2005 AMC 10B Problem 10

Question

In

\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;

its foot is the midpoint of

AB

The altitude has length

\sqrt{7^2-1^2},

shared by right triangles

CHB

and

CHD

Let

H

be the foot of the altitude from

C

to line

AB.

Since

AC = BC,

H

is the midpoint of

AB,

so

BH = 1

and

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

Applying the Pythagorean theorem in

\triangle CHD,

where

HD = BH + BD = 1 + BD,

gives

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

so

(1 + BD)^2 = 16.

Then

1 + BD = 4,

so

BD = 3.

Thus, A is the correct answer.

Problem 10 in other years