AMC 10 · 2013 · #18
Grade 8 geometry-2dLet points A=(0,0), B=(1,2), C=(3,3), and D=(4,0). Quadrilateral ABCD is cut into equal area pieces by a line passing through A. This line intersects CD at point (qp,sr), where these fractions are in lowest terms. What is p+q+r+s?
Pick an answer.
AMC 10 2013 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: Four points $A=(0,0)$, $B=(1,2)$, $C=(3,3)$, $D=(4,0)$ form quadrilateral $ABCD$. A line through $A$ splits it into two pieces of equal area. That line meets side $\overline{CD}$ at a point $\left(\tfrac{p}{q},\tfrac{r}{s}\right)$ in lowest terms. Find $p+q+r+s$.
Givens: Vertices $A=(0,0)$, $B=(1,2)$, $C=(3,3)$, $D=(4,0)$ in order; A cutting line passes through vertex $A$; The line divides $ABCD$ into two equal-area pieces; The line crosses side $\overline{CD}$ at $\left(\tfrac{p}{q},\tfrac{r}{s}\right)$, fractions in lowest terms; Answer choices: (A) 54, (B) 58, (C) 62, (D) 70, (E) 75
Unknowns: The intersection point on $\overline{CD}$ and the sum $p+q+r+s$ of its reduced-fraction parts
Understand
Restated: Four points $A=(0,0)$, $B=(1,2)$, $C=(3,3)$, $D=(4,0)$ form quadrilateral $ABCD$. A line through $A$ splits it into two pieces of equal area. That line meets side $\overline{CD}$ at a point $\left(\tfrac{p}{q},\tfrac{r}{s}\right)$ in lowest terms. Find $p+q+r+s$.
Givens: Vertices $A=(0,0)$, $B=(1,2)$, $C=(3,3)$, $D=(4,0)$ in order; A cutting line passes through vertex $A$; The line divides $ABCD$ into two equal-area pieces; The line crosses side $\overline{CD}$ at $\left(\tfrac{p}{q},\tfrac{r}{s}\right)$, fractions in lowest terms; Answer choices: (A) 54, (B) 58, (C) 62, (D) 70, (E) 75
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #7 Identify Subproblems, #13 Convert to Algebra
The heart of the problem is the unknown crossing point on $\overline{CD}$, so Tool #4 (Introduce a Variable) is primary: call the point $E=(x,y)$ and turn "equal areas" into an equation for it. Tool #1 (Draw a Diagram) is used first to see that the line from $A$ to a point on $\overline{CD}$ cuts off triangle $AED$ sitting on the x-axis. Tool #7 (Identify Subproblems) splits the work into two clean pieces — find the whole area, then match half of it. Tool #13 (Convert to Algebra) writes side $\overline{CD}$ as a line equation so the point's $x$-coordinate can be solved once its height is known.
Execute — Answer: B
6.G.A.3 Step 1 Plot the points and the cut
- Place the four points.
- $A$ and $D$ both sit on the x-axis, so side $\overline{AD}$ is flat along the bottom with length $4$.
- A line drawn from $A$ to a point $E$ on side $\overline{CD}$ cuts the quadrilateral into two parts: triangle $AED$ hugging the bottom-right, and the four-sided piece $ABCE$ above it.
- Making those two areas equal is the whole task.
💡 Seeing which two pieces the line makes tells you exactly which area to control.
6.G.A.1 Step 2 Find the whole area
- Use the shoelace formula on $A,B,C,D$ in order.
- Multiply each $x$ by the next $y$ and subtract each next $x$ times $y$: $(0\cdot2-1\cdot0)+(1\cdot3-3\cdot2)+(3\cdot0-4\cdot3)+(4\cdot0-0\cdot0)=0-3-12+0=-15$.
- Take half the absolute value: the area is $\tfrac{15}{2}$.
- So each equal piece must have area $\tfrac12\cdot\tfrac{15}{2}=\tfrac{15}{4}$.
💡 The shoelace formula reads the area straight off the corner coordinates.
6.EE.B.7 Step 3 Set the triangle's area to half
- Let the crossing point be $E=(x,y)$.
- Triangle $AED$ has base $\overline{AD}=4$ lying on the x-axis, so its height is just the height of $E$ above that axis, namely $y$.
- Its area is $\tfrac12\cdot4\cdot y=2y$.
- This triangle must be one of the two equal halves, so set $2y=\tfrac{15}{4}$, giving $y=\tfrac{15}{8}$.
- That is the $y$-coordinate of $E$.
💡 Because the base sits on the axis, the point's height alone sets the triangle's area.
8.EE.B.6 Step 4 Locate the point on side CD
- Now find $x$ from the fact that $E$ lies on $\overline{CD}$.
- The segment runs from $C=(3,3)$ to $D=(4,0)$: its slope is $\tfrac{0-3}{4-3}=-3$, so the line is $y=-3x+12$.
- Substitute the known height $y=\tfrac{15}{8}$: $\tfrac{15}{8}=-3x+12$, so $3x=12-\tfrac{15}{8}=\tfrac{81}{8}$ and $x=\tfrac{27}{8}$.
- Thus $E=\left(\tfrac{27}{8},\tfrac{15}{8}\right)$, and $\tfrac{27}{8}$ sits between $3$ and $4$ as required.
💡 The point must obey the line of $\overline{CD}$, so its known height pins down its $x$.
6.NS.C.6 Step 5 Add the reduced parts
- The point is $\left(\tfrac{27}{8},\tfrac{15}{8}\right)$.
- Both fractions are already in lowest terms: $27$ and $8$ share no factor, and $15$ and $8$ share none.
- So $p=27,\ q=8,\ r=15,\ s=8$, and $p+q+r+s=27+8+15+8=58$.
- The answer is $\textbf{(B)}$.
💡 Once the fractions are fully reduced, just add the four whole numbers.
6.G.A.3 Place the four points. $A$ and $D$ both sit on the x-axis, so side $\overline{AD 6.G.A.1 Use the shoelace formula on $A,B,C,D$ in order. Multiply each $x$ by the next $y 6.EE.B.7 Let the crossing point be $E=(x,y)$. Triangle $AED$ has base $\overline{AD}=4$ l 8.EE.B.6 Now find $x$ from the fact that $E$ lies on $\overline{CD}$. The segment runs fr 6.NS.C.6 The point is $\left(\tfrac{27}{8},\tfrac{15}{8}\right)$. Both fractions are alre Review
Reasonableness: Check the height is possible: side $\overline{CD}$ falls from $y=3$ at $C$ down to $y=0$ at $D$, and $\tfrac{15}{8}=1.875$ lies between them, so $E$ really is on the segment (its $x=\tfrac{27}{8}=3.375$ is between $3$ and $4$ too). Confirm equal areas: triangle $AED$ has area $2\cdot\tfrac{15}{8}=\tfrac{15}{4}$, exactly half of $\tfrac{15}{2}$, so the other piece is the matching half. The parts sum to $58$, choice (B); the other choices don't come from any reduced form of this point.
Alternative: Instead of a line equation, use proportion along $\overline{CD}$. Going from $C$ to $D$ the height drops by $3$ (from $3$ to $0$) while $x$ rises by $1$. The needed drop is $3-\tfrac{15}{8}=\tfrac{9}{8}$, which is $\tfrac{9/8}{3}=\tfrac38$ of the way down, so $x=3+\tfrac38=\tfrac{27}{8}$ — the same point, giving $58$ again.
CCSS standards used (min grade 8)
6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Plotting the four vertices to see that AD lies on the x-axis and the line from A cuts off triangle AED.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Using the shoelace formula to get the quadrilateral's area of 15/2, so each half is 15/4.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Solving 2y = 15/4 for the height y = 15/8 of the intersection point.)8.EE.B.6Use similar triangles to explain why the slope is the same between any two points (Writing side CD as the line y = -3x + 12 and solving for x = 27/8 at the known height.)6.NS.C.6Understand a rational number as a point on the number line (Reading the reduced fraction coordinates (27/8, 15/8) and summing the parts to 58.)
⭐ Since AD lies flat on the x-axis, the cut's triangle has area just $2\times$ its height, so make that equal half the total, then ride the line of CD to find the exact point.
⭐ Since AD lies flat on the x-axis, the cut's triangle has area just $2\times$ its height, so make that equal half the total, then ride the line of CD to find the exact point.
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