Inspired by Sandeep Bhardwaj

Geometry Level 5

How many lines are there in the space, such that they are equidistant from three non-collinear points?

This is an easier version of this problem .

The answer is 4.

2 solutions

Niranjan Khanderia
Nov 10, 2015

Three non-collinear points lie in a plane. A line perpendicular to the plane through the circumcenter of the triangle formed by the three points is one such line. Another is parallel to the line joining two points, and equidistance from this line and third point. There will naturally be three such lines in the same plane. So total 4 lines.

I don't see why it cannot be infinite lines.

Let the points be $A$ , $B$ and $C$ .

Let the midpoint of $AB$ be $M$ and the distance between $AB$ be $2n$

Any line that passes through point $M$ and another point $D$ which is distance $n$ away from point $C$ and lies perpendicular to the plane that $A$ $B$ and $C$ lies on will satisfy the question.

Since point $D$ can lie anywhere on a circle, there should be an infinite number of lines.

Julian Poon - 5 years, 7 months ago

If I understood what you tried to explain, any line which passes through the midpoint of two of the points which distance is $n$ and through any point on a sphere with center in the third point and radious equal to $n$ is going to be equidistant to the three points, right? In that case the fact is that everytime you change the slope of the line, the distance through the first two points will fail, because we define the distance from a point to a line as the shorter or perpendicular distance.

Hjalmar Orellana Soto - 5 years, 7 months ago

If finding the perpendicular distance is what we want, the answer should be 10 lines.

Suppose we have another line $k$ which is perpendicular to the plane which A, B and C lies on, and this line passes through point $C$ . We have point $D$ which lies on line $k$ and is distance $n$ away from point $C$ . If a line passes through point $M$ , or midpoint of points $AB$ , and also passes through point $D$ , it would also be equidistant from all 3 points. There are $6$ of such points. Adding to your solution, the answer should be $10$ points.

Julian Poon - 5 years, 7 months ago

@Julian Poon – A point on line through M will be equidistance from A and B only if it is perpendicular to the plane with A, B, C. But its distance from A and B will not be n, except for point M.

Niranjan Khanderia - 5 years, 7 months ago

There is only one line perpendicular to the plain passing through M. Unless C is located at a distance of n from that line our condition is. Not fulfilled. Please rethink what you have said.

Niranjan Khanderia - 5 years, 7 months ago

The line will be lying perpendicular to line $AB$ , passing through point $M$ . What I'm saying, is that there might be $2$ such lines that will be distance $n$ away from point $C$ if the distance from point $C$ and the line that is perpendicular to line $AB$ and also lies on the plane which which $A$ $B$ and $C$ sits on is less than or equal to $n$ .

Ok, I agree the answer would be $4$ if the question asks for the minimum number of such lines.

Julian Poon - 5 years, 7 months ago

@Julian Poon – Any line in the plane perpendicular to AB at M will be equiditance from A and B. But there is only two location of C that will be at a distance of AB/2 from The plane, and hence from a line from M. C at other location would NOT satisfy the condition..

Niranjan Khanderia - 5 years, 7 months ago

@Niranjan Khanderia – If the distance from $C$ and that plane is less than $\frac{AB}{2}$ , there will be still $2$ lines that satisfy the condition. But you are getting my point: The question has to state that it is finding the least number of lines, of if not the answer would not necessarily be $4$ .

Julian Poon - 5 years, 7 months ago

@Julian Poon – You are absolutely correct. But these are special cases. The wording of the problem can be changed. But it is not my problem. I can not do it.

Niranjan Khanderia - 5 years, 7 months ago

@Niranjan Khanderia – @Hjalmar Orellana Soto I'll summon the god of this problem

Julian Poon - 5 years, 7 months ago

@Julian Poon – I'm going to try to explain you my solution, with a blackboard I could explain it in 15 seconds...Ok, you have a picture of my next explanation in "View Reports" in advance. Let A, B, C be the three no-collinear points in the space $\mathbb{R}^3$ . They are in a plane, the plane ABC. Take the line AB passing through A and B. From C we can trace an only perpendicular line to AB cutting to this at the point D. Take the midpoint E between A and D and take the only plane P perpendicular a ABC passing through E with a parallel vector to AB, then all the lines in P parallel to AB equidistant of A.B,C. The end.

Guillermo Templado - 5 years, 7 months ago

@Julian Poon – Well, this answer, $4$ , is completely wrong. @Guillermo Templado has the reason in his solution, there are infinite lines, and are different lines you are talking about. Imagine three arbitrary points $A$ , $B$ and $C$ and call the plane which contains it $\pi _{0}$ . We will have a line $L_{1}$ parallel to $\overline{AB}$ (we'll call it $L_{0}$ passing through $C$ . The line $L_{2}$ which is parallel to $L_{0}$ and is equidistant from $L_{0}$ and $L_{1}$ is equidistant from the three points. If we draw or think in a plane $\pi _{1}$ which passes through $L_{2}$ and is perpendicular to $\pi _{0}$ we'll have that every line which belongs to $\pi _{1}$ and are parallel to $\pi _{0}$ are equidistant from the three points, so there are infinite lines, because it makes a lot of isosceles triangles.... Is somehting here wrong?

Hjalmar Orellana Soto - 5 years, 7 months ago

@Hjalmar Orellana Soto – Muchas gracias Hjalmar por darme la razón,con todo mi respeto hacia ti voy a intentar explicárselo a Julian,es que no son necesariamente triángulos isósceles(son triángulos rectángulos),pido disculpas de antemano por si me equivoco o te corrijo sin tener razón, muchas gracias de todas formas.

Guillermo Templado - 5 years, 7 months ago

@Guillermo Templado – u are thanking him in other language , right ?

A Former Brilliant Member - 4 years, 3 months ago

@A Former Brilliant Member – Right, the answer is $\infty$ lines... (if we are talking in 3D-Euclidean space).

Guillermo Templado - 4 years, 2 months ago

Why did u consider only the perpendicular to the plane, many lines which are not perpendicular to the plane can pass through the circumcentre.

Debmeet Banerjee - 5 years, 7 months ago

Because only the perpendicular to the plane is equidistant from the three points, the other lines which passes through the circumcentre are not equidistant (the distance is not from the points to the circumcenter, but to the line, this menas perpendicular distance to the line)

Hjalmar Orellana Soto - 5 years, 7 months ago

OK! So we have to keep it equidistant at every point?

Debmeet Banerjee - 5 years, 6 months ago

Michael Mendrin
Nov 24, 2015

I agree with the objections being raised that there are actually infinitely many lines in 3D space that can be equidistant from 3 non-colinear points. In 2D space, there are just 3 such lines, but in 3D, any lines parallel to those 3 lines but translated in the new axis are also equidistant from those same 3 non-colinear points.

I think this problem needs some extra wording to avoid this confusion. Like, for example, "A line in 3D space is equidistant from 3 non-colinear points. How many different directions can it have?" Then the answer would be 4.

Inspired by Sandeep Bhardwaj

This is an easier version of this problem .

The answer is 4.

2 solutions

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