AMC 10 · 2009 · #17
Grade 6 geometry-2dFive unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from (c,0) to (3,3), divides the entire region into two regions of equal area. What is c?
Pick an answer.
AMC 10 2009 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: Five unit squares are placed in the coordinate plane with the lower-left corner at the origin, forming the figure shown. A slanted line runs from the point $(c,0)$ on the $x$-axis up to $(3,3)$, and this line splits the whole figure into two pieces of equal area. Find the value of $c$.
Givens: The figure is made of five unit squares, so its total area is $5$.; From the figure, the squares occupy the cells $[0,1]\times[0,1]$, $[1,2]\times[0,1]$, $[1,2]\times[1,2]$, $[2,3]\times[1,2]$, and $[2,3]\times[2,3]$.; The cutting line goes from $(c,0)$ to $(3,3)$.; The line divides the figure into two regions of equal area.; Answer choices: (A) $\frac12$, (B) $\frac35$, (C) $\frac23$, (D) $\frac34$, (E) $\frac45$.
Unknowns: The $x$-coordinate $c$ where the slanted line meets the $x$-axis.
Understand
Restated: Five unit squares are placed in the coordinate plane with the lower-left corner at the origin, forming the figure shown. A slanted line runs from the point $(c,0)$ on the $x$-axis up to $(3,3)$, and this line splits the whole figure into two pieces of equal area. Find the value of $c$.
Givens: The figure is made of five unit squares, so its total area is $5$.; From the figure, the squares occupy the cells $[0,1]\times[0,1]$, $[1,2]\times[0,1]$, $[1,2]\times[1,2]$, $[2,3]\times[1,2]$, and $[2,3]\times[2,3]$.; The cutting line goes from $(c,0)$ to $(3,3)$.; The line divides the figure into two regions of equal area.; Answer choices: (A) $\frac12$, (B) $\frac35$, (C) $\frac23$, (D) $\frac34$, (E) $\frac45$.
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #4 Introduce a Variable
The shaded region has an awkward staircase edge, but it becomes easy once broken into pieces. Tool #7 (Identify Subproblems) is primary: fill in the one missing unit square at the bottom-right so the shaded part plus that square is a single clean triangle, then the shaded area is just (triangle) minus (one square). Tool #1 (Draw a Diagram) reads the exact corner coordinates off the figure so the triangle's base and height are known. Tool #4 (Introduce a Variable) keeps $c$ as the unknown, writes the triangle's area in terms of $c$, and turns the equal-area condition into one linear equation to solve.
Execute — Answer: C
3.MD.C.6 Step 1 Find each region's target area
- The figure is five unit squares, so its total area is $5$.
- The slanted line splits it into two equal pieces, so each piece must have area $\frac52=2.5$.
- In particular the shaded piece (the part below-right of the line) must equal $2.5$.
💡 Cutting an area in half means each side is exactly half of the whole.
3.MD.C.7 Step 2 Complete the shaded part to a triangle
- The shaded region sits below the line from $(c,0)$ to $(3,3)$, bounded on the right by $x=3$ and along the bottom by the $x$-axis, but it has a notch: the cell $[2,3]\times[0,1]$ is empty because no square is drawn there.
- If we imagine filling that one unit square in, the shaded part together with it becomes the full triangle with corners $(c,0)$, $(3,0)$, and $(3,3)$.
- So the shaded area equals the triangle's area minus $1$.
💡 Adding the one missing square straightens the jagged edge into a plain triangle, and area is additive so we just subtract it back.
6.G.A.1 Step 3 Write the triangle's area with c
- The triangle with corners $(c,0)$, $(3,0)$, $(3,3)$ is a right triangle.
- Its base lies on the $x$-axis from $x=c$ to $x=3$, so the base length is $3-c$.
- Its height is the vertical side from $(3,0)$ up to $(3,3)$, so the height is $3$.
- Using area $=\frac12\cdot\text{base}\cdot\text{height}$, the triangle's area is $\frac12(3-c)(3)$.
💡 A right triangle is just half of the rectangle built on its two perpendicular sides.
6.EE.B.7 Step 4 Set up the equation and solve
- The shaded area must be $2.5$, and it equals the triangle minus $1$, so the triangle must have area $3.5$.
- Set $\frac12(3-c)(3)=3.5$.
- Multiplying both sides by $2$ gives $3(3-c)=7$, so $3-c=\frac73$.
- Then $c=3-\frac73=\frac93-\frac73=\frac23$.
- This matches choice $\textbf{(C)}$.
💡 Once the equal-area rule fixes the triangle's area, one equation pins down where the line must start.
3.MD.C.6 The figure is five unit squares, so its total area is $5$. The slanted line spli 3.MD.C.7 The shaded region sits below the line from $(c,0)$ to $(3,3)$, bounded on the ri 6.G.A.1 The triangle with corners $(c,0)$, $(3,0)$, $(3,3)$ is a right triangle. Its bas 6.EE.B.7 The shaded area must be $2.5$, and it equals the triangle minus $1$, so the tria Review
Reasonableness: Plug $c=\frac23$ back in. The triangle base is $3-\frac23=\frac73$ and height $3$, giving area $\frac12\cdot\frac73\cdot3=\frac72=3.5$; subtracting the one empty square leaves shaded area $2.5$, exactly half of $5$. As a second check, the shaded polygon $(\frac23,0),(3,3),(3,1),(2,1),(2,0)$ has area $2.5$ by the shoelace formula, agreeing. The value $\frac23$ is between $0$ and $3$, so the line really does start on the figure's bottom edge, which is sensible.
Alternative: Instead of completing to a triangle, use the shoelace (coordinate) formula directly on the shaded polygon with vertices $(c,0),(3,3),(3,1),(2,1),(2,0)$, set its area equal to $2.5$, and solve the resulting linear equation for $c$; it again gives $c=\frac23$.
CCSS standards used (min grade 6)
3.MD.C.6Measure areas by counting unit squares (Reading the total area as five unit squares $=5$ and halving it to get the $2.5$ target for each region.)3.MD.C.7Relate area to multiplication and addition operations (Using that area is additive to write the shaded region as a full triangle minus the one missing unit square.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the right-triangle area $\frac12(3-c)(3)$ from its base $3-c$ and height $3$.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Solving $\frac12(3-c)(3)=3.5$ for the unknown starting point $c=\frac23$.)
⭐ Fill in the missing square to turn a jagged shape into a clean triangle, find its area, then subtract the square back.
⭐ Fill in the missing square to turn a jagged shape into a clean triangle, find its area, then subtract the square back.
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