AMC 10 · 2007 · #20

Grade 9 geometry-2dnumber-theory

Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Vertices then Area

coordinate-geometryslope-interceptoptimization extreme-principleconvert-to-algebra ↑ Prerequisites: coordinate-geometrysystems-of-equations

📏 Long solution 💡 4 insights

Problem

Two parallelograms cut out by pairs of parallel lines have known areas. All four constants are positive whole numbers. Find the smallest possible sum of the four.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Extreme Principle

The question asks for a smallest value, so Tool #14 governs the ending: a minimum claim has two halves, a lower bound that no legal quadruple can beat and a witness that reaches it, and skipping either half leaves the answer unproved. Everything before that is setup. The two figures are built the same way out of the same two slopes, so solve one subproblem once (Tool #7) — the area of a parallelogram cut by y=ax+p₁, y=ax+p₂, y=bx+q₁, y=bx+q₂ — and reuse it, rather than analysing two pictures. To get that formula, sketch the figure (Tool #1) and notice the one piece of structure that makes it cheap: the two lines sharing an intercept meet on the y-axis, so a whole diagonal lies along the y-axis and the parallelogram splits into two triangles with a common base. With the formula in hand the geometry is gone and only integers remain (Tool #13); naming the smaller of c,d and the gap |a-b| (Tool #4) turns both area conditions into one Diophantine equation whose full solution set can be written down, which is what makes an honest minimisation possible.

1STEP 1

Find where the corners sit

Solving the line pairs locates the corners.

ax+c=bx+d ⟹ x=(d-c)/(a-b), ax+d=bx+c ⟹ x=(c-d)/(a-b)

2STEP 2

Turn that diagonal into an area

Each area becomes one clean expression in the constants.

Area=2·1/2·|c-d|·(|c-d|)/(|a-b|)=((c-d)²)/(|a-b|)

3STEP 3

Compare the two areas

Comparing them forces a fixed ratio between two constants.

((c-d)²)/m=18, ((c+d)²)/m=72 ⟹ (3d-c)(3c-d)=0 ⟹ {c,d}={t,3t}, 2t²=9m

4STEP 4

Force a factor of three

A divisibility argument forces one of them to be a multiple of three.

2t²=9m, gcd(2,9)=1 ⟹ 9 ∣ t² ⟹ 3 ∣ t; t=3k ⟹ m=2k²

5STEP 5

Prove nothing smaller can happen

Bounding every part gives a floor of 16.

S=(a+b)+(c+d)=(2s+2k²)+12k ≥ 2+2k²+12k ≥ 2+2+12=16

6STEP 6

Exhibit a quadruple reaching 16

An explicit quadruple reaches it, so the answer is 16, choice (D).

(a,b,c,d)=(3,1,9,3): ((9-3)²)/(|3-1|)=36/2=18, ((9+3)²)/(|3-1|)=144/2=72, a+b+c+d=16

Answer

A parity check kills three of the five choices outright. The numbers a+b and a-b always have the same parity, and a-b=± 2k² is even, so a+b is even; the intercepts give c+d=4t, also even. So a+b+c+d is even for every legal quadruple, and 13, 15, 17 can never occur at all. That leaves 14 and 16. The value 14 is exactly what (a,b,c,d)=(2,0,9,3) would give: the same picture with one family of lines horizontal. It is the trap for anyone who lets b drop to 0 after reading that the numbers are positive. Since b ≥ 1 and |a-b|=2, the pair {1,3} is forced and 16 stands. The nearby legal sums confirm the gap: staying at k=1 but taking {a,b}={2,4} gives 18 and {3,5} gives 20, while k=2 forces {c,d}={6,18} and |a-b|=8, so at least 2+8+24=34. The geometry checks out on the winning quadruple too: with a=3,b=1,c=9,d=3 the first parallelogram has corners (0,9),(3,12),(0,3),(-3,0), whose y-axis diagonal has length 6 and whose far corners sit 3 units to either side, giving 2·1/2 · 6 · 3=18; the second has corners (0,9),(6,15),(0,-3),(-6,-9), with diagonal 12 and corners 6 units out, giving 2·1/2 · 12 · 6=72.

💡Key takeaway

Derive one area formula you can use on both pictures, then finish a "smallest possible" question twice: prove nothing smaller can happen, and build one example that hits the number you claim.

Find where the corners sit
Turn that diagonal into an area
Compare the two areas
Force a factor of three
Prove nothing smaller can happen
Exhibit a quadruple reaching 16