2008 AMC 12B Problem 17

Question

Let

A,

B

and

C

be three distinct points on the graph of

y = x^2

such that line

AB

is parallel to the

x

-axis and

\triangle ABC

is a right triangle with area

2008.

What is the sum of the digits of the

y

-coordinate of

C?

16

17

18

19

20

Answer: C

Concepts:parabola slope triangle area digits

Difficulty rating: 1800

Small hint:

Write

A = (a, a^2)

and

B = (-a, a^2);

since

A, B, C

have distinct

x

-coordinates, the right angle is at

C

Big hint:

Perpendicular legs

CA

and

CB

force

a^2 - c^2 = 1,

the height above

\overline{AB};

get

|a|

from the area, then

c^2

Solution:

Since

AB

is horizontal, take

A = (a, a^2)

and

B = (-a, a^2),

and let

C = (c, c^2).

The right angle cannot be at

A

or

B

(that would need

c = \pm a

), so it is at

C.

Put

d = a^2-c^2.

Taking the dot product of

\overrightarrow{CA}

and

\overrightarrow{CB}

gives

d(d-1)=0.

The points are distinct, so

d \ne 0;

hence

a^2-c^2=1.

This value is the height of the triangle above

\overline{AB}.

The area is

\tfrac12 \cdot AB \cdot \text{height}

= \tfrac12 (2|a|)(1)

= |a| = 2008,

so

a^2 = 2008^2 = 4{,}032{,}064

and the

y