Question

Triangle $ABC$ is a right triangle with $\angle ACB$ as its right angle, $m\angle ABC = 60^\circ,$ and $AB = 10.$ Let $P$ be randomly chosen inside $\triangle ABC,$ and extend $\overline{BP}$ to meet $\overline{AC}$ at $D.$ What is the probability that $BD \gt 5\sqrt{2}?$

$\dfrac{2 - \sqrt{2}}{2}$

$\dfrac{1}{3}$

$\dfrac{3 - \sqrt{3}}{3}$

$\dfrac{1}{2}$

$\dfrac{5 - \sqrt{5}}{5}$

Solution:

Since $AB = 10$ and $\angle ABC = 60^\circ,$ the $30$ - $60$ - $90$ triangle has $BC = 5$ and $AC = 5\sqrt3.$

Place $E$ on $\overline{AC}$ with $CE = 5;$ then $BE = \sqrt{5^2 + 5^2} = 5\sqrt2.$ As $D$ moves along $\overline{AC},$ $BD = \sqrt{25 + CD^2}$ exceeds $5\sqrt2$ exactly when $CD \gt 5,$ i.e. when $D$ lies beyond $E,$ which happens iff $P$ is inside $\triangle ABE.$

The probability is $\dfrac{[ABE]}{[ABC]} = \dfrac{EA}{CA} = \dfrac{5\sqrt3 - 5}{5\sqrt3} = \dfrac{3 - \sqrt3}{3}.$

Thus, the correct answer is C.

2002 AMC 12A Problem 22

Solution:

Problem 22 in Other Years