2004 AMC 12B Problem 24

Question

In

\triangle ABC,

AB = BC,

and

\overline{BD}

is an altitude. Point

E

is on the extension of

\overline{AC}

such that

BE = 10.

The values of

\tan \angle CBE,

\tan \angle DBE,

and

\tan \angle ABE

form a geometric progression, and the values of

\cot \angle DBE,

\cot \angle CBE,

\cot \angle DBC

form an arithmetic progression. What is the area of

\triangle ABC?

16

\dfrac{50}{3}

10\sqrt{3}

8\sqrt{5}

18

Answer: B

Concepts:trigonometric identity geometric sequence isosceles triangle

Difficulty rating: 2390

Small hint:

Let

\angle DBE = \alpha

and

\angle DBC = \beta;

then

\angle CBE = \alpha - \beta

and

\angle ABE = \alpha + \beta

Big hint:

The geometric progression gives

\tan(\alpha - \beta)\tan(\alpha + \beta) = \tan^2\alpha,

which forces

\alpha = 45^\circ

Solution:

Let

\angle DBE = \alpha

and

\angle DBC = \beta.

Since

\overline{BD}

is the altitude of the isosceles triangle,

\angle CBE = \alpha - \beta

and

\angle ABE = \alpha + \beta.

The geometric progression gives

\tan(\alpha - \beta)\tan(\alpha + \beta) = \tan^2\alpha,

which simplifies to

\tan^2\beta(\tan^4\alpha - 1) = 0,

so

\tan\alpha = 1

and

\alpha = 45^\circ.

Writing

DC = a

and

BD = b,

the arithmetic progression

\cot\angle DBE,

\cot\angle CBE,

\cot\angle DBC

becomes

1, \dfrac{b + a}{b - a}, \dfrac{b}{a},

forcing

b = 3a.

With

BE = 10

and

\angle DBE = 45^\circ,

we get

b = \dfrac{BE}{\sqrt2} = 5\sqrt2,

so

a = \dfrac{5\sqrt2}{3}.

The area of

\triangle ABC

is

\tfrac12 (AC)(BD) = ab

= 5\sqrt2 \cdot \dfrac{5\sqrt2}{3}

= \dfrac{50}{3}.

Thus, the correct answer is B.