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

m
Line 41: Line 41:
 
as in Solution 2.  Note that <math>x, y, z, \in (0, 1)</math> because <math>K, L, M</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.
 
are in the interior of the respective sides.
 +
 +
[[File:Prob_1966_6.png|400px]]
 +
 +
Using the fact that 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
 +
<math>\mathbf{area} AML = x(1 - z) \mathbf{area} ABC,
 +
\mathbf{area} BKM = y(1 - x) \mathbf{area} ABC,
 +
\mathbf{area} CLK = z(1 - y) \mathbf{area} ABC</math>.
 +
 +
Now the problem has nothing to do with geometry anymore: we just have
 +
to show that given three numbers <math>x, y, z, \in (0, 1)</math>, at least one
 +
of <math>x(1 - z), y(1 - x), z(1 - y)</math> is <math>\le \frac{1}{4}</math>.
 +
 +
If <math>y(1 - x) \le \frac{1}{4}</math>, we are done.  Otherwise, we have
 +
<math>y(1 - x) > \frac{1}{4}</math>.  It follows that <math>y > \frac{1}{4(1 - x)}</math>
 +
(recall that <math>0 < x, y, z < 1</math>).
 +
 +
If <math>z(1 - y) \le \frac{1}{4}</math>, we are done.  Otherwise, we have
 +
<math>z(1 - y) > \frac{1}{4}</math>.  Using the inequality on <math>y</math> from the
 +
previous paragraph, we have
 +
<math>z \left( 1 - \frac{1}{4(1 - x)} \right) > \frac{1}{4}</math>, or
 +
<math>z \cdot \frac{3 - 4x}{1 - x} > 1</math>.
 +
 +
ttt
  
  
  
 +
After some computations and simplifications, we get that
 +
<math>z > \frac{1 - x}{3 - 4x}</math>.
  
−
Using the fact thar the
+
Now consider <math>x(1 - z)</math>.  Using the inequality on <math>z</math> from the previous
−
area of a triangle is half of the product  of two sides and <math>\sin</math> of
+
paragraph, we have that
−
the angle between them (like in the first Solution), we have that
+
<math>x(1 - z) < x \left( 1 - \frac{1 - x}{3 - 4x} \right) </math>.  To finish the
−
$\bathbf{area}  
+
solution to the problem, it is enough to show that
 +
<math>x \left( 1 - \frac{1 - x}{3 - 4x} \right) \le \frac{1}{4}</math>.
  
−
[[File:Prob_1966_6.png|400px]]
+
After some easy computations, this becomes <math>3(4x^2 - 4x + 1) \ge 0</math>,
 +
which is obvious.
  
  

Revision as of 15:01, 22 September 2024

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 than one 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.

Solution 2

Let $AR : AB = x, BP : BC = y, CQ : CA = z$. Then it is clear that the ratio of areas of $AQR, BPR, CPQ$ to that of $ABC$ equals $x(1-y), y(1-z), z(1-x)$, respectively. Suppose all three quantities exceed $\frac{1}{4}$. Then their product also exceeds $\frac{1}{64}$. However, it is clear by AM-GM that $x(1-x) \le \frac{1}{4}$, and so the product of all three quantities cannot exceed $\frac{1}{64}$ (by the associative property of multiplication), a contradiction. Hence, at least one area is less than or equal to $\frac{1}{4} [ABC]$.


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 $\triangle ABC$ and $K, L, M$ be as in the problem. Denote $x = \frac{AM}{AB}, y = \frac{BK}{BC}, z = \frac{CL}{CA}$ as in Solution 2. Note that $x, y, z, \in (0, 1)$ because $K, L, M$ are in the interior of the respective sides.

Prob 1966 6.png

Using the fact that the area of a triangle is half of the product of two sides and $\sin$ of the angle between them (like in the first Solution), we have that $\mathbf{area} AML = x(1 - z) \mathbf{area} ABC, \mathbf{area} BKM = y(1 - x) \mathbf{area} ABC, \mathbf{area} CLK = z(1 - y) \mathbf{area} ABC$.

Now the problem has nothing to do with geometry anymore: we just have to show that given three numbers $x, y, z, \in (0, 1)$, at least one of $x(1 - z), y(1 - x), z(1 - y)$ is $\le \frac{1}{4}$.

If $y(1 - x) \le \frac{1}{4}$, we are done. Otherwise, we have $y(1 - x) > \frac{1}{4}$. It follows that $y > \frac{1}{4(1 - x)}$ (recall that $0 < x, y, z < 1$).

If $z(1 - y) \le \frac{1}{4}$, we are done. Otherwise, we have $z(1 - y) > \frac{1}{4}$. Using the inequality on $y$ from the previous paragraph, we have $z \left( 1 - \frac{1}{4(1 - x)} \right) > \frac{1}{4}$, or $z \cdot \frac{3 - 4x}{1 - x} > 1$.

ttt


After some computations and simplifications, we get that $z > \frac{1 - x}{3 - 4x}$.

Now consider $x(1 - z)$. Using the inequality on $z$ from the previous paragraph, we have that $x(1 - z) < x \left( 1 - \frac{1 - x}{3 - 4x} \right)$. To finish the solution to the problem, it is enough to show that $x \left( 1 - \frac{1 - x}{3 - 4x} \right) \le \frac{1}{4}$.

After some easy computations, this becomes $3(4x^2 - 4x + 1) \ge 0$, which is obvious.


(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
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