2024 AMC 12A Problema 20

Question

Los puntos

P

y

Q

se eligen de manera uniforme e independiente al azar sobre los lados

\overline{AB}

y

\overline{AC},

respectivamente, del triángulo equilátero

\triangle ABC.

¿Cuál de los siguientes intervalos contiene la probabilidad de que el área de

\triangle APQ

sea menor que la mitad del área de

\triangle ABC

?

Points

P

and

Q

are chosen uniformly and independently at random on sides

\overline{AB}

and

\overline{AC},

respectively, of equilateral triangle

\triangle ABC.

Which of the following intervals contains the probability that the area of

\triangle APQ

is less than half the area of

\triangle ABC?

\left[\tfrac38,\tfrac12\right]

\left(\tfrac12,\tfrac23\right]

\left(\tfrac23,\tfrac34\right]

\left(\tfrac34,\tfrac78\right]

\left(\tfrac78,1\right]

Respuesta: D

Conceptos:probabilidad geométrica razón de áreas

Nivel de dificultad: 2100

Pista pequeña:

Sean

x=\dfrac{AP}{AB}

y

y=\dfrac{AQ}{AC}

uniformes en

[0,1];

la razón de áreas es

xy

Let

x=\dfrac{AP}{AB}

and

y=\dfrac{AQ}{AC}

be uniform on

[0,1];

the area ratio is

xy

Pista grande:

Calcula

P(xy\ge\tfrac12)=\displaystyle\int_{\frac{1}{2}}^{1}\left(1-\tfrac{1}{2x}\right)dx

y réstalo de

1

Compute

P(xy\ge\tfrac12)=\displaystyle\int_{\frac{1}{2}}^{1}\left(1-\tfrac{1}{2x}\right)dx

and subtract from

1

Solución:

Con

x=\tfrac{AP}{AB}

y

y=\tfrac{AQ}{AC}

uniformes en

[0,1],

la razón de áreas es

\tfrac{[APQ]}{[ABC]}=xy.

El evento complementario

xy\ge\tfrac12

exige

x\ge\tfrac12

y

y\in[\tfrac{1}{2x},1],

y tiene probabilidad

\begin{aligned} &\int_{\frac{1}{2}}^{1}\left(1-\frac{1}{2x}\right)dx \\ &=\frac12-\frac{\ln2}{2}\approx0.153. \end{aligned}

Por tanto,

P(xy\lt\tfrac12)\approx1-0.153=0.847,

que pertenece a

\left(\tfrac34,\tfrac78\right].

Así, D es la respuesta correcta.

With

x=\tfrac{AP}{AB}

and

y=\tfrac{AQ}{AC}

uniform on

[0,1],

the area ratio

\tfrac{[APQ]}{[ABC]}=xy.

The complementary event

xy\ge\tfrac12

requires

x\ge\tfrac12

and

y\in[\tfrac{1}{2x},1],

with probability

\begin{aligned} &\int_{\frac{1}{2}}^{1}\left(1-\frac{1}{2x}\right)dx \\ &=\frac12-\frac{\ln2}{2}\approx0.153. \end{aligned}

Therefore

P(xy\lt\tfrac12)\approx1-0.153=0.847,

which lies in

\left(\tfrac34,\tfrac78\right].

Thus, the correct answer is D.