2005 AMC 12B Problem 6

Attempt Problem 6 of the 2005 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 2005 AMC 12B solutions, or check the answer key.

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\triangle ABC,

we have

AC = BC = 7

and

AB = 2.

Suppose that

D

is a point on line

AB

such that

B

lies between

A

and

D

and

CD = 8.

What is

BD?

3

2\sqrt{3}

4

5

4\sqrt{2}

Drop the altitude from

C

to line

AB

; it meets

AB

at its midpoint

H

CH^2 = 7^2 - 1^2,

and

CD^2 = CH^2 + HD^2

with

HD = 1 + BD

Let

H

be the foot of the altitude from

C

to line

AB.

Since

\triangle ABC

is isosceles with

AC = BC,

H

is the midpoint of

AB,

AH = HB = 1.

Then

CH^2 = 7^2 - 1^2 = 48.

Applying the Pythagorean Theorem to

\triangle CHD

with

HD = HB + BD = 1 + BD

gives

8^2 = 48 + (1 + BD)^2,

(1 + BD)^2 = 16.

Therefore

1 + BD = 4,

which means

BD = 3.

Thus, the correct answer is A.

Problem 6 in other years