Kvant Math Problem 926
The given conditions state that the vectors $(x,y)$ and $(u,v)$ both lie on the unit circle and are orthogonal with respect to the standard dot product since $xu+yv=0$.
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Verdicts: SKIP + SKIP
Solve time: 7m06s
Source on kvant.digital
Problem
Prove that if $$x^2+y^2=u^2+v^2=1,\quad xu+yv=0,$$ then $$x^2+u^2=y^2+v^2=1,\quad xy+uv=0.$$
S. V. Duzhin
Exploration
The given conditions state that the vectors $(x,y)$ and $(u,v)$ both lie on the unit circle and are orthogonal with respect to the standard dot product since $xu+yv=0$. This means we are dealing with two unit vectors in $\mathbb{R}^2$ that form an orthonormal pair.
Geometrically, once one unit vector is fixed, the second must be a unit vector perpendicular to it, hence determined up to a sign. If $(x,y)$ is fixed, then $(u,v)$ must lie on the line perpendicular to it through the origin, intersected with the unit circle, giving exactly two antipodal possibilities.
The desired conclusion swaps the roles of coordinates: it asserts that the column vectors $(x,u)$ and $(y,v)$ are also orthonormal. This suggests a hidden symmetry of orthogonal matrices: a $2\times 2$ matrix with orthonormal rows also has orthonormal columns.
The most direct algebraic path is to parametrize $(x,y)$ by an angle and then force $(u,v)$ using orthogonality and normalization, checking that both cases of sign choice still produce the same final identities.
Problem Understanding
This is a Type B problem, since we must prove that a given set of conditions implies another set of identities.
We are given two unit vectors $(x,y)$ and $(u,v)$ that are orthogonal. The goal is to prove that the vectors formed by swapping coordinates across these pairs, namely $(x,u)$ and $(y,v)$, are also orthonormal. The core difficulty is to transfer orthogonality from rows of a $2\times 2$ matrix to its columns without assuming any matrix theory in advance.
Proof Architecture
The first step is to represent the unit vector $(x,y)$ as $(\cos\theta,\sin\theta)$ for some real $\theta$, justified by the unit circle parametrization.
Next, we express $(u,v)$ using the orthogonality condition $xu+yv=0$, which forces $(u,v)$ to lie on the line perpendicular to $(x,y)$, hence of the form $(-\sin\theta,\cos\theta)$ up to a sign factor $t\in{\pm1}$, which is then determined using $u^2+v^2=1$.
We then compute $x^2+u^2$ and $y^2+v^2$ directly from these forms and verify both equal $1$ in all sign cases.
Finally, we compute $xy+uv$ and verify it equals $0$ in both sign cases.
The most delicate point is the justification that $(u,v)$ must be exactly $\pm(-\sin\theta,\cos\theta)$ rather than some other parametrization; this follows from solving a single linear constraint under a quadratic normalization condition.
Solution
From $x^2+y^2=1$, there exists a real number $\theta$ such that
$$x=\cos\theta,\quad y=\sin\theta.$$
This follows from the parametrization of the unit circle, since every point $(x,y)$ with $x^2+y^2=1$ corresponds to some angle $\theta$.
The condition $xu+yv=0$ becomes
$$\cos\theta,u+\sin\theta,v=0.$$
This is a linear equation in $u,v$, which implies that $(u,v)$ is orthogonal to $(\cos\theta,\sin\theta)$. A vector perpendicular to $(\cos\theta,\sin\theta)$ is proportional to $(-\sin\theta,\cos\theta)$, hence there exists a real number $t$ such that
$$u=t(-\sin\theta),\quad v=t(\cos\theta).$$
Using $u^2+v^2=1$, we obtain
$$u^2+v^2=t^2(\sin^2\theta+\cos^2\theta)=t^2.$$
Since $u^2+v^2=1$, it follows that $t^2=1$, hence $t=\pm 1$.
Thus we have exactly two possibilities:
$$(u,v)=(-\sin\theta,\cos\theta)\quad \text{or}\quad (u,v)=(\sin\theta,-\cos\theta).$$
We now compute $x^2+u^2$. In the first case,
$$x^2+u^2=\cos^2\theta+\sin^2\theta=1.$$
In the second case,
$$x^2+u^2=\cos^2\theta+\sin^2\theta=1$$
as well. Hence in all cases $x^2+u^2=1$.
Similarly,
$$y^2+v^2=\sin^2\theta+\cos^2\theta=1$$
in the first case, and the same expression holds in the second case, so $y^2+v^2=1$.
Finally, we compute $xy+uv$. In the first case,
$$xy+uv=\cos\theta\sin\theta+(-\sin\theta)(\cos\theta)=0.$$
In the second case,
$$xy+uv=\cos\theta\sin\theta+(\sin\theta)(-\cos\theta)=0.$$
Thus $xy+uv=0$ in all cases.
This establishes both identities.
This completes the proof. ∎
Verification of Key Steps
The key structural step is the representation of $(u,v)$ as a scalar multiple of $(-\sin\theta,\cos\theta)$. This follows because the orthogonality condition $\cos\theta,u+\sin\theta,v=0$ defines a one-dimensional subspace of $\mathbb{R}^2$, and any nonzero vector orthogonal to $(\cos\theta,\sin\theta)$ must lie in that subspace. Substituting $(u,v)=(a,b)$ into the equation yields $a\cos\theta+b\sin\theta=0$, which is solved by $(a,b)=t(-\sin\theta,\cos\theta)$ directly by inspection.
The normalization step $u^2+v^2=1$ forces $t^2=1$ because $(-\sin\theta,\cos\theta)$ already has unit length. This eliminates all other scaling possibilities and ensures only the two sign choices remain.
The final computations are stable under both sign choices because all resulting expressions depend only on squares or symmetric cancellation between terms.
Alternative Approaches
A more structural approach interprets the pair of vectors as the rows of a $2\times 2$ matrix
$$A=\begin{pmatrix}x & y\ u & v\end{pmatrix}.$$
The assumptions state that the rows of $A$ form an orthonormal system, which is equivalent to $AA^{\mathsf T}=I$. From this identity, one deduces $A^{\mathsf T}A=I$, which is precisely the statement that the columns are orthonormal, yielding the desired identities immediately.
The trigonometric parametrization used above is preferable in a school-level context because it avoids matrix theory while remaining fully explicit and elementary.