AMC 10 · 2003 · #22
Grade 8 geometry-2dIn rectangle ABCD, we have AB=8, BC=9, H is on BC with BH=6, E is on AD with DE=4, line EC intersects line AH at G, and F is on line AD with GF⊥AF. Find the length of GF.
Pick an answer.
AMC 10 2003 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: In rectangle $ABCD$ with $AB=8$ and $BC=9$, point $H$ lies on $BC$ with $BH=6$ and point $E$ lies on $AD$ with $DE=4$. Line $EC$ and line $AH$ cross at $G$, and $F$ is the point on line $AD$ for which $GF\perp AF$. Find the length $GF$.
Givens: Rectangle $ABCD$ with $AB=8$ and $BC=9$, so $AD=9$ and $CD=8$; $H$ is on $BC$ with $BH=6$; $E$ is on $AD$ with $DE=4$; Line $EC$ and line $AH$ meet at $G$; $F$ is on line $AD$ with $GF\perp AF$; Answer choices: (A) 16, (B) 20, (C) 24, (D) 28, (E) 30
Unknowns: The length of the segment $GF$
Understand
Restated: In rectangle $ABCD$ with $AB=8$ and $BC=9$, point $H$ lies on $BC$ with $BH=6$ and point $E$ lies on $AD$ with $DE=4$. Line $EC$ and line $AH$ cross at $G$, and $F$ is the point on line $AD$ for which $GF\perp AF$. Find the length $GF$.
Givens: Rectangle $ABCD$ with $AB=8$ and $BC=9$, so $AD=9$ and $CD=8$; $H$ is on $BC$ with $BH=6$; $E$ is on $AD$ with $DE=4$; Line $EC$ and line $AH$ meet at $G$; $F$ is on line $AD$ with $GF\perp AF$; Answer choices: (A) 16, (B) 20, (C) 24, (D) 28, (E) 30
Plan
Primary tool: #1 Draw a Diagram
Secondary: #16 Change Focus / Count the Complement, #13 Convert to Algebra, #7 Identify Subproblems
The figure is nothing but straight lines, so Tool #1 (Draw a Diagram) is strongest when the diagram is a coordinate grid: drop $D$ at the origin and lay line $AD$ along the $x$-axis. That single choice pays off through Tool #16 (Change Focus): since $F$ is on line $AD$ (the $x$-axis) and $GF\perp AF$, the segment $GF$ points straight up, so its length is simply how high $G$ sits above the $x$-axis — the $y$-coordinate of $G$. We never have to locate $F$ at all. From there Tool #13 (Convert to Algebra) turns each of the two given lines into an equation, and Tool #7 (Identify Subproblems) isolates the real task: find where those two lines cross and read off its height.
Execute — Answer: B
6.NS.C.8 Step 1 Put the figure on a coordinate grid
- Place $D$ at the origin with line $AD$ along the $x$-axis.
- Reading the side lengths off the rectangle gives $D=(0,0)$, $A=(9,0)$, $B=(9,8)$, and $C=(0,8)$.
- Because $BH=6$ measured from $B$ toward $C$, we get $H=(3,8)$; because $DE=4$ measured from $D$ toward $A$, we get $E=(4,0)$.
- Now notice what $F$ is: it sits on line $AD$, which is the $x$-axis, and $GF\perp AF$ forces $GF$ to be vertical.
- So $GF$ is just how high $G$ stands above the $x$-axis — the $y$-coordinate of $G$.
- The whole problem collapses to one question: what is the $y$-coordinate of the point where lines $EC$ and $AH$ cross?
💡 When the base line is the $x$-axis, the length of a vertical segment down to it is just the point's height.
8.EE.B.6 Step 2 Write the equation of line $EC$
- Line $EC$ passes through $E=(4,0)$ and $C=(0,8)$.
- Its slope is $\frac{8-0}{0-4}=-2$.
- Point $C=(0,8)$ is already on the $y$-axis, so it is the $y$-intercept, and the line is $y=-2x+8$.
💡 Slope is the rise over the run between two known points, and the intercept is where the line meets the $y$-axis.
8.EE.B.6 Step 3 Write the equation of line $AH$
- Line $AH$ passes through $A=(9,0)$ and $H=(3,8)$.
- Its slope is $\frac{8-0}{3-9}=\frac{8}{-6}=-\frac{4}{3}$.
- Using point $A=(9,0)$ in point-slope form, $y-0=-\frac{4}{3}(x-9)$, which simplifies to $y=-\frac{4}{3}x+12$.
💡 Two points fix a line: the slope sets its tilt and one point pins it in place.
8.EE.C.8 Step 4 Cross the two lines and read the height
- The point $G$ lies on both lines, so set the two right-hand sides equal: $-2x+8=-\frac{4}{3}x+12$.
- Multiply every term by $3$ to clear the fraction: $-6x+24=-4x+36$.
- Collecting terms gives $-2x=12$, so $x=-6$.
- Substitute back into the simpler line $y=-2x+8$: $y=-2(-6)+8=20$.
- So the lines cross at $G=(-6,20)$, and its height above line $AD$ is $20$.
- Since $GF$ equals that height, $GF=20$, which is choice (B).
💡 The single point on both lines is the one $(x,y)$ pair that satisfies both equations at the same time.
6.NS.C.8 Place $D$ at the origin with line $AD$ along the $x$-axis. Reading the side leng 8.EE.B.6 Line $EC$ passes through $E=(4,0)$ and $C=(0,8)$. Its slope is $\frac{8-0}{0-4}= 8.EE.B.6 Line $AH$ passes through $A=(9,0)$ and $H=(3,8)$. Its slope is $\frac{8-0}{3-9}= 8.EE.C.8 The point $G$ lies on both lines, so set the two right-hand sides equal: $-2x+8= Review
Reasonableness: The crossing point $G=(-6,20)$ lands well to the upper-left of the rectangle, which matches the picture: the two lines both lean up-and-to-the-left and meet far outside the figure. The value $20$ is one of the listed choices. A quick coordinate-free check confirms it: segment $CH$ (from $C=(0,8)$ to $H=(3,8)$) is horizontal with length $3$, and segment $EA$ (from $E=(4,0)$ to $A=(9,0)$) is horizontal with length $5$, so $CH\parallel EA$. Since $C$ and $H$ lie on the two lines through $G$, triangle $GCH$ is similar to triangle $GEA$ with ratio $\frac{CH}{EA}=\frac{3}{5}$. If $GF=h$ is the height of $G$ above $EA$, then $h-8$ is its height above $CH$, so $\frac{h-8}{h}=\frac{3}{5}$, giving $5(h-8)=3h$ and $h=20$ — the same answer.
Alternative: Skip coordinates entirely and lean on similar triangles. Because $C$ lies on line $EC$ (through $G$) and $H$ lies on line $AH$ (through $G$), and $CH$ is parallel to $EA$ (both are horizontal in the rectangle), triangles $GCH$ and $GEA$ are similar. Their base ratio is $\frac{CH}{EA}=\frac{3}{5}$, where $CH=BC-BH-\dots$ is found from the rectangle ($CH=3$) and $EA=AD-DE=9-4=5$. The heights of $G$ above the two parallel bases are in the same ratio $\frac{3}{5}$; since $CH$ sits $8$ above $EA$, writing $\frac{h-8}{h}=\frac{3}{5}$ and solving gives $h=GF=20$ in a single equation, with no slopes or intercepts.
CCSS standards used (min grade 8)
6.NS.C.8Solve real-world problems by graphing points in all four quadrants (Placing the rectangle on a coordinate grid ($D$ at the origin, line $AD$ on the $x$-axis) and reading off $A,B,C,H,E$, with $G$ landing in the second quadrant.)8.EE.B.6Use similar triangles to explain why the slope is the same between any two points (Computing the slopes of lines $EC$ and $AH$ from two points each and writing their equations $y=-2x+8$ and $y=-\frac{4}{3}x+12$.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Setting the two line equations equal to find their intersection $G=(-6,20)$, whose height gives $GF=20$.)
⭐ To find a vertical distance to a line, make that line the $x$-axis — the distance becomes the point's height, and the problem turns into finding where two lines cross.
⭐ To find a vertical distance to a line, make that line the $x$-axis — the distance becomes the point's height, and the problem turns into finding where two lines cross.
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