Question

En el cuadrado unitario $ABCD,$ la circunferencia inscrita $\omega$ corta a $\overline{CD}$ en $M,$ y $\overline{AM}$ corta a $\omega$ en un punto $P$ distinto de $M.$ ¿Cuánto vale $AP$ ?

In unit square $ABCD,$ the inscribed circle $\omega$ intersects $\overline{CD}$ at $M,$ and $\overline{AM}$ intersects $\omega$ at a point $P$ different from $M.$ What is $AP?$

$\dfrac{\sqrt{5}}{12}$

$\dfrac{\sqrt{5}}{10}$

$\dfrac{\sqrt{5}}{9}$

$\dfrac{\sqrt{5}}{8}$

$\dfrac{2\sqrt{5}}{15}$

Solución:

Sea $A = (0, 0),$ $B = (1, 0),$ $C = (1, 1),$ $D = (0, 1).$ La circunferencia inscrita tiene centro $\left(\tfrac12, \tfrac12\right)$ y radio $\tfrac12,$ y toca $\overline{CD}$ en $M = \left(\tfrac12, 1\right).$

La recta $AM$ es $y = 2x.$ Sustituyendo en $\left(x - \tfrac12\right)^2 + \left(y - \tfrac12\right)^2 = \tfrac14$ se obtiene $20x^2 - 12x + 1 = 0,$ con raíces $x = \tfrac12$ (punto $M$ ) y $x = \tfrac{1}{10}$ (punto $P$ ).

Así que $P = \left(\tfrac{1}{10}, \tfrac15\right)$ y $AP = \sqrt{\left(\tfrac{1}{10}\right)^2 + \left(\tfrac15\right)^2} = \frac{\sqrt5}{10}.$

Por lo tanto, la respuesta correcta es B.

Let $A = (0, 0),$ $B = (1, 0),$ $C = (1, 1),$ $D = (0, 1).$ The inscribed circle has center $\left(\tfrac12, \tfrac12\right)$ and radius $\tfrac12,$ touching $\overline{CD}$ at $M = \left(\tfrac12, 1\right).$

Line $AM$ is $y = 2x.$ Substituting into $\left(x - \tfrac12\right)^2 + \left(y - \tfrac12\right)^2 = \tfrac14$ gives $20x^2 - 12x + 1 = 0,$ with roots $x = \tfrac12$ (point $M$ ) and $x = \tfrac{1}{10}$ (point $P$ ).

So $P = \left(\tfrac{1}{10}, \tfrac15\right)$ and $AP = \sqrt{\left(\tfrac{1}{10}\right)^2 + \left(\tfrac15\right)^2} = \frac{\sqrt5}{10}.$

Thus, the correct answer is B.

2020 AMC 12B Problema 10

Solución:

El Problema 10 en otros años