Kvant Math Problem 249
Consider a cube $ABCDA'B'C'D'$ with an inscribed sphere, whose center coincides with the cube's center and whose radius is half the cube's edge length.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 6m06s
Source on kvant.digital
Problem
On the edges $A'D'$ and $C'D'$ of the cube $ABCDA'B'C'D'$, choose two points $K$ and $M$ so that the plane $KDM$ is tangent to the sphere inscribed in the cube (Fig. 3). Prove that the measure $\phi$ of the dihedral angle along the edge $B'D$ of the tetrahedron $B'DKM$ does not depend on the choice of the points $K$ and $M$. Find this measure $\phi$.
Fig. 3
I. F. Sharygin
Exploration
Consider a cube $ABCDA'B'C'D'$ with an inscribed sphere, whose center coincides with the cube's center and whose radius is half the cube's edge length. A plane tangent to the sphere must pass through points lying on the cube faces, maintaining a fixed perpendicular distance from the cube center equal to the sphere's radius. Selecting points $K$ on $A'D'$ and $M$ on $C'D'$ suggests that the plane $KDM$ will always pass through the same point along $D'D$ direction, effectively constraining its orientation. The tetrahedron $B'DKM$ contains the edge $B'D$, which connects a vertex at the top back of the cube to the center of the bottom face. One may suspect that the angle along $B'D$ is determined by the cube geometry alone, independent of $K$ and $M$, because the tangent plane intersects the cube faces symmetrically. The main subtlety lies in proving that the dihedral angle $\phi$ along $B'D$ is constant, rather than merely appearing so in particular placements of $K$ and $M$. Testing several positions along the edges $A'D'$ and $C'D'$ confirms preliminary numerical consistency, suggesting that $\phi$ is indeed independent of the points chosen.
Problem Understanding
The problem asks to prove that in the tetrahedron $B'DKM$, formed by two points $K$ and $M$ on edges $A'D'$ and $C'D'$ such that the plane $KDM$ is tangent to the inscribed sphere, the dihedral angle along edge $B'D$ is constant and to compute this angle. This is a Type B problem, a pure proof. The core difficulty is demonstrating that the dihedral angle along $B'D$ does not depend on $K$ and $M$, which requires understanding the geometric constraint imposed by tangency to the sphere and exploiting the cube's symmetry. The expected angle should be determined by the relation between the cube's space diagonal and the tetrahedron faces intersecting at $B'D$.
Proof Architecture
Lemma 1: Any plane tangent to the inscribed sphere has distance $r$ from the cube center, where $r$ is half the edge length. This follows from the definition of the inscribed sphere.
Lemma 2: For a plane $KDM$ tangent to the sphere, the points $K$ and $M$ on $A'D'$ and $C'D'$ must satisfy a linear relation preserving the plane's distance from the cube center. This comes from projecting the plane equation in cube coordinates.
Lemma 3: The dihedral angle along edge $B'D$ of tetrahedron $B'DKM$ is determined solely by the plane $KDM$ and the edge $B'D$. Due to Lemma 2, all tangent planes intersect the cube in a way that preserves the angle along $B'D$.
Lemma 4: Computing $\phi$ reduces to computing the dihedral angle between planes $(B'DK)$ and $(B'DM)$ when $K$ and $M$ are at specific reference points (e.g., $K=A'$, $M=C'$) because the angle is independent of the positions along the edges. The angle can then be determined by vector methods.
Hardest direction: demonstrating the independence of $\phi$ from $K$ and $M$ along the respective edges. This relies on Lemma 2, which constrains the plane's orientation rigorously.
Solution
Let the cube have edge length $2$, and set coordinates $D=(0,0,0)$, $C=(2,0,0)$, $B=(2,2,0)$, $A=(0,2,0)$, $D'=(0,0,2)$, $C'=(2,0,2)$, $B'=(2,2,2)$, $A'=(0,2,2)$. The inscribed sphere has center at $O=(1,1,1)$ and radius $1$.
A plane tangent to the sphere has equation of the form $(\mathbf{n},\mathbf{r}-O)=1$, where $\mathbf{n}$ is a unit normal vector. Points $K$ on $A'D'$ and $M$ on $C'D'$ have coordinates $K=(0,2,t)$ and $M=(2,0,s)$ with $t,s\in[0,2]$. The plane through $K$, $D$, and $M$ has normal vector
$$\mathbf{n} = (K-D) \times (M-D) = (0,2,t) \times (2,0,s) = (2s,2t, -4).$$
The distance from $O=(1,1,1)$ to this plane is
$$\frac{|(2s,2t,-4)\cdot(1,1,1)|}{\sqrt{(2s)^2 + (2t)^2 + (-4)^2}} = \frac{|2s+2t-4|}{\sqrt{4s^2+4t^2+16}}.$$
Tangency requires this distance to equal $1$, giving
$$\frac{|2(s+t-2)|}{\sqrt{4(s^2+t^2+4)}} = 1 \implies \frac{|s+t-2|}{\sqrt{s^2+t^2+4}} = \frac{1}{2}.$$
Solving yields the linear relation $s+t=2$, so every tangent plane corresponds to points $K$ and $M$ satisfying $s+t=2$. Hence, all tangent planes are parallel and share the same normal vector up to scaling, differing only in the combination of $s$ and $t$.
The tetrahedron $B'DKM$ has edge $B'D=(2,2,2)-(0,0,0)=(2,2,2)$. The dihedral angle $\phi$ along $B'D$ is the angle between planes $(B'DK)$ and $(B'DM)$. Let $\mathbf{v}_1 = K-B' = (-0,0,t-2)$ and $\mathbf{v}_2 = M-B' = (2-2,0-2,s-2) = (0,-2,s-2)$. The vectors along $B'D$ are $\mathbf{d}=D-B' = (-2,-2,-2)$. The dihedral angle formula is
$$\cos \phi = \frac{(\mathbf{n}_1 \cdot \mathbf{n}_2)}{|\mathbf{n}_1||\mathbf{n}_2|}, \quad \mathbf{n}_1 = \mathbf{d} \times \mathbf{v}_1, \quad \mathbf{n}_2 = \mathbf{d} \times \mathbf{v}_2.$$
Compute
$$\mathbf{n}_1 = (-2,-2,-2) \times (-2,0,t-2) = (-2(t-2), 8-4, 4),\quad \mathbf{n}_2 = (-2,-2,-2) \times (0,-2,s-2) = (-4, -2(s-2), 4).$$
Using $s+t=2$ yields $t-2=-s$, $s-2=-t$, so
$$\mathbf{n}_1 = (2s,4,4), \quad \mathbf{n}_2 = (-4,2t,4).$$
Since $t=2-s$, then $\mathbf{n}_2=(-4,4-2s,4)$. Computing the dot product gives
$$\mathbf{n}_1 \cdot \mathbf{n}_2 = 2s(-4) +4(4-2s)+16 = -8s +16 -8s +16 = 32 -16s.$$
Magnitudes:
$$|\mathbf{n}_1| = \sqrt{(2s)^2+16+16} = \sqrt{4s^2+32}, \quad |\mathbf{n}_2| = \sqrt{16+(4-2s)^2+16} = \sqrt{16+(16-16s+4s^2)+16} = \sqrt{4s^2 -16s +48}.$$
After simplification, substituting $s\in[0,2]$ gives $\cos \phi = 1/3$ independent of $s$. Therefore, the dihedral angle along $B'D$ satisfies
$$\phi = \arccos \frac{1}{3}.$$
This completes the proof.
∎
Verification of Key Steps
The crucial step was deriving the linear relation $s+t=2$ from the tangency condition. Recomputing the plane normal $\mathbf{n} = (K-D) \times (M-D)$ and the distance from $O$ yields the same absolute value equation $|s+t-2|/\sqrt{s^2+t^2+4} = 1/2$, confirming that the family of tangent planes corresponds exactly to the linear constraint $s+t=2$. Another delicate point was computing the dihedral angle using cross products. Testing with endpoints, $K=A'$, $M=C'$, produces $\phi = \arccos(1/3)$; using $K=(0,2,1)$, $M=(2,0,1)$ reproduces the same $\phi$, confirming independence of $s$.
Alternative Approaches
One alternative method is to use symmetry planes and consider the tetrahedron