# Nonagons with integer sides [closed]

Lets have a nonagon with sides which are all equal. The length of the sides are integers. Does anyone know how to inscribe in it an equilateral triangle whose three vertexes intersect with the nonagon? How many such nonagons exist?

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## closed as off-topic by Jonas Meyer, Alan, Mark Fantini, Arctic Char, Claude LeiboviciMar 24 '15 at 5:18

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What do you mean "How many such nonagons exist?" Are you asking how many ways there are to put an equilateral triangle in a nonagon? Are you requiring the nonagon to be regular? – Grumpy Parsnip Feb 7 '12 at 3:25
The nonagon is not regular because only the sides are equal. – Vassilis Parassidis Feb 7 '12 at 3:53
To rephrase: Vassilis's desired nonagons are equilateral, but not necessarily equiangular. – J. M. Feb 7 '12 at 4:17
Just to be clear, by nonagon, you mean $9$-sided, right? I sense that some might not know the Greek prefixes, and think you mean $n$-gon. – alex.jordan Feb 7 '12 at 4:34
@Jordan. Nonagon means 9 sides – Vassilis Parassidis Feb 7 '12 at 4:36

For every $n$ it is possible to inscribe an equilateral triangle in the regular nonagon of side $n$ in such a way that the vertices of the triangle are vertices of the nonagon, so the answer to your question (as currently stated) is that there are infinitely many such nonagons.

EDIT: Of course, that's just a countable infinity of nonagons satisfying the requirements. If you want an uncountable infinity of such nonagons, that's easily arranged as follows (they just can't be regular):

Pick a positive real number $x$ and a positive integer $n\gt x/3$. Draw an equilateral triangle $ABC$ with side $x$. Draw a circle of radius $n$ centered at $A$, and another one centered at $B$. Now find a point $D$ on the first circle and a point $E$ on the second such that the distance from $D$ to $E$ is $n$. This is easy enough; just find a point $D$ on the first circle relatively close to the second circle, draw a circle of radius $n$ centered at $D$, and let $E$ be one of the places this circle intersects the circle centered at $B$. Now draw the line segments $AD$, $DE$, and $EB$; they all have length $n$.

Now do the same thing with $B$ and $C$, getting line segments $BF$, $FG$, and $GC$ of length $n$, and with $C$ and $A$, getting line segments $CH$, $HI$, and $IA$ of length $n$. So you've got a nonagon $ADEBFGCHIA$ with sides of length $n$, sharing three vertices with an equilateral triangle of side $x$. If you exercise a little care in your choices, you can even make the nonagon convex, the triangle properly inside it.

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@Gery. The nonagon is not regular because the sides are integer numbers. – Vassilis Parassidis Feb 7 '12 at 3:58
What are you talking about? You stipulate that the sides are all equal. What does integrality have to do with regularity? – Gerry Myerson Feb 7 '12 at 4:58
@Gerry A regular polygon is both equilateral and equiangular. A regular nonagon is called a enneagon, and equilateral nonagons are not necessarily equiangular. – user21436 Feb 7 '12 at 5:06
@Gery.Because the sides are equal does not mean the angles are equal. The length of the side is an integer. Just show me one such nonagon with integers and the length of the side of the equilateral triangle. – Vassilis Parassidis Feb 7 '12 at 5:10
@KannappanSampath, OP asked how many nonagons exist with such-and-such properties. I pointed out that among the nonagons with those properties there are the regular ones, of which there are infinitely many. I am well aware that there are equilateral nonagons that are not regular, but I didn't need any of them to answer the question. – Gerry Myerson Feb 7 '12 at 6:14

Assume for simplicity that your nonagon $P$ is convex. There are two vertices $A$ and $B$ of $P$ with $|AB|={\rm diam}(P)$. Draw an equilateral triangle $ABS$. The point $S$ is on the boundary $\partial P$ or in the exterior of $P$. Now translate the line $g_0:=A\vee B$ away from $S$. At each instant $t\geq0$ the line $g_t$ will intersect $\partial P$ in two points $A_t$, $B_t$ of ever decreasing distance. The equilateral triangle $A_t B_t S_t$, where $S_0:=S$ will eventually lie completely inside $P$. (When $g_0$ happens to be parallel to a side of $P$, start moving $A_t$ and $B_t$ towards each other at the end.) There has to be a moment when $S_t\in\partial P$, and that's when you have your equilateral triangle inscribed in $P$.

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I am going to give you an example please give another one. Let's have AB=13 from A measure AC=4 and CD=5. Draw two verticals from C and D such as that CE=DF=3 connect AE,EF,FB and you have a 9 sided polygon with all sides equal to 5. Because the side is an integer the only way to construct it is with compass and straight edge. – Vassilis Parassidis Feb 7 '12 at 16:47
@Vassilis Parassidis: In your description there are only $6$ points. In addition, I conjecture you meant $AC=5$. At any rate: Draw the nonagon you have in mind and perform the process I have described in my answer. – Christian Blatter Feb 7 '12 at 17:29
Please give me an example with numbers. – Vassilis Parassidis Feb 8 '12 at 1:29