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Difference between revisions of "2001 USAMO Problems/Problem 6"
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== Solution ==
 
== Solution ==
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{{solution}}
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We label each upper case point with the corresponding lower case letter as its assigned number. The key step is the following lemma.
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'''Lemma''': If <math>ABCD</math> is an isosceles trapezoid, then <math>a + c = b + d</math>.
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''Proof'': Assume without loss of generality that <math>BC\parallel AD</math>, and that rays <math>AB</math> and <math>DC</math> meet at <math>P</math>. Let <math>I</math> be the incenter of triangle <math>PAC</math>, and let line <math>l</math> bisect <math>\angle APD</math>. Then <math>I</math> is on <math>l</math>, so reflecting everything across line <math>l</math> shows that <math>I</math> is also the incenter of triangle <math>PDB</math>. Therefore,
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<cmath>\frac{p + a + c}{3} = i = \frac{p + b + d}{3}.</cmath>
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Hence <math>a + c = b + d</math>, as desired.<math>\blacksquare</math>
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<center>[[File:2001usamo6-1.png]]</center>
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For any two distinct points <math>A_1</math> and <math>A_2</math> in the plane, we construct a regular pentagon <math>A_1A_2A_3A_4A_5</math>. Applying the lemma to isosceles trapezoids <math>A_1A_3A_4A_5</math> and <math>A_2A_3A_4A_5</math> yields
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<cmath>a_1 + a_4 = a_3 + a_5\quad\text{and}\quad a_2 + a_4 = a_3 + a_5.</cmath>
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Hence <math>a_1 = a_2</math>. Since <math>A_1</math> and <math>A_2</math> were arbitrary, all points in the plane are assigned the same number.
  
 
== See also ==
 
== See also ==

Latest revision as of 17:11, 22 February 2016

Problem

Each point in the plane is assigned a real number such that, for any triangle, the number at the center of its inscribed circle is equal to the arithmetic mean of the three numbers at its vertices. Prove that all points in the plane are assigned the same number.

Solution

We label each upper case point with the corresponding lower case letter as its assigned number. The key step is the following lemma.

Lemma: If $ABCD$ is an isosceles trapezoid, then $a + c = b + d$.

Proof: Assume without loss of generality that $BC\parallel AD$, and that rays $AB$ and $DC$ meet at $P$. Let $I$ be the incenter of triangle $PAC$, and let line $l$ bisect $\angle APD$. Then $I$ is on $l$, so reflecting everything across line $l$ shows that $I$ is also the incenter of triangle $PDB$. Therefore, \[\frac{p + a + c}{3} = i = \frac{p + b + d}{3}.\] Hence $a + c = b + d$, as desired.$\blacksquare$

2001usamo6-1.png

For any two distinct points $A_1$ and $A_2$ in the plane, we construct a regular pentagon $A_1A_2A_3A_4A_5$. Applying the lemma to isosceles trapezoids $A_1A_3A_4A_5$ and $A_2A_3A_4A_5$ yields \[a_1 + a_4 = a_3 + a_5\quad\text{and}\quad a_2 + a_4 = a_3 + a_5.\] Hence $a_1 = a_2$. Since $A_1$ and $A_2$ were arbitrary, all points in the plane are assigned the same number.

See also

2001 USAMO (Problems • Resources)
Preceded by
Problem 5 		Followed by
Last question
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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