2006 AMC 12A Problema 15

Question

Supón que

\cos x = 0

y

\cos(x + z) = \tfrac{1}{2}.

¿Cuál es el menor valor positivo posible de

z

?

Suppose

\cos x = 0

and

\cos(x + z) = \tfrac{1}{2}.

What is the smallest possible positive value of

z?

\dfrac{\pi}{6}

\dfrac{\pi}{3}

\dfrac{\pi}{2}

\dfrac{5\pi}{6}

\dfrac{7\pi}{6}

Respuesta: A

Conceptos:trigonometría

Nivel de dificultad: 1590

Pista pequeña:

\cos x = 0

significa que

x

es un múltiplo impar de

\tfrac{\pi}{2}

.

\cos x = 0

means

x

is an odd multiple of

\tfrac{\pi}{2}

Pista grande:

\cos(x + z) = \tfrac{1}{2}

significa que

x + z = 2n\pi \pm \tfrac{\pi}{3};

minimiza

z = (x+z) - x

.

\cos(x + z) = \tfrac{1}{2}

means

x + z = 2n\pi \pm \tfrac{\pi}{3};

minimize

z = (x+z) - x

Solución:

Como

\cos x = 0,

tenemos

x = \tfrac{\pi}{2} + k\pi.

Como

\cos(x + z) = \tfrac{1}{2},

tenemos

x + z = 2n\pi \pm \tfrac{\pi}{3}.

En el círculo unitario, la menor separación angular positiva entre un múltiplo impar de

\tfrac{\pi}{2}

y un ángulo congruente con

\pm\tfrac{\pi}{3}

es

\tfrac{\pi}{6}.

Se alcanza tomando

x = -\tfrac{\pi}{2}

y

x + z = -\tfrac{\pi}{3},

lo que da

z = -\tfrac{\pi}{3} + \tfrac{\pi}{2} = \tfrac{\pi}{6}.

Por lo tanto, la respuesta correcta es A.

Since

\cos x = 0,

we have

x = \tfrac{\pi}{2} + k\pi.

Since

\cos(x + z) = \tfrac{1}{2},

we have

x + z = 2n\pi \pm \tfrac{\pi}{3}.

On the unit circle, the smallest positive angular separation between an odd multiple of

\tfrac{\pi}{2}

and an angle congruent to

\pm\tfrac{\pi}{3}

is

\tfrac{\pi}{6}.

It is attained by taking

x = -\tfrac{\pi}{2}

and

x + z = -\tfrac{\pi}{3},

which gives

z = -\tfrac{\pi}{3} + \tfrac{\pi}{2} = \tfrac{\pi}{6}.

Thus, the correct answer is A.