It is a long time since my last geometry class. I have a question about triangles that have constant perimeter.
If I have the triangle $ABC$. For the moment let's say that $A$ is at the origin of an $XY$ plane. $B$ is the apex and $C$ is $n$ units to the right of $A$ on the $X$ axis. Then $AC$ is the base and that base has length $n$. $A$ and $C$ are fixed in place and can not move.
If I have a bit of string and can form a loop that perfectly encloses $ABC$ and doesn't sag then the loop, and so the triangle, has a perimeter of $p$.
If I translate the triangle along the $X$ axis so that the origin bisects the base $AC$ and the apex of the triangle is on the $Y$ axis then each side $AB$ and $BC$ have the same length (they are mirrored around the $Y$ axis). I can find the lengths of $AB$ and $BC$ (I think) by subtracting the base length $n$ from the perimeter $p$ and then dividing that by $2$. I can then pull out Pythagoras to find the height or use Herron's formula (thanks google).
Ok, in this configuration, with the base bisected by the origin, if I move the apex one unit to the right in the $X$ direction the lengths of $AB$ and $BC$ will change as will the internal angles. $AC$ remains the same and the perimeter remains the same.
Is there a method or formula to determine any of:
-
The new $Y$ value for $B$? We know the new $X$ value.
-
The new $AB$, $BC$ lengths
-
The internal angles?
My goal is to find the new triangle height after the apex move.
Best Answer
$A$, $C$ are your focal points. Every formula can be found here: https://en.wikipedia.org/wiki/Ellipse