Difference between revisions of "1977 USAMO Problems"
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==Problem 3== | ==Problem 3== | ||
| − | If <math> a</math> and <math> b</math> are two of the roots of <math> x^4 | + | If <math> a</math> and <math> b</math> are two of the roots of <math> x^4+x^3-1=0</math>, prove that <math> ab</math> is a root of <math> x^6+x^4+x^3-x^2-1=0</math>. |
[[1977 USAMO Problems/Problem 3 | Solution]] | [[1977 USAMO Problems/Problem 3 | Solution]] | ||
Revision as of 21:55, 13 March 2015
Problems from the 1977 USAMO.
Problem 1
Determine all pairs of positive integers 
such that
is divisible by 
.
Problem 2
and 
are two triangles in the same plane such that the lines 
are mutually parallel. Let ![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
denote the area of triangle with an appropriate 
sign, etc.; prove that
![\[3([ABC] + [A'B'C']) = [AB'C'] + [BC'A'] + [CA'B'] + [A'BC] + [B'CA] + [C'AB].\]](http://latex.artofproblemsolving.com/f/d/a/fda9d7733fb8e4454a4b3637dc8a0977ea38e7df.png)
Problem 3
If 
and 
are two of the roots of 
, prove that 
is a root of 
.
Problem 4
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.
Problem 5
If 
are positive numbers bounded by 
and 
, i.e, if they lie in ![$[p,q], 0 < p$](http://latex.artofproblemsolving.com/b/7/d/b7db0d1c44b17ab521242db4599465126d5ad617.png)
, prove that
![\[(a + b + c + d + e)(\frac {1}{a} + \frac {1}{b} + \frac {1}{c} + \frac {1}{d} + \frac {1}{e}) \le 25 + 6\left(\sqrt {\frac {p}{q}} - \sqrt {\frac {q}{p}}\right)^2\]](http://latex.artofproblemsolving.com/6/3/3/6335c58fcf19184291f5c4449812dc094c8b65bf.png)
and determine when there is equality.
See Also
| 1977 USAMO (Problems • Resources) | ||
| Preceded by 1976 USAMO |
Followed by 1978 USAMO | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
