Question

Sea $ABCD$ un trapecio isósceles con $AD = BC$ y $AB \lt CD.$ Suponga que las distancias de $A$ a las rectas $BC,$ $CD,$ y $BD$ son $15,$ $18,$ y $10,$ respectivamente. Sea $K$ el área de $ABCD.$ Halle $\sqrt{2} \cdot K.$

Let $ABCD$ be an isosceles trapezoid with $AD = BC$ and $AB \lt CD.$ Suppose that the distances from $A$ to the lines $BC,$ $CD,$ and $BD$ are $15,$ $18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt{2} \cdot K.$

Respuesta

Respuesta: 567

Conceptos:trapecio geometría analítica fórmula de la distancia

Nivel de dificultad: 2990

Pista pequeña:

La distancia $18$ a $CD$ es la altura, así que ponga $D = (0, 0),$ $C = (c, 0),$ $A = (a, 18),$ $B = (c - a, 18)$

The distance $18$ to $CD$ is the height, so set $D = (0, 0),$ $C = (c, 0),$ $A = (a, 18),$ $B = (c - a, 18)$

Pista grande:

Las distancias a $BC$ y $BD$ tienen ambas numerador $18 \cdot AB$ en la fórmula punto-recta; al compararlas se obtiene $a = \frac{2}{5} AB,$ y luego se despeja $AB$

The distances to $BC$ and $BD$ both have numerator $18 \cdot AB$ in the point-line formula; comparing them gives $a = \frac{2}{5} AB,$ then solve for $AB$

Solución:

Como $AB \parallel CD,$ la distancia $18$ de $A$ a $CD$ es la altura. Ponga $D = (0, 0),$ $C = (c, 0),$ $A = (a, 18),$ $B = (c - a, 18),$ y sea $u = AB = c - 2a \gt 0.$ Las fórmulas de distancia punto-recta dan $\begin{aligned} \frac{18u}{\sqrt{324 + a^2}} &= 15, \\ \small \frac{18u}{\sqrt{324 + (u + a)^2}} \\ &= 10. \end{aligned}$ Por lo tanto, $324 + a^2 = \left(\frac{6u}{5}\right)^2$ y $324 + (u + a)^2 = \left(\frac{9u}{5}\right)^2.$

Restando, $u(u + 2a) = \frac{81 - 36}{25}u^2 = \frac{9}{5}u^2,$ por lo que $u + 2a = \frac{9}{5}u$ y $a = \frac{2}{5}u.$ Sustituyendo de nuevo, $324 = \frac{36u^2}{25} - \frac{4u^2}{25} = \frac{32u^2}{25},$ así que $u^2 = \frac{2025}{8}$ y $u = \frac{45\sqrt{2}}{4}.$

Entonces $CD = c = u + 2a = \frac{9}{5}u$ $= \frac{81\sqrt{2}}{4},$ y $\begin{aligned} K &= \frac{AB + CD}{2} \cdot 18 \\ &= 9\left(\frac{45\sqrt{2}}{4} + \frac{81\sqrt{2}}{4}\right) \\ &= \frac{567\sqrt{2}}{2}, \end{aligned}$ así que $\sqrt{2} \cdot K = 567.$

Since $AB \parallel CD,$ the distance $18$ from $A$ to $CD$ is the height. Put $D = (0, 0),$ $C = (c, 0),$ $A = (a, 18),$ $B = (c - a, 18),$ and let $u = AB = c - 2a \gt 0.$ The point-to-line distance formulas give $\begin{aligned} \frac{18u}{\sqrt{324 + a^2}} &= 15, \\ \small \frac{18u}{\sqrt{324 + (u + a)^2}} \\ &= 10. \end{aligned}$ so $324 + a^2 = \left(\frac{6u}{5}\right)^2$ and $324 + (u + a)^2 = \left(\frac{9u}{5}\right)^2.$

Subtracting, $u(u + 2a) = \frac{81 - 36}{25}u^2 = \frac{9}{5}u^2,$ hence $u + 2a = \frac{9}{5}u$ and $a = \frac{2}{5}u.$ Substituting back, $324 = \frac{36u^2}{25} - \frac{4u^2}{25} = \frac{32u^2}{25},$ so $u^2 = \frac{2025}{8}$ and $u = \frac{45\sqrt{2}}{4}.$

Then $CD = c = u + 2a = \frac{9}{5}u$ $= \frac{81\sqrt{2}}{4},$ and $\begin{aligned} K &= \frac{AB + CD}{2} \cdot 18 \\ &= 9\left(\frac{45\sqrt{2}}{4} + \frac{81\sqrt{2}}{4}\right) \\ &= \frac{567\sqrt{2}}{2}, \end{aligned}$ so $\sqrt{2} \cdot K = 567.$

2021 AIME I Problema 9

Respuesta: 567

Nivel de dificultad: 2990

Pista pequeña:

Pista grande:

Solución:

El Problema 9 en otros años