2004 AMC 12B Problem 14

Attempt Problem 14 of the 2004 AMC 12B 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 2004 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

14.

In

\triangle ABC,

AB = 13,

AC = 5

and

BC = 12.

Points

M

and

N

lie on

\overline{AC}

and

\overline{BC},

respectively, with

CM = CN = 4.

Points

J

and

K

are on

\overline{AB}

so that

\overline{MJ}

and

\overline{NK}

are perpendicular to

\overline{AB}.

What is the area of pentagon

CMJKN?

15

\dfrac{81}{5}

\dfrac{205}{12}

\dfrac{240}{13}

20

Answer: D

Concepts:similarity right triangle area decomposition

Difficulty rating: 1680

Small hint:

\triangle ABC

is right-angled at

C

since

5^2 + 12^2 = 13^2

Big hint:

\triangle AMJ

and

\triangle NBK

are similar to

\triangle ABC

with hypotenuses

AM = 1

and

BN = 8

Solution:

Since

5^2 + 12^2 = 13^2,

\triangle ABC

is right-angled at

C

with area

\tfrac12 (5)(12) = 30.

The small right triangles

\triangle AMJ

and

\triangle NBK

are each similar to

\triangle ABC,

with hypotenuses

AM = 5 - 4 = 1

and

BN = 12 - 4 = 8.

Their areas are

\left(\dfrac{1}{13}\right)^2 (30)

and

\left(\dfrac{8}{13}\right)^2 (30).

The pentagon is what remains:

\begin{gathered} \left(1 - \dfrac{1}{169} - \dfrac{64}{169}\right)(30) \\ {}= \dfrac{104}{169}(30) = \dfrac{240}{13}. \end{gathered}

Thus, the correct answer is D.

Problem 14 in other years