<p>If three different circles are drawn on a piece of paper, at most how many points can be common to all three??</p>
<p>I know the answer is 2, but my question is: can we generalize and say that the maximum number of points of intersection of any number of circles is 2?</p>
<p>Is it true for all 2D figures?</p>
<p>Why is the answer 2 ?</p>
<p>Isn’t it 6 ?</p>
<p>No because note: “common to all three”. You can have six where the points are common only to “two” circles.</p>
<p>Since two circles can’t intersect in more than two points, more than two circles can’t possibly intersect in more than two points.</p>
<p>And if you have n circles to draw, there is a way to draw them so that they all contain those two points of intersection. (Just draw a line perpendicular to the segment connecting the two points, and locate the centers of the various circles at different points along that line.)</p>
<p>So yes, you can generalize and say that.</p>
<p>But you can’t generalize on all 2D figures, right?</p>
<p>For example, can we say that since 2 triangles will have 6 “common” maximum number of intersection points (e.g star of david), 4, 5 or 100 triangles will have the same maximum number of “common” intersection points?
My guess: we can.</p>
<p>The problem of generalizing becomes more difficult. Also, you can draw triangles that have infinitely many points in common if you allow one side of a triangle to coincide with another side.</p>
<p>For example, you can draw infinitely many squares on a sheet of paper, in such a way that there are four points common to every square (no two sides coincide). Can you figure out a way?</p>
<p>You can’t have four points in common with two or more squares. You’ll have a maximum of 2 common points. And you can conclude that an infinite number of squares will have 2 max common points of intersection.</p>
<p>2 or more squares can have infinitely (in fact, uncountably) many points in common if they each have a side that lies on the same line. In fact you can have infinitely many distinct squares that intersect in infinitely many points.</p>
<p>I’ll assume that no two sides of any two squares lie on the same line. </p>
<p>And yes, you can in fact draw uncountably many squares on a plane such that every square passes through the same set of four points, and no two sides lie on the same line.</p>
<p>MITer94, I don’t think you can have two squares intersecting at 4 points. With a square and a rectangle it could work, but I don’t think it’d work with two squares.</p>
<p>@ccuser here’s an example. Draw a square. Construct a smaller square whose vertices are midpoints of the sides of the first square.</p>
<p>Yeh, but that’s just two squares. How’ll you have an infinite number of squares with 4 common intersection points?</p>
<p>Given a square ABCD, construct four congruent right triangles with hypotenuses AB, BC, CD, DA (outside the square) so that a larger square is formed. We can construct infinitely more squares from ABCD, by varying the angles of the right triangle. All of these squares have four common intersection points (A, B, C, D).</p>
<p>Two squares can intersect at a maximum of 8 points. Using what you did for 4 points (previous post), only two squares, not an infinite number of squares, can intersect at 8 points. Am I correct?</p>