Difference between revisions of "1966 IMO Problems/Problem 6"
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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>. | |
− | == See | + | 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 than one 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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+ | ==Solution 2== | ||
+ | Let <math>AR : AB = x, BP : BC = y, CQ : CA = z</math>. Then it is clear that the ratio of areas of <math>AQR, BPR, CPQ</math> to that of <math>ABC</math> equals <math>x(1-y), y(1-z), z(1-x)</math>, respectively. Suppose all three quantities exceed <math>\frac{1}{4}</math>. Then their product also exceeds <math>\frac{1}{64}</math>. However, it is clear by AM-GM that <math>x(1-x) \le \frac{1}{4}</math>, and so the product of all three quantities cannot exceed <math>\frac{1}{64}</math> (by the associative property of multiplication), a contradiction. Hence, at least one area is less than or equal to <math>\frac{1}{4} [ABC]</math>. | ||
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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]] |
Latest revision as of 23:37, 16 May 2015
Contents
Problem
In the interior of sides of triangle , any points , respectively, are selected. Prove that the area of at least one of the triangles is less than or equal to one quarter of the area of triangle .
Solution
Let the lengths of sides , , and be , , and , respectively. Let , , and .
Now assume for the sake of contradiction that the areas of , , and are all at greater than one fourth of that of . Therefore
In other words, , or . Similarly, and . Multiplying these three inequalities together yields
We also have that , , and from the Arithmetic Mean-Geometric Mean Inequality. Multiplying these three inequalities together yields
This is a contradiction, which shows that our assumption must have been false in the first place. This proves the desired result.
Solution 2
Let . Then it is clear that the ratio of areas of to that of equals , respectively. Suppose all three quantities exceed . Then their product also exceeds . However, it is clear by AM-GM that , and so the product of all three quantities cannot exceed (by the associative property of multiplication), a contradiction. Hence, at least one area is less than or equal to .
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 |