Difference between revisions of "2001 USAMO Problems/Problem 4"
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== Solution == | == Solution == | ||
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+ | We know that <math>PB^2+PC^2 < PA^2</math> and we wish to prove that <math>AB^2 + AC^2 > BC^2</math>. | ||
+ | It would be sufficient to prove that | ||
+ | <cmath>PB^2+PC^2+AB^2+AC^2 \geq PA^2 + BC^2.</cmath> | ||
+ | Set <math>A(0,0)</math>, <math>B(1,0)</math>, <math>C(x,y)</math>, <math>P(p,q)</math>. | ||
+ | Then, we wish to show | ||
+ | |||
+ | <cmath>\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 \ | ||
+ | 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*}</cmath> | ||
+ | |||
+ | which is true by the trivial inequality. | ||
== See also == | == See also == |
Revision as of 21:30, 18 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 Set , , , . Then, we wish to show
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 |