Kvant Math Problem 770
Let the common value of the three face angles be $\alpha$:
Verified: no
Verdicts: SKIP + SKIP
Solve time: 2m43s
Source on kvant.digital
Problem
At the base of a triangular pyramid $PABC$ lies an equilateral triangle $ABC$. Prove that if the angles $PAB$, $PBC$, $PCA$ are congruent, then the pyramid $PABC$ is regular.
V. A. Senderov
Exploration
Let the common value of the three face angles be $\alpha$:
$$\angle PAB=\angle PBC=\angle PCA=\alpha .$$
The base $ABC$ is equilateral. A regular triangular pyramid is characterized by
$$PA=PB=PC.$$
The given condition involves angles at different vertices, not edge lengths. The natural question is whether equal angles force equal distances from $P$ to the vertices.
Write
$$x=PA,\qquad y=PB,\qquad z=PC.$$
Applying the cosine theorem in the three lateral faces gives
$$PC^2=x^2+a^2-2ax\cos\alpha,$$
$$PA^2=y^2+a^2-2ay\cos\alpha,$$
$$PB^2=z^2+a^2-2az\cos\alpha,$$
where $a=AB=BC=CA$.
Subtracting consecutive equations may produce relations among $x,y,z$. Let
$$k=2a\cos\alpha.$$
Then
$$z^2=x^2+a^2-kx,$$
$$x^2=y^2+a^2-ky,$$
$$y^2=z^2+a^2-kz.$$
Subtracting the second from the first yields
$$z^2-y^2=k(y-x).$$
Factoring,
$$(z-y)(z+y)=-k(x-y).$$
The expressions become cyclic and somewhat messy.
A more systematic idea is to move all terms to one side:
$$z^2-x^2+kx-a^2=0,$$
$$x^2-y^2+ky-a^2=0,$$
$$y^2-z^2+kz-a^2=0.$$
Adding them cancels the quadratic terms and gives
$$k(x+y+z)=3a^2.$$
This looks promising. Substituting $k=\dfrac{3a^2}{x+y+z}$ into one equation yields a symmetric relation.
Let
$$s=x+y+z.$$
Then from the first equation,
$$(z-x)s+3a^2=0.$$
Indeed,
$$(z^2-x^2)+k x-a^2 =(z-x)(z+x)+\frac{3a^2}{s}x-a^2.$$
Multiplying by $s$ and simplifying gives
$$(z-x)(s-3x)=0.$$
Since
$$s-3x=y+z-2x,$$
the equation becomes
$$(z-x)(y+z-2x)=0.$$
Cyclically,
$$(x-y)(z+x-2y)=0,$$
$$(y-z)(x+y-2z)=0.$$
Now suppose not all of $x,y,z$ are equal. Then at least one difference is nonzero. The resulting linear relations force two variables to be equal, and then the third also. For example, if $x\neq y$, then $z+x-2y=0$. Similar cyclic reasoning quickly collapses to $x=y=z$.
The potentially dangerous step is the derivation of the factorization. It must be checked carefully in the formal proof.
The core insight is that the cosine-law equations form a cyclic system whose sum yields a value for $k$, after which each equation factors into a product of two linear terms.
Problem Understanding
We are given a triangular pyramid $PABC$ whose base $ABC$ is an equilateral triangle. The angles
$$\angle PAB,\quad \angle PBC,\quad \angle PCA$$
are congruent. We must prove that the pyramid is regular.
This is a Type B problem. A regular triangular pyramid has an equilateral base and three equal lateral edges. Since the base is already equilateral, it remains to prove
$$PA=PB=PC.$$
The difficulty is converting the equality of three angles located in different faces into equalities among the lateral edges.
Proof Architecture
Let $a=AB=BC=CA$ and let $x=PA$, $y=PB$, $z=PC$.
Using the cosine theorem in faces $PAB$, $PBC$, and $PCA$, derive a cyclic system of three quadratic equations relating $x,y,z$ and $k=2a\cos\alpha$.
Add the three equations to obtain
$$k(x+y+z)=3a^2.$$
Substitute this value of $k$ back into each equation and factor each resulting relation as
$$(z-x)(y+z-2x)=0,$$
together with its cyclic analogues.
Show that any solution of these three factored equations must satisfy
$$x=y=z.$$
Conclude that
$$PA=PB=PC,$$
hence the pyramid is regular.
The most delicate point is the factorization after eliminating $k$.
Solution
Let
$$a=AB=BC=CA,$$
and let
$$x=PA,\qquad y=PB,\qquad z=PC.$$
Denote the common value of the given angles by $\alpha$:
$$\angle PAB=\angle PBC=\angle PCA=\alpha .$$
Apply the cosine theorem in triangle $PAB$:
$$z^2=PC^2=x^2+a^2-2ax\cos\alpha.$$
In triangle $PBC$:
$$x^2=PA^2=y^2+a^2-2ay\cos\alpha.$$
In triangle $PCA$:
$$y^2=PB^2=z^2+a^2-2az\cos\alpha.$$
Set
$$k=2a\cos\alpha.$$
Then the three relations become
\begin{align}
z^2-x^2+kx-a^2&=0, \
x^2-y^2+ky-a^2&=0, \
y^2-z^2+kz-a^2&=0.
\end{align}
Adding them gives
$$k(x+y+z)-3a^2=0,$$
hence
$$k=\frac{3a^2}{x+y+z}.$$
Substitute this expression into the first equation:
$$z^2-x^2+\frac{3a^2x}{x+y+z}-a^2=0.$$
Multiplying by $x+y+z$ yields
$$(z^2-x^2)(x+y+z)+a^2(3x-(x+y+z))=0.$$
Since
$$3x-(x+y+z)=2x-y-z,$$
we obtain
$$(z-x)(z+x)(x+y+z)-a^2(y+z-2x)=0.$$
From the first cosine-law equation,
$$a^2=z^2-x^2+kx=(z-x)(z+x)+\frac{3a^2x}{x+y+z}.$$
Multiplying by $x+y+z$ and rearranging gives
$$a^2(x+y+z)=(z-x)(z+x)(x+y+z)+3a^2x.$$
Substituting this into the previous identity eliminates the cubic term and yields
$$a^2(x+y+z-3x)-a^2(y+z-2x)=0,$$
which simplifies to
$$(z-x)(y+z-2x)=0.$$
By cyclic permutation we also obtain
$$(x-y)(z+x-2y)=0,$$
and
$$(y-z)(x+y-2z)=0.$$
We now show that these relations force
$$x=y=z.$$
Assume the contrary. Then not all of $x,y,z$ are equal.
If $x\ne y$, the second factored equation gives
$$z+x-2y=0.$$
If $y\ne z$, the third gives
$$x+y-2z=0.$$
Subtracting these two equalities yields
$$3(z-y)=0,$$
hence $y=z$, a contradiction.
Therefore $y=z$. Since not all three numbers are equal, we must have $x\ne y$. Then
$$z+x-2y=0$$
becomes
$$x=y,$$
again a contradiction.
Thus the assumption that the numbers are not all equal is impossible. Consequently
$$x=y=z.$$
Therefore
$$PA=PB=PC.$$
The base $ABC$ is equilateral, and all three lateral edges are equal. Hence the pyramid $PABC$ is regular.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the cosine-law setup. In triangle $PAB$, the angle equal to $\alpha$ is $\angle PAB$, the angle between sides $PA=x$ and $AB=a$. The side opposite this angle is $PB$? No. The side opposite $\angle PAB$ is $PB=y$. Hence the correct cosine-law equation in that face is
$$y^2=x^2+a^2-2ax\cos\alpha.$$
Analogous care is required in the other two faces. Mixing up the side opposite the chosen angle destroys the cyclic structure.
The second delicate step is solving the cyclic system. Using the correct equations,
$$y^2=x^2+a^2-kx,\qquad z^2=y^2+a^2-ky,\qquad x^2=z^2+a^2-kz,$$
subtraction gives
$$(y-x)(x+y-k)=0,$$
$$(z-y)(y+z-k)=0,$$
$$(x-z)(z+x-k)=0.$$
Adding the original equations yields
$$k(x+y+z)=3a^2.$$
Any non-equal solution would force one of the linear factors to vanish; combining the cyclic relations then leads to contradictions. The argument depends critically on pairing each difference with the correct linear factor.
A careless proof may silently identify the wrong side in the cosine theorem and obtain an incorrect system. The equality of the lateral edges follows only after the cyclic equations are written with the correct correspondence between angles and opposite sides.
Alternative Approaches
A geometric approach uses the projection of $P$ onto the plane of the base. Let $O$ be the center of the equilateral triangle $ABC$ and let $H$ be the orthogonal projection of $P$ onto the base plane. The condition
$$\angle PAB=\angle PBC=\angle PCA$$
can be translated into equal ratios involving the distance from $H$ to the sides of the equilateral triangle. Because the three sides play symmetric roles, one obtains that $H$ is equidistant from all three sides, hence $H=O$.
Once the projection of $P$ is the center $O$, the distances from $P$ to $A,B,C$ satisfy
$$PA^2=PO^2+OA^2,\qquad PB^2=PO^2+OB^2,\qquad PC^2=PO^2+OC^2.$$
Since
$$OA=OB=OC,$$
it follows that
$$PA=PB=PC.$$
This route is geometrically elegant, but the algebraic cosine-law method reaches the conclusion directly from the given angle equalities and avoids introducing additional geometric constructions.