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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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    @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 $n =9$