AMC 10 · 2025 · #6
Grade 8 geometry-2dThe line y=31x+1 divides the square region defined by 0≤x≤2 and 0≤y≤2 into an upper region and a lower region. The line x=a divides the lower region into two regions of equal area. Then a can be written as s−t, where s and t are positive integers. What is s+t?
Pick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Inside the square with $0 \le x \le 2$ and $0 \le y \le 2$, the line $y = \frac{1}{3}x + 1$ cuts off a lower region (the part of the square below the line). A vertical line $x = a$ splits that lower region into two pieces of equal area. Writing $a = \sqrt{s} - t$ with $s$ and $t$ positive integers, find $s + t$.
Givens: The square is $0 \le x \le 2$ and $0 \le y \le 2$; The dividing line is $y = \frac{1}{3}x + 1$; The vertical line $x = a$ cuts the lower region into two equal areas; $a = \sqrt{s} - t$ with $s, t$ positive integers
Unknowns: The value of $s + t$
Understand
Restated: Inside the square with $0 \le x \le 2$ and $0 \le y \le 2$, the line $y = \frac{1}{3}x + 1$ cuts off a lower region (the part of the square below the line). A vertical line $x = a$ splits that lower region into two pieces of equal area. Writing $a = \sqrt{s} - t$ with $s$ and $t$ positive integers, find $s + t$.
Givens: The square is $0 \le x \le 2$ and $0 \le y \le 2$; The dividing line is $y = \frac{1}{3}x + 1$; The vertical line $x = a$ cuts the lower region into two equal areas; $a = \sqrt{s} - t$ with $s, t$ positive integers
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
The unknown $a$ is exactly the number to name and chase. The plan is: first draw the picture to see that the lower region is a trapezoid, then find its total area, then write the area of the left piece (from $x=0$ to $x=a$) as an expression in $a$, set that equal to half the total, and solve the resulting equation. Naming $a$ turns 'cut the area in half' into an equation we can solve, and the answer comes out in the $\sqrt{s} - t$ form the problem asks for.
Execute — Answer: C
6.G.A.3 Step 1 Draw the region and name its corners
- Plot the line $y = \frac{1}{3}x + 1$ on the square.
- At $x = 0$ it is at height $1$, and at $x = 2$ it is at height $\frac{1}{3}(2) + 1 = \frac{5}{3}$.
- Both heights are below the top of the square, so the line runs from $(0,1)$ across to $(2, \frac{5}{3})$.
- The lower region sits below the line and above the $x$-axis, with corners $(0,0)$, $(2,0)$, $(2, \frac{5}{3})$, and $(0,1)$.
- Its left and right edges are vertical and parallel, so the region is a trapezoid.
💡 Drawing the line inside the square shows the lower region is just a trapezoid standing on the $x$-axis.
6.G.A.1 Step 2 Find the total lower area
- A trapezoid's area is the average of its two parallel sides times the distance between them.
- Here the two vertical (parallel) sides have lengths $1$ (the left edge, from $y=0$ to $y=1$) and $\frac{5}{3}$ (the right edge, from $y=0$ to $y=\frac{5}{3}$), and they are $2$ apart.
- So the whole lower region has area $\frac{1}{2}\left(1 + \frac{5}{3}\right)(2) = \frac{8}{3}$.
- Half of that area is $\frac{4}{3}$, which is the target for the left piece.
💡 The area of a trapezoid is just its average height times its width.
6.EE.B.6 Step 3 Write the left piece's area in terms of a
- The vertical line $x = a$ cuts off a smaller trapezoid on the left, with corners $(0,0)$, $(a,0)$, $(a, \frac{1}{3}a + 1)$, and $(0,1)$.
- Its two parallel vertical sides have lengths $1$ (at $x=0$) and $\frac{1}{3}a + 1$ (at $x=a$), and they are $a$ apart.
- So its area is $\frac{1}{2}\left(1 + \left(\frac{1}{3}a + 1\right)\right)a$.
- Set this equal to the target $\frac{4}{3}$.
💡 Using the same trapezoid rule, but with the movable edge at $x=a$, makes the area a formula in $a$.
7.EE.B.4 Step 4 Turn it into a clean equation
- Expand the left side: $\frac{1}{2}\left(2 + \frac{a}{3}\right)a = a + \frac{a^2}{6}$.
- So $a + \frac{a^2}{6} = \frac{4}{3}$.
- Multiply every term by $6$ to clear the fractions, which gives $6a + a^2 = 8$.
- Move everything to one side to get a tidy equation in $a$.
💡 Clearing fractions turns the area condition into a plain equation with no denominators.
8.EE.A.2 Step 5 Solve for a and read off the answer
- Complete the square.
- From $a^2 + 6a = 8$, add $9$ to both sides so the left side becomes a perfect square: $a^2 + 6a + 9 = 17$, that is $(a+3)^2 = 17$.
- Taking the square root gives $a + 3 = \sqrt{17}$ (the positive root, since $a$ must be between $0$ and $2$), so $a = \sqrt{17} - 3 \approx 1.12$, which indeed lies in the region.
- Matching $a = \sqrt{s} - t$ gives $s = 17$ and $t = 3$, so $s + t = 20$.
- This is choice (C).
💡 Completing the square repackages the equation as $(a+3)^2 = 17$, so a single square root unlocks $a$.
6.G.A.3 Plot the line $y = \frac{1}{3}x + 1$ on the square. At $x = 0$ it is at height $ 6.G.A.1 A trapezoid's area is the average of its two parallel sides times the distance b 6.EE.B.6 The vertical line $x = a$ cuts off a smaller trapezoid on the left, with corners 7.EE.B.4 Expand the left side: $\frac{1}{2}\left(2 + \frac{a}{3}\right)a = a + \frac{a^2} 8.EE.A.2 Complete the square. From $a^2 + 6a = 8$, add $9$ to both sides so the left side Review
Reasonableness: Check that $a = \sqrt{17} - 3 \approx 1.12$ sits between $0$ and $2$: it does, so the vertical cut really falls inside the region. It also lands just past the middle, which makes sense because the region is slightly taller on the right, so the left half has to be a bit wider to hold the same area. Plugging back, the left area is $\frac{1}{2}(2 + \frac{1.12}{3})(1.12) \approx 1.33 = \frac{4}{3}$, exactly half of $\frac{8}{3}$. With $s = 17$ and $t = 3$, $s + t = 20$ is choice (C).
Alternative: Use calculus: the lower area up to $x = a$ is $\int_0^a \left(\frac{1}{3}x + 1\right)dx = \frac{1}{6}a^2 + a$. Set it equal to half the total, $\frac{4}{3}$, giving $\frac{1}{6}a^2 + a = \frac{4}{3}$, the same equation $a^2 + 6a - 8 = 0$, and the same answer $a = \sqrt{17} - 3$.
CCSS standards used (min grade 8)
6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Plotting the line inside the square and finding the corner points of the lower region to see it is a trapezoid.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Using the trapezoid area rule to compute the total lower area $\frac{8}{3}$ and its half $\frac{4}{3}$.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming the cut position $a$ and writing the left piece's area as an expression in $a$.)7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Setting the left area equal to $\frac{4}{3}$ and clearing fractions to get $a^2 + 6a - 8 = 0$.)8.EE.A.2Use square root and cube root symbols to represent solutions (Completing the square to $(a+3)^2 = 17$ and taking a square root to get $a = \sqrt{17} - 3$.)
⭐ Turn 'cut the area in half' into an equation by naming the cut $a$, write the left trapezoid's area, set it to half the total, and solve to get $a = \sqrt{17} - 3$.
⭐ Turn 'cut the area in half' into an equation by naming the cut $a$, write the left trapezoid's area, set it to half the total, and solve to get $a = \sqrt{17} - 3$.
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