AMC 10 · 2012 · #20
Grade 9 geometry-2dPick an answer.
The phrase "all the possible areas" makes this a completeness problem, so tool #2 (Make a Systematic List) is the backbone: the list must contain every trapezoid and no duplicates. Two things have to be nailed down before any area is computed. First, how many cases there really are — naively 4!=24 orderings, but tool #2 plus a mirror-symmetry observation collapses this to the C(4, 2)=6 choices of which pair is parallel. Second, and this is where the problem is actually decided, a test for when a chosen arrangement exists at all. Tools #4 (Introduce a Variable) and #13 (Convert to Algebra) set up coordinates and solve for the height, and tool #9 (Solve an Easier Related Problem) converts "does this trapezoid exist?" into the far easier "do these three lengths form a triangle?" — as an if-and-only-if, so it both rules cases out and certifies the survivors. Tool #3 (Eliminate Possibilities) then screens the six candidates by arithmetic alone, tool #7 (Identify Subproblems) computes the three heights and areas, and tool #15 (Organize Information in More Ways) matches the total to the prescribed r₁√(n₁)+r₂√(n₂)+r₃ shape and checks that the match is forced, so the requested sum is well defined.
Only the parallel pair matters
Only the parallel pair distinguishes the arrangements.
Swapping the two slanted sides only flips the picture left-to-right, and a mirror image has the same area.
8.G.A.2Make A Systematic ListPut the trapezoid on coordinates
Coordinates turn the shape into two unknowns.
Once both parallel sides sit on horizontal lines, each slanted side is a hypotenuse and the height is the shared vertical leg.
8.G.B.7Introduce A VariableSolve for the offset and the height
Solving gives the offset and the height.
The difference of the two Pythagorean equations is linear in the offsets, so both offsets drop out in one move.
8.EE.C.8Convert To AlgebraExists exactly when a triangle exists
Existence reduces to a triangle inequality.
The two legs must be long enough to bridge the overhang c-a but not so lopsided that one alone outreaches the other two — the same balance that lets three lengths close into a triangle.
The trapezoid exists exactly when the two slanted sides and the overhang could form a triangle.
▸ Why?
Any two of those lengths together must reach further than the third, or the ends never meet.
▸ Why?
Each slanted side is the hypotenuse over the shared height, so the height only exists when that reach is possible.
Screen the six candidates
Only three of the six arrangements survive.
A tiny overhang forces the two legs to be nearly equal, so pairs of close lengths like 3,5 or 5,7 cannot be the parallel sides here.
7.G.A.2Eliminate PossibilitiesCompute the three areas
Each survivor gives a clean area.
Height times the average of the two parallel sides is the area, and the coordinate formula delivers that height directly.
7.G.B.6Identify SubproblemsMatch the form and take the floor
Matching the form and rounding down gives 63.
The two irrational pieces cannot merge or hide inside the rational piece, so the way the total is written is essentially forced.
8.NS.A.1Organize Information In More WaysOnly the choice of which two sides are parallel matters, and such a trapezoid exists exactly when the two legs and the overhang c-a can form a triangle — three of the six choices pass, and their areas add to 35√3/2+32√5/3+27.
- Only the parallel pair matters
- Put the trapezoid on coordinates
- Solve for the offset and the height
- Exists exactly when a triangle exists
- Screen the six candidates
- Compute the three areas
- Match the form and take the floor
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