2008 AMC 12B Problem 17

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Concepts:parabolaslopetriangle areadigits

Difficulty rating: 1800

17.

Let A,A, BB and CC be three distinct points on the graph of y=x2y = x^2 such that line ABAB is parallel to the xx-axis and ABC\triangle ABC is a right triangle with area 2008.2008. What is the sum of the digits of the yy-coordinate of C?C?

1616

1717

1818

1919

2020

Solution:

Since ABAB is horizontal, take A=(a,a2)A = (a, a^2) and B=(a,a2),B = (-a, a^2), and let C=(c,c2).C = (c, c^2). The right angle cannot be at AA or BB (that would need c=±ac = \pm a), so it is at C.C.

Then CACBCA \perp CB gives (c+a)(ca)=1,(c + a)(c - a) = -1, so a2c2=1.a^2 - c^2 = 1. This value is the height of the triangle above AB.\overline{AB}.

The area is 12ABheight=12(2a)(1)=a=2008,\tfrac12 \cdot AB \cdot \text{height} = \tfrac12 (2|a|)(1) = |a| = 2008, so a2=20082=4,032,064a^2 = 2008^2 = 4{,}032{,}064 and the yy-coordinate of CC is c2=a21=4,032,063.c^2 = a^2 - 1 = 4{,}032{,}063.

Its digit sum is 4+0+3+2+0+6+3=18.4 + 0 + 3 + 2 + 0 + 6 + 3 = 18.

Thus, the correct answer is C.

Problem 17 in Other Years