Kvant Math Problem 368
Choose coordinates so that the three cylinder axes are parallel to the coordinate axes.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 6m32s
Source on kvant.digital
Problem
Prove that the intersection of three right circular cylinders of radius 1 whose axes are pairwise perpendicular (but do not necessarily intersect) is contained in some sphere of radius $\sqrt\dfrac32$.
S. V. Fomin
Exploration
Choose coordinates so that the three cylinder axes are parallel to the coordinate axes. Since the axes are pairwise perpendicular but need not intersect, they can be written as
$$L_x={(x,a,b)}, \qquad L_y={(c,y,d)}, \qquad L_z={(e,f,z)}.$$
A point $(x,y,z)$ belongs to the three cylinders of radius $1$ exactly when
$$(y-a)^2+(z-b)^2\le 1,$$
$$(x-c)^2+(z-d)^2\le 1,$$
$$(x-e)^2+(y-f)^2\le 1.$$
The problem asks for a sphere of radius $\sqrt{3/2}$ containing the whole intersection.
For the classical Steinmetz solid, when all three axes pass through the origin, the intersection satisfies
$$x^2+y^2\le 1,\quad x^2+z^2\le 1,\quad y^2+z^2\le 1.$$
Adding these inequalities gives
$$2(x^2+y^2+z^2)\le 3,$$
hence the solid lies in the sphere of radius $\sqrt{3/2}$ centered at the origin. The present problem differs only by translations of the three axis lines.
A natural attempt is to translate the coordinates by a suitable point $(u,v,w)$ and obtain three inequalities of the same form. Let
$$X=x-\alpha,\qquad Y=y-\beta,\qquad Z=z-\gamma.$$
The cylinder conditions become
$$(Y+\beta-a)^2+(Z+\gamma-b)^2\le 1,$$
$$(X+\alpha-c)^2+(Z+\gamma-d)^2\le 1,$$
$$(X+\alpha-e)^2+(Y+\beta-f)^2\le 1.$$
To make the paired variables appear symmetrically, choose
$$\beta-a=\gamma-b=:p,\qquad \alpha-c=\gamma-d=:q,\qquad \alpha-e=\beta-f=:r.$$
Then the inequalities become
$$(Y+p)^2+(Z+p)^2\le 1,$$
$$(X+q)^2+(Z+q)^2\le 1,$$
$$(X+r)^2+(Y+r)^2\le 1.$$
The system for $\alpha,\beta,\gamma$ is
$$\beta-\gamma=a-b,\qquad \alpha-\gamma=c-d,\qquad \alpha-\beta=e-f.$$
These equations are compatible because
$$(\alpha-\gamma)-(\alpha-\beta) = \beta-\gamma,$$
which reduces exactly to
$$(c-d)-(e-f)=a-b,$$
or
$$a+d+e=b+c+f.$$
This identity need not hold, so that approach is too restrictive.
A better idea is to choose the center directly. Let
$$O=(\alpha,\beta,\gamma)$$
and define
$$A=\beta-a,\quad B=\gamma-b,$$
$$C=\alpha-c,\quad D=\gamma-d,$$
$$E=\alpha-e,\quad F=\beta-f.$$
The cylinder inequalities become
$$(Y+A)^2+(Z+B)^2\le1,$$
$$(X+C)^2+(Z+D)^2\le1,$$
$$(X+E)^2+(Y+F)^2\le1.$$
After summing, the mixed linear terms are
$$2X(C+E)+2Y(A+F)+2Z(B+D).$$
If we choose $O$ so that
$$C+E=0,\qquad A+F=0,\qquad B+D=0,$$
the linear terms disappear. These equations are
$$2\alpha=c+e,\qquad 2\beta=a+f,\qquad 2\gamma=b+d.$$
This always has a solution. Then the sum of the three inequalities yields
$$2(X^2+Y^2+Z^2) + (A^2+B^2+C^2+D^2+E^2+F^2) \le 3.$$
Since the extra constant term is nonnegative,
$$2(X^2+Y^2+Z^2)\le 3.$$
Hence every point of the intersection lies in the sphere centered at $O$ of radius $\sqrt{3/2}$. The crucial point is choosing the center so that the linear terms cancel simultaneously.
Problem Understanding
The three cylinders have radius $1$. Their axes are pairwise perpendicular, but the three axes need not meet at one point. We must prove that the common intersection of the cylinders is contained in some sphere of radius
$$\sqrt{\frac32}.$$
This is a Type B problem.
The core difficulty is that the cylinder axes may be displaced from one another. For the standard Steinmetz solid, the estimate follows immediately by adding three inequalities. Here one must find a suitable center of a sphere so that, after translating coordinates to that center, the same summation argument survives.
Proof Architecture
Let the cylinder axes be represented by three mutually perpendicular lines parallel respectively to the coordinate axes, and write the cylinder inequalities in coordinates.
Choose a point $O=(\alpha,\beta,\gamma)$ whose coordinates are the midpoints of the corresponding axis offsets, namely
$$\alpha=\frac{c+e}{2},\quad \beta=\frac{a+f}{2},\quad \gamma=\frac{b+d}{2}.$$
This choice makes the coefficients of the linear terms vanish when the three cylinder inequalities are added.
After translating coordinates by $O$, sum the three inequalities. The left-hand side becomes
$$2(X^2+Y^2+Z^2) + \text{nonnegative constant}.$$
Since the right-hand side equals $3$, deduce
$$X^2+Y^2+Z^2\le \frac32.$$
The most delicate step is verifying that the chosen center indeed cancels all linear terms simultaneously.
Solution
Choose Cartesian coordinates so that the three cylinder axes are parallel respectively to the coordinate axes.
Let
$$L_x={(x,a,b):x\in\mathbb R},$$
$$L_y={(c,y,d):y\in\mathbb R},$$
$$L_z={(e,f,z):z\in\mathbb R}.$$
A point $(x,y,z)$ belongs to the three cylinders of radius $1$ precisely when
$$(y-a)^2+(z-b)^2\le1, \tag{1}$$
$$(x-c)^2+(z-d)^2\le1, \tag{2}$$
$$(x-e)^2+(y-f)^2\le1. \tag{3}$$
Define
$$\alpha=\frac{c+e}{2},\qquad \beta=\frac{a+f}{2},\qquad \gamma=\frac{b+d}{2},$$
and introduce translated coordinates
$$X=x-\alpha,\qquad Y=y-\beta,\qquad Z=z-\gamma.$$
Set
$$A=\beta-a,\qquad B=\gamma-b,$$
$$C=\alpha-c,\qquad D=\gamma-d,$$
$$E=\alpha-e,\qquad F=\beta-f.$$
Then (1), (2), (3) become
$$(Y+A)^2+(Z+B)^2\le1, \tag{4}$$
$$(X+C)^2+(Z+D)^2\le1, \tag{5}$$
$$(X+E)^2+(Y+F)^2\le1. \tag{6}$$
Because of the definitions of $\alpha,\beta,\gamma$,
$$C+E=(\alpha-c)+(\alpha-e)=2\alpha-(c+e)=0,$$
$$A+F=(\beta-a)+(\beta-f)=2\beta-(a+f)=0,$$
$$B+D=(\gamma-b)+(\gamma-d)=2\gamma-(b+d)=0.$$
Adding (4), (5), and (6), we obtain
$$\begin{aligned} &2X^2+2Y^2+2Z^2 \ &\quad +2X(C+E)+2Y(A+F)+2Z(B+D) \ &\quad +(A^2+B^2+C^2+D^2+E^2+F^2) \le 3 . \end{aligned}$$
The linear terms vanish, hence
$$2(X^2+Y^2+Z^2) + (A^2+B^2+C^2+D^2+E^2+F^2) \le 3.$$
Since the constant term is nonnegative,
$$2(X^2+Y^2+Z^2)\le3.$$
Therefore
$$X^2+Y^2+Z^2\le\frac32.$$
Recalling that
$$X=x-\alpha,\qquad Y=y-\beta,\qquad Z=z-\gamma,$$
every point of the intersection of the three cylinders lies inside the sphere centered at
$$\left(\frac{c+e}{2},\frac{a+f}{2},\frac{b+d}{2}\right)$$
with radius
$$\sqrt{\frac32}.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the representation of the cylinders. Since the axes are pairwise perpendicular, a rigid motion of space sends them to lines parallel to the coordinate axes. The distance from $(x,y,z)$ to the line $L_x$ equals
$$\sqrt{(y-a)^2+(z-b)^2},$$
which gives inequality (1), and similarly for the other two cylinders.
The second delicate step is the cancellation of linear terms. After translation, the coefficient of $X$ in the sum of the three inequalities is
$$2(C+E).$$
Substituting
$$C=\alpha-c,\qquad E=\alpha-e,$$
gives
$$C+E=2\alpha-(c+e).$$
Choosing
$$\alpha=\frac{c+e}{2}$$
makes this coefficient exactly zero. The same computation for $Y$ and $Z$ yields the other two cancellations. Any error in these midpoint choices would leave residual linear terms and destroy the argument.
The final delicate step is extracting the radius bound. After summation,
$$2(X^2+Y^2+Z^2)+K\le3,$$
where
$$K=A^2+B^2+C^2+D^2+E^2+F^2\ge0.$$
Discarding the nonnegative quantity $K$ gives
$$2(X^2+Y^2+Z^2)\le3.$$
This implication is valid because removing a nonnegative term from the left-hand side weakens the inequality.
Alternative Approaches
A more geometric proof interprets each cylinder inequality as a bound on the squared distance from a point to one of the coordinate directions. After choosing the same center
$$O=\left(\frac{c+e}{2},\frac{a+f}{2},\frac{b+d}{2}\right),$$
one can express the three inequalities directly in terms of the vector $\mathbf r=(X,Y,Z)$. Summing the squared distances to the three coordinate axes yields
$$2|\mathbf r|^2$$
plus a nonnegative correction depending only on the relative positions of the axes. The same estimate
$$2|\mathbf r|^2\le3$$
follows immediately.
The coordinate computation is preferable because it makes the cancellation mechanism completely explicit and reduces the proof to a single algebraic summation.