Question

Sean $a, b, x,$ y $y$ números reales con $a \gt 4$ y $b \gt 1$ tales que $\begin{aligned} \frac{x^2}{a^2} + \frac{y^2}{a^2 - 16} &= \frac{(x - 20)^2}{b^2 - 1} \\ &\quad {}+ \frac{(y - 11)^2}{b^2} \\ &= 1. \end{aligned}$ Halla el menor valor posible de $a + b.$

Let $a, b, x,$ and $y$ be real numbers with $a \gt 4$ and $b \gt 1$ such that $\begin{aligned} \frac{x^2}{a^2} + \frac{y^2}{a^2 - 16} &= \frac{(x - 20)^2}{b^2 - 1} \\ &\quad {}+ \frac{(y - 11)^2}{b^2} \\ &= 1. \end{aligned}$ Find the least possible value of $a + b.$

Respuesta

Solución:

La primera elipse tiene $c^2 = a^2 - (a^2 - 16) = 16,$ por lo que sus focos son $F_1 = (-4, 0)$ y $F_2 = (4, 0),$ con suma de distancias $2a.$ La segunda está centrada en $(20, 11)$ con eje mayor vertical y $c^2 = b^2 - (b^2 - 1) = 1,$ por lo que sus focos son $G_1 = (20, 10)$ y $G_2 = (20, 12),$ con suma de distancias $2b.$ Si $P = (x, y)$ está en ambas, entonces $\begin{aligned} 2a + 2b \\ &= (PF_1 + PG_1) \\ &\quad {}+ (PF_2 + PG_2) \\ &\ge F_1G_1 + F_2G_2 \\ &= \sqrt{24^2 + 10^2} \\ &\quad {}+ \sqrt{16^2 + 12^2} \\ &= 26 + 20 \\ &= 46, \end{aligned}$ así que $a + b \ge 23.$

La igualdad requiere que $P$ esté sobre ambos segmentos $\overline{F_1G_1}$ y $\overline{F_2G_2}.$ Estos segmentos sí se cortan, en $P = \left(14, \frac{15}{2}\right):$ entonces $PF_1 = \frac{39}{2},$ $PF_2 = \frac{25}{2},$ así que $2a = 32,$ $a = 16 \gt 4;$ y $PG_1 = \frac{13}{2},$ $PG_2 = \frac{15}{2},$ así que $2b = 14,$ $b = 7 \gt 1.$

Por lo tanto el menor valor posible de $a + b$ es $16 + 7 = 23.$

The first ellipse has $c^2 = a^2 - (a^2 - 16) = 16,$ hence foci $F_1 = (-4, 0)$ and $F_2 = (4, 0),$ with distance sum $2a.$ The second is centered at $(20, 11)$ with vertical major axis and $c^2 = b^2 - (b^2 - 1) = 1,$ hence foci $G_1 = (20, 10)$ and $G_2 = (20, 12),$ with distance sum $2b.$ If $P = (x, y)$ lies on both, then $\begin{aligned} 2a + 2b \\ &= (PF_1 + PG_1) \\ &\quad {}+ (PF_2 + PG_2) \\ &\ge F_1G_1 + F_2G_2 \\ &= \sqrt{24^2 + 10^2} \\ &\quad {}+ \sqrt{16^2 + 12^2} \\ &= 26 + 20 \\ &= 46, \end{aligned}$ so $a + b \ge 23.$

Equality requires $P$ to lie on both segments $\overline{F_1G_1}$ and $\overline{F_2G_2}.$ These segments do intersect, at $P = \left(14, \frac{15}{2}\right):$ then $PF_1 = \frac{39}{2},$ $PF_2 = \frac{25}{2},$ so $2a = 32,$ $a = 16 \gt 4;$ and $PG_1 = \frac{13}{2},$ $PG_2 = \frac{15}{2},$ so $2b = 14,$ $b = 7 \gt 1.$

Hence the least possible value of $a + b$ is $16 + 7 = 23.$

2022 AIME II Problema 12

Solución:

El Problema 12 en otros años