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Here is the question: The difference between any two consecutive interior angles of a polygon is 5 degrees. If the smallest angle is 120 degrees. Find the number of sides the polygon has.

I think the answer is 9. But I thought it will be an even number since only then the figure can be made(I am not sure about this since I got 9)

Can you help me with this question?

2 Answers 2

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It seems to me there is no solution. Proof:

The sum of the exterior angles of any polygon is $360^\circ$, so we need to find a set of numbers $\theta_0,\theta_1,\dots,\theta_k=\theta_0$ such that $|\theta_i-\theta_{i-1}|=5^\circ$ and $\sum_1^k\theta_i=360^\circ$. We also know that $\min\{180^\circ-\theta_i\}=120^\circ$, so $\theta_i\leq 60^\circ$ and $(\exists i)[\theta_i=60^\circ]$. WLOG, we can let $\theta_0=60^\circ$. Noting that $\theta_i/5^\circ=n_i\in\mathbb Z$, the constraint $|n_i-n_{i-1}|=1$ means that if $n_i$ is even, $n_{i+1}$ is odd and vice-versa so that $n_i$ is even iff $i$ is even since $n_0=12$ by induction.

Thus, if $k$ is odd, then $n_k=n_0$ is odd and even, a contradiction. Thus $k$ is even. Additionally, since $n_0+n_1,n_2+n_3,\dots$ are all odd, $\sum_1^k n_i$ will be odd if $k/2$ is odd, which is a contradiction since $\sum_1^k n_i=72$. Thus $k$ is a multiple of $4$.

Now $$n_i\leq 12\Rightarrow\sum_1^k n_i=72\leq12k\Rightarrow k\geq 6,$$ and $|n_i-n_{i-1}|=1\Rightarrow n_i\geq12-\min(i,n-i)\Rightarrow$

$$72=\sum_1^k n_i=n_0+n_{k/2}+\sum_1^{k/2-1} (n_i+n_{n-i})\geq 24-k/2+2\sum_1^{k/2-1} (12-i)=12k-(k/2)^2,$$

so $k\leq12(2-\sqrt2)\approx7.02$. There are no multiples of 4 in the range $k\in[6,7.02]$, so this has no solution.

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    with no offense, I just want to say that even though I cannot properly understand your solution ....but then again...I dont see any problem with the solution given below... So if maybe you can show that the solution given below is wrong ...then it will be really good...!!2016-01-26
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    I had no intention of restarting this really old question....but then this is a very popular sort of question one can say ....and almost every site...on the internet that give its solution....say that answer is $9$...I also disagree with this because I have not able to geometrically make such a polygon... But since I on my own can't prove it....that such condition is impossible to be satisfied...so maybe if i/or you can help me to prove that the below solution is wrong....then my purpose will be solved ....any help is appreciated..2016-01-26
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    Actually this question is from the book ( NCERT for class 11th Mathematics) used in India...And the book says answer is $9$...which makes me doubt your and my claim...that this is not possible.2016-01-26
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    @Freelancer The easiest way to see the answer cannot be 9 is to look at the part where I prove that $k$ is even. Let $n_i$ be the exterior angle of each vertex, divided by $5^\circ$. One of the vertices, which we take as vertex zero, is said to have $120^\circ$ on the inside, so $60^\circ$ on the outside hence $n_0=12$. As you go around the circle, the $n_i$ go either up or down by $1$, for example $12,11,10,11,10,9,\dots$. Note the numbers are always alternately odd and even. You get back to the starting point after $k$ steps, so you have to land on $12$, but if $k$ is odd that is impossible.2016-01-26
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    @Freelancer The other answer does not prove that there is such a polygon, only that if there is a solution it must have $9$ or $16$ sides. A proof of existence would have to show this by a construction. It is also wrong; the angles are not in arithmetic progression because they might go up or down from vertex to vertex.2016-01-26
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    If there is a book which says this problem has a solution, then either it is wrong or some part of the assumptions has been mis-stated (for example the difference in angles is at least $5^\circ$ instead of exactly $5^\circ$).2016-01-26
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    Looking up the problem on the internet, I find the following statement: "The smallest angle of a polygon is 120 degrees and the angles are in an arithmetic progression with difference 5 degrees. How many sides?" The commonly quoted 9-side solution has exterior angles 60,55,50,45,40,35,30,25,20 which you can verify adds to 360. However, obviously the arithmetic progression is broken as you keep going around the figure since $60\ne 20-5$. I assume that the statement applies to *all* consecutive angles, not just all but the last, so my analysis doesn't consider this a proper solution.2016-01-26
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    ....sorry for bugging you but just wanted to say that the the only thing mis-stated in this question is that the polygon is **Concave**,but then... I don't think that makes any difference to your claim that such a polygon is not possible to make....right ??2016-01-27
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    @Freelancer Actually the $9$-sided polygon in question is convex (there is also a similar $16$-sided polygon that is concave). The mis-stating is in not mentioning that each angle is 5 degrees more than the previous one (so that an AP is created instead of just a random walk of $+5,-5$ degrees), and in stating that the relation does not apply between the adjacent largest and smallest angles (the other online versions also omit this constraint, but at least if you use the term AP it is obvious that you mean a finite AP because there are no non-constant infinite periodic APs).2016-01-27
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    OK..so do you mean to say that If we add the above instructions in the question ..it will have the solution $9$ right...??2016-01-27
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    Also ..if I do not state that the relation does not apply between the adjacent largest and smallest angles....it will always be false...right ??2016-01-27
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    Moreover... you say that this thing is not mentioned in question here... *each angle is 5 degrees more than the previous one (so that an AP is created instead of just a random walk of +5,−5 degrees*)2016-01-27
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    This means that if we ignore the second correction(*about smallest/largest angle*).....and if i am not wrong that you were considering the case that a random walk of +5,-5 ...was there...but...even if you were considering this....and (you proved in your answer that this is not possible)...but still among one of these random walks one case was there in which they will form an increasing A.P so this means that you proved that this is also not possible(that the question is incorrect even with the first correction)...right???2016-01-27
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    @Freelancer That is correct. The AP requirement is not necessary, as long as you say that one pair of adjacent angles is excluded there are many solutions, such as 60,55,50,45,40,35,40,35 which is 8 sides, but after adding the AP requirement you get the "desired" solution method shown in @${}$SushruthS answer. You can even get arbitrarily many sides with 60,55,50,...,0,-5,-10,-5,-10 (17 sides) followed by -5,0,5,0 repeated $n$ times, to get $17+4n$ sides.2016-01-27
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The sum of all angles in a polygon is given by the formula....$180(n-2)$....also we know that angles are in A.P( according to question) so again using the formula for that $\frac{n}{2}(2a + (n-1)d )$.....equating these we get.... $$180(n-2) = n/2( 120 + 120 + 5(n-1))$$ If you simplify, you get \begin{align} 5n^2 -125n +720 &=0\\ n^2 -25n +144 &=0\\ (n -16)(n-9) &=0 \end{align} Thereofore, $n =16$ or $n =9$