Solución 019

::: Sea ∠BAC = α → ∠CDB = 4α ∧ ∠ACB = 5α
::: → ∠CBD = α
::: ΔABD ~ ΔBCD
::: (x + CD)/BD = BD/CD
::: x = BD²/CD - CD . . . . . (I)

::: BD²/CD = (100/3)·[√3/(20 - 10√3)]
::: BD²/CD = (10/3)·[√3/(2 - √3)]
::: BD²/CD = (10/3)·(2√3 + 3) = (20/√3) + 10

Finalmente en (I)
::: x = (20/√3) + 10 - [(20/√3) - 10]
::: ⇒ x = 20

VOLVER A LA PREGUNTA ::: UNI 2010-I: MATEMATICA