Interesting Combinatorics Problem :: Help

There are 5 Points in a plane from each points Perpendicular's are drawn to the Lines Joining other Points. what is maximum number of point of Intersection of these perpendiculars ?

Explain it to Me Please ! Thanks

Answer is 315315

Note by Karan Shekhawat
5 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

Sort by:

Top Newest

Just for the sake of convenience, let us name the points as A,B,C,DA,B,C,D and EE. Consider any point as a reference point and WLOG take it to be AA. A total of six lines can be made among the remaining points. So there would be six perpendiculars drawn from AA. Thus there would be a total of of thirty lines. Thus the maximum number of points of intersection are (302)=435\displaystyle \binom{30}{2} = 435 .

As stated before there would be six lines through each of the points and we have counted the intersection points of these lines separately. Therefore we have counted55 points as 5×(62)=75\displaystyle 5 \times \binom{6}{2} = 75. Also all the perpendiculars from to a line would be parallel and would not have a point of intersection. A total of (52)=10\displaystyle \binom{5}{2} =10 lines can be drawn between any two given points and there would a set of three parallel lines on each of these. Therefore we have counted an extra of 3030 points. Consider any of the (53)=10\displaystyle \binom{5}{3} =10 triangles formed by these points. Just by a simple analysis, it is not a difficult task to realize that the orthocentres of these triangles would lie on these lines. We have counted each of these points thrice i.e. the intersection of three lines thereby counting an additional of 2020 points. Thus we have counted a total of 755+30+20=120 75 -5 +30 +20 =120 points extra. Therefore, the total number of maximum possible points of intersections are 435120=315 435-120 = \boxed{315} .

Sudeep Salgia - 5 years, 4 months ago

Log in to reply

Thanks ! Many-Many Time To You "The great Sudeep " :)

Karan Shekhawat - 5 years, 4 months ago

Log in to reply

Assuming no trhee points are collinear and the pentagon is not regular, the number of Perpendiculars is 5(42)=30 5 \cdot {4 \choose 2} = 30 . Hence, the maximum number of intersection point is at most 30292=435 \frac{30\cdot 29}{2} = 435

But we have to notice that:

  • if the perpendiculars are the altitudes of a triangle, then they are concurrent. Therefore the total number must be reduced by 2(53)=20 2 \cdot {5 \choose 3} = 20

  • if the perpendiculars are drawn from three different points, and perpendicular to the two remaining, then they don't intersect each other (because they are parallel to each other). Therefore the total number must be reduced by 3(53)=30 3 \cdot {5 \choose 3} = 30

  • if the perpendiculars are drawn all from the same point, they do not intersect each other in any other point (otherwise two sides would be parallel). Therefore the total number must be reduced by ((62)1)5=70 \left( {6 \choose 2} -1 \right) \cdot 5 = 70

These considerations reduce the total number by, respectively, 2020, 3030 and 7070 points. Hence the maximum number of points is N435203070=315 N \leq 435-20-30-70 = 315. Now we should show that it's possible to achive it... Maybe with a diagram...

Andrea Gallese - 5 years, 4 months ago

Log in to reply

Thanks a lot ! But sorry I did not understand The Third Point (70) . Can You Please explain it more in word's Thanks

Karan Shekhawat - 5 years, 4 months ago

Log in to reply

If we pick one of the five point, let's say AA, the perpendiculars we draw from AA are (62) {6 \choose 2} and since two lines cannot intersect in 2 different points only, alle these perpendiculars intersect in A A only. They could intersect in infinitely many points, but we can assume this is not happening. Therefore, we have to subtract the points of intersection of 6 lines with themselves, but one (that is AA ). Repeating the process for other 4 point we obtain the result above.

Andrea Gallese - 5 years, 4 months ago

Log in to reply

Suppose, those five points are (A,B,C,D,E)(A, B, C, D, E). Now, we want to create some special structure. Let, we take the line BCBC and draw a perpendicular from AA on BCBC, andd call it P1P_1. We can do this set up in (51)(42)=30\binom{5}{1}\binom{4}{2}=30 ways. There will 3030 such PiP_i s.

Now we will try to find the number of other perpendiculars which intersect P1P_1.

See, can draw perpendiculars from BB and CC to other lines( we haven't counted the perpendicular from BB to ACAC and perpendicular from CC on ABAB , as they intersect P1P_1 at the same point) in 55 ways for each. So, total 1010 ways. Now, 55 perpendiculars from each DD and EE on the other lines except on BCBC( because in this case teh perpendiculars from DD and EE will be parallel to P1P_1 , and so shall not intersect). So,total 1010 cases. So, total 10+10=2010+10=20 cases. From, these two cases we get P1P_1 will be intersected at 5×6×(5+5+5+5)=6005×6×(5+5+5+5)=600 ways. But, as we have passed this algorithm over all the five points, we have counted each intersection points twice. So, there are total 6002=300\frac{600}{2}=300 ways. Now, as we had excluded the orthocentres, we have to add now. There are total (53)=10\binom{5}{3}=10 orthocentres. Also we should add those vertices as these are also point of intersection of silimar perpendiculars, there are $5$ such.

So, total ways 300+10+5=315300+10+5=315.

Alapan Das - 3 months, 1 week ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...