2016 AMC 10B Problem 9

P1-10~2 minPrint

Coordinate Geometry Quadratics Triangles: Area & Pythagorean Set up the equation/formula and compute; no special trick needed Exploit symmetry to reduce work or pair up objects

All three vertices of $\bigtriangleup ABC$ are lying on the parabola defined by $y=x^2$ , with $A$ at the origin and $\overline{BC}$ parallel to the $x$ -axis. The area of the triangle is $64$ . What is the length of $BC$ ?

Show solution without answering

Key insight

The parabola is symmetric about the y-axis, so B and C are (-b, b^2) and (b, b^2). The area is (1/2)(2b)(b^2) = b^3 = 64.

Solution

$\overline{BC}$ is flat (parallel to the $x$ -axis), so $B$ and $C$ are at the same height.
The graph of $y = x^2$ is the same on the left and right of the $y$ -axis. For example, $(-3, 9)$ and $(3, 9)$ are both on it. So two points at the same height are mirror images. Call them $B = (-b, b^2)$ and $C = (b, b^2)$ , with $b$ positive.
The base $BC$ goes from $x = -b$ to $x = b$ , so its length is $2b$ .
The height is the vertical distance from $A = (0,0)$ up to the line $BC$ , which is $b^2$ .
Area $= \frac12 \cdot \text{base} \cdot \text{height} = \frac12 \cdot 2b \cdot b^2 = b^3$ .
Set $b^3 = 64$ . Then $b = 4$ , and $BC = 2b = 8$ .

Check: the points are $(-4, 16)$ and $(4, 16)$ , the base is $8$ , the height is $16$ , and the area is $\frac12 \cdot 8 \cdot 16 = 64$ .

The answer is $\boxed{\textbf{(C)}\ 8}$ .

Why this works

When a shape sits on a graph, use the graph's symmetry to name the points with as few letters as possible. Here one letter, $b$ , describes the whole triangle, and the area formula gives one equation for it.

Alternative approach

Test the choices. If $BC = 8$ , then $B$ and $C$ are at $x = \pm 4$ , so they sit at height $16$ . The area is $\frac12 \cdot 8 \cdot 16 = 64$ , which matches. (For example, $BC = 6$ gives only $\frac12 \cdot 6 \cdot 9 = 27$ .)

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Since $\overline{BC}$ is horizontal, $B$ and $C$ have the same $y$ -coordinate. On the parabola $y = x^2$ , two points with the same height are mirror images across the $y$ -axis, so let $B = (-b, b^2)$ and $C = (b, b^2)$ with $b > 0$ .

Then $BC = 2b$ , and the height from $A = (0,0)$ to line $BC$ is $b^2$ . The area is
$\frac12 \cdot 2b \cdot b^2 = b^3 = 64,$
so $b = 4$ and $BC = 2b = 8$ .

The answer is $\boxed{\textbf{(C)}\ 8}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Algebra, Chapter 10: Quadratic Equations – Part 1 (§10.2, §10.3, §10.4)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)