Tangent lines to 2 circles, tangent planes to 3 spheres, and so on. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:25:37Z http://mathoverflow.net/feeds/question/87047 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87047/tangent-lines-to-2-circles-tangent-planes-to-3-spheres-and-so-on Tangent lines to 2 circles, tangent planes to 3 spheres, and so on. Rogelio Fernández-Alonso 2012-01-30T17:46:07Z 2012-01-31T06:33:53Z <p>Although it is known the solution to the first two questions, somebody may have different nice answers, so I include them:</p> <ol> <li><p>Given two circles in the plane, there is (at least) a line which is tangent to both of them.</p></li> <li><p>Given three spheres in the space, there is a plane which is tangent to all of them.</p></li> <li><p>In general, given $n$ n-spheres in the n-dimensional space, is there a hyperplane which is tangent to all of them?</p></li> <li><p>What other generalizations does this problem admit?</p></li> </ol> <p>EDIT: As @Noam kindly remarked below, the existence of the tangent objects is not always true. I think that in #2 the hypothesis must be: one of the spheres not in the cone determined by the other two. In #1 the "cone" determined by one circle is the circle itself. So in #3 we need a suitable definition for the "cone" determined by $n-1$ n-spheres.</p> http://mathoverflow.net/questions/87047/tangent-lines-to-2-circles-tangent-planes-to-3-spheres-and-so-on/87111#87111 Answer by Noam D. Elkies for Tangent lines to 2 circles, tangent planes to 3 spheres, and so on. Noam D. Elkies 2012-01-31T06:33:53Z 2012-01-31T06:33:53Z <p>This answers expands on my comment on the original question.</p> <p><strong>1.</strong> <em>Given two circles in the plane, there is (at least) a line which is tangent to both of them</em>: this is not true unless we allow lines with complex coefficients (and even then there's an exception, see below). In the real plane, two circles have:</p> <p>$\bullet$ no common tangents if one is contained in the other's interior,</p> <p>$\bullet$ two common tangents if they intersect at two points, and </p> <p>$\bullet$ four common tangents if they're disjoint and neither's interior contains the other.</p> <p>At the boundaries between these cases, the circles can have 1 or 3 common tangents if they're tangent internally or externally. There's also a generate case where the two circles coincide and thus have infinitely many common tangents.</p> <p><strong>2.</strong> <em>Given three spheres in the space, there is a plane which is tangent to all of them</em>: again not true, and and with a wider variety of counterexamples, and a wider variety of degenerate cases where there are infinitely many common tangents. Aside of these degenerate examples, the maximum number of common tangent planes is $8$. (See below for how to compute them.) A simple configuration with infinitely many tangent planes is three identical spheres with collinear centers. More generally, three spheres with collinear centers that have one tangent plane have infinitely many, because we can rotate the plane about the line $l$ joining the centers. But make one sphere a bit larger or smaller about the same center, and there's no common tangent plane at all, and then this stays true if we move the centers a bit off $l$.</p> <p><strong>3.</strong> <em>In general, given $n$ spheres in $n$-dimensional space, is there a hyperplane which is tangent to all of them?</em> [The proposer wrote "$n$ n-spheres", but current terminology uses "n-sphere" for a sphere in ${\bf R}^{n+1}$.] Again, not necessarily; the number can range from $0$ to $2^n$, or be infinitely large, and there are various ways for the number to be zero, most simply when one sphere is contained in another's interior.</p> <p>The hyperplanes in ${\bf R}^n$ constitute an $n$-dimensional space, more precisely a projective space ${\bf RP}^n$ with a point removed (the hyperplane at infinity, which is not relevant to us because it is not tangent to any sphere). The hyperplanes tangent to a given sphere constitute a degree-$2$ hypersurface in this projective space. Generically $n$ such surfaces in $n$-dimensional space meet in $2^n$ points, but it is possible for none of them to have real coordinates; and there are also special configurations that intersect in a positive-dimensional variety, most simply when they are linearly dependent (but this is far from the only reason, as noted in <a href="http://mathoverflow.net/questions/84297" rel="nofollow">this mathoverflow answer</a>).</p> <p><strong>4.</strong> <em>What other generalizations does this problem admit?</em> I wrote that one gets the same $2^n$ enumeration with $n$ arbitrary quadrics in place of spheres, but Mariano Su&aacute;rez-Alvarez suggested a much more complete answer:</p> <blockquote> <p>The ultimate generalization (or, at least, one of them) is, I guess, intersection theory and enumerative geometry.</p> </blockquote> <p>Still, there are special features of the $n$-sphere problem that do not survive generalization even to $n$ quadrics. Most notably, the $2^n$ tangent planes can be given in closed form, each requiring only the extraction of a single square root. Suppose the spheres have centers $v_i \in {\bf R}^n$ and radii $r_i \in (0,\infty)$. A hyperplane $a \cdot x = b$ is tangent to the sphere $|x-v_i|=r$ iff $a \cdot v_i - b = \pm r_i |a|$. Now for each of $2^n$ choices of $\pm$ signs we get $n$ simultaneous linear equations in $n+1$ variables: the coordinates of $a$, and the length $|a|$. Generically there is a line of solutions, and then the additional condition that $|a|$ really be the length of the vector $a$ gives a quadratic equation on that line which usually has $2$ solutions, either real or complex conjugate. [This gives $2^n$ hyperplanes, not twice that number, because switching <em>all</em> the $\pm$ signs yields the same solution.] But there are also non-generic cases where the equations are dependent, and so give a higher-dimensional space of solutions (not necessarily defined over ${\bf R}$), or inconsistent, and so give no solutions at all. If I did this right, one of these happens <strong>iff</strong> the centers of the spheres lie on a linear subspace of dimension at most $n-2$. So for $n=2$ we're back to concentric circles, with infinitely many solutions if they coincide, and none otherwise, not even over ${\bf C}$. Likewise for <code>$n=3$</code> we get the configurations of three spheres with collinear centers.</p>