AMC 10 · 2011 · #24
Grade 11 geometry-2dPick an answer.
Tool #14 (Extreme Principle): the question asks for a maximum, and a maximum claim has two halves that must both be delivered — a ceiling that no shape can pass, and one actual shape that touches the ceiling. Skipping the second half is the classic trap here: it is tempting to declare 'obviously the circle touches all four sides and the quadrilateral is cyclic' and compute, but that assumption is exactly what needs earning. Tool #7 (Identify Subproblems): cutting the quadrilateral into four triangles from the circle's center converts 'how big is the circle' into 'how much area is available', which is the move that produces the ceiling. Tool #4 (Introduce a Variable): naming one tangent length t turns the touching circle into a one-knob family, and one equation fixes the knob. Tool #1 (Draw a Diagram): the final existence check is a picture question — do the four corner wedges wrap around the center exactly once?
An inside circle costs area
An inner circle costs area proportional to the perimeter.
A circle of radius r demands at least r of clearance from every wall, and buying that clearance along all 42 units of perimeter costs area.
6.G.A.1Identify SubproblemsCaved-in shapes lose at once
Caved-in shapes lose at once.
A dented quadrilateral is trapped inside the triangle formed by its two longest walls, and that triangle is too thin to hold a big circle.
11.G-SRT.D.9Extreme PrincipleCap the quadrilateral's area
The area itself has its own ceiling.
Flexing only moves the subtracted cosine term, so the frame spreads to its widest when opposite angles add to a straight angle.
Flexing only moves the cosine term, so the frame spreads widest when opposite angles add to a straight angle.
▸ Why?
With the sides fixed, the diagonal and hence the area depend on the corner angles alone.
▸ Why?
Opposite angles adding to a straight angle is exactly the condition that puts all four corners on one circle.
Chain the two ceilings
Chaining the two bounds gives a single number.
Circle size is capped by area and area is capped by the cyclic shape, so the two ceilings stack into one number.
9.A-CED.A.1Extreme PrincipleEquality demands two things at once
Equality needs opposite sides to balance.
Two ceilings can only be reached together if one shape satisfies both, and equal sums of opposite sides is the signal that a touching circle is even possible.
10.G-C.A.2Introduce A VariableSolve for the touch points
The touch points solve one small equation.
Sliding the four touch points is a single free knob, and demanding opposite angles be supplementary turns it to exactly one setting.
9.A-REI.B.3Introduce A VariableCheck the wedges close up
The shape really closes, so the radius is 2√6.
Naming a radius is not enough — the four corner wedges have to fit around the center exactly once, and here they do.
10.G-SRT.C.8Draw A DiagramA circle of radius r needs r of clearance along every wall, so r can never beat area divided by half the perimeter — then the work is finding the one shape that hits that ceiling exactly.
- An inside circle costs area
- Caved-in shapes lose at once
- Cap the quadrilateral's area
- Chain the two ceilings
- Equality demands two things at once
- Solve for the touch points
- Check the wedges close up
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