Moving Legs of an Equilateral Triangle

Geometry Level 5

Let $ABC$ be an equilateral triangle in the plane with side length of $1$ with vertices labeled in counterclockwise order. Let point $A'$ coincide with $A$ , $B'$ coincide with $B$ , and $C'$ coincide with $C$ . Suppose the leg $AB'$ rotates about $A$ counterclockwise, $CA'$ rotates about $C$ counterclockwise, and $BC'$ rotates about $B$ counterclockwise all at the same rate until the three legs intersect at one point. Each leg intersects another leg at a point. These $3$ intersection points trace out paths as the legs rotate. Find the total length of these paths. Note: The diagram is not completely accurate or to scale.

The answer is 1.813.

1 solution

Ujjwal Rane
Aug 21, 2016

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Since the sides are rotating CCW (Counter Clock Wise) at the same rate, the angle they make at the point of intersection (120°) does not change. This is akin to an angle subtended by a chord (the side of the triangle here) at the circumference of a circle. Hence the path must be an arc of a circle. The radius of the circle will be $\frac{1/2}{\sin 60} = 0.577$ and the path will be a 60° arc of this circle. So three such arcs will have the length $3 \times \frac{\pi}{3} 0.577 = \textbf{1.813799}$

Please guide me where I am wrong.
The point the three lines meet is the CG of ABC. So radius that rotates is R=0.577.
Take AB'rotating about A, and say D along AB', AD=0.577.
So it is this point D that is tracing the path of radius 0.577 about A. (I think my understanding is at fault here.) D traces an angle ArcCos{(1/2)/0.577}=0.522546.
Total path traced =.522546 0.577 3=0.94528.

Niranjan Khanderia - 4 years, 1 month ago

Sir, the circular arc of a path will pass through two of the given triangle vertices and the centroid. So the center of the arc will be outside the triangle. I have added a figure to the solution above, showing one such path from vertex A to centroid O.

Ujjwal Rane - 4 years, 1 month ago

Moving Legs of an Equilateral Triangle

The answer is 1.813.

1 solution

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