AMC 10 · 2009 · #16

Grade 10 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
similar-trianglesisosceles-triangleangle-sum-triangleratio-proportion identify-subproblemsconvert-to-algebra ↑ Prerequisites: similar-trianglesratio-proportion
📏 Long solution 💡 3 insights
Problem
A trapezoid has one diagonal of known length, two known angles at its ends, and a known ratio of parallel sides. Find the length of one leg.

Pick an answer.

(A)
$\frac 79$
(B)
$\frac 45$
(C)
$\frac {13}{15}$
(D)
$\frac 89$
(E)
$\frac {14}{15}$
How to solve
Strategy Draw a Diagram

Tool #1 (Draw a Diagram): the given angles live at opposite ends of the diagonal and face opposite legs, so nothing in the figure as drawn ties them together. Extending the two legs until they meet adds the one point that puts both angles into a single triangle — this is the move the whole problem is built around, and picking the right auxiliary point is a drawing decision, not a computation. Tool #15 (Organize Information in More Ways): BC:AD = 9:5 compares two segments that sit at opposite ends of the figure; the same fact re-read through similar triangles becomes a comparison of two pieces of one straight line, where lengths simply add. Tool #7 (Identify Subproblems): after the construction, the work splits into two independent small questions — how long is DE (an angle-chase inside one triangle) and what is ED:EC (a similarity) — which are answered separately and then multiplied together. Tool #4 (Introduce a Variable): naming CD = x turns the final relation into one linear equation. Tool #11 (Work Backwards): everything up to that point proves only that IF such a trapezoid exists THEN CD has one value; running the construction in reverse from the isosceles triangle builds a figure meeting all four conditions and shows the value is actually attained.

1STEP 1

The legs cannot be parallel

The two legs cannot be parallel, so they meet at a point.

AB ∥ DC → ∠ ABD = ∠ BDC → 23° = 46°, false. So AB and DC meet at a point E.
2STEP 2

Pin down where E sits

Similar triangles locate that meeting point.

△ EAD ∼ △ EBC (AA) → ED/EC = AD/BC = 5/9 ＜ 1, so the order is E, A, B and E, D, C.
3STEP 3

The doubling makes DE = DB

The doubling makes a triangle isosceles.

∠ BED = ∠ BDC - ∠ DBE = 46° - 23° = 23° = ∠ DBE → DE = DB = 1
4STEP 4

Add along the line and solve

Adding along that line gives the leg as 4/5.

EC/ED = (1 + x)/1 = 9/5 → x = 9/5 - 1 = 4/5
5STEP 5

Build the trapezoid to prove it exists

Building the trapezoid proves it exists, choice (C).

DB = DE = 1, ∠ BDE = 134°; EC = 9/5, EA = 5/9EB → AD ∥ BC, BC/AD = 9/5, BD = 1, ∠ DBA = 23°, ∠ BDC = 46°, CD = 4/5
Answer
4/5
First a crude size check that uses no construction: CD and BD are two sides of △ BCD meeting at D in an angle of 46°, so CD should be the same order of magnitude as BD = 1, and every choice lies in [0.78, 0.93]. That is the problem's design — the choices are packed too close to separate by feel, so a second exact route is needed rather than an estimate. Solve the trapezoid directly instead. Write φ = ∠ BCD; since AD ∥ BC, the diagonal gives ∠ ADB = ∠ DBC = 134° - φ, and the Law of Sines in the two triangles gives AD = (sin 23°)/(sin(23° + φ)), BC = (sin 46°)/(sin φ), CD = (sin(134° - φ))/(sin φ). Forcing BC/AD = 9/5 gives φ ≈ 81.67°, hence ∠ DBC = ∠ ADB ≈ 52.33°, ∠ BAD ≈ 104.67°, ∠ ADC ≈ 98.33° — all positive, with ∠ BAD + ∠ ABC ≈ 104.67° + 75.33° = 180° and ∠ ADC + ∠ BCD ≈ 98.33° + 81.67° = 180°, exactly as co-interior angles on the parallel sides must. So the figure is a real convex trapezoid. Its measurements come out AD ≈ 0.40389, BC ≈ 0.72700 (ratio 1.80000), and CD ≈ 0.80000, matching 4/5 to five decimals by a route that never mentions E. Choice check: since CD = BC/AD - 1, each wrong choice corresponds to a different side ratio — 7/9 would need 16/9, 13/15 would need 28/15, 8/9 would need 17/9, 14/15 would need 29/15 — and none of those is 9:5. Finally a boundary check on the relation itself: as BC/AD → 1 the formula sends CD → 0, and indeed a trapezoid whose parallel sides become equal is a parallelogram, where C has collapsed onto D and the legs no longer meet. The formula fails in exactly the case Step 1 had to exclude, which is the right place for it to fail.
💡Key takeaway

When one given angle is exactly double the other, extend the trapezoid's slanted sides to the point where they meet: the doubling collapses into an isosceles triangle, and the side ratio then reads off a single straight line.

  • The legs cannot be parallel
  • Pin down where E sits
  • The doubling makes DE = DB
  • Add along the line and solve
  • Build the trapezoid to prove it exists

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