Kvant Math Problem 1172
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Problem
What is the greatest possible angle between the segments $OA$ and $OB$, drawn from the origin $O$ of a rectangular coordinate system in space, if the point $A$ has coordinates $(x, y, z)$ and the point $B$ has coordinates $(y, z, x)$?
S. N. Bychkov
Exploration
Let
$$\vec a=(x,y,z),\qquad \vec b=(y,z,x).$$
The angle between $OA$ and $OB$ is the angle between $\vec a$ and $\vec b$.
Its cosine equals
$$\cos\theta=\frac{\vec a\cdot\vec b}{|\vec a|,|\vec b|}.$$
Since
$$|\vec a|^2=x^2+y^2+z^2=|\vec b|^2,$$
we have
$$\cos\theta=\frac{xy+yz+zx}{x^2+y^2+z^2}.$$
To maximize the angle, we must minimize this quotient.
A first guess is that the quotient might become arbitrarily close to $-1$. Testing examples quickly shows otherwise. For $(1,-1,0)$ we obtain
$$\frac{-1}{2}.$$
For $(1,1,-2)$ we obtain
$$\frac{-3}{6}=-\frac12.$$
For $(1,-2,1)$ we again obtain $-\frac12$.
This suggests that perhaps
$$xy+yz+zx\ge -\frac12(x^2+y^2+z^2).$$
Checking,
$$(x-y)^2+(y-z)^2+(z-x)^2 =2(x^2+y^2+z^2)-2(xy+yz+zx).$$
Hence
$$3(x^2+y^2+z^2) = (x-y)^2+(y-z)^2+(z-x)^2 +2(x^2+y^2+z^2)+2(xy+yz+zx),$$
which is not the desired inequality directly.
A better identity is
$$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\ge0.$$
Therefore
$$xy+yz+zx\ge-\frac12(x^2+y^2+z^2).$$
The bound is attained when
$$x+y+z=0.$$
Then
$$\cos\theta\ge-\frac12,$$
so
$$\theta\le120^\circ.$$
Equality occurs whenever $x+y+z=0$ and $(x,y,z)\ne(0,0,0)$, for example $(1,-1,0)$.
The crucial point is proving that the quotient cannot be smaller than $-1/2$ and that equality is attainable.
Problem Understanding
We are given two vectors in space,
$$\vec a=(x,y,z),\qquad \vec b=(y,z,x),$$
where $\vec b$ is obtained from $\vec a$ by a cyclic permutation of coordinates. We must determine the largest possible angle between these vectors.
This is a Type C problem. We must find the maximum value of the angle and prove that no larger angle can occur.
The answer should be $120^\circ$. The reason is that the cosine of the angle simplifies to
$$\frac{xy+yz+zx}{x^2+y^2+z^2},$$
and the identity
$$(x+y+z)^2\ge0$$
forces this quantity to be at least $-1/2$.
Proof Architecture
Lemma 1. The cosine of the angle between $OA$ and $OB$ equals
$$\frac{xy+yz+zx}{x^2+y^2+z^2}.$$
This follows from the scalar product formula and the fact that the two vectors have the same length.
Lemma 2. For all real $x,y,z$,
$$xy+yz+zx\ge-\frac12(x^2+y^2+z^2).$$
This is equivalent to the nonnegativity of $(x+y+z)^2$.
Lemma 3. Equality in Lemma 2 holds exactly when
$$x+y+z=0.$$
This follows directly from the equality condition in $(x+y+z)^2\ge0$.
Using Lemmas 1 and 2, we obtain
$$\cos\theta\ge-\frac12.$$
Since cosine decreases on $[0,\pi]$, the angle satisfies $\theta\le120^\circ$. Lemma 3 provides examples attaining equality.
The most delicate point is verifying that the lower bound for the cosine is exactly $-1/2$ and that it is attained.
Solution
Let
$$\vec a=(x,y,z),\qquad \vec b=(y,z,x).$$
The angle $\theta$ between the segments $OA$ and $OB$ is the angle between the vectors $\vec a$ and $\vec b$.
Their scalar product is
$$\vec a\cdot\vec b=xy+yz+zx.$$
Also,
$$|\vec a|^2=x^2+y^2+z^2,$$
and
$$|\vec b|^2=y^2+z^2+x^2=x^2+y^2+z^2.$$
Hence $|\vec a|=|\vec b|$, and therefore
$$\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a|,|\vec b|} = \frac{xy+yz+zx}{x^2+y^2+z^2}.$$
Now
$$(x+y+z)^2\ge0.$$
Expanding gives
$$x^2+y^2+z^2+2(xy+yz+zx)\ge0.$$
Therefore
$$xy+yz+zx\ge-\frac12(x^2+y^2+z^2).$$
Substituting into the expression for $\cos\theta$, we obtain
$$\cos\theta = \frac{xy+yz+zx}{x^2+y^2+z^2} \ge -\frac12.$$
Since $\theta\in[0,\pi]$ and the cosine function is decreasing on this interval,
$$\theta\le\arccos!\left(-\frac12\right)=120^\circ.$$
It remains to show that this value is attained. Equality in the inequality above holds when
$$x+y+z=0.$$
Take, for example,
$$(x,y,z)=(1,-1,0).$$
Then
$$\vec a=(1,-1,0),\qquad \vec b=(-1,0,1),$$
and
$$\cos\theta=-\frac12.$$
Hence
$$\theta=120^\circ.$$
Thus the greatest possible angle is
$$\boxed{120^\circ}.$$
Equality holds whenever $(x,y,z)\ne(0,0,0)$ and $x+y+z=0$.
Verification of Key Steps
The first delicate step is the computation of the cosine. Since
$$|\vec a|^2=x^2+y^2+z^2, \qquad |\vec b|^2=y^2+z^2+x^2,$$
the lengths are equal. Forgetting this fact would leave an unnecessary denominator
$$\sqrt{x^2+y^2+z^2}\sqrt{y^2+z^2+x^2},$$
but these two factors are identical.
The second delicate step is the lower bound
$$xy+yz+zx\ge-\frac12(x^2+y^2+z^2).$$
Starting from
$$(x+y+z)^2\ge0$$
gives
$$x^2+y^2+z^2+2(xy+yz+zx)\ge0,$$
and rearrangement yields the desired inequality. The equality condition is exactly
$$x+y+z=0.$$
A numerical check confirms optimality. For $(1,-1,0)$,
$$\cos\theta=\frac{-1}{2},$$
so $\theta=120^\circ$. For $(1,1,-2)$,
$$\cos\theta=\frac{-3}{6}=-\frac12,$$
again giving $120^\circ$. These examples show that the bound is attained and cannot be improved.
Alternative Approaches
Using the identity
$$(x-y)^2+(y-z)^2+(z-x)^2\ge0,$$
we obtain
$$2(x^2+y^2+z^2)-2(xy+yz+zx)\ge0,$$
hence
$$xy+yz+zx\le x^2+y^2+z^2.$$
Combining this with
$$(x+y+z)^2\ge0$$
places the quotient
$$\frac{xy+yz+zx}{x^2+y^2+z^2}$$
in the interval
$$\left[-\frac12,1\right].$$
The lower endpoint again yields the maximal angle $120^\circ$.
Another viewpoint is to regard the cyclic permutation
$$(x,y,z)\mapsto(y,z,x)$$
as a linear transformation. The vector $(1,1,1)$ is fixed, while every vector orthogonal to $(1,1,1)$ is rotated by $120^\circ$ in the plane $x+y+z=0$. Choosing $\vec a$ in that plane gives an angle of exactly $120^\circ$ between $\vec a$ and its image, which immediately identifies the maximum angle. The coordinate computation above is more elementary and requires only standard algebraic identities.