1962 IMO Problems/Problem 3


Consider the cube $ABCDA'B'C'D'$($ABCD$ and $A'B'C'D'$ are the upper and lower bases, respectively, and edges $AA'$, $BB'$, $CC'$, $DD'$ are parallel). The point $X$ moves at constant speed along the perimeter of the square $ABCD$ in the direction $ABCDA$, and the point $Y$ moves at the same rate along the perimeter of the square $B'C'CB$ in the direction $B'C'CBB'$. Points $X$ and $Y$ begin their motion at the same instant from the starting positions $A$ and $B'$, respectively. Determine and draw the locus of the midpoints of the segments $XY$.


First we prove a small lemma: If the particles $X$ and $Y$ move along straight lines at constant velocities, then the locus of the midpoint of $XY$ is also a line. This is rather trivial, since all lines in 3D space may take the parametric form $(at+n, bt+m, ct+p)$, with $t$ being time, and the average of two such lines must also have a linear parametric form.

The locus clearly starts at the midpoint of $AB'$, or the center of face $ABB'A'$. As $X$ moves from $A$ to $B$, and $Y$ moves from $C'$ to $C$, both $X$ and $Y$ move along straight lines, so the midpoint of $XY$ traces out a line segment, starting at the midpoint of $AB'$ and ending at the midpoint of $BC'$, or the center of face $BCC'B'$. This concludes the first phase of motion. A quick check reveals that the locus goes nowhere during the second and fourth phases of motion, and only moves backward on the third. Thus the locus is just the segment connecting the centers of sides $ABB'A'$ and $BCC'B'$.

solution is actually wrong, it's not segment it's parallelogram- author mixed the movement of points (even if he clearly understands what he is doing- just a little mistake) Here is correct soluton:

Answer: the rhombus CUVW, where U is the center of ABCD, V is the center of ABB'A, and W is the center of BCC'B'.

Take rectangular coordinates with A as (0, 0, 0) and C' as (1, 1, 1). Let M be the midpoint of XY. Whilst X is on AB and Y on B'C', X is (x, 0, 0) and Y is (1, x, 1), so M is (x/2 + 1/2, x/2, 1/2) = x (1, 1/2, 1/2) + (1-x) (1/2, 0, 1/2) = x W + (1-x) V, so M traces out the line VW.

Whilst X is on BC and Y is on C'C, X is (1, x, 0) and Y is ( 1, 1, 1-x), so M is (1, x/2+1/2, 1/2 - x/2) = x (1, 1, 0) + (1-x) (1, 1/2, 1/2) = x C + (1-x) W, so M traces out the line WC.

Whilst X is on CD and Y is on CB, X is (1-x, 1, 0) and Y is (1, 1-x, 0), so M is (1-x/2, 1-x/2, 0) = x (1, 1, 0) + (1-x) (1/2, 1/2, 0) = x C + (1-x) U, so M traces out the line CU.

Whilst X is on DA and Y is on BB', X is (0, 1-x, 0) and Y is (1, 0, x), so M is (1/2, 1/2 - x/2, x/2) = x (1/2, 0, 1/2) + (1-x) (1/2, 1/2, 0) = x V + (1-x) U, so M traces out the line UV.

See Also

1962 IMO (Problems) • Resources
Preceded by
Problem 2
1 2 3 4 5 6 Followed by
Problem 4
All IMO Problems and Solutions