Kvant Math Problem 462

Let the apex of the regular square pyramid be $S$, and let the base square be $ABCD$ with center $O$.

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Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 1m46s
Source on kvant.digital

Problem

A plane intersects the lateral edges of a regular square pyramid at points located at distances $a$, $b$, $c$, and $d$ from the apex (see Fig. 1). Prove that $$\dfrac1a+\dfrac1c=\dfrac1b+\dfrac1d.$$

Figure 1

K. Shvartsman

Exploration

Let the apex of the regular square pyramid be $S$, and let the base square be $ABCD$ with center $O$. The lateral edges are the segments $SA$, $SB$, $SC$, $SD$. A plane intersects these edges at points $A_1,B_1,C_1,D_1$ such that $SA_1=a$, $SB_1=b$, $SC_1=c$, $SD_1=d$.

Because the pyramid is regular, all lateral edges have the same length, so each intersection point divides its edge in a proportional parameter from $S$. This suggests writing each point as an affine combination of $S$ and the corresponding base vertex.

The key structure is that all four intersection points lie in a single plane, so a single linear functional vanishes on all of them. This converts geometric collinearity into linear relations among the parameters along the four edges. The symmetry of the square should force a cancellation between opposite vertices.

The most delicate point is to express the condition of coplanarity in a way that makes the symmetry between opposite vertices $A,C$ and $B,D$ manifest, rather than relying on coordinate computation.

Problem Understanding

This is a Type B problem. One must prove the identity

$\frac{1}{a}+\frac{1}{c}=\frac{1}{b}+\frac{1}{d}$

for the distances from the apex of a regular square pyramid to the intersection points of a plane with its four lateral edges.

The essential difficulty is to translate a geometric coplanarity condition into a linear relation among reciprocals of segment lengths along symmetric edges. The symmetry of the square base relative to its center is expected to enforce the equality between sums over opposite vertices.

Proof Architecture

Let $S$ be the apex, and let $A,B,C,D$ be the vertices of the square base with center $O$.

First lemma establishes that each intersection point on a lateral edge can be written uniquely as $S+tX$ where $X\in{A,B,C,D}$ and $t$ is proportional to the given distance from $S$.

Second lemma expresses coplanarity of $A_1,B_1,C_1,D_1$ as the existence of a linear functional $f$ such that $f(S+t_X X)=0$ for each vertex $X$.

Third lemma reduces this condition to a linear relation among the values $f(A),f(B),f(C),f(D)$.

Fourth lemma uses the geometry of a square centered at $O$ to prove $f(A)+f(C)=f(B)+f(D)$.

The hardest point is the transition from coplanarity to the linear relation on vertex values and ensuring the correct handling of affine shifts from $S$.

Solution

Let $S$ be the apex and let $A,B,C,D$ be the vertices of the square base, ordered cyclically so that $AC$ and $BD$ are diagonals. Let $O$ be the center of the square.

Because the pyramid is regular, all lateral edges have equal length. Denote this common length by $L$. For each vertex $X\in{A,B,C,D}$, the point on $SX$ at distance $x\in{a,b,c,d}$ from $S$ is given by

$X_1 = S + \frac{x}{L}(X-S).$

Let $t_A=\frac{a}{L}$, $t_B=\frac{b}{L}$, $t_C=\frac{c}{L}$, $t_D=\frac{d}{L}$. Then

$A_1 = (1-t_A)S+t_A A,\quad B_1=(1-t_B)S+t_B B,\quad C_1=(1-t_C)S+t_C C,\quad D_1=(1-t_D)S+t_D D.$

All four points lie in one plane. Therefore there exists a nonzero linear functional $f$ on space such that

$f(A_1)=f(B_1)=f(C_1)=f(D_1).$

Since only equality matters, we may subtract this common value and assume

$f(A_1)=f(B_1)=f(C_1)=f(D_1)=0.$

Expanding using linearity gives

$t_A f(A)+(1-t_A)f(S)=0,$

and similarly for $B,C,D$. Rearranging yields

$\frac{1}{t_A}f(S)+f(A)=0,\quad \frac{1}{t_B}f(S)+f(B)=0,\quad \frac{1}{t_C}f(S)+f(C)=0,\quad \frac{1}{t_D}f(S)+f(D)=0.$

Assuming $f(S)\neq 0$ (otherwise all four $f(X)=0$ would force degeneracy of the plane, impossible since it intersects edges in distinct interior points), we obtain

$\frac{1}{t_A}=-\frac{f(A)}{f(S)},\quad \frac{1}{t_B}=-\frac{f(B)}{f(S)},\quad \frac{1}{t_C}=-\frac{f(C)}{f(S)},\quad \frac{1}{t_D}=-\frac{f(D)}{f(S)}.$

Thus

$\frac{1}{a}=\frac{1}{Lt_A}=-\frac{f(A)}{L f(S)},\quad \frac{1}{b}=-\frac{f(B)}{L f(S)},\quad \frac{1}{c}=-\frac{f(C)}{L f(S)},\quad \frac{1}{d}=-\frac{f(D)}{L f(S)}.$

Multiplying the desired identity by $-Lf(S)$ reduces it to

$f(A)+f(C)=f(B)+f(D).$

Now express vertices relative to the center $O$. Write

$A=O+\vec{u},\quad B=O+\vec{v},\quad C=O-\vec{u},\quad D=O-\vec{v}.$

Applying linearity of $f$ gives

$f(A)=f(O)+f(\vec{u}),\quad f(C)=f(O)-f(\vec{u}),$

so

$f(A)+f(C)=2f(O).$

Similarly,

$f(B)+f(D)=f(O)+f(\vec{v})+f(O)-f(\vec{v})=2f(O).$

Hence

$f(A)+f(C)=f(B)+f(D),$

which implies

$\frac{1}{a}+\frac{1}{c}=\frac{1}{b}+\frac{1}{d}.$

This completes the proof. ∎

Verification of Key Steps

The first delicate step is the reduction from coplanarity of $A_1,B_1,C_1,D_1$ to the existence of a linear functional $f$ vanishing on all four points after normalization. This holds because any plane in affine space is the zero set of an affine linear functional, and subtracting a constant shifts it to a linear functional on the vector space of position vectors.

The second point is the conversion from $f(A_1)=0$ to a relation involving $1/t_A$. The equation $f((1-t_A)S+t_AA)=0$ expands linearly to $(1-t_A)f(S)+t_A f(A)=0$, and dividing by $t_A$ is legitimate since the intersection points lie strictly inside edges, so $t_A\neq 0$.

The third point is the symmetry identity $f(A)+f(C)=f(B)+f(D)$. This depends only on the decomposition of opposite vertices as reflections through the center $O$, which guarantees cancellation of opposite displacement vectors under any linear functional.

Alternative Approaches

A coordinate approach places the base square in the plane $z=0$ with center at the origin and apex at $(0,0,h)$, then expresses intersection points using barycentric coordinates along each edge. The plane equation becomes $px+qy+rz=s$, leading directly to proportionality relations for reciprocals of parameters. This method is computational but obscures the geometric origin of the symmetry $A+C=B+D$ in vector form, which is the structural reason the identity holds.