1977 USAMO Problems/Problem 4

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Problem

Prove that if the opposite sides of a skew (non-planar) quadrilateral are congruent, then the line joining the midpoints of the two diagonals is perpendicular to these diagonals, and conversely, if the line joining the midpoints of the two diagonals of a skew quadrilateral is perpendicular to these diagonals, then the opposite sides of the quadrilateral are congruent.

Hint

Skew quadrilateral, perpendicular? Use vectors!



Solution

We first prove that if the opposite sides are congruent, then the line joining the midpoints of the two diagonals is perpendicular to these diagonals.

Let the vertices of the quadrilateral be A, B, C, D, in that order. Thus, the opposite sides congruent condition translates to (after squaring):

\[AB^2 = CD^2, BC = AD^2\]

Let the position vectors of A, B, C, D be a, b, c, d with respect to an arbitrary origin $O$. Define the square of a vector $a^2 = a * a$. Thus, $(a-b)^2 = (c-d)^2$ and so \[a^2 - 2ab + b^2 = c^2 - 2cd + d^2.\] Similarly, \[b^2 - 2bc + c^2 = a^2 - 2ad + d^2.\] Subtracting the second equation from the first eventually results in \[a^2 - ab + cd + bc - ad - c^2 = 0,\] which factors as \[(a - b + c - d)(a - c) = 0.\]

Let M and N be the midpoints of $AC$ and $BD$, with position vectors $m = \frac{a+c}{2} and n = \frac{b+d}{2}$, respectively. Then notice that the perpendicularity condition for vectors gives $MN$ perpendicular to $AC$. Similarly (by adding the equations), we find that $MN$ is perpendicular to $BD$, completing the first part of the proof.

Proving the converse is straightforward also: indeed, $MN$ perpendicular to both diagonals gives

\[a^2 - ab + cd + bc - ad - c^2 = 0\] \[b^2 - ab + cd - bc + ad - d^2 = 0\]

Adding and subtracting the equations eventually simplifies to

\[(a-b)^2 = (c-d)^2\] \[(a-d)^2 = (b-c)^2,\]

or $AB^2 = CD^2$ and $AD^2 = BC^2$. This reduces to $AB = CD$ and $AD = BC$, as desired. $\blacksquare$

Solution 2

Let $A, B, C, D$ be position vectors rather than the specified points. We're given that $|A-B| = |C-D| \rightarrow |A-B|^2=|C-D|^2 \rightarrow (A-B) \cdot (A-B) = (C-D) \cdot (C-D) \rightarrow \\ A \cdot A - 2A \cdot B + B \cdot B - C \cdot C + 2C \cdot D - D \cdot D=0$. We're also given that $|A-D| = |C-B| \rightarrow |A-D|^2=|C-B|^2 \rightarrow (A-D) \cdot (A-D) = (C-B) \cdot (C-B) \rightarrow \\ A \cdot A - 2A \cdot D + D \cdot D - C \cdot C + 2C \cdot B - B \cdot B=0$. Adding the two equations gives $2A \cdot A - 2A \cdot B - 2A \cdot D - 2 C \cdot C + 2C \cdot D + 2C \cdot B = 0$. Dividing by 4 and rearranging gives $\frac{A+C-B-D}{2} \cdot (A-C) = 0$. On the left hand side of this equation, one of the two factors is the vector between the midpoints of the two diagonals, and the other factor is on of the two diagonals. Their dot product is $0$, so they are perpendicular. The same methods can be used to show that the vector between the two midpoints is perpendicular to the other diagonal. We can employ this process in reverse to get the converse (this isn't completely simple, but I'm lazy so it's left as an exercise i guess).

See Also

1977 USAMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5
All USAMO Problems and Solutions

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