AMC 10 · 2012 · #21
Grade 8 geometry-2dPick an answer.
AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The problem is entirely about positions and distances, so Tool #1 (Draw a Diagram) is the natural driver: the answer falls out once the four points are pinned down on paper. The single long segment 2a is the pressure point. Tool #3 (Eliminate Possibilities) uses the triangle inequality to rule out any real triangle with sides a,a,2a, which forces three of the points onto one straight line. Tool #7 (Identify Subproblems) then splits the six segments into two clean pieces — the three collinear segments a,a,2a and the three segments from the fourth point. Tool #8 (Analyze the Units) keeps every length written as a multiple of a, so the final answer comes out as a pure ratio with the a dividing away.
Zoom in on the long segment
The longest segment is PQ = 2a. If a third point R lies a from both P and Q, triangle PQR would have sides a, a, 2a.
The one length that is twice as long as all the others is what controls the whole shape, so start there.
7.G.A.2Draw A DiagramRule out the flat triangle
But a + a = 2a is not greater than 2a, so PQR cannot be a real triangle: P, R, Q lie on one line with R the midpoint of PQ.
When two sides add up to exactly the third, the triangle collapses flat onto a line.
When two sides add up to exactly the third, the triangle collapses flat onto a line.
▸ Why?
Two sides must together outreach the third, or the ends only just meet without enclosing anything.
▸ Why?
That single flat case is enough to throw the arrangement out and force the other one.
Put the three collinear points on a number line
Put the line on the x-axis: P=(0,0), R=(a,0), Q=(2a,0). That uses a, a, 2a; the other three segments run to a fourth point D.
Coordinates turn distance questions into arithmetic you can just compute.
5.G.A.2Identify SubproblemsLocate the fourth point
D's three distances are a, a, b. The two a's cannot reach P and Q, so DP=DR=a: D caps an equilateral triangle at D=(a/2, √(3)a/2).
Two equal a-legs on a base of length a build the familiar equilateral triangle.
7.G.A.2Eliminate PossibilitiesMeasure the last segment
The last unnamed segment is DQ = b. Horizontal gap 3a/2, vertical gap √(3)a/2, so b² = 9a²/4 + 3a²/4 = 3a² and b = √(3) a.
The distance between two plotted points is just the Pythagorean theorem on the horizontal and vertical gaps.
8.G.B.8Draw A DiagramForm the ratio
All six lengths check out: PR, RQ, DP, DR are a, PQ = 2a, DQ = b. So b:a = √(3), choice (A).
Writing both lengths as multiples of a lets the a cancel, leaving a pure number for the ratio.
6.RP.A.1Analyze The UnitsThe length 2a is too long to be a triangle side with two a-sides, so three points must lie flat on a line; the fourth point makes an equilateral triangle, and the last segment measures √(3) a, giving b/a=√(3).
- Zoom in on the long segment
- Rule out the flat triangle
- Put the three collinear points on a number line
- Locate the fourth point
- Measure the last segment
- Form the ratio