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, 1993

Geometry Difficulty 5.5 AIME, harder Prove it Baltic Way

Problem:

An equilateral triangle $ABC$ is divided into $100$ congruent equilateral triangles. What is the greatest number of vertices of small triangles that can be chosen so that no two of them lie on a line that is parallel to any of the sides of the triangle $ABC$ ?

Solution

Solution:

Figure 2

An example for $7$ vertices is shown in Figure 2. Now assume we have chosen $8$ vertices satisfying the conditions of the problem. Let the height of each small triangle be equal to $1$ and denote by $a_{i}, b_{i}, c_{i}$ the distance of the $i$ th point from the three sides of the big triangle. For any $i=1,2, \ldots, 8$ we then have $a_{i}, b_{i}, c_{i} \geq 0$ and $a_{i}+b_{i}+c_{i}=10$ . Thus, $\left(a_{1}+a_{2}+\cdots+a_{8}\right)+\left(b_{1}+b_{2}+\cdots+b_{8}\right)+\left(c_{1}+c_{2}+\cdots+c_{8}\right)=80$ . On the other hand, each of the sums in the brackets is not less than $0+1+\cdots+7=28$ , but $3 \cdot 28=84>80$ , a contradiction.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.