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2016 AMC 10B Problem 9

2016 AMC 10B Problem 9AMC 10 Step by Step

Name: Date: September 13, 2026

Coordinate Geometry · difficulty 2 of 5 · about 2 min · position P1-10

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 $BC$ ?

A)
$4$
B)
$6$
C)
$8$
D)
$10$
E)
$16$

Answer

Key insight

A horizontal chord of y = x^2 has endpoints (-b, b^2) and (b, b^2), so base 2b, height b^2, and area b^3 = 64 gives b = 4.

Solution

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 $\boxed{\textbf{(C)}\ 8}$ .

Why this works

The parabola $y = x^2$ is symmetric about the $y$ -axis, so a horizontal chord is centered on that axis and a single parameter $b$ describes the whole triangle. Whenever a figure sits on a graph, name the coordinates with as few unknowns as the symmetry allows and let the area formula produce the equation.

Alternative approach

Test the choices: if $BC = 8$ then $b = 4$ , the chord sits at height $16$ , and the area is $\frac12 \cdot 8 \cdot 16 = 64$ . Only this choice works.

The trap

Setting the area to (1/2)(b)(b^2) with base b instead of 2b, which gives b^3 = 128 and no clean answer.

Common mistakes

Setting the area to $\frac12 \cdot b \cdot b^2$ with base $b$ instead of $2b$ , which gives $b^3 = 128$ and no clean answer.
Reporting $b = 4$ (half the chord) instead of the full length $BC = 8$ .

Techniques

Set up the equation/formula and compute; no special trick needed · Exploit symmetry to reduce work or pair up objects