Kvant Math Problem 578
Introduce
Verified: no
Verdicts: SKIP + SKIP
Solve time: 9m12s
Source on kvant.digital
Problem
Find $x$ and $y$ from the system of equations $$\left{\begin{array}{l} \dfrac{x-y\sqrt{x^2-y^2}}{\sqrt{1-x^2+y^2}}=a,\ \dfrac{y-x\sqrt{x^2-y^2}}{\sqrt{1-x^2+y^2}}=b \end{array}\right.$$ ($a$ and $b$ are given numbers).
V. L. Gutenmakher
All-Union Mathematical Olympiad for School Students (1979, Grades 8 and 10)
Exploration
Introduce
$$s=\sqrt{x^2-y^2},\qquad t=\sqrt{,1-x^2+y^2,}.$$
Then $s,t\ge0$ and
$$s^2+t^2=1.$$
The system becomes
$$x-ys=at,\qquad y-xs=bt.$$
This is a linear system in $x$ and $y$.
Writing it as
$$\begin{pmatrix} 1&-s\ -s&1 \end{pmatrix} \begin{pmatrix} x\y \end{pmatrix} = t \begin{pmatrix} a\b \end{pmatrix},$$
its determinant equals $1-s^2=t^2$. Since the original denominators contain $t$, we must have $t>0$. Hence the matrix is invertible.
Solving,
$$x=\frac{a+bs}{t},\qquad y=\frac{b+as}{t}.$$
The unknowns have now been expressed through the single parameter $s$. The next step is to use the defining relation $s^2=x^2-y^2$. Substituting the formulas for $x,y$,
$$x^2-y^2 = \frac{(a+bs)^2-(b+as)^2}{t^2} = \frac{(a^2-b^2)(1-s^2)}{t^2}.$$
Since $t^2=1-s^2$,
$$x^2-y^2=a^2-b^2.$$
But $x^2-y^2=s^2$, so
$$s^2=a^2-b^2.$$
This is the crucial point. Once $s\ge0$ is fixed by
$$s=\sqrt{a^2-b^2},$$
the relation $t^2=1-s^2$ gives
$$t=\sqrt{1-a^2+b^2}.$$
Substituting into the formulas for $x,y$ yields a candidate solution. The remaining task is to verify that it indeed satisfies all defining relations and to determine when real solutions exist.
The existence conditions are forced by $s\ge0$ and $t>0$:
$$a^2-b^2\ge0,\qquad 1-a^2+b^2>0.$$
Equivalently,
$$0\le a^2-b^2<1.$$
Problem Understanding
We must solve explicitly the system
$$\frac{x-y\sqrt{x^2-y^2}}{\sqrt{1-x^2+y^2}}=a, \qquad \frac{y-x\sqrt{x^2-y^2}}{\sqrt{1-x^2+y^2}}=b,$$
where $a$ and $b$ are given real numbers.
This is a Type A problem. We must determine all real solutions $(x,y)$ and prove that no others exist.
The main difficulty is that the square roots contain the unknowns. The natural substitution
$$s=\sqrt{x^2-y^2},\qquad t=\sqrt{1-x^2+y^2}$$
turns the system into linear equations in $x$ and $y$, after which the defining identity $s^2=x^2-y^2$ determines $s$.
The expected answer is that solutions exist exactly when
$$0\le a^2-b^2<1,$$
and then $x$ and $y$ are given by explicit formulas obtained from the linear system.
Proof Architecture
First, set
$$s=\sqrt{x^2-y^2},\qquad t=\sqrt{1-x^2+y^2}.$$
Then $s,t\ge0$, $t>0$, and $s^2+t^2=1$.
Next, rewrite the original equations as a linear system
$$x-ys=at,\qquad y-xs=bt.$$
Then solve this system to obtain
$$x=\frac{a+bs}{t},\qquad y=\frac{b+as}{t}.$$
Substitute these expressions into $x^2-y^2=s^2$. This yields
$$s^2=a^2-b^2.$$
Use $s\ge0$ to conclude
$$s=\sqrt{a^2-b^2}.$$
Use $s^2+t^2=1$ and $t>0$ to obtain
$$t=\sqrt{1-a^2+b^2}.$$
Substitute these values into the formulas for $x$ and $y$, obtaining a candidate solution.
Finally, verify directly that the obtained pair satisfies the original system and determine the necessary and sufficient conditions for reality of all square roots.
The most delicate step is deriving $s^2=a^2-b^2$ from the substituted formulas for $x$ and $y$.
Solution
Let
$$s=\sqrt{x^2-y^2},\qquad t=\sqrt{1-x^2+y^2}.$$
Since the denominators in the original system are defined and nonzero,
$$t>0.$$
Also,
$$s\ge0, \qquad s^2+t^2=(x^2-y^2)+(1-x^2+y^2)=1.$$
Multiplying both equations of the system by $t$, we obtain
$$x-ys=at, \qquad y-xs=bt.$$
Equivalently,
$$\begin{pmatrix} 1&-s\ -s&1 \end{pmatrix} \begin{pmatrix} x\y \end{pmatrix} = t \begin{pmatrix} a\b \end{pmatrix}.$$
The determinant of the coefficient matrix is
$$1-s^2=t^2.$$
Since $t>0$, this determinant is positive. Hence
$$x=\frac{t(a+bs)}{t^2} =\frac{a+bs}{t}, \qquad y=\frac{t(b+as)}{t^2} =\frac{b+as}{t}.$$
Now compute $x^2-y^2$:
$$x^2-y^2 = \frac{(a+bs)^2-(b+as)^2}{t^2}.$$
Expanding,
$$(a+bs)^2-(b+as)^2 = (a^2-b^2)+s^2(b^2-a^2) = (a^2-b^2)(1-s^2).$$
Since $t^2=1-s^2$,
$$x^2-y^2=a^2-b^2.$$
By definition, $x^2-y^2=s^2$. Therefore
$$s^2=a^2-b^2.$$
Because $s\ge0$,
$$s=\sqrt{a^2-b^2}.$$
Using $s^2+t^2=1$,
$$t=\sqrt{1-s^2} = \sqrt{1-a^2+b^2}.$$
Since $t>0$, we must have
$$1-a^2+b^2>0.$$
Substituting the values of $s$ and $t$ into the formulas for $x$ and $y$, we obtain
$$x= \frac{a+b\sqrt{a^2-b^2}} {\sqrt{,1-a^2+b^2,}},$$
$$y= \frac{b+a\sqrt{a^2-b^2}} {\sqrt{,1-a^2+b^2,}}.$$
These expressions are real precisely when
$$a^2-b^2\ge0, \qquad 1-a^2+b^2>0.$$
Equivalently,
$$0\le a^2-b^2<1.$$
It remains to verify that the obtained pair satisfies the system. For this pair,
$$s=\sqrt{a^2-b^2}, \qquad t=\sqrt{1-a^2+b^2},$$
and the formulas for $x$ and $y$ were derived from
$$x-ys=at, \qquad y-xs=bt.$$
Furthermore,
$$x^2-y^2=s^2, \qquad 1-x^2+y^2=t^2.$$
Hence
$$\frac{x-ys}{t}=a, \qquad \frac{y-xs}{t}=b,$$
which is exactly the original system.
Thus a real solution exists if and only if
$$0\le a^2-b^2<1,$$
and in that case it is unique and equals
$$x= \frac{a+b\sqrt{a^2-b^2}} {\sqrt{,1-a^2+b^2,}}, \qquad y= \frac{b+a\sqrt{a^2-b^2}} {\sqrt{,1-a^2+b^2,}}.$$
$$\boxed{ \begin{aligned} x&=\frac{a+b\sqrt{a^2-b^2}} {\sqrt{,1-a^2+b^2,}},\[2mm] y&=\frac{b+a\sqrt{a^2-b^2}} {\sqrt{,1-a^2+b^2,}}, \end{aligned} \qquad 0\le a^2-b^2<1 }$$
Verification of Key Steps
The first delicate step is the computation of $x^2-y^2$. Starting from
$$x=\frac{a+bs}{t},\qquad y=\frac{b+as}{t},$$
we have
$$x^2-y^2 = \frac{(a+bs)^2-(b+as)^2}{t^2}.$$
The numerator equals
$$a^2+2abs+b^2s^2-b^2-2abs-a^2s^2,$$
which simplifies to
$$(a^2-b^2)(1-s^2).$$
Division by $t^2=1-s^2$ gives
$$x^2-y^2=a^2-b^2.$$
A sign error in this expansion would produce an incorrect value of $s$.
The second delicate step is the determinant. The linear system is solvable only because
$$\det \begin{pmatrix} 1&-s\ -s&1 \end{pmatrix} = 1-s^2=t^2.$$
The original equations require $t>0$, so $t^2>0$. If one allowed $t=0$, the original system would be undefined.
The third delicate step is the existence condition. The square root
$$\sqrt{a^2-b^2}$$
requires
$$a^2-b^2\ge0,$$
while the denominator
$$\sqrt{1-a^2+b^2}$$
must be strictly positive, not merely nonnegative. Hence
$$a^2-b^2<1.$$
Replacing the strict inequality by $\le$ would admit values for which the original denominators vanish.
Alternative Approaches
Instead of solving the linear system by matrix inversion, one may add and subtract the equations
$$x-ys=at, \qquad y-xs=bt.$$
This gives
$$(x+y)(1-s)=t(a+b), \qquad (x-y)(1+s)=t(a-b).$$
Using
$$1-s^2=t^2,$$
one obtains
$$x+y=\frac{t(a+b)}{1-s}, \qquad x-y=\frac{t(a-b)}{1+s},$$
and then reconstructs $x$ and $y$. Substitution into $x^2-y^2=s^2$ again yields
$$s^2=a^2-b^2.$$
The main approach is preferable because the linear structure becomes transparent immediately. The formulas for $x$ and $y$ emerge directly, and the relation $s^2=a^2-b^2$ follows from a single computation.