AMC 10 · 2005 · #23

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation
area-trapezoidratio-proportionlinear-equations-one-var convert-to-algebraidentify-subproblems ↑ Prerequisites: ratio-proportionlinear-equations-one-var
📏 Medium solution 💡 2 insights
Problem
In trapezoid ABCD the top side AB is parallel to the bottom side DC. Point E is the midpoint of leg BC and point F is the midpoint of leg DA, so segment FE cuts the trapezoid into an upper piece ABEF and a lower piece FECD. The upper piece ABEF has twice the area of the lower piece FECD. Find the ratio AB/DC.

Pick an answer.

(A)
2
(B)
3
(C)
5
(D)
6
(E)
8

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The picture has no numbers, only two unknown bases and a middle line, so Tool #4 (Introduce a Variable) is the natural start: call the top base a=AB and the bottom base b=DC and hunt for one equation relating them. Tool #1 (Draw a Diagram) makes the key feature visible — the segment FE joining the two leg-midpoints is the midsegment, which is parallel to both bases, equals their average, and sits exactly halfway up. Tool #7 (Identify Subproblems) then breaks the job into two clean pieces: first pin down the length of FE and the fact that the two smaller trapezoids have the same height, then turn the 'twice the area' condition into a single linear equation in a and b and read off the ratio.

1STEP 1

Name the two bases

Let a = AB be the top base and b = DC the bottom base; the goal a/b needs just one equation linking a and b.

a = AB, b = DC, goal: a/b
2STEP 2

The middle line is the average, halfway up

FE joins the leg midpoints, so it is the midsegment: FE = (a+b)/2, parallel to both bases and halfway up, giving each piece height h/2.

FE = (a+b)/2, each small height = h/2
3STEP 3

Equal heights turn area into a sum of bases

Equal heights make the areas proportional to the sums of parallel sides: ABEF gives (3a+b)/2 and FECD gives (a+3b)/2.

[ABEF] ∝ a+(a+b)/2=(3a+b)/2, [FECD] ∝ (a+b)/2+b=(a+3b)/2
4STEP 4

Use 'twice as big' and solve

Set (3a+b)/2 = 2·(a+3b)/2, so 3a+b = 2a+6b and a = 5b; dividing by b gives AB/DC = 5, choice (C).

3a+b=2(a+3b) → 3a+b=2a+6b → a=5b → AB/DC=5 → (C)
Answer
5
A quick sanity test: with a=5 and b=1 the midsegment is (5+1)/2=3. The upper piece then behaves like 5+3=8 and the lower like 3+1=4, and 8 really is twice 4 — the condition checks out and gives AB/DC=5. The ratio also makes sense in the picture: the top must be much wider than the bottom for its half of the trapezoid to carry twice the area. A common trap is to forget that the midsegment sits halfway and instead compare only the bases a:b, or to set the lower piece to be the doubled one; both mis-steps land on the decoy answers like 2 or 3.
💡Key takeaway

The line joining the midpoints of a trapezoid's slanted sides is the average of the two bases and sits halfway up, so the two halves have the same height and their areas compare just by the length of their parallel sides.

  • Name the two bases
  • The middle line is the average, halfway up
  • Equal heights turn area into a sum of bases
  • Use 'twice as big' and solve

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