Polygon is not possible

[sanju09]

How many sides are there in a convex polygon whose consecutive interior angles, cycle-wise, are ascending by 5º each; with the smallest interior angle being 120º?

A. 12
B. 16
C. 9
D. 15
E. Polygon is not possible

[DanaJ]

The starting point is that formula for the sum of angles of a convex polygon = 180(n - 2), where n is the number of sides. The number of interior angles will also be n, with the following formula:
angle A: 120 (first angle)
angle B: 120 + 1*5 (second angle)
angle C: 120 + 2*5 (third angle)
....
angle N: 120 + (n-1)*5 (n-th angle)
So we have a general formula here:
measurement of n-th angle will be 120 + (n-1)*5.
The sum of interior angles will be 120 + 120 + 1*5 + 120 + 2*5 + ... 120 + (n-1)*5 = n*120 + 5(1 + 2 + ... + n - 2 + n - 1) = 120n + 5(n - 1)n/2.
This will have to equal 180(n - 2), so you get that:
120n + 5(n - 1)n/2 = 180(n - 2). Eliminate 5 from each side and you get:
24n + (n - 1)n/2 = 36n - 72. Multiply everything by 2 and get:
48n + n^2 - n = 72n - 144
n^2 - 25n + 144 = 0.
(n - 16)(n - 9) = 0.
I get two values for n: 16 and 9. Where am I wrong?

[sanju09]

The starting point is that formula for the sum of angles of a convex polygon = 180(n - 2), where n is the number of sides. The number of interior angles will also be n, with the following formula:
angle A: 120 (first angle)
angle B: 120 + 1*5 (second angle)
angle C: 120 + 2*5 (third angle)
....
angle N: 120 + (n-1)*5 (n-th angle)
So we have a general formula here:
measurement of n-th angle will be 120 + (n-1)*5.
The sum of interior angles will be 120 + 120 + 1*5 + 120 + 2*5 + ... 120 + (n-1)*5 = n*120 + 5(1 + 2 + ... + n - 2 + n - 1) = 120n + 5(n - 1)n/2.
This will have to equal 180(n - 2), so you get that:
120n + 5(n - 1)n/2 = 180(n - 2). Eliminate 5 from each side and you get:
24n + (n - 1)n/2 = 36n - 72. Multiply everything by 2 and get:
48n + n^2 - n = 72n - 144
n^2 - 25n + 144 = 0.
(n - 16)(n - 9) = 0.
I get two values for n: 16 and 9. Where am I wrong?

What is a convex polygon?

[Ian Stewart]

EDIT - this post actually refers to a similar question posted in another thread! My mistake.

I made the same mistake the first time I saw this question - angles N and A are also adjacent angles, so they must differ by 5 degrees! When the question was posted on this forum before, there was no answer choice 'no such polygon is possible', so I'm pretty sure the question designer intended the answer to be 9, and the wording of the question is bad. At first glance, I don't think it's at all easy to prove that no such polygon is possible, since there are so many possibilities - the angles could be 120, 125, 130, 125, 130, 125, 130, 135, etc for example - though I haven't given this much thought.

[sanju09]

The starting point is that formula for the sum of angles of a convex polygon = 180(n - 2), where n is the number of sides. The number of interior angles will also be n, with the following formula:
angle A: 120 (first angle)
angle B: 120 + 1*5 (second angle)
angle C: 120 + 2*5 (third angle)
....
angle N: 120 + (n-1)*5 (n-th angle)

Where am I wrong?

I made the same mistake the first time I saw this question - angles N and A are also adjacent angles, so they must differ by 5 degrees! When the question was posted on this forum before, there was no answer choice 'no such polygon is possible', so I'm pretty sure the question designer intended the answer to be 9, and the wording of the question is bad. At first glance, I don't think it's at all easy to prove that no such polygon is possible, since there are so many possibilities - the angles could be 120, 125, 130, 125, 130, 125, 130, 135, etc for example - though I haven't given this much thought.

OA is not E Ian! and the wording of the question is bad only because it's not openly suggesting that the interior angles (clock/counter-clock) form an arithmetic progression, I agree. But if it was so given, then I think DanaJ did it right till the threshold. Let's help her out now:

A convex polygon has all its interior angles between 0 and 180 exclusive. Now if you take 16 as correct choice, the 16th angle according to your assumptions DanaJ, will be 120 + 15*5 = 195 > 180. This is not possible with a convex polygon. Although the other root of your quadratic, 9, fits in the requirement; see the 9th angle your way, will be 120 + 8*5 = 160 < 180. Very much welcome choice.

Back to you Ian

I am not getting any satisfactory response for my this option-less question posted in this forum few days back. Will you please help us out?

[spoiler]3 people are to be selected out of 8 people. Out of the three, Only Jim and not Jill needs to be selected. What is the probability of this?[/spoiler]

Thanks

[DanaJ]

Thank you, sanju09. Now I understand my mistake.

[krisraam]

3 people are to be selected out of 8 people. Out of the three, Only Jim and not Jill needs to be selected. What is the probability of this?

No of ways 3 people selected out of 8 = 8C3 = 56

No of ways of selection Jim and 2 other members from the remaining 6 other than Jill = 1* 6C2 = 15

Probabilty = 15/56

Thanks
raama

[Ian Stewart]

Wow, my last post probably didn't make any sense. A nearly identical problem was posted earlier:

www.beatthegmat.com/sides-of-polygon-t21590.html

That's the problem which had the faulty wording I was referring to, and after glancing at the OP, I assumed the question was identical. It's not; Sanju's post changes the wording of the question (though the other details are the same), so my response didn't refer back to his question, but rather to a question in a different thread!

Still, it's impossible for the angles to 'ascend cycle-wise' in a polygon, since eventually you must come back (descend) to the smallest angle.