Equilateral Triangle in Cube

Geometry Level 4

$A$ , $B$ and $C$ are points inside a cube of side length 1 (including faces of the cube) such that $\triangle ABC$ is an equilateral triangle. If the maximum area of $\triangle ABC$ can be represented as $\frac{\sqrt{a}}{b}$ , where $a$ and $b$ are positive integers such that $a$ is square-free, find $a+b$ .

The answer is 5.

2 solutions

Marta Reece
Jun 3, 2017

The largest equilateral triangle inside a cube is made of face diagonals as shown, with a side equal to $\sqrt 2$ .

The area of the triangle is $\dfrac{\sqrt{3}}{4}\times( \sqrt 2)^2=\dfrac{\sqrt{3}}{2}$

$a+b=3+2=\boxed 5$

Can you prove rigorously this is the maximum?

Sharky Kesa - 4 years ago

Niranjan Khanderia
Jun 6, 2017

Longest line that can be inside a cub is between opposite corners = $\sqrt3$ , as a diagonal of section that is a rectangle sides $1~by~x~\sqrt2$ and it is the section with largest area. It can be proved that the largest equilateral triangle in this rectangle has sides $<\sqrt2$ .
The largest equilateral triangle can only be obtained by a section that itself is an equilateral triangle. This is diagonals on the faces and as shown by Mr. Marta Reece.

What about the line from a vertex to the midpoint of the edge defined by the vertex opposite the original vertex through the cube and the vertex opposite the vertex on a face? That line has length $1.5>\sqrt{2}$ . This solution is invalid.

Sharky Kesa - 4 years ago

Not only midpoint, all point on that edge will have length $~\geq \sqrt2$ , but there are no matching points in the same plane that can give an equivalent triangle greater than that with sides $>\sqrt2$ . Your comment is correct. I was wrong in my statement. Please see my corrected solution, awiting your comment on it.

Niranjan Khanderia - 4 years ago

Equilateral Triangle in Cube

The answer is 5.

2 solutions

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