2001 USAMO Problems/Problem 4: Difference between revisions
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Then, we wish to show | Then, we wish to show | ||
< | <center><math>\begin{align*} | ||
(p-1)^2 + q^2 + (p-x)^2 + (q-y)^2 + 1 + x^2 + y^2 &\geq p^2 + q^2 + (x-1)^2 + y^2 \\ | (p-1)^2 + q^2 + (p-x)^2 + (q-y)^2 + 1 + x^2 + y^2 &\geq p^2 + q^2 + (x-1)^2 + y^2 \\ | ||
2p^2 + 2q^2 + 2x^2 + 2y^2 - 2p - 2px - 2qy + 2 &\geq p^2 + q^2 + x^2 + y^2 - 2x + 1 \\ | 2p^2 + 2q^2 + 2x^2 + 2y^2 - 2p - 2px - 2qy + 2 &\geq p^2 + q^2 + x^2 + y^2 - 2x + 1 \\ | ||
| Line 17: | Line 17: | ||
(x-p)^2 + (q-y)^2 + 2(x-p) + 1 &\geq 0 \\ | (x-p)^2 + (q-y)^2 + 2(x-p) + 1 &\geq 0 \\ | ||
(x-p+1)^2 + (q-y)^2 &\geq 0, | (x-p+1)^2 + (q-y)^2 &\geq 0, | ||
\end{align*}</ | \end{align*}</math></center> | ||
which is true by the trivial inequality. | which is true by the trivial inequality. | ||
Revision as of 18:00, 19 April 2010
Problem
Let 
be a point in the plane of triangle 
such that the segments 
, 
, and 
are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to . Prove that 
is acute.
Solution
We know that 
and we wish to prove that 
.
It would be sufficient to prove that
![\[PB^2+PC^2+AB^2+AC^2 \geq PA^2 + BC^2.\]](http://latex.artofproblemsolving.com/1/3/0/130788f8cf9a54d97df4b71ffc5d21849aa7cd63.png)
Set 
, 
, 
, 
.
Then, we wish to show
(p-1)^2 + q^2 + (p-x)^2 + (q-y)^2 + 1 + x^2 + y^2 &\geq p^2 + q^2 + (x-1)^2 + y^2 \\ 2p^2 + 2q^2 + 2x^2 + 2y^2 - 2p - 2px - 2qy + 2 &\geq p^2 + q^2 + x^2 + y^2 - 2x + 1 \\ p^2 + q^2 + x^2 + y^2 + 2x - 2p - 2px - 2qy + 1 &\geq 0 \\ (x-p)^2 + (q-y)^2 + 2(x-p) + 1 &\geq 0 \\ (x-p+1)^2 + (q-y)^2 &\geq 0,
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)which is true by the trivial inequality.
See also
| 2001 USAMO (Problems • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAMO Problems and Solutions | ||
