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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 ==
+
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
 +
 
 +
<cmath>\frac{AM\cdot AL\sin{\angle BAC}}{2}>\frac{AB\cdot AC\sin{\angle BAC}}{8}</cmath>
 +
 
 +
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
 +
 
 +
<cmath>def(a-d)(b-e)(c-f)>\frac{a^2b^2c^2}{64}</cmath>
 +
 
 +
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
 +
 
 +
<cmath>def(a-d)(b-e)(c-f)\leq\frac{a^2b^2c^2}{64}</cmath>
 +
 
 +
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 <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.
 +
 
 +
[[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>
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(recall that <math>0 < x, y, z < 1</math>).  In particular, it follows that
 +
<math>\frac{1}{4(1 - x)} < 1</math>, which implies <math>3 - 4x > 0</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 after
 +
a few computations, <math>z \cdot \frac{3 - 4x}{1 - x} > 1</math>.  Using the
 +
observation about <math>3 - 4x</math> from the preceding paragraph, we get
 +
<math>z > \frac{1 - x}{3 - 4x}</math>.
 +
 
 +
Now consider <math>x(1 - z)</math>.  Using the inequality on <math>z</math> from the previous
 +
paragraph, we have that
 +
<math>x(1 - z) < x \left( 1 - \frac{1 - x}{3 - 4x} \right) </math>.  To finish the
 +
solution to the problem, it is enough to show that
 +
<math>x \cdot \left( 1 - \frac{1 - x}{3 - 4x} \right) \le \frac{1}{4}</math>.
 +
 
 +
After some easy computations (and using again that <math>3 - 4x > 0</math>), this
 +
becomes <math>3(4x^2 - 4x + 1) \ge 0</math>, which is true because.
 +
<math>4x^2 - 4x + 1 = (2x - 1)^2</math>.
 +
 
 +
(Solution by pf02, September 2024)
 +
 
 +
 
 +
== 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 16:06, 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 $0 < x, y, z < 1$). In particular, it follows that $\frac{1}{4(1 - x)} < 1$, which implies $3 - 4x > 0$.

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 after a few computations, $z \cdot \frac{3 - 4x}{1 - x} > 1$. Using the observation about $3 - 4x$ from the preceding paragraph, we get $z > \frac{1 - x}{3 - 4x}$.

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 $x \cdot \left( 1 - \frac{1 - x}{3 - 4x} \right) \le \frac{1}{4}$.

After some easy computations (and using again that $3 - 4x > 0$), this becomes $3(4x^2 - 4x + 1) \ge 0$, which is true because. $4x^2 - 4x + 1 = (2x - 1)^2$.

(Solution by pf02, September 2024)


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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