For every $n$-sided equiangular polygon, define $f(n)$ to be the maximum number of consecutive sides whose lengths you can choose without uniquely determining the polygon. Find the last two digits of:$$\sum_{n=3}^{2019} f(n)$$

Can someone verify this ? I think $(n-3)$ works, and I posted it in AOPS, but the guy is saying that it is wrong.

Here's the solution:

Answer : $36$, $f(n)=n-3 $

clearly for any $n , n-2$ doesn't satisfy because then equiangular property fixes the polygon.

For $n=3, 4$ , clearly $f(n)=n-3$ works . For $n\ge 5$ , consider the following construction , which i took for $n=7$ ,( but one can understand what's the construction) . It's just extending $2$ sides and making a parallel side . So $n-3$ consecutives doesn't fix.

enter image description here

So $n-3$ works .

And answer is $\frac{2016\cdot 2017}{2}=2033136 \equiv 36 \pmod {100}$

Here's the other guy's solution, which many people are saying it to be true : Equilateral triangle which is regular, square but we can have a rectangle, pentagon where all it's sides are equal and so on. Then I see a pattern. This was the crucial observation.

In Euclidean Geometry, if we have an equiangular polygon with odd sides then it's all the sides will be equal, i.e it will be a regular polygon. So, we have the regular polygons with side lengths $3,5,\cdots$ and so on. Here, in this case we can choose only one of it's side length and all the side lengths will be automatically chosen since it is an equiangular and hence regular polygon. Therefore, $f(n)=\boxed{1}$, in the case when we have an equiangular polygon with odd sides. Now, we see what happens when we choose an equiangular polygon with odd sides, so we can choose half of the sides of the polygon(these needs to be consecutive sides). Therefore $f(n)=\frac{n}{2}$ in the case where $n=\text{even}$. $\text{Sum}=(1+2)+(1+3)+(1+4)+\cdots+(1+1009)+1=510553 $.

Therefore Last two digits of the sum$\boxed{53}$.

Is there a flaw in the solution , I wrote ?

  • 3
    $\begingroup$ Your solution is flawless. The other person's claim is clearly wrong as you have shown. $\endgroup$
    – cosmo5
    Dec 24 '20 at 8:41
  • 1
    $\begingroup$ In the original question, does it include the "consecutive" requirement? If you drop that requirement, for even $n$, it is possible to specify the length of $n-2$ sides without fixing the polygon (fixing the length of all sides except for a pair of parallel sides) $\endgroup$ Dec 24 '20 at 15:42
  • $\begingroup$ @achillehui yes it was specified .. $\endgroup$ Dec 24 '20 at 15:44
  • $\begingroup$ @SunainaPati okay, then the other guy is wrong. $\endgroup$ Dec 24 '20 at 15:45

Your solution is correct.

You can have an equiangular pentagon that is not regular. Take the regular pentagon and extend any two of its non-adjacent alternate sides. You can now shift the intermediate side parallel to itself such that it remains between the extensions.

You now have an equiangular non-regular pentagon.



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