2017 AMC 12B Problema 24

Question

El cuadrilátero

ABCD

tiene ángulos rectos en

B

y

C,

\triangle ABC \sim \triangle BCD,

y

AB \gt BC.

Hay un punto

E

en el interior de

ABCD

tal que

\triangle ABC \sim \triangle CEB

y el área de

\triangle AED

es

17

veces el área de

\triangle CEB.

¿Cuánto vale

\dfrac{AB}{BC}

?

Quadrilateral

ABCD

has right angles at

B

and

C,

\triangle ABC \sim \triangle BCD,

and

AB \gt BC.

There is a point

E

in the interior of

ABCD

such that

\triangle ABC \sim \triangle CEB

and the area of

\triangle AED

is

17

times the area of

\triangle CEB.

What is

\dfrac{AB}{BC}?

1 + \sqrt{2}

2 + \sqrt{2}

\sqrt{17}

2 + \sqrt{5}

1 + 2\sqrt{3}

Respuesta: D

Conceptos:semejanza geometría analítica fórmula del cordón

Nivel de dificultad: 2550

Pista pequeña:

Pon

BC = 1,

AB = r,

y sitúa

C = (0,0),

B = (0,1),

A = (r,1),

D = (\tfrac1r, 0)

Set

BC = 1,

AB = r,

and place

C = (0,0),

B = (0,1),

A = (r,1),

D = (\tfrac1r, 0)

Pista grande:

Halla

E

a partir de

\triangle ABC \sim \triangle CEB,

calcula ambas áreas, y plantea

[\triangle AED] = 17[\triangle CEB]

Find

E

from

\triangle ABC \sim \triangle CEB,

compute both areas, and set

[\triangle AED] = 17[\triangle CEB]

Solución:

Pon

BC = 1

y

AB = r \gt 1.

La semejanza

\triangle ABC \sim \triangle BCD

con los ángulos rectos sitúa la figura en

C = (0,0),

B = (0,1),

A = (r,1),

D = \bigl(\tfrac1r, 0\bigr).

Sea

E = (x, y)

con

x, y \gt 0.

De

\triangle ABC \sim \triangle CEB

obtenemos

\dfrac{x}{y} = \tan(\angle ECB)

= \tan(\angle BAC)

= \dfrac1r

y

x^2 + y^2 = \dfrac{r^2}{1 + r^2},

de modo que

x = \dfrac{r}{1+r^2},

y = \dfrac{r^2}{1+r^2}.

Las dos áreas relevantes son

\begin{aligned} [\triangle CEB]&=\dfrac{r}{2(1+r^2)},\\ [\triangle AED]&=\dfrac{r^4-r^2+1}{2r(1+r^2)}. \end{aligned}

Igualar la segunda a

17

veces la primera da

r^4 - 18r^2 + 1 = 0.

Entonces

r^2 = 9 + 4\sqrt5 = (2 + \sqrt5)^2,

así que

r = 2 + \sqrt5.

Por lo tanto, la respuesta correcta es D.

Set

BC = 1

and

AB = r \gt 1.

The similarity

\triangle ABC \sim \triangle BCD

with the right angles places the figure at

C = (0,0),

B = (0,1),

A = (r,1),

D = \bigl(\tfrac1r, 0\bigr).

Let

E = (x, y)

with

x, y \gt 0.

From

\triangle ABC \sim \triangle CEB

we get

\dfrac{x}{y} = \tan(\angle ECB)

= \tan(\angle BAC)

= \dfrac1r

and

x^2 + y^2 = \dfrac{r^2}{1 + r^2},

so

x = \dfrac{r}{1+r^2},

y = \dfrac{r^2}{1+r^2}.

The two relevant areas are

\begin{aligned} [\triangle CEB]&=\dfrac{r}{2(1+r^2)},\\ [\triangle AED]&=\dfrac{r^4-r^2+1}{2r(1+r^2)}. \end{aligned}

Setting the second equal to

17

times the first gives

r^4 - 18r^2 + 1 = 0.

Then

r^2 = 9 + 4\sqrt5 = (2 + \sqrt5)^2,

so

r = 2 + \sqrt5.

Thus, the correct answer is D.