Reply for the maths problem that introduced by Calvin Lin

Calvin Lin: You have introduced a problem that describes about a minimum number of sets of lines.

If every line is perpendicular at most one line, and since the lines have placed arbitrary: then within the the possible space volume, the angle theta should obey 0<theta< pi/2 and the azimuthal angle should obey 0<azimuthal angle< pi/2 as usual notations in spherical polar coordinates. Otherwise there is a possibility to be at least 2 lines perpendicular to one line(for 3-D case) Then there is a contradiction to the the statement: every line is perpendicular at most one line.(since lines have placed arbitrary). But within above constrains for angles, no any two lines perpendicular to each other. If we divide the set of lines into N groups according to the criteria: every line is perpendicular at most one line in its group, we have to divide the set of lines into 2014 groups. We can't think about 2-D case. Because in one plane the maximum number of lines is 3.

#Geometry

Note by Geerasee Wijesuriya
7 years, 4 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

It is best to make your comments in the solution discussion.

The statement is true for any set of 2014 lines (through the origin). It is possible for 2 of those lines to be perpendicular, which would give θ=π/2 \theta = \pi / 2 . It is also possible for 1 line to be perpendicular to 2 other lines, like in the case of the x, y and z axis.

It is possible for 2 lines in the same group to be perpendicular to each other. For example, consider the set of 3 lines which are the x, y and z axis, where each of these lines is perpendicular to the other. 2 groups are sufficient - in group 1, we have the x and y axis, and in group 2 we have the z axis. You can check that this satisfies the condition.

Calvin Lin Staff - 7 years, 4 months ago

Log in to reply

You are correct. I meant that if 0<theta<= pi/2 or 0<azimuthal angle<= pi/2, then there are infinite number of lines exist those are perpendicular to one line. Since you divide the sets according to the constrain, the problem is correct. But, I couldn't noted the constrain of dividing the lines into sets.

Geerasee Wijesuriya - 7 years, 4 months ago
×

Problem Loading...

Note Loading...

Set Loading...