Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Find the answer

7. Given five points in a plane, where no two lines connecting any two points are parallel, perpendicular, or coincident, draw perpendiculars from one of the points to the lines connecting the other points. How many intersection points can these perpendiculars have, excluding the original five points, at most?

保留源文本的换行和格式,所以翻译结果如下:

7. Given five points in a plane, where no two lines connecting any two points are parallel, perpendicular, or coincident, draw perpendiculars from one of the points to the lines connecting the other points. How many intersection points can these perpendiculars have, excluding the original five points, at most?

A number or a short expression. Spacing and $ signs are ignored.

Solution

7. For five points, the number of lines connecting any two points is C52=10\mathrm{C}_{5}^{2}=10, and the number of triangles is C53=10\mathrm{C}_{5}^{3}=10.

Among any four points, the number of lines connecting them is C42=6\mathrm{C}_{4}^{2}=6, and the fifth point can draw 6 perpendicular lines to these lines, resulting in a total of 30 perpendicular lines. If we do not consider non-intersecting and repeated lines, we can get C302=435\mathrm{C}_{30}^{2}=435 intersection points.

For the five points, there are 10 connecting lines, and each line has three parallel perpendicular lines, so we need to subtract 30 intersection points.
In the 10 triangles, the three altitudes of each triangle intersect at one point, so we should subtract (C321)×10=\left(\mathrm{C}_{3}^{2}-1\right) \times 10= 20 intersection points.

Furthermore, for each point PiP_{i}, the perpendicular lines drawn to C42=6\mathrm{C}_{4}^{2}=6 lines intersect at the same point PiP_{i} (excluding PiP_{i} itself), so we should subtract C62×5=75\mathrm{C}_{6}^{2} \times 5=75, leaving us with 435302075=310435-30-20-75=310 intersection points.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.