AMC 10 · 2002 · #24
Grade 8 geometry-2dPick an answer.
The sides are exactly what is not given, so tool #16 (Change Focus) says stop looking at sides and look at the diagonals — those are the segments the four given distances actually control. That produces an upper bound on the area, and the given area turns out to equal that bound exactly. Tool #14 (Extreme Principle) is then the engine of the whole solution: when a quantity is handed to you at its ceiling, every inequality used to build the ceiling has to be an equality, and each of those equalities is a geometric fact about the picture. Tool #1 (Draw a Diagram) is needed twice — once to see the heights onto a diagonal, and once to build the extremal quadrilateral and check that it really exists, because a bound with nothing attaining it proves nothing. Tool #7 (Identify Subproblems) closes the job: once the diagonals are known to cross at right angles at P, each side is a separate small Pythagorean computation.
Cut along the diagonal AC
A diagonal splits the quadrilateral, giving area as half the diagonal times the two heights.
The sides are unknown, but the diagonals are the only segments the distances from P say anything about, so build the area out of a diagonal.
6.G.A.1Change Focus Count The ComplementTrade the two heights for BD
Those heights are capped by the other diagonal, with equality only when the diagonals are perpendicular.
A slanted segment is always longer than its shadow, so two heights measured off a line can never add up to more than the segment joining them.
A slanted segment is never shorter than its shadow, so two heights measured off a line cannot exceed the segment itself.
▸ Why?
Going from one point to another by way of a third is never shorter than going straight there.
▸ Why?
Each height is a leg of a right triangle whose hypotenuse is the slanted piece, so the leg can never beat the hypotenuse.
Bound both diagonals through P
The triangle inequality through P caps the diagonals at 52 and 77.
The only thing the distances from P can say about a diagonal is how far apart its two ends are allowed to be.
7.G.A.2Extreme PrincipleClose the squeeze
Both ends of the chain equal 2002, so every step is an equality and the diagonals cross at P at right angles.
When the value you are given equals the largest value the setup allows, nothing was wasted anywhere, so every inequality along the way has to be tight.
7.EE.B.4Extreme PrincipleCheck the shape really exists
Building that figure explicitly confirms its four right triangles total 2002.
Bounding is only half the job — the shape that reaches the bound still has to be drawable, and here it is.
6.G.A.1Draw A DiagramPythagoras at the four corners
Pythagoras at each corner gives sides 40, √(1808), 53, 51.
Right angles at P turn every side of the quadrilateral into a hypotenuse whose two legs are already written in the problem.
8.G.B.7Identify SubproblemsAdd the four sides
Adding and simplifying the radical gives 4(36+√(113)), choice (E).
Three of the four sides land on whole numbers from Pythagorean triples, so only one radical survives, and it factors into the printed form.
8.EE.A.2Identify SubproblemsWhen the number a problem hands you is exactly the biggest the setup allows, every inequality you used to reach that ceiling has to be tight — and each tight inequality hands back a piece of the shape.
- Cut along the diagonal AC
- Trade the two heights for BD
- Bound both diagonals through P
- Close the squeeze
- Check the shape really exists
- Pythagoras at the four corners
- Add the four sides
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