AMC 10 · 2003 · #22

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Point as the Answer
coordinate-geometryslope-interceptsystems-of-equations convert-to-algebra ↑ Prerequisites: coordinate-geometry
📏 Long solution 💡 3 insights 📊 Diagram
Problem
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 ⊥ AF. Find the length GF.

Pick an answer.

(A)
16
(B)
20
(C)
24
(D)
28
(E)
30

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

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 ⊥ 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.

1STEP 1

Put the figure on a coordinate grid

Put D at the origin with line AD on the x-axis, so H=(3,8) and E=(4,0) — and GF is just the height of G.

D=(0,0), A=(9,0), B=(9,8), C=(0,8), H=(3,8), E=(4,0)
2STEP 2

Write the equation of line EC

Line EC runs through E=(4,0) and C=(0,8): slope -2, y-intercept 8, so y=-2x+8.

slope=(8-0)/(0-4)=-2 → y=-2x+8
3STEP 3

Write the equation of line AH

Line AH runs through A=(9,0) and H=(3,8): slope −43-\frac{4}{3}, so point-slope gives y=−43x+12y=-\frac{4}{3}x+12.

slope=8/(3-9)=-4/3 → y=-4/3x+12
4STEP 4

Cross the two lines and read the height

Solving -2x+8=−43x+12-\frac{4}{3}x+12 gives x=-6, so y=-2(-6)+8=20 — that height is GF, choice (B).

-2x+8=-4/3x+12 → x=-6, y=20 → GF=20 (B)
Answer
20
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 ∥ EA. Since C and H lie on the two lines through G, triangle GCH is similar to triangle GEA with ratio CH/EA=3/5. If GF=h is the height of G above EA, then h-8 is its height above CH, so (h-8)/h=3/5, giving 5(h-8)=3h and h=20 — the same answer.
💡Key takeaway

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.

  • Put the figure on a coordinate grid
  • Write the equation of line EC
  • Write the equation of line AH
  • Cross the two lines and read the height

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