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Difference between revisions of "2001 IMO Shortlist Problems/G3"

(New page: == Problem == Let <math>ABC</math> be a triangle with centroid <math>G</math>. Determine, with proof, the position of the point <math>P</math> in the plane of <math>ABC</math> such that <m...)
 
 
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== Solution ==
 
== Solution ==
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{{solution}}
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We claim that the expression is minimized at <math>P=G</math>, resulting it having a value of <math>(a^2+b^2+c^2)/3</math> (<math>a,b,c</math> being the side lengths of <math>ABC</math>).
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 +
We will use vectors, with <math>G=\vec{0}</math> (meaning that <math>\vec A+\vec B+\vec C=\vec 0</math>). Note that by Cauchy-Schwarz,
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<cmath>\|\vec A\| \|\vec A-\vec P \|+\|\vec B\|\|\vec B-\vec P \|+\|\vec C\| \|\vec C-\vec P \|</cmath>                    <cmath>\ge \vec A\cdot (\vec A-\vec P) +\vec B\cdot (\vec B-\vec P)+\vec C\cdot (\vec C-\vec P) </cmath>
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<cmath>=\|\vec A\|^2+\|\vec B\|^2+\|\vec C\|^2-\vec P\cdot (\vec A+\vec B+\vec C)=\|\vec A\|^2+\|\vec B\|^2+\|\vec C\|^2, </cmath>
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and this bound is clearly reached by <math>\vec P=\vec 0</math>. Furthermore, equality is only reached when <math>\vec A</math>, <math>\vec B</math>, <math>\vec C</math> are scalar multiples of <math>\vec A-\vec P</math>, <math>\vec B-\vec P</math>, <math>\vec C-\vec P</math>, respectively. This means that <math>\vec P</math> is a scalar multiple of <math>\vec A</math>, <math>\vec B</math>, and <math>\vec C</math>, so <math>\vec P=\vec 0</math>. (Note that <math>\vec A</math> and <math>\vec B</math> are linearly independent, since the centroid is not on <math>AB</math>.)
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Now all that remains is to calculate <math>\|\vec A\|^2+\|\vec B\|^2+\|\vec C\|^2=AG^2+BG^2+CG^2</math>. To calculate <math>AG</math>, first let <math>D</math> be the midpoint of <math>BC</math>. Then by Stewart's theorem,
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<cmath>AD^2\cdot a+\frac{a^3}{4}=\frac{b^2a}{2}+\frac{c^2a}{2} </cmath>
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<cmath>AD^2=\frac{b^2}{2}+\frac{c^2}{2}-\frac{a^2}{4}=\frac{2b^2+2c^2-a^2}{4}. </cmath>
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Furthermore, <math>AG=2/3\cdot AD</math>, so
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<cmath>AG^2=\frac{2b^2+2c^2-a^2}{9}. </cmath>
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By similar reasoning, we can calculate <math>BG^2=(2c^2+2a^2-b^2)/9</math> and <math>CG^2=(2a^2+2b^2-c^2)/9</math>, so
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<cmath>AG^2+BG^2+CG^2=\frac{a^2+b^2+c^2}{3}. </cmath>
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<math>\blacksquare</math>
  
 
== Resources ==
 
== Resources ==

Latest revision as of 17:13, 29 January 2019

Problem

Let $ABC$ be a triangle with centroid $G$. Determine, with proof, the position of the point $P$ in the plane of $ABC$ such that $AP{\cdot}AG + BP{\cdot}BG + CP{\cdot}CG$ is a minimum, and express this minimum value in terms of the side lengths of $ABC$.

Solution

We claim that the expression is minimized at $P=G$, resulting it having a value of $(a^2+b^2+c^2)/3$ ($a,b,c$ being the side lengths of $ABC$).

We will use vectors, with $G=\vec{0}$ (meaning that $\vec A+\vec B+\vec C=\vec 0$). Note that by Cauchy-Schwarz, \[\|\vec A\| \|\vec A-\vec P \|+\|\vec B\|\|\vec B-\vec P \|+\|\vec C\| \|\vec C-\vec P \|\] \[\ge \vec A\cdot (\vec A-\vec P) +\vec B\cdot (\vec B-\vec P)+\vec C\cdot (\vec C-\vec P)\] \[=\|\vec A\|^2+\|\vec B\|^2+\|\vec C\|^2-\vec P\cdot (\vec A+\vec B+\vec C)=\|\vec A\|^2+\|\vec B\|^2+\|\vec C\|^2,\] and this bound is clearly reached by $\vec P=\vec 0$. Furthermore, equality is only reached when $\vec A$, $\vec B$, $\vec C$ are scalar multiples of $\vec A-\vec P$, $\vec B-\vec P$, $\vec C-\vec P$, respectively. This means that $\vec P$ is a scalar multiple of $\vec A$, $\vec B$, and $\vec C$, so $\vec P=\vec 0$. (Note that $\vec A$ and $\vec B$ are linearly independent, since the centroid is not on $AB$.)

Now all that remains is to calculate $\|\vec A\|^2+\|\vec B\|^2+\|\vec C\|^2=AG^2+BG^2+CG^2$. To calculate $AG$, first let $D$ be the midpoint of $BC$. Then by Stewart's theorem, \[AD^2\cdot a+\frac{a^3}{4}=\frac{b^2a}{2}+\frac{c^2a}{2}\] \[AD^2=\frac{b^2}{2}+\frac{c^2}{2}-\frac{a^2}{4}=\frac{2b^2+2c^2-a^2}{4}.\]

Furthermore, $AG=2/3\cdot AD$, so \[AG^2=\frac{2b^2+2c^2-a^2}{9}.\] By similar reasoning, we can calculate $BG^2=(2c^2+2a^2-b^2)/9$ and $CG^2=(2a^2+2b^2-c^2)/9$, so \[AG^2+BG^2+CG^2=\frac{a^2+b^2+c^2}{3}.\]

$\blacksquare$

Resources

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