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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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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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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