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Difference between revisions of "1960 IMO Problems/Problem 5"

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[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Revision as of 18:03, 15 September 2012

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 $ZY=2XZ$.

See Also

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