2013 AMC 10A Problem 18

P16-20~5 minPrint

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

Let 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 $\overline{CD}$ at point $\bigg(\frac{p}{q}, \frac{r}{s}\bigg)$ , where these fractions are in lowest terms. What is $p+q+r+s$ ?

Show solution without answering

Key insight

The piece next to D is a triangle with base AD = 4 on the x-axis. Half the area fixes the point's height; line CD gives x.

Solution

1. Total area. Split $ABCD$ with vertical lines at $x = 1$ and $x = 3$ into three simple pieces:
- Triangle from $x = 0$ to $1$ : base $1$ , height $2$ , area $1$ .
- Trapezoid from $x = 1$ to $3$ : parallel sides $2$ and $3$ , width $2$ , area $\tfrac{2 + 3}{2} \times 2 = 5$ .
- Triangle from $x = 3$ to $4$ : base $1$ , height $3$ , area $\tfrac32$ .

Total: $1 + 5 + \tfrac32 = \tfrac{15}{2}$ . Each half must be $\tfrac{15}{4}$ .
2. Look at the piece near $D$ . Let the line from $A$ hit $CD$ at the point $P$ . The piece next to $D$ is the triangle $A\,P\,D$ . Its base $AD$ lies on the $x$ -axis and has length $4$ . Its height is just the $y$ -coordinate of $P$ .
3. Find the height. $\tfrac12 \times 4 \times y = \tfrac{15}{4}$ , so $2y = \tfrac{15}{4}$ and $y = \tfrac{15}{8}$ .
4. Find $x$ using the side $CD$ . From $C = (3, 3)$ to $D = (4, 0)$ , $y$ drops $3$ while $x$ grows $1$ . Starting at $D$ and going back up to height $\tfrac{15}{8}$ , $x$ goes back by $\tfrac{15}{8} \div 3 = \tfrac{5}{8}$ . So
$x = 4 - \frac58 = \frac{27}{8}.$
5. Answer. $P = \left(\tfrac{27}{8}, \tfrac{15}{8}\right)$ . Both fractions are in lowest terms (the tops are odd, the bottom is $8$ ). So $p + q + r + s = 27 + 8 + 15 + 8 = 58$ .

The answer is $\boxed{\textbf{(B)}\ 58}$ .

Why this works

A line through a corner that cuts the area in half cuts off a triangle on one side. If that triangle has its base on an axis, its area depends only on how high the unknown point is. So one area equation gives one coordinate, and the side the point lies on gives the other. Choosing the piece with the easy base ( $AD$ on the $x$ -axis) keeps the work short.

Concise solution

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

First find the area of $ABCD$ by the shoelace formula on $(0,0), (1,2), (3,3), (4,0)$ :
$\frac12\left|0\cdot2 - 1\cdot0 + 1\cdot3 - 3\cdot2 + 3\cdot0 - 4\cdot3 + 4\cdot0 - 0\cdot0\right| = \frac12\left|0 - 3 - 12\right| = \frac{15}{2}.$
Each piece must have area $\frac{15}{4}$ .

Let the cutting line meet $\overline{CD}$ at $P$ . The piece on the $D$ side is triangle $APD$ , whose base $AD$ lies on the $x$ -axis with length $4$ . Its area is $\frac12 \cdot 4 \cdot y_P = 2y_P$ , so $2y_P = \frac{15}{4}$ and $y_P = \frac{15}{8}$ .

Now use that $P$ is on line $CD$ . From $C = (3,3)$ to $D = (4,0)$ the slope is $-3$ , so the line is $y = -3(x - 4)$ , i.e. $x = 4 - \frac{y}{3}$ . With $y = \frac{15}{8}$ :
$x = 4 - \frac{15}{24} = 4 - \frac{5}{8} = \frac{27}{8}.$

So $P = \left(\frac{27}{8}, \frac{15}{8}\right)$ , both fractions in lowest terms, and $p + q + r + s = 27 + 8 + 15 + 8 = 58$ .

The answer is $\boxed{\textbf{(B)}\ 58}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2, §4.3)
Introduction to Algebra, Chapter 8: Graphing Lines