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Difference between revisions of "1966 IMO Problems/Problem 6"

(Created page with "== Problem == In the interior of sides <math>BC, CA, AB</math> of triangle <math>ABC</math>, any points <math>K, L,M</math>, respectively, are selected. Prove that the area of at...")
 
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== Solution ==
 
== Solution ==
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{{solution}}
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Let the lengths of sides <math>BC</math>, <math>CA</math>, and <math>AB</math> be <math>a</math>, <math>b</math>, and <math>c</math>, respectively. Let <math>BK=d</math>, <math>CL=e</math>, and <math>AM=f</math>.
  
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== See also ==
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Now assume for the sake of contradiction that the areas of <math>\Delta AML</math>, <math>\Delta BKM</math>, and <math>\Delta CLK</math> are all at greater thanone fourth of that of <math>\Delta ABC</math>. Therefore
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<cmath>\frac{AM\cdot AL\sin{\angle BAC}}{2}>\frac{AB\cdot AC\sin{\angle BAC}}{8}</cmath>
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In other words, <math>AM\cdot AL>\frac{1}{4}AB\cdot AC</math>, or <math>f(b-e)>\frac{bc}{4}</math>. Similarly, <math>d(c-f)>\frac{ac}{4}</math> and <math>e(a-d)>\frac{ab}{4}</math>. Multiplying these three inequalities together yields
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<cmath>def(a-d)(b-e)(c-f)>\frac{a^2b^2c^2}{64}</cmath>
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We also have that <math>d(a-d)\leq \frac{a^2}{4}</math>, <math>e(b-e)\leq \frac{b^2}{4}</math>, and <math>f(c-f)\leq \frac{c^2}{4}</math> from the [[Arithmetic Mean-Geometric Mean Inequality]]. Multiplying these three inequalities together yields
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<cmath>def(a-d)(b-e)(c-f)\leq\frac{a^2b^2c^2}{64}</cmath>
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This is a contradiction, which shows that our assumption must have been false in the first place. This proves the desired result.
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== See Also ==
 
{{IMO box|year=1966|num-b=5|after=Last Problem}}
 
{{IMO box|year=1966|num-b=5|after=Last Problem}}
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Revision as of 12:00, 16 May 2012

Problem

In the interior of sides $BC, CA, AB$ of triangle $ABC$, any points $K, L,M$, respectively, are selected. Prove that the area of at least one of the triangles $AML, BKM, CLK$ is less than or equal to one quarter of the area of triangle $ABC$.

Solution

Let the lengths of sides $BC$, $CA$, and $AB$ be $a$, $b$, and $c$, respectively. Let $BK=d$, $CL=e$, and $AM=f$.

Now assume for the sake of contradiction that the areas of $\Delta AML$, $\Delta BKM$, and $\Delta CLK$ are all at greater thanone fourth of that of $\Delta ABC$. Therefore

\[\frac{AM\cdot AL\sin{\angle BAC}}{2}>\frac{AB\cdot AC\sin{\angle BAC}}{8}\]

In other words, $AM\cdot AL>\frac{1}{4}AB\cdot AC$, or $f(b-e)>\frac{bc}{4}$. Similarly, $d(c-f)>\frac{ac}{4}$ and $e(a-d)>\frac{ab}{4}$. Multiplying these three inequalities together yields

\[def(a-d)(b-e)(c-f)>\frac{a^2b^2c^2}{64}\]

We also have that $d(a-d)\leq \frac{a^2}{4}$, $e(b-e)\leq \frac{b^2}{4}$, and $f(c-f)\leq \frac{c^2}{4}$ from the Arithmetic Mean-Geometric Mean Inequality. Multiplying these three inequalities together yields

\[def(a-d)(b-e)(c-f)\leq\frac{a^2b^2c^2}{64}\]

This is a contradiction, which shows that our assumption must have been false in the first place. This proves the desired result.

See Also

1966 IMO (Problems) • Resources
Preceded by
Problem 5 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Last Problem
All IMO Problems and Solutions
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