AMC 10 · 2021 · #24
Grade 11 geometry-2dPick an answer.
The shape has three unknown sides but only one equation tying them together, so the work has to be organised before it is computed. Tool #4 (Introduce a Variable) does that twice over: first it names the three unknown sides, and then — the move that makes the problem finite — it names the common difference d of the progression, which shrinks three unknowns down to one. Tool #1 (Draw a Diagram) supplies the missing equation: sliding the short parallel side across turns the trapezoid into a single triangle carrying the 60° angle, where the Law of Cosines applies. Tool #13 (Convert to Algebra) then makes that geometric relation a formula, and after substituting the progression the constant and the d-free terms cancel, leaving a plain linear equation in d. Tool #2 (Make a Systematic List) is needed because the problem says the sides form a progression 'in some order' — there are exactly six ways to attach the three smaller terms to the three unnamed sides, and each must be tested. Tool #3 (Eliminate Possibilities) finishes the job, since most of those six produce a side of length 0 and have to be thrown out.
Name the three unknown sides
Name the three unknown sides.
Naming the unknowns is what lets the one given angle and the one given length reach the rest of the figure.
9.A-CED.A.2Introduce A VariableSlide the short side onto AB
Slide it over to form a parallelogram.
Opposite sides of a parallelogram are equal, so sliding CD down onto AB moves BC into a triangle without changing its length.
10.G-CO.C.11Draw A DiagramOne equation from the Law of Cosines
The law of cosines gives one equation.
A 60° angle is the friendliest case of the Law of Cosines, because cos 60° = 1/2 cancels the factor of 2 and leaves whole-number coefficients.
With two sides and the angle between them known, the third side is fixed by those three alone.
▸ Why?
Two sides and the enclosed angle determine the remaining side completely.
▸ Why?
Opposite sides of the figure keep a constant gap and equal length, so sliding one into the triangle changes nothing.
Write the progression around 18
Write the progression around eighteen.
An arithmetic progression is fixed by its largest term and its gap, so one new letter d replaces three unknown lengths.
9.F-IF.A.3Introduce A VariableCheck the flat case d = 0
Check the flat case with zero difference.
Zero is a legal common difference, and it is the case most likely to be forgotten because it makes the progression invisible.
11.G-SRT.D.11Make A Systematic ListSix orderings, one linear equation
Six orderings give one linear equation each.
The quadratic looks frightening until the constants cancel and a common factor of d comes out, turning six hard cases into six one-line divisions.
9.A-SSE.A.2Convert To AlgebraRun the six cases, drop the dead ones
Drop the ones giving a zero length.
An equation can hand back a number that no picture can match, so every solution has to be walked back to the figure before it is kept.
9.A-REI.B.3Eliminate PossibilitiesAdd up the values a can take
Adding them all gives 84.
The question asks for the values a can take, not for one quadrilateral, so the answer is a pooled list across every figure that survives.
9.A-CED.A.3Eliminate PossibilitiesSlide the short parallel side onto the long one and the trapezoid becomes a single triangle with a 60° angle; then write the four sides as 18, 18-d, 18-2d, 18-3d, test all six ways of matching them to the figure, and keep only the ones where no side shrinks to zero.
- Name the three unknown sides
- Slide the short side onto AB
- One equation from the Law of Cosines
- Write the progression around 18
- Check the flat case d = 0
- Six orderings, one linear equation
- Run the six cases, drop the dead ones
- Add up the values a can take