1960 IMO Problems/Problem 5

Problem

Consider the cube $ABCDA'B'C'D'$ (with face $ABCD$ directly above face $A'B'C'D'$ ).

a) Find the locus of the midpoints of the segments $XY$ , where $X$ is any point of $AC$ and $Y$ is any point of $B'D'$ ;

b) Find the locus of points $Z$ which lie on the segment $XY$ of part a) with $ZY = 2XZ$ .

Solution

Let $A=(0,0,2)$ , $B=(2,0,2)$ , $C=(2,0,0)$ , $D=(0,0,0)$ , $A'=(0,2,2)$ , $B'=(2,2,2)$ , $C'=(2,2,0)$ , and $D'=(0,2,0)$ . Then there exist real $x$ and $y$ in the closed interval $[0,2]$ such that $X=(x,0,2-x)$ and $Y=(y,2,y)$ .

The midpoint of $XY$ has coordinates $((x+y)/2, 1, (2-x+y)/2)$ . Let $a$ and $b$ be the $x$ - and $z$ -coordinates of the midpoint of $XY$ , respectively. We then have that $a+b=y+1$ and $a-b=x-1$ , so $a+b\in [1,3]$ and $a-b\in [-1,1]$ . The region of points that satisfy these inequalities is the closed square with vertices at $(1,1,2)$ , $(2,1,1)$ , $(1,1,0)$ , and $(0,1,1)$ . For every point $P$ in this region, there exist unique points $X$ and $Y$ such that $P$ is the midpoint of $XY$ .

If $Z\in XY$ and $ZY=2XZ$ , then $Z$ has coordinates $((2x+y)/3, 2/3, (4-2x+y)/3)$ . Let $a$ and $b$ be the $x$ - and $z$ - coordinates of $Z$ . We then have that $a+b=(4/3)+(2/3)y$ and $a-b=(4x-4)/3$ , and $a\in (4/3,8/3)$ and $b\in (-4/3,4/3)$ . The region of points that satisfy these inequalities is the closed rectangle with vertices at $(0,2/3,4/3)$ , $(2/3,2/3,1)$ , $(1,2/3,2/3)$ , and $(4/3,2/3,0)$ . For every point $Z$ in this region, there exist unique points $X$ and $Y$ such that $Z\in XY$ and $ZY=2XZ$ .

1960 IMO (Problems)
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6 • 7	Followed by Problem 6

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1960 IMO Problems/Problem 5

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