AMC 10 · 2021 · #17
Grade 9 geometry-2d
Pick an answer.
Tool #4 (Introduce a Variable) — nothing in the picture has a number attached except the four areas, so name what matters: let a = AB, b = CD, and let h₁, h₂ be the distances from P up to CD and down to AB. Tool #16 (Change Focus) — the two triangles that sit on the parallel bases are the useful ones, because each has a base that is one of the two unknowns and a height that is one of h₁, h₂. The two slanted triangles have no clean base or height, so do not chase them individually; use them only through the total area 14. Tool #7 (Identify Subproblems) supplies the link that closes the system: h₁ + h₂ is exactly the trapezoid's height. Tool #13 (Convert to Algebra) turns the three area facts into one equation in a and b. Tool #15 (Organize Information in More Ways) then rewrites that equation in the single quantity actually asked for, r = a/b, and Tool #3 (Eliminate Possibilities) throws out the root that violates AB ＞ CD.
Label the four areas
Label the four areas on the picture.
Four pieces that fit together with no gaps and no overlaps have areas that simply add.
6.G.A.1Draw A DiagramName the bases and heights
Name the bases and the two heights.
A triangle's area only cares about one base and the perpendicular height to it, so those are the only lengths worth naming.
9.A-CED.A.2Introduce A VariableTurn base triangles into heights
The two triangles give the heights.
Each of these triangles has a full base of the trapezoid as its base, so its area hands you its height for free.
7.G.B.6Convert To AlgebraThe heights fill the trapezoid
Together they make the full height.
Between two parallel lines, every interior point splits the gap into two pieces that add back to the full gap.
Between two parallel lines, every interior point splits the gap into two pieces that add back to the full gap.
▸ Why?
Parallel lines keep a constant gap everywhere, so that full gap is the same wherever it is measured.
▸ Why?
The gap is exactly its two pieces put together, so the two heights add to it with nothing left over.
Get the height from the total area
The total area gives the height again.
The two slanted triangles have no convenient base or height, so use them only as part of the total.
6.G.A.1Change Focus Count The ComplementSet the two equal
Setting them equal makes an equation.
One quantity computed two different ways is an equation waiting to be written down.
9.A-CED.A.2Convert To AlgebraRewrite in the ratio
Reduce it to the ratio alone.
When every term has the same degree, only the ratio of the variables can matter — so make the ratio the variable.
9.A-SSE.A.2Organize Information In More WaysSolve the quadratic
Solve the quadratic.
A quadratic with a positive discriminant gives two candidates; the geometry decides which one is real.
9.A-REI.B.4Introduce A VariableReject the impossible root
What remains is two plus root two.
Algebra produces both roots because it never heard the words AB ＞ CD; you have to supply that filter yourself.
9.A-CED.A.3Eliminate PossibilitiesThe two triangles sitting on the parallel bases are the only ones worth measuring: each one's area tells you how far the point is from that base, and those two distances add up to the trapezoid's height. Set that against the trapezoid's own area formula and you get a² - 4ab + 2b² = 0; divide by b² and the answer is the root of r² - 4r + 2 = 0 bigger than 1, namely 2+√(2).
- Label the four areas on the picture
- Name the bases and the two heights
- Turn the base triangles into height formulas
- The two heights fill the trapezoid
- Use the total area to get h again
- Set the two expressions for h equal
- Rewrite in the ratio r = a/b
- Solve the quadratic
- Reject the impossible root
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