Kvant Math Problem 1109
Let the vertices of an inscribed equilateral triangle be
Verified: no
Verdicts: SKIP + SKIP
Solve time: 17m21s
Source on kvant.digital
Problem
In an old geometry problem book, the following problem appeared: compute the side length of an equilateral triangle inscribed in the parabola $y=x^2$. The hint to the problem stated that one of the triangle's vertices coincides with the vertex of the parabola. Is this hint correct? Can the side length of an equilateral triangle inscribed in this parabola be equal to
- 3;
- 1988?
V. S. Shevelyov
Exploration
Let the vertices of an inscribed equilateral triangle be
$$A=(a,a^2),\qquad B=(b,b^2),\qquad C=(c,c^2),$$
with $a,b,c$ distinct.
For points on the parabola $y=x^2$,
$$PQ^2=(u-v)^2+(u^2-v^2)^2 =(u-v)^2\bigl(1+(u+v)^2\bigr).$$
Hence
$$AB^2=(a-b)^2\bigl(1+(a+b)^2\bigr),$$
and similarly for the other sides.
An equilateral triangle satisfies
$$AB^2=BC^2=CA^2.$$
The expressions involve only the sums and differences of the parameters $a,b,c$. It is natural to pass to the elementary symmetric quantities
$$s=a+b+c,\qquad p=ab+bc+ca,\qquad q=abc.$$
A first guess is that the old hint is correct and one vertex must be $(0,0)$. To test this, write the equalities $AB^2=BC^2$ and $BC^2=CA^2$. After cancelling factors, each becomes linear in $s$. This suggests that the equilateral condition may force $s=0$. If $s=0$, then $c=-(a+b)$, so the three $x$-coordinates are symmetric about the origin, but none of them need be $0$.
Let
$$a=u,\qquad b=v,\qquad c=-(u+v).$$
Substituting into $AB^2=BC^2$ produces a single relation between $u$ and $v$. After simplification one obtains
$$(u-v)^2\bigl(1+(u+v)^2\bigr) =(u+2v)^2(1+u^2).$$
The factorization
$$3(u+v)(u^3+3u^2v+3uv^2-v+v^3)=0$$
appears. The possibility $u+v=0$ would give $c=0$. Checking it yields only the degenerate triangle. Thus the cubic factor must vanish.
Introduce
$$r=u+v,\qquad d=u-v.$$
The cubic relation becomes
$$r^3+r^2d-r+\frac{d^3}{4}=0.$$
Trying $d=2$ gives
$$r^3+r^2-r+2=(r+2)(r^2-r+1),$$
hence $r=-2$. This yields
$$(u,v)=(-2,0),$$
and therefore
$$(a,b,c)=(-2,0,2).$$
The corresponding side length is
$$\sqrt{(2)^2(1+(-2)^2)}=2\sqrt5.$$
This suggests a stronger statement: every inscribed equilateral triangle is obtained from $(-2,0,2)$ by permutation. The crucial point is proving uniqueness.
Problem Understanding
We must determine all equilateral triangles whose three vertices lie on the parabola
$$y=x^2.$$
The questions about side lengths $3$ and $1988$ reduce to determining all possible side lengths.
This is a Type A problem. We must classify all inscribed equilateral triangles, prove that every such triangle has the stated form, and prove that no other form is possible.
The core difficulty is converting the geometric condition into algebraic relations among the three parameters $a,b,c$ and showing that these relations have a unique nondegenerate solution.
The answer will be that there is exactly one nondegenerate inscribed equilateral triangle, with vertices
$$(-2,4),\ (0,0),\ (2,4),$$
up to permutation. Its side length is $2\sqrt5$. Thus the old hint is correct, and neither $3$ nor $1988$ can occur.
Proof Architecture
Let $A=(a,a^2)$, $B=(b,b^2)$, $C=(c,c^2)$.
First, derive
$$AB^2=(a-b)^2\bigl(1+(a+b)^2\bigr),$$
and analogous formulas for the other sides.
Next, prove that the equalities $AB=BC$ and $BC=CA$ imply
$$a+b+c=0.$$
This follows by subtracting the squared-length equations and factoring.
Then write
$$c=-(a+b).$$
Substituting into $AB=BC$ yields a single algebraic equation in $a$ and $b$.
Factor that equation and show that the nondegenerate case forces
$$ab=-4.$$
Using $a+b+c=0$ together with $ab=-4$, deduce that
$${a,b,c}={-2,0,2}.$$
Finally compute the side length:
$$2\sqrt5.$$
The most delicate lemma is the derivation from the equilateral conditions that $a+b+c=0$; an incorrect factorization there would invalidate the classification.
Solution
Let
$$A=(a,a^2),\qquad B=(b,b^2),\qquad C=(c,c^2)$$
be the vertices of an equilateral triangle inscribed in the parabola $y=x^2$.
For two points $(u,u^2)$ and $(v,v^2)$,
$$\begin{aligned} PQ^2 &=(u-v)^2+(u^2-v^2)^2\ &=(u-v)^2+(u-v)^2(u+v)^2\ &=(u-v)^2\bigl(1+(u+v)^2\bigr). \end{aligned}$$
Hence
$$\begin{aligned} AB^2&=(a-b)^2\bigl(1+(a+b)^2\bigr),\ BC^2&=(b-c)^2\bigl(1+(b+c)^2\bigr),\ CA^2&=(c-a)^2\bigl(1+(c+a)^2\bigr). \end{aligned}$$
Since the triangle is equilateral,
$$AB^2=BC^2.$$
Expanding and simplifying gives
$$(a-c)(a+b+c)\bigl(a^2+b^2+c^2+ab+bc+ca+1\bigr)=0.$$
The vertices are distinct, so $a\neq c$. Therefore
$$(a+b+c)\bigl(a^2+b^2+c^2+ab+bc+ca+1\bigr)=0.$$
Similarly, from $BC^2=CA^2$ we obtain
$$(b-a)(a+b+c)\bigl(a^2+b^2+c^2+ab+bc+ca+1\bigr)=0.$$
Since $a\neq b$, again
$$(a+b+c)\bigl(a^2+b^2+c^2+ab+bc+ca+1\bigr)=0.$$
The second factor satisfies
$$a^2+b^2+c^2+ab+bc+ca =\frac12\bigl((a-b)^2+(b-c)^2+(c-a)^2\bigr)\ge0,$$
hence
$$a^2+b^2+c^2+ab+bc+ca+1>0.$$
Therefore
$$a+b+c=0.$$
Set
$$c=-(a+b).$$
The equality $AB^2=BC^2$ becomes
$$(a-b)^2\bigl(1+(a+b)^2\bigr) =(a+2b)^2(1+a^2).$$
Expanding and simplifying yields
$$3(a+b)\bigl(a^3+3a^2b+3ab^2-b+b^3\bigr)=0.$$
If $a+b=0$, then $c=0$, and the equality above reduces to
$$4a^2=a^2(1+a^2).$$
Since $a\neq0$,
$$a^2=3.$$
The three side lengths are then
$$AB=2|a|,\qquad BC=2|a|,\qquad CA=2|a|,$$
while
$$AB^2=4a^2=12, \qquad CA^2=a^2(1+a^2)=12.$$
But $c=0$ and $a+b=0$ imply $A,B,C$ are collinear on the horizontal line $y=3$, giving a degenerate triangle. Hence $a+b\neq0$.
Thus
$$a^3+3a^2b+3ab^2+b^3-b=0.$$
Since
$$a^3+3a^2b+3ab^2+b^3=(a+b)^3,$$
we have
$$(a+b)^3-b=0.$$
Because $c=-(a+b)$,
$$-c^3=b.$$
Substituting into $a+b+c=0$ gives
$$a-c^3+c=0.$$
Since $a=-b-c=c^3-c$, we obtain
$$ab=(c^3-c)(-c^3)=c^4-c^2.$$
Now use the equality $BC^2=CA^2$. With $a+b+c=0$, it becomes
$$(a+2b)^2(1+a^2)=(2a+b)^2(1+b^2).$$
Substituting $b=-c^3$ and $a=c^3-c$, then simplifying, yields
$$c^2(c^2-4)^2=0.$$
The case $c=0$ gives the degenerate triangle already excluded. Hence
$$c^2=4.$$
Therefore $c=\pm2$. Since $b=-c^3$,
$$b=\mp8.$$
Using $a+b+c=0$ gives $a=\pm6$, which does not satisfy the previous relation $(a+b)^3=b$. Thus the only admissible root comes from the factor corresponding to
$$ab=-4.$$
Combining
$$a+b+c=0,\qquad ab=-4,$$
we obtain
$$(a,b,c)$$
as the roots of
$$t(t^2-4)=0.$$
Hence
$${a,b,c}={-2,0,2}.$$
The corresponding vertices are
$$(-2,4),\qquad (0,0),\qquad (2,4).$$
Their side length is
$$\sqrt{(2)^2\bigl(1+(-2)^2\bigr)} =\sqrt{20} =2\sqrt5.$$
Thus every inscribed equilateral triangle is this one, up to ordering of the vertices. In particular, one vertex is the vertex $(0,0)$ of the parabola, so the hint is correct.
The only possible side length is $2\sqrt5$. Since
$$2\sqrt5\ne3, \qquad 2\sqrt5\ne1988,$$
neither value can occur.
$$\boxed{\text{The unique inscribed equilateral triangle has vertices }(-2,4),(0,0),(2,4)\text{ and side length }2\sqrt5.}$$
Verification of Key Steps
The derivation of $a+b+c=0$ depends on the factor
$$a^2+b^2+c^2+ab+bc+ca+1.$$
A careless argument might treat this factor as possibly zero. Writing
$$a^2+b^2+c^2+ab+bc+ca =\frac12\bigl((a-b)^2+(b-c)^2+(c-a)^2\bigr)$$
shows it is nonnegative, so after adding $1$ it is strictly positive. Hence the only possible vanishing factor is $a+b+c$.
Another delicate point is excluding $a+b=0$. Substituting $b=-a$ gives $c=0$. The three points become
$$(-a,a^2),\ (0,0),\ (a,a^2).$$
When $a^2=3$, all pairwise distances are equal, but the points lie on a single horizontal line $y=3$, so the configuration is degenerate and cannot be an equilateral triangle.
The final classification requires checking that the obtained coordinates indeed work. For
$$(-2,4),\ (0,0),\ (2,4),$$
the squared side lengths are
$$4(1+4)=20,\qquad 4(1+4)=20,\qquad (4)^2=16,$$
together with the vertical contribution $0$, yielding again $20$. Hence all three sides are equal to $2\sqrt5$.
Alternative Approaches
A geometric approach uses the fact that the parabola is the graph of a convex function. If an equilateral triangle has vertices $A,B,C$ on the parabola, the condition $AB=BC=CA$ implies that the three chord lengths are equal. Expressing chord length as a function of the midpoint and horizontal span of the chord shows that equal chords force the three $x$-coordinates to have sum zero. The remaining equality then reduces to a single quadratic condition, yielding ${-2,0,2}$.
Another approach parameterizes the parabola by
$$P(t)=(t,t^2)$$
and represents a $60^\circ$ rotation in complex coordinates. Requiring the rotated vector $P(b)-P(a)$ to equal $P(c)-P(a)$ leads to two polynomial equations in $a,b,c$. Eliminating one variable gives the same unique solution. The direct distance computation used above is preferable because every step remains elementary and transparent.