This post's format will be a bit different from the usual. In this post, I will state a problem and then present my solution. More often than not (unfortunately) my solutions may have mistakes. If you find any, let me know!
Problem statement:
Let $[ABC] $ be an equilateral triangle of side length $1$. We now define a transformation of the triangle which moves one vertex and leaves the other two unchanged.
To apply a transformation, start by picking the vertex you want to move. Assume it was $A $. Consider the line that goes through $A $ which is parallel to $[BC] $, and pick any point $A' $ in it. Your transformed triangle is $[A'BC] $. You can repeat this process as many times as you want.
Q: Can the original triangle be transformed into an equilateral triangle of side length $2$? How/why?
Try figuring this for yourself.
Solution:
The answer to the question is no, one cannot apply the transformation a given number of times and get to an equilateral triangle with side length $2$. To see why, we will show that the area of the triangle remains unchanged when you apply the transformation. If we do that and if we consider that an equilateral triangle of side length $1$ has a smaller area than an equilateral triangle of side length $2$, then it is obvious that the answer is no.
Say we have a triangle $[ABC]$ and, without loss of generality, we are going to apply the transformation on $A$. Furthermore, we can assume $[BC]$ is horizontal (this is not needed, just makes it even simpler to visualize the proof). We know that $A$ is going to be moved to a point $A'$ in a line that is parallel to $[BC]$. We know that the area of a triangle is $\frac{b \times h}{2}$ where $b$ is the length of the base and $h$ is the height. If $[BC]$ is seen as the base (its length doesn't change when $A$ moves to $A'$) we see that the height $h$ is the length of the line segment that goes from $A$ to $BC$ and is perpendicular to $BC$. But the line segment that goes from $A'$ to $BC$ and is perpendicular to it has exactly the same length, so the area of the triangle didn't change. QED.
Bonus question: if $a,b,c$ are the side lengths of a triangle of area $\frac{\sqrt{3}}{4}$ (the area of an equilateral triangle of side length $1$) is there a sequence of moves that transforms our equilateral triangle into that with side lengths $a,b,c$?
-RGS
Problem statement:
Let $[ABC] $ be an equilateral triangle of side length $1$. We now define a transformation of the triangle which moves one vertex and leaves the other two unchanged.
To apply a transformation, start by picking the vertex you want to move. Assume it was $A $. Consider the line that goes through $A $ which is parallel to $[BC] $, and pick any point $A' $ in it. Your transformed triangle is $[A'BC] $. You can repeat this process as many times as you want.
Q: Can the original triangle be transformed into an equilateral triangle of side length $2$? How/why?
Try figuring this for yourself.
Solution:
The answer to the question is no, one cannot apply the transformation a given number of times and get to an equilateral triangle with side length $2$. To see why, we will show that the area of the triangle remains unchanged when you apply the transformation. If we do that and if we consider that an equilateral triangle of side length $1$ has a smaller area than an equilateral triangle of side length $2$, then it is obvious that the answer is no.
Say we have a triangle $[ABC]$ and, without loss of generality, we are going to apply the transformation on $A$. Furthermore, we can assume $[BC]$ is horizontal (this is not needed, just makes it even simpler to visualize the proof). We know that $A$ is going to be moved to a point $A'$ in a line that is parallel to $[BC]$. We know that the area of a triangle is $\frac{b \times h}{2}$ where $b$ is the length of the base and $h$ is the height. If $[BC]$ is seen as the base (its length doesn't change when $A$ moves to $A'$) we see that the height $h$ is the length of the line segment that goes from $A$ to $BC$ and is perpendicular to $BC$. But the line segment that goes from $A'$ to $BC$ and is perpendicular to it has exactly the same length, so the area of the triangle didn't change. QED.
Bonus question: if $a,b,c$ are the side lengths of a triangle of area $\frac{\sqrt{3}}{4}$ (the area of an equilateral triangle of side length $1$) is there a sequence of moves that transforms our equilateral triangle into that with side lengths $a,b,c$?
-RGS
Problem #01 - a dancing triangle
![Problem #01 - a dancing triangle]()
Reviewed by Unknown
on
November 05, 2017
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