Question

Sea $\begin{aligned} p(x, y) &= a_0 + a_1x + a_2y \\ &\quad {}+ a_3x^2 + a_4xy + a_5y^2 \\ &\quad {}+ a_6x^3 + a_7x^2y \\ &\quad {}+ a_8xy^2 + a_9y^3. \end{aligned}$ Supongamos que $\begin{aligned} p(0, 0) &= p(1, 0) = p(-1, 0) \\ &= p(0, 1) = p(0, -1) \\ &= p(1, 1) = p(1, -1) \\ &= p(2, 2) = 0. \end{aligned}$

Existe un punto $\left(\frac{a}{c}, \frac{b}{c}\right)$ para el cual $p\left(\frac{a}{c}, \frac{b}{c}\right) = 0$ para todos esos polinomios, donde $a,$ $b,$ y $c$ son enteros positivos, $a$ y $c$ son primos entre sí, y $c \gt 1.$ Halla $a + b + c.$

Let $\begin{aligned} p(x, y) &= a_0 + a_1x + a_2y \\ &\quad {}+ a_3x^2 + a_4xy + a_5y^2 \\ &\quad {}+ a_6x^3 + a_7x^2y \\ &\quad {}+ a_8xy^2 + a_9y^3. \end{aligned}$ Suppose that $\begin{aligned} p(0, 0) &= p(1, 0) = p(-1, 0) \\ &= p(0, 1) = p(0, -1) \\ &= p(1, 1) = p(1, -1) \\ &= p(2, 2) = 0. \end{aligned}$

There is a point $\left(\frac{a}{c}, \frac{b}{c}\right)$ for which $p\left(\frac{a}{c}, \frac{b}{c}\right) = 0$ for all such polynomials, where $a,$ $b,$ and $c$ are positive integers, $a$ and $c$ are relatively prime, and $c \gt 1.$ Find $a + b + c.$

Respuesta

Solución:

De $p(0,0) = 0$ obtenemos $a_0 = 0.$ Sumando y restando $p(1,0) = p(-1,0) = 0$ se obtiene $a_3 = 0$ y $a_6 = -a_1;$ de manera similar $p(0,\pm 1) = 0$ dan $a_5 = 0$ y $a_9 = -a_2.$ Luego $p(1,1) = 0$ y $p(1,-1) = 0$ se reducen a $a_4 + a_7 + a_8 = 0$ y $-a_4 - a_7 + a_8 = 0,$ así que $a_8 = 0$ y $a_7 = -a_4.$ Ahora $p = a_1(x - x^3)$ $+ a_2(y - y^3)$ $+ a_4(xy - x^2y),$ y $p(2,2) = 0$ da $-6a_1 - 6a_2 - 4a_4 = 0,$ es decir $a_4 = -\frac{3}{2}(a_1 + a_2).$

Por lo tanto $\begin{aligned} p &= a_1\left[x - x^3 - \tfrac{3}{2}xy(1 - x)\right] \\ &\quad {}+ a_2\left[y - y^3 - \tfrac{3}{2}xy(1 - x)\right], \end{aligned}$ y un punto $(r, s)$ que es un cero para toda elección de $a_1, a_2$ debe anular ambos corchetes. El primer corchete se factoriza como $r(1 - r)\left(1 + r - \tfrac{3}{2}s\right),$ así que para un nuevo punto (con $r \ne 0, 1$ ) necesitamos $s = \tfrac{2}{3}(r + 1).$ El segundo corchete es $\tfrac{1}{2}s(2 - 2s^2 - 3r + 3r^2);$ sustituyendo $s^2 = \tfrac{4}{9}(r + 1)^2$ convierte $2 - 2s^2 - 3r + 3r^2 = 0$ en $\frac{19r^2 - 43r + 10}{9} = 0,$ cuyas raíces son $r = 2$ y $r = \frac{5}{19}.$

La raíz $r = 2$ reproduce el punto dado $(2, 2),$ así que el nuevo punto tiene $r = \frac{5}{19}$ y $s = \frac{2}{3} \cdot \frac{24}{19} = \frac{16}{19}.$ Por lo tanto $(a, b, c) = (5, 16, 19)$ y $a + b + c = 40.$

From $p(0,0) = 0$ we get $a_0 = 0.$ Adding and subtracting $p(1,0) = p(-1,0) = 0$ gives $a_3 = 0$ and $a_6 = -a_1;$ similarly $p(0,\pm 1) = 0$ give $a_5 = 0$ and $a_9 = -a_2.$ Then $p(1,1) = 0$ and $p(1,-1) = 0$ reduce to $a_4 + a_7 + a_8 = 0$ and $-a_4 - a_7 + a_8 = 0,$ so $a_8 = 0$ and $a_7 = -a_4.$ Now $p = a_1(x - x^3)$ $+ a_2(y - y^3)$ $+ a_4(xy - x^2y),$ and $p(2,2) = 0$ gives $-6a_1 - 6a_2 - 4a_4 = 0,$ i.e. $a_4 = -\frac{3}{2}(a_1 + a_2).$

Therefore $\begin{aligned} p &= a_1\left[x - x^3 - \tfrac{3}{2}xy(1 - x)\right] \\ &\quad {}+ a_2\left[y - y^3 - \tfrac{3}{2}xy(1 - x)\right], \end{aligned}$ and a point $(r, s)$ that is a zero for every choice of $a_1, a_2$ must kill both brackets. The first bracket factors as $r(1 - r)\left(1 + r - \tfrac{3}{2}s\right),$ so for a new point (with $r \ne 0, 1$ ) we need $s = \tfrac{2}{3}(r + 1).$ The second bracket is $\tfrac{1}{2}s(2 - 2s^2 - 3r + 3r^2);$ substituting $s^2 = \tfrac{4}{9}(r + 1)^2$ turns $2 - 2s^2 - 3r + 3r^2 = 0$ into $\frac{19r^2 - 43r + 10}{9} = 0,$ whose roots are $r = 2$ and $r = \frac{5}{19}.$

The root $r = 2$ reproduces the given point $(2, 2),$ so the new point has $r = \frac{5}{19}$ and $s = \frac{2}{3} \cdot \frac{24}{19} = \frac{16}{19}.$ Thus $(a, b, c) = (5, 16, 19)$ and $a + b + c = 40.$

2008 AIME I Problema 13

Solución:

El Problema 13 en otros años