Difference between revisions of "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 $\frac{1}{4} [ABC]$.