Difference between revisions of "1966 IMO Problems/Problem 6"
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==Solution 2== | ==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>. | 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>. | ||
+ | |||
+ | |||
+ | ==Remarks (added by pf02, September 2024)== | ||
+ | |||
+ | Solution 2 is written in a very sloppy way. However, an interested | ||
+ | reader can make sense of it. More importantly, the two solutions | ||
+ | are identical. If it wasn't for the sloppy writing, Solution 2 could | ||
+ | be obtained from the first Solution after applying a word by word | ||
+ | translation which replaces line segments by ratios. | ||
+ | |||
+ | Below I will give another solution. It is formally different from | ||
+ | the previous solutions, even if not at a deep level. | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Let <math>\triangle ABC</math> and <math>K, L, M</math> be as in the problem. | ||
+ | Denote <math>x = \frac{AM}{AB}, y = \frac{BK}{BC}, z = \frac{CL}{CA}</math> | ||
+ | as in Solution 2. Note that <math>x, y, z, \in (0, 1)</math> because <math>K, L, M</math> | ||
+ | are in the interior of the respective sides. | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | Using the fact thar the | ||
+ | area of a triangle is half of the product of two sides and <math>\sin</math> of | ||
+ | the angle between them (like in the first Solution), we have that | ||
+ | $\bathbf{area} | ||
+ | |||
+ | [[File:Prob_1966_6.png|400px]] | ||
+ | |||
+ | |||
+ | |||
+ | (Solution by pf02, September 2024) | ||
+ | |||
+ | TO BE CONTINUED. SAVING MID WAY SO I DON'T LOOSE WORK DONE SO FAR | ||
+ | |||
== See Also == | == 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 13:00, 22 September 2024
Contents
[hide]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 .
Remarks (added by pf02, September 2024)
Solution 2 is written in a very sloppy way. However, an interested reader can make sense of it. More importantly, the two solutions are identical. If it wasn't for the sloppy writing, Solution 2 could be obtained from the first Solution after applying a word by word translation which replaces line segments by ratios.
Below I will give another solution. It is formally different from the previous solutions, even if not at a deep level.
Solution 3
Let and be as in the problem. Denote as in Solution 2. Note that because are in the interior of the respective sides.
Using the fact thar the
area of a triangle is half of the product of two sides and of
the angle between them (like in the first Solution), we have that
$\bathbf{area}
(Solution by pf02, September 2024)
TO BE CONTINUED. SAVING MID WAY SO I DON'T LOOSE WORK DONE SO FAR
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 |