For an equilateral triangle $ABC$ which has edge-length $1$, let $D,E,F$ be a point of contact of the inscribed circle and an edge $BC, CA, AB$ respectively.
Then, let us call the region surrounded by lines $AE,AF$ and a minor arc $EF$ region $A^\prime$, the region surrounded by lines $BD, BF$ and a minor arc $DF$ region $B^\prime$, the region surrounded by lines $CD, CE$ and a minor arc $DE$ region $C^\prime$.
Also, letting $P,Q,R$ be a point in the region $A^\prime,B^\prime, C^\prime$ respectively, let $I, O, G$ be the incenter, the circumcenter, the centroid of a triangle $PQR$ respectively.
Then, here is my question.
Question : Can we draw the region where each of $I,O,G$ can exist? Also, can we find the area of the region where each of $I, O, G$ can exist?
Note that the latter question is based on my assumption that each region would be closed.
I've been thinking about these questions, but I'm facing difficulty. I would like to know what the shape of each region is. Can anyone help?
This question comes from the following question :
"Prove that the region where $I$ can exist is included in the inscribed circle of $ABC$."
After solving this question, I can prove that the region where $I$ can exist is included in a hexagon $LFMDNE$ where $L,M,N$ is an intersection of the inscribed circle of $ABC$ and a line segment $IA, IB, IC$ respectively. However, this is all I've got.
for circumcenter,it is same as incenter, the area is a hex.
for centroid, it is a equilateral triangle which inscribe the circle and the position is similar to ABC.
to find the area, you can find fix two points on the special points and move one point along the side which you specified.
if you can find incenter area, then the other two should be easy for you.
edit: just a example to explain the method
you can always specify a area, such as the red one which created by fixing $Q=B$, moving $P$ on the arc(black) and $R$ on the side $CD$. Assume $I$ is on the red area, we can make a circle (orange)to tangent BC , then to make a line from $Q=B$ to tangent the circle .The line intersect the black arc to get $P$. From $P$, we can make a tangent line to get $R$. So for any specified area, we can find $P,Q,R$ for a $X$.
Now you can fix $R=C$ and move other two points to get another area. Next to fix $Q=D$, and move other two points, etc. you will have 6 areas which cross each other and become a big fan area(red line). With symmetry, it is a hex if you have same way to put $P,Q$ on the $AB$ and $P,R$ on the $AC$.
But there is hole in the middle of the Hex. Imaging the red are is moving to $A$,so the new area will cover part of the hole.All the points in this new area can be find with $Q$ on $BF$. We can also to change $Q$ moving on $BF$ but $R$ on $CD$ to get some new area to cover the hole.
Finally we can cover the whole hex area with limited specified area by a specified method. So for any $X$, wherever it falls in hex,we can always have at least one solution to find $P,Q,R$ with the specified area.
These are the key points.
edit2: for centroid, it is much easy:
Fixed $P=F$, when you move $Q,R$ on $BC$, it is a line for centroid.(green) The two terminal are the centroid of $\triangle FBD $ and $\triangle FCD $.Simply moving the line toward $A$, you will get a small equilateral triangle.(consider $P$ is on $E$ also)
For any $G$ on the green line, let $P=F$,connect $PG$ and get $M$, flip $D$ to $T$. find any point $Q$ on $BT$,(this is to make sure $P$ is on $CD$) flip $Q$ to $P$. that are $(P,Q,R)$ which we wanted.