Algebraic Transformations in Problems

We have \(x^2-8x=(x-4)^2-16\) and \(y^2+2y=(y+1)^2-1\). Hence the expression equals \((x-4)^2+(y+1)^2+3\).

A sum of squares is nonnegative, so the minimum is \(3\), attained at \(x=4\), \(y=-1\).

From \((a+b+c)^2=0\) we get \(a^2+b^2+c^2+2(ab+bc+ca)=0\), which gives the first equality.

Also, \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\). The right-hand side is zero, hence \(a^3+b^3+c^3=3abc\).

\(x^2+y^2=(x+y)^2-2xy=49-20=29\).

Next, \(x^3+y^3=(x+y)^3-3xy(x+y)=343-210=133\).

\(\frac{(x+y)^2}{xy}=\frac{x^2+2xy+y^2}{xy}=\frac{x}{y}+2+\frac{y}{x}=3+2=5\).

Since the expression is homogeneous, the factor \(t\) cancels. We get \(\frac{4t^2+9t^2+25t^2}{6t^2+15t^2+10t^2}=\frac{38}{31}\).

We get \(s+p=7\) and \(s^2-2p=10\). Since \(p=7-s\), it follows that \(s^2+2s-24=0\), so \(s=4\) or \(s=-6\).

If \(s=4\), then \(p=3\), and \(x,y\) are the roots of \(t^2-4t+3=0\), hence \(\{x,y\}=\{1,3\}\). If \(s=-6\), then \(p=13\), but the discriminant \(s^2-4p=36-52<0\), so no real \(x,y\) exist.

Answer: \((1,3)\) and \((3,1)\).

From \((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\) we get \(36=a^2+b^2+c^2+22\), so \(a^2+b^2+c^2=14\).

Then \((a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2-ab-bc-ca)=2(14-11)=6\).

From \((p+q+r)^2=0\) we get \(18+2(pq+qr+rp)=0\), so \(pq+qr+rp=-9\).

Therefore \(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{pq+qr+rp}{pqr}=\frac{-9}{6}=-\frac{3}{2}\).

The left-hand side equals \((x-2)^2+4(y-1)^2+5\). This is always at least \(5\), so it cannot be zero.

After dividing numerator and denominator by \(b\), we get \(\frac{t-1}{t+1}=\frac{1}{3}\). Thus \(3t-3=t+1\), so \(2t=4\) and \(t=2\).

The left-hand side equals \(2(a^2+b^2+c^2-ab-bc-ca)\). Since \(a+b+c=0\), we have \(ab+bc+ca=-\frac{a^2+b^2+c^2}{2}\).

Substitution gives \(2\left(a^2+b^2+c^2+\frac{a^2+b^2+c^2}{2}\right)=3(a^2+b^2+c^2)\).

From the conditions we get \(xy+yz+zx=-1\). Hence \(x,y,z\) are roots of the polynomial \(t^3+(xy+yz+zx)t-xyz=t^3-t-xyz\).

Therefore each of \(x,y,z\) satisfies \(t^3-t=xyz\). Thus \(x^3-x=y^3-y=z^3-z=xyz\).

Let \(s=x+y=3\), \(p=xy\). Then \(x^2+y^2+xy=s^2-p\). We get \(9-p=7\), hence \(p=2\).

The numbers \(x,y\) are roots of \(t^2-3t+2=0\), so \(\{x,y\}=\{1,2\}\). Answer: \((1,2)\), \((2,1)\).

Since \(x+y+z=0\), we have \(y+z=-x\), \(z+x=-y\), \(x+y=-z\). Therefore the sum is \(\frac{x}{-x}+\frac{y}{-y}+\frac{z}{-z}=-1-1-1=-3\).

\(\left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)\). Hence \(27=x^3+\frac{1}{x^3}+9\), so the desired value is \(18\).

We have \((a-1)^2+(b-1)^2+(c-1)^2=a^2+b^2+c^2-2(a+b+c)+3=3-6+3=0\).

A sum of squares is zero only when each square is zero. Hence \(a=b=c=1\).

From the zero sum we get \(xy+yz+zx=-3\). Then \((xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)\), and the second term is zero. Hence \(x^2y^2+y^2z^2+z^2x^2=9\).

Now \((x^2+y^2+z^2)^2=x^4+y^4+z^4+2(x^2y^2+y^2z^2+z^2x^2)\). Therefore \(36=x^4+y^4+z^4+18\), so \(x^4+y^4+z^4=18\).

From the first two conditions, \(ab+bc+ca=-3\). From \(a^3+b^3+c^3=3abc\) we get \(abc=2\).

Thus \(a,b,c\) are roots of \(t^3-3t-2=0\). The factorization \(t^3-3t-2=(t-2)(t+1)^2\) gives the roots \(2,-1,-1\). Hence all solutions are permutations of \((2,-1,-1)\).

Since \(\frac{x^2+y^2}{xy}=\frac{s^2-2p}{p}=\frac{5}{2}\), we have \(\frac{s^2}{p}=\frac{9}{2}\).

Then \(\left(\frac{x-y}{x+y}\right)^2=\frac{s^2-4p}{s^2}=1-\frac{4p}{s^2}=1-\frac{8}{9}=\frac{1}{9}\). Thus the possible values are \(\frac{1}{3}\) and \(-\frac{1}{3}\). Both are attained, for example, by the ratios \(x:y=2:1\) and \(1:2\).

From the conditions we get \(ab+bc+ca=-1\). Let \(q=abc\). Then \(a,b,c\) are three real roots of \(F(t)=t^3-t-q\).

If the cubic \(t^3-t-q\) has three real roots, then \(q\) must lie between the local maximum and local minimum values of the function \(t^3-t\). These extrema occur at \(3t^2-1=0\), so \(t=\pm\frac{1}{\sqrt{3}}\).

We get \(\left(\frac{1}{\sqrt{3}}\right)^3-\frac{1}{\sqrt{3}}=-\frac{2}{3\sqrt{3}}\) and \(\left(-\frac{1}{\sqrt{3}}\right)^3+\frac{1}{\sqrt{3}}=\frac{2}{3\sqrt{3}}\). Therefore \(-\frac{2}{3\sqrt{3}}\le q\le\frac{2}{3\sqrt{3}}\), as required.

Answer: no.

Since the graphs are lines and the coefficients \(v_i\) are distinct, any two graphs intersect at most once. Consider the order of the graphs by height for a very large \(t\), after all positive intersections.

The graph \(f_1\) was the lowest at \(t=0\). Each of its meetings moves it one position upward in the current order. If it has exactly four meetings, then finally it is fifth from the bottom.

Similarly, \(f_9\) was the highest at \(t=0\). Each of its meetings moves it one position downward. If it has exactly four meetings, then finally it is also fifth from the bottom. Two different graphs cannot occupy the same position. Contradiction.

Take any two given numbers \(x\) and \(y\). Consider two choices of four numbers that differ only in this: in one choice \(x\) is in the square group and \(y\) is in the other group, while in the second choice they are swapped. The other five contributions are the same.

By the condition, the sums are equal, so \(x^2+(2y-y^2)=y^2+(2x-x^2)\). Moving everything to one side gives \(2x^2-2y^2-2x+2y=0\), that is \((x-y)(x+y-1)=0\).

Thus any two numbers in the set are either equal or have sum \(1\). Fix one number \(x\). Then every other number is either \(x\) or \(1-x\). Hence there are at most two distinct values among the seven numbers. By the pigeonhole principle, one of them occurs at least four times.

Since \(S\) and \(C\) are rational, \(S^2+C^2\) is rational. Expanding, \[S^2+C^2=2+2(\sin48t\sin49t+\cos48t\cos49t)=2+2\cos t.\] Hence \(\cos t\) is rational.

For every positive integer \(n\), \(\cos nt\) is a polynomial with integer coefficients in \(\cos t\). Therefore \(\cos48t\) is rational. Since \(C=\cos48t+\cos49t\) is rational, \(\cos49t=C-\cos48t\) is rational as well.

We have \(\cos y=A-\sin x\) and \(\sin y=\cos x-B\). Square and add: \[(A-\sin x)^2+(\cos x-B)^2=1.\]

After expanding and using \(\sin^2x+\cos^2x=1\), we get \(A^2+B^2-2A\sin x-2B\cos x=0\), that is \(2A\sin x+2B\cos x=A^2+B^2\).

Let \(N\) be a common denominator of the rational numbers \(A\) and \(B\). Then \(m=2AN^2\) and \(n=2BN^2\) are positive integers, and \((A^2+B^2)N^2\) is a positive integer. Multiplying the equality by \(N^2\) gives the result.

Answer: no.

For neighboring numbers \(u\neq v\), \[\left(\frac{uv}{u-v}\right)^{-1}=\frac{u-v}{uv}=\frac{1}{v}-\frac{1}{u}.\] Hence \(\frac{uv}{u-v}\) is rational if and only if the difference \(\frac{1}{v}-\frac{1}{u}\) is rational.

Assume that all values are rational except possibly the value for the pair \(a_{2028},a_1\). Then the differences \(\frac{1}{a_2}-\frac{1}{a_1}\), \(\frac{1}{a_3}-\frac{1}{a_2}\), ..., \(\frac{1}{a_{2028}}-\frac{1}{a_{2027}}\) are rational. Their sum is \(\frac{1}{a_{2028}}-\frac{1}{a_1}\), so it is rational as well.

Therefore \(\frac{a_{2028}a_1}{a_{2028}-a_1}\) is rational. This contradicts the assumption that exactly one value is irrational.