AMC 10 · 2015 · #21
Grade 11 geometry-2dPick an answer.
"All circles through both foci" sounds like a two-dimensional search, but Tool #4 (Introduce a Variable) collapses it: the center must sit on the y-axis, so a single height n names the whole family and r=√(n²+15) comes free. Tool #1 (Draw a Diagram) keeps the picture honest — the ellipse is 8 wide and only 2 tall, so the interesting failure happens at the top and bottom vertices, not at the ends. Tool #7 (Identify Subproblems) splits the job into: where are the foci, which circles qualify, when are there exactly four crossings, and what radii those circles have. Tool #14 (Extreme Principle) then finds the boundary case where the fourth crossing is about to be lost, which is precisely what makes the interval closed at one end and open at the other. The reason for going algebraic rather than eyeballing the picture: "exactly four" is a counting claim, and only the root count of an explicit quadratic proves it in every case instead of the ones we happen to sketch.
Locate the two foci
The two foci are located immediately.
The foci always sit on the long axis, and how far out they land is set by the gap between the two squared semi-axes.
11.G-GPE.A.3Draw A DiagramPut every candidate center on the y-axis
Every candidate centre sits on one line.
Equidistant from two fixed points means sitting on their perpendicular bisector, which shrinks a hunt for a center in the plane down to sliding a point along one line.
Being the same distance from two fixed points puts a centre on the line that folds one onto the other.
▸ Why?
Those equidistant points make up exactly that fold line and nothing else.
▸ Why?
A circle through both points has them both one radius away, which is exactly that equal-distance condition.
Trade two curves for one quadratic in y
Combining the curves gives one quadratic.
Both curves use x only as x², so subtracting one from the other erases x and leaves just the heights at which they can meet.
11.A-REI.C.7Identify SubproblemsForce both heights strictly inside the band
Four crossings means both heights stay inside.
The ellipse is only two units tall, so a crossing height pays for two points only while it stays strictly between -1 and 1.
9.A-CED.A.1Extreme PrincipleConvert the range of n into the range of r
Converting gives ends summing to choice (D).
A continuous increasing function on a half-open interval sweeps out a half-open interval: it hits everything up to the endpoint value but never the value itself.
9.F-IF.B.4Extreme PrincipleWhen two curves both use x only as x², subtract them: the crossings collapse into one quadratic in y, and counting how many of its roots stay strictly inside the shape's height band tells you exactly how many intersection points there are.
- Locate the two foci
- Put every candidate center on the y-axis
- Trade two curves for one quadratic in y
- Force both heights strictly inside the band
- Convert the range of n into the range of r