2002 AMC 12A Problem 22

Attempt Problem 22 of the 2002 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12A solutions, or check the answer key.

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22.

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}

Answer: C

Concepts:geometric probability area ratio special right triangle

Difficulty rating: 1990

Small hint:

\triangle ABC

is

30

-

60

-

90

with

BC = 5

and

AC = 5\sqrt{3}

Big hint:

Choose

E

on

\overline{AC}

with

CE = 5,

so

BE = 5\sqrt{2};

then

BD \gt 5\sqrt{2}

iff

P

lies inside

\triangle ABE

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.

Problem 22 in other years