Question

En el cuadrilátero convexo $ABCD,$ $\angle A \cong \angle C,$ $AB = CD = 180,$ y $AD \ne BC.$ El perímetro de $ABCD$ es $640.$ Halla $\lfloor 1000 \cos A \rfloor.$ (La notación $\lfloor x \rfloor$ significa el mayor entero que es menor o igual que $x.$ )

In convex quadrilateral $ABCD,$ $\angle A \cong \angle C,$ $AB = CD = 180,$ and $AD \ne BC.$ The perimeter of $ABCD$ is $640.$ Find $\lfloor 1000 \cos A \rfloor.$ (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x.$ )

Respuesta

Solución:

Sean $\angle A = \angle C = \alpha,$ $AD = x,$ y $BC = y.$ Aplicando la ley de los cosenos a la diagonal $BD$ en los triángulos $ABD$ y $CDB,$ $\begin{aligned} BD^2 &= x^2 + 180^2 \\ &\quad {}- 2 \cdot 180x\cos\alpha \\ &= y^2 + 180^2 \\ &\quad {}- 2 \cdot 180y\cos\alpha. \end{aligned}$

Reordenando se obtiene $x^2 - y^2 = 2 \cdot 180(x - y)\cos\alpha,$ y como $x \ne y$ podemos dividir entre $x - y:$ $\begin{aligned} \cos\alpha &= \frac{x + y}{360} \\ &= \frac{640 - 2 \cdot 180}{360} \\ &= \frac{280}{360} = \frac{7}{9}. \end{aligned}$

Entonces $1000\cos A = \frac{7000}{9} = 777.7\ldots,$ así que $\lfloor 1000\cos A \rfloor = 777.$

Let $\angle A = \angle C = \alpha,$ $AD = x,$ and $BC = y.$ Applying the Law of Cosines to diagonal $BD$ in triangles $ABD$ and $CDB,$ $\begin{aligned} BD^2 &= x^2 + 180^2 \\ &\quad {}- 2 \cdot 180x\cos\alpha \\ &= y^2 + 180^2 \\ &\quad {}- 2 \cdot 180y\cos\alpha. \end{aligned}$

Rearranging gives $x^2 - y^2 = 2 \cdot 180(x - y)\cos\alpha,$ and since $x \ne y$ we may divide by $x - y:$ $\begin{aligned} \cos\alpha &= \frac{x + y}{360} \\ &= \frac{640 - 2 \cdot 180}{360} \\ &= \frac{280}{360} = \frac{7}{9}. \end{aligned}$

Then $1000\cos A = \frac{7000}{9} = 777.7\ldots,$ so $\lfloor 1000\cos A \rfloor = 777.$

2003 AIME I Problema 12

Solución:

El Problema 12 en otros años