2000 AMC 10 Problem 7

Attempt Problem 7 of the 2000 AMC 10 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 2000 AMC 10 solutions, or check the answer key.

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

7.

In rectangle

ABCD,

AD = 1,

P

is on

\overline{AB},

and

\overline{DB}

and

\overline{DP}

trisect

\angle ADC.

What is the perimeter of

\triangle BDP?

3 + \dfrac{\sqrt3}{3}

2 + \dfrac{4\sqrt3}{3}

2 + 2\sqrt2

\dfrac{3 + 3\sqrt5}{2}

2 + \dfrac{5\sqrt3}{3}

Answer: B

Concepts:special right triangle rectangle perimeter

Difficulty rating: 1390

Small hint:

Angle

ADC = 90^\circ

is split into three

30^\circ

angles

Big hint:

Triangles

ADP

and

ADB

are both

30

-

60

-

90

right triangles with leg

AD = 1

Solution:

The right angle

\angle ADC = 90^\circ

is trisected into three

30^\circ

angles, so

\angle ADP = 30^\circ

and

\angle ADB = 60^\circ.

In right triangle

ADP,

with

AD = 1,

we get

DP = \dfrac{1}{\cos 30^\circ} = \dfrac{2\sqrt3}{3}

and

AP = \tan 30^\circ = \dfrac{\sqrt3}{3}.

In right triangle

ADB,

with

AD = 1,

we get

DB = \dfrac{1}{\cos 60^\circ} = 2

and

AB = \tan 60^\circ = \sqrt3.

Then

\begin{aligned} PB = AB - AP &= \sqrt3 - \dfrac{\sqrt3}{3} \\ &= \dfrac{2\sqrt3}{3}. \end{aligned}

The perimeter of

\triangle BDP

is

\begin{gathered} DP + PB + DB \\ = \dfrac{2\sqrt3}{3} + \dfrac{2\sqrt3}{3} + 2 \\ = 2 + \dfrac{4\sqrt3}{3}. \end{gathered}

Thus, the correct answer is B.