2025 AMC 10B Problem 6

P1-10~3 minPrint

Coordinate Geometry Quadratics Quadrilaterals & Polygon Areas Set up the equation/formula and compute; no special trick needed Cut the figure into known shapes (triangles, rectangles, sectors)

The line $y = \frac{1}{3}x + 1$ divides the square region defined by $0 \leq x \leq 2$ and $0 \leq y \leq 2$ into an upper and lower region. The line $x=a$ divides the lower region into two regions of equal area. Then $a$ can be written as $\sqrt{s} - t$ , where $s$ and $t$ are positive integers. What is $s+t$ ?

Show solution without answering

Key insight

The line crosses both side edges, so the lower region is a trapezoid of area 8/3. The piece left of x = a must have area 4/3.

Solution

See where the line goes. At $x = 0$ the line is at height $y = 1$ . At $x = 2$ it is at height $y = \frac{2}{3} + 1 = \frac53$ . Both heights are between $0$ and $2$ , so the line goes in through the left side of the square and out through the right side.

2. Find the area of the lower region. Below the line is a trapezoid with vertical sides of lengths $1$ and $\frac53$ and width $2$ :
$\text{area} = \frac{1 + \frac53}{2} \times 2 = \frac83.$
Each half must have area $\frac83 \div 2 = \frac43$ .

3. Find the area to the left of $x = a$ . This piece is also a trapezoid. Its left side has height $1$ , its right side has height $\frac{a}{3} + 1$ , and its width is $a$ :
$\text{area} = \frac{1 + \left(\frac a3 + 1\right)}{2} \times a = \left(1 + \frac{a}{6}\right) a = a + \frac{a^2}{6}.$

4. Set it equal to $\frac43$ and solve. Multiply both sides of $a + \frac{a^2}{6} = \frac43$ by $6$ :
$a^2 + 6a = 8.$
The answer is supposed to look like $\sqrt{s} - t$ , which hints at completing the square. Add $9$ to both sides:
$a^2 + 6a + 9 = 17, \qquad (a + 3)^2 = 17.$
Since $a$ is positive, $a + 3 = \sqrt{17}$ , so $a = \sqrt{17} - 3 \approx 1.12$ . That is between $0$ and $2$ , as it should be.

So $s = 17$ , $t = 3$ , and $s + t = 20$ .

The answer is $\boxed{\textbf{(C)}\ 20}$ .

Why this works

Checking the line's height at both edges tells you the shape of the region, and a trapezoid has an easy area formula. Cutting it with a vertical line gives a smaller trapezoid whose area is a quadratic in $a$ . When an answer is supposed to be of the form $\sqrt{s} - t$ , completing the square is almost always the intended finish.

Concise solution

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

First locate the line inside the square. At $x = 0$ it is at $y = 1$ ; at $x = 2$ it is at $y = \tfrac53$ . Both heights lie strictly between $0$ and $2$ , so the line enters through the left edge and leaves through the right edge, and the lower region is the trapezoid with vertices $(0,0)$ , $(2,0)$ , $\left(2, \tfrac53\right)$ , $(0,1)$ . Its area is
$\frac{1 + \tfrac53}{2}\cdot 2 = \frac{8}{3},$
so each half must have area $\tfrac43$ .

The part of the lower region with $0 \le x \le a$ is again a trapezoid: its parallel sides are the heights of the line at $x = 0$ and $x = a$ , namely $1$ and $\tfrac{a}{3} + 1$ , and its width is $a$ . Hence
$\frac{1 + \left(\tfrac{a}{3} + 1\right)}{2}\cdot a = a + \frac{a^2}{6}.$

Set this equal to $\tfrac43$ and multiply through by $6$ :
$a^2 + 6a = 8 \quad\Longrightarrow\quad a^2 + 6a + 9 = 17 \quad\Longrightarrow\quad (a+3)^2 = 17.$
Since $a > 0$ , $a = \sqrt{17} - 3 \approx 1.12$ , which indeed lies in $[0,2]$ . So $s = 17$ , $t = 3$ , and $s + t = 20$ .

The answer is $\boxed{\textbf{(C)}\ 20}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Algebra, Chapter 10: Quadratic Equations – Part 1 (§10.2, §10.3, §10.4)
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2, §4.3)