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Difference between revisions of "2001 USAMO Problems/Problem 4"
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== Solution ==
 
== Solution ==
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{{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>.
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It would be sufficient to prove that
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<cmath>PB^2+PC^2+AB^2+AC^2 \geq PA^2 + BC^2.</cmath>
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Set <math>A(0,0)</math>, <math>B(1,0)</math>, <math>C(x,y)</math>, <math>P(p,q)</math>.
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Then, we wish to show
 +
 
 +
<cmath>\begin{align*}
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(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 \\
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2p^2 + 2q^2 + 2x^2 + 2y^2 - 2p - 2px - 2qy + 2 &\geq p^2 + q^2 + x^2 + y^2 - 2x + 1 \\
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p^2 + q^2 + x^2 + y^2 + 2x - 2p - 2px - 2qy + 1 &\geq 0 \\
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(x-p)^2 + (q-y)^2 + 2(x-p) + 1 &\geq 0 \\
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(x-p+1)^2 + (q-y)^2 &\geq 0,
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\end{align*}</cmath>
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which is true by the trivial inequality.
  
 
== See also ==
 
== See also ==

Revision as of 21:30, 18 April 2010

Problem

Let $P$ be a point in the plane of triangle $ABC$ such that the segments $PA$, $PB$, and $PC$ are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to $PA$. Prove that $\angle BAC$ is acute.

Solution

We know that $PB^2+PC^2 < PA^2$ and we wish to prove that $AB^2 + AC^2 > BC^2$. It would be sufficient to prove that \[PB^2+PC^2+AB^2+AC^2 \geq PA^2 + BC^2.\] Set $A(0,0)$, $B(1,0)$, $C(x,y)$, $P(p,q)$. Then, we wish to show

\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*}

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
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