Thinking Binary in Geometry

Geometry Level 5

$S = \big\{(0, 0, 0), (n, 0, 0), \dots, (n, n, n)\big\}$ is the set of coordinates of the vertices of a cube with side length $n$ .

$P$ is a plane such that the distances of the points in $S$ to $P$ are $0, 1, 2, \dots, 7$ (not necessarily in that order), and its Cartesian equation is $ax+by+cz=0$ , where $a, b, c$ are positive integers that statisfy $\gcd(a,b,c)=1.$

Find $a+b+c +n^2.$

The answer is 28.

1 solution

Daniel Xiang
Feb 17, 2018

Think of the coordinates of the vertices as binary numbers.

$(0, 0, 0) \Rightarrow 000_2 = 0 \\ (0, 0, n) \Rightarrow 001_2 = 1 \\ (0, n, 0) \Rightarrow 010_2 = 2\\ (0, n, n) \Rightarrow 011_2 = 3\\ \vdots\\ (n, n, n) \Rightarrow 111_2 = 7 \\$

And now we can easily determine the equation of the plane

$2^2x+2^1 y+2^0 z=0 \quad\Rightarrow\quad 4x+2y+z=0$

Because this is the way we turn binary numbers into denary numbers.

For example, $110_2 = 1\times 2^2 + 1\times 2^1 + 0\times 2^0 = 6$

And by symmetry, There are five more solutions:

$4x+y+2z=0 \\ 2x+y+4z=0 \\ 2x+4y+z=0 \\ x+2y+4z=0 \\ x+4y+2z= 0 \\$

And now we find the side length of the cube $n = \sqrt{a^2+b^2+c^2} = \sqrt{1+4+16} = \sqrt{21}$

Finally we have $a+b+c+n^2 = 4+2+1+21 = \boxed{28}$

Note: the distance of a point $(x_0, y_0, z_0)$ to the plane $ax+by+cz=d$ is $\displaystyle \frac{ax_0+by_0+cz_0 - d}{\sqrt{a^2+b^2+c^2}}$

I like this problem but I was a little disappointed to see my solution refused. Indeed I believe it to be correct.

Take $(a,b,c) = (7,-1,-2)$ for example and $n=\sqrt{a^2+b^2+c^2}=\sqrt{54}$ . We still have the set of distances from the plan of equation $7x-y-2z=0$ to each of the $8$ vertices of the cube equal to $\{0,1,2,3,4,5,6,7\}$ . Hence $a+b+c+n^2 = 7-1-2+54 = 58$ should be a valid solution.

Romain Bouchard - 3 years, 3 months ago

Thinking Binary in Geometry

The answer is 28.

1 solution

0 pending reports