11
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Re: HOT Competition 28 Aug/8AM: If a convex polygon has 88 more diagonals [#permalink]
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If a convex polygon has 88 more diagonals than sides, how many vertices does it have?

A. 9
B. 11
C. 12
D. 16
E. 18

Solution:

Number of Diagonals: nc2 - n ; for a n sided convex polygon
As per the problem: (nc2-n)-n = 88
nc2 - 2n = 88
n(n-1)/2 - 2n = 88
n^2 - n - 4n = 176
n^2 - 5n - 176 = 0
(n - 16) ( n + 11 ) = 0
n = 16 or -11 ( -11 is not feasible)
Number of vertices = 16 ; Hence D is the correct answer

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Re: HOT Competition 28 Aug/8AM: If a convex polygon has 88 more diagonals [#permalink]
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IMO Ans is D=16

If a convex polygon has 88 more diagonals than sides, how many vertices does it have

No of Vertices for Convex Polygon = No. of Sides of Convex Polygon

eg:- Square =4 Side = 4 Vertices, Hexagon= 6 Side= 6 Vertices.

For Convex Polygon with n vertices or n side

Total no. of Diagonal = nC2 - n (Combinaton of all 2 Points - No. of Side)

Total No. of Diagonal = ((n!) / (2!*(n-2)!)) -n

=> Total No. of Diagonal = n(n-3)/2---------------eq (1)

Let Convex Polygon with side n

=> n = (n(n-3)/2)- 88(From Question & eq(1)

=>2n+176=n^2-3n

=>n^2-5n-176=0

SOlving this Quadratic equation, we will have roots as 16, -11

Only Value of n=16 is Possible.

Convex Polygon will have 16 Vertices

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Re: HOT Competition 28 Aug/8AM: If a convex polygon has 88 more diagonals [#permalink]
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it's simple figure out the equation nc2 (no of ways of chosing diagonal ) -n(colinear lines concave ) = 88 +n(what the question indicating ) the equation indicating n^2 - 5n -176 , which gives us n=16 imo D
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Re: HOT Competition 28 Aug/8AM: If a convex polygon has 88 more diagonals [#permalink]
Bunuel wrote:
If a convex polygon has 88 more diagonals than sides, how many vertices does it have?

A. 9
B. 11
C. 12
D. 16
E. 18

 This question was provided by Veritas Prep for the Heroes of Timers Competition

No of ways of chosing a diagonal = nc2 - n(collinear points )
Given :
nc2 -n =88 + n
=> n^2 - 5n -176

Upon solving we get n=16 , -11

Therefore IMO D
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Re: HOT Competition 28 Aug/8AM: If a convex polygon has 88 more diagonals [#permalink]
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Everybody is solving in the route of quadratic formula. I wanted to show my different approach.

Since number of sides is equal to number of vertices, I set up the formulae for sides.

((n)(n - 2))/2 = n + 88.

Always choose C which gives us the result after working out

72 = 88 + 12. Wrong.

Eliminate A and B because you need a larger number for this to work.

Choose D.

104 = 16 + 88.

Don't need to check E.
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HOT Competition 28 Aug/8AM: If a convex polygon has 88 more diagonals [#permalink]
I think for certain problems from geometry, it is better to remember the basic formulas than to come up with cases in such a limited time. Here we can use the formula to find diagonals of a n-sided polygon: (n(n-3))/2. From the question, (n(n-3))/2=n+88. Solving for n, we get n=16. The number of vertices for any polygon=number of sides therefore 16 vertices are there. Option D is the answer.
HOT Competition 28 Aug/8AM: If a convex polygon has 88 more diagonals [#permalink]
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