Difference between revisions of "1977 USAMO Problems/Problem 2"

 
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<cmath> 3([ABC]+ [A'B'C']) = [AB'C'] + [BC'A'] + [CA'B']+ [A'BC]+[B'CA] + [C'AB].</cmath>
 
<cmath> 3([ABC]+ [A'B'C']) = [AB'C'] + [BC'A'] + [CA'B']+ [A'BC]+[B'CA] + [C'AB].</cmath>
  
== Solution ==
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== Hint ==
{{solution}}
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Let the parallel lines <math>AA', BB', CC'</math> be parallel to the <math>x-axis</math>, and choose arbitrary origin. Then we can define <math>A(x_1, a), A'(x_2, a), B(y_1, b), B'(y_2, b), C(z_1, c), C'(z_2, c),</math> and so, by the area of a triangle formula, it suffices to prove an algebraic statement that is readily shown to be true.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 22:24, 19 May 2015

Problem

$ABC$ and $A'B'C'$ are two triangles in the same plane such that the lines $AA',BB',CC'$ are mutually parallel. Let $[ABC]$ denote the area of triangle $ABC$ with an appropriate $\pm$ sign, etc.; prove that \[3([ABC]+ [A'B'C']) = [AB'C'] + [BC'A'] + [CA'B']+ [A'BC]+[B'CA] + [C'AB].\]

Hint

Let the parallel lines $AA', BB', CC'$ be parallel to the $x-axis$, and choose arbitrary origin. Then we can define $A(x_1, a), A'(x_2, a), B(y_1, b), B'(y_2, b), C(z_1, c), C'(z_2, c),$ and so, by the area of a triangle formula, it suffices to prove an algebraic statement that is readily shown to be true.

See Also

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

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