1966 IMO Problems/Problem 6

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.

Using the fact thar 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 $\bathbf{area}

(Solution by pf02, September 2024)

TO BE CONTINUED. SAVING MID WAY SO I DON'T LOOSE WORK DONE SO FAR

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