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Difference between revisions of "1977 USAMO Problems/Problem 4"

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or <math>AB^2 = CD^2</math> and <math>AD^2 = BC^2</math>. This reduces to <math>AB = CD</math> and <math>AD = BC</math>, as desired. <math>\blacksquare</math>
 
or <math>AB^2 = CD^2</math> and <math>AD^2 = BC^2</math>. This reduces to <math>AB = CD</math> and <math>AD = BC</math>, as desired. <math>\blacksquare</math>
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==Solution 2==
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== See Also ==
 
== See Also ==

Revision as of 09:48, 16 October 2017

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

See Also

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

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