Kvant Math Problem 587

Verified: no
Verdicts: SKIP + SKIP
Solve time: 3m28s
Source on kvant.digital

Problem

A triple of numbers 2, $\sqrt2$, $\dfrac1{\sqrt2}$ is given. It is allowed to replace any two of them with two numbers defined as follows: their sum divided by $\sqrt2$ and their difference divided by $\sqrt2$. Is it possible, by performing this procedure several times, to obtain the triple of numbers 1, $\sqrt2$, $1+\sqrt2$?

All-Russian School Mathematics Olympiad (1979, 9th grade)

Exploration

The operation replaces two numbers $x,y$ by

$$\frac{x+y}{\sqrt2},\qquad \frac{x-y}{\sqrt2}.$$

This is the standard orthogonal transformation given by the matrix

$$\frac1{\sqrt2} \begin{pmatrix} 1&1\ 1&-1 \end{pmatrix}.$$

A first invariant is the sum of squares:

$$\left(\frac{x+y}{\sqrt2}\right)^2+\left(\frac{x-y}{\sqrt2}\right)^2 = x^2+y^2.$$

For the initial triple,

$$2^2+(\sqrt2)^2+\left(\frac1{\sqrt2}\right)^2 =4+2+\frac12=\frac{13}2.$$

For the target triple,

$$1^2+(\sqrt2)^2+(1+\sqrt2)^2 =1+2+(3+2\sqrt2) =6+2\sqrt2.$$

These values are different, so the target cannot be reached.

Before accepting this, it is worth checking that the operation indeed leaves the third number unchanged and preserves the total sum of squares of the whole triple. Since only two coordinates are transformed and the identity above is exact, the invariant is global.

The only potentially hidden error would be using an invariant that is preserved for the chosen pair but not for the entire triple. The computation above avoids that issue.

Problem Understanding

We start with the triple

$$\left(2,\sqrt2,\frac1{\sqrt2}\right).$$

At each step we choose any two entries $x,y$ and replace them by

$$\frac{x+y}{\sqrt2},\qquad \frac{x-y}{\sqrt2}.$$

The question is whether, after finitely many such operations, we can obtain the triple

$$(1,\sqrt2,1+\sqrt2).$$

This is a Type B problem. We must prove or disprove the possibility of reaching the specified triple.

The core difficulty is to find a quantity that remains unchanged under every allowed operation and then compare its value for the initial and target triples.

Proof Architecture

The proof uses one lemma.

Lemma. For any real numbers $x,y$,

$$\left(\frac{x+y}{\sqrt2}\right)^2+\left(\frac{x-y}{\sqrt2}\right)^2 =x^2+y^2.$$

Sketch. Expand both squares and simplify.

From the lemma, the sum of the squares of all three numbers in the triple is invariant under every operation, because only two coordinates are changed.

The hardest point is verifying that this invariant is indeed preserved for the entire triple after each move.

Solution

Let

$$S=a^2+b^2+c^2$$

denote the sum of the squares of the three numbers currently present.

Suppose an operation is applied to the numbers $x$ and $y$. They are replaced by

$$x'=\frac{x+y}{\sqrt2},\qquad y'=\frac{x-y}{\sqrt2}.$$

Then

$$\begin{aligned} (x')^2+(y')^2 &= \left(\frac{x+y}{\sqrt2}\right)^2 + \left(\frac{x-y}{\sqrt2}\right)^2 \ &= \frac{(x+y)^2+(x-y)^2}{2} \ &= \frac{x^2+2xy+y^2+x^2-2xy+y^2}{2} \ &= x^2+y^2. \end{aligned}$$

Hence the contribution of the chosen pair to $S$ remains unchanged. The third number is not altered, so the entire quantity

$$S=a^2+b^2+c^2$$

is preserved by every allowed operation.

For the initial triple,

$$S_0 = 2^2+(\sqrt2)^2+\left(\frac1{\sqrt2}\right)^2 = 4+2+\frac12 = \frac{13}{2}.$$

For the proposed final triple,

$$\begin{aligned} S_1 &= 1^2+(\sqrt2)^2+(1+\sqrt2)^2 \ &= 1+2+\bigl(1+2\sqrt2+2\bigr) \ &= 6+2\sqrt2. \end{aligned}$$

Since

$$\frac{13}{2}\neq 6+2\sqrt2,$$

the invariant $S$ has different values for the initial and target triples.

Therefore the target triple cannot be obtained from the initial one by any sequence of the allowed operations.

This completes the proof.

∎

Verification of Key Steps

The invariant computation can be checked independently by writing

$$(x+y)^2+(x-y)^2=2x^2+2y^2.$$

Division by $2$ yields exactly $x^2+y^2$. The mixed terms cancel, which is the essential reason the invariant works.

To ensure that the invariant applies to the whole triple, write the triple as $(x,y,z)$. After the move it becomes

$$\left(\frac{x+y}{\sqrt2},\frac{x-y}{\sqrt2},z\right).$$

The new sum of squares equals

$$x^2+y^2+z^2,$$

so the unchanged third coordinate contributes identically before and after the move.

The final comparison also deserves a direct check:

$$(1+\sqrt2)^2=1+2\sqrt2+2=3+2\sqrt2,$$

hence

$$1+2+(3+2\sqrt2)=6+2\sqrt2.$$

Since $6+2\sqrt2$ is irrational while $\frac{13}{2}$ is rational, equality is impossible.

Alternative Approaches

The operation on a chosen pair is an orthogonal transformation. Indeed,

$$\frac1{\sqrt2} \begin{pmatrix} 1&1\ 1&-1 \end{pmatrix}$$

has orthonormal rows. Orthogonal transformations preserve Euclidean length, so the squared norm of the whole triple remains unchanged. The rest of the argument is the same comparison of norms.

Another viewpoint is to interpret the triple as a vector in $\mathbb R^3$. Each move applies an orthogonal transformation in one coordinate plane while leaving the third coordinate fixed. Since orthogonal transformations preserve the quantity $a^2+b^2+c^2$, the initial and target vectors must have equal squared lengths if one is reachable from the other. Their squared lengths differ, so reachability is impossible.