2002 AMC 12A Problem 22

Problem 22 of 25HarderGeometry Counting & Probability

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}?

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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

\begin{aligned} \dfrac{[ABE]}{[ABC]} &= \dfrac{EA}{CA} \\ &= \dfrac{5\sqrt3 - 5}{5\sqrt3} \\ &= \dfrac{3 - \sqrt3}{3}. \end{aligned}

Thus, the correct answer is C.

More practice

Concepts: geometric probability · area ratio · special right triangle

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.