AMC 10 · 2007 · #19

Grade 8 geometry-2dalgebra
Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Point as the Answer
coordinate-geometryarea-trianglesabsolute-valueslope-intercept convert-to-algebracaseworksymmetry-argument ↑ Prerequisites: coordinate-geometryarea-triangles
📏 Long solution 💡 4 insights
Problem
A point must give one triangle a fixed area and a second triangle another fixed area at the same time. Find the total of the first coordinates of every such point.

Pick an answer.

(A)
282
(B)
300
(C)
600
(D)
900
(E)
1200
How to solve
Strategy Convert to Algebra

There is no picture to stare at, because A is not one point — it is an unknown set of points, and the answer is a sum over that set. So Tool #4 (Introduce a Variable) comes first: write A=(x,y) and treat the two area statements as two conditions on the pair (x,y). Tool #13 (Convert to Algebra) is then the spine of the whole solution, and the exact form of the conversion is what everything turns on. Each area condition must be turned into an equation that is equivalent to it, not merely implied by it — an area of 2007 has to become a statement that is true for a point exactly when the area is 2007. Done that way, each condition turns out to describe a pair of parallel lines, and the solution set is an intersection of two such pairs, which can be counted with certainty instead of guessed at. Tool #7 (Identify Subproblems) supplies the one geometric fact that makes the second area computable by hand: a triangle with two vertices on a slanted line can be split along a vertical segment, which replaces an ugly perpendicular distance with a plain difference of y-values. Tool #15 (Organize Information in More Ways) then re-reads the finished solution set through its symmetry and recovers the same sum by a completely different route — one that never computes either area — which is the real cross-check. Tool #3 (Eliminate Possibilities) finishes by matching each wrong choice to the slip that produces it.

1STEP 1

Name the unknown point, aim at a set

Name the point and aim at the whole solution set.

A=(x,y); find S={(x,y) : [ABC]=2007 and [ADE]=7002}, then compute Σ_(x,y)∈ S x
2STEP 2

The first area pins y to ± 18

The first area fixes the height to two values.

[ABC]=1/2 · 223·|y| = 2007 ⇔ |y| = 4014/223 = 18 ⇔ y = 18 or y = -18
3STEP 3

Line DE, and the vertical gap to it

The second triangle's line gives a vertical gap to measure with.

slope(DE)=(389-380)/(689-680)=1, DE: y=x-300, A'=(x, x-300), v=|AA'|=|y-x+300|
4STEP 4

Split along AA': the second area is 9/2v

Splitting along that segment turns the second area into two more lines.

[ADE]=9/2 |y-x+300| = 7002 ⇔ |x-y-300| = 14004/9 = 1556 ⇔ x = y+300± 1556
5STEP 5

Two pairs of lines cross in exactly four points

Two pairs of lines cross in exactly four points.

S={(1874,18), (-1238,18), (1838,-18), (-1274,-18)}
6STEP 6

Re-read the four points as one symmetry

A symmetry pairs them up, so the total comes out at once.

σ(x,y)=(600-x,-y); σ(S)=S, P=(300,0)∉ S ⟹ Σ x = 2 · 600 = 4 · 300
7STEP 7

Total, and what each wrong choice is

Adding gives 1200, choice (E).

1874-1238+1838-1274 = 1200 ⟹ (E)
Answer
1200
Check each of the four points directly instead of trusting the derivation. Every one has |y|=18, so [ABC]=1/2 · 223 · 18 = 2007 for all four. For [ADE] use 9/2|y-x+300|: at (1874,18) it is 9/2|18-1874+300| = 9/2 · 1556 = 7002; at (-1238,18), 9/2|18+1238+300| = 9/2 · 1556 = 7002; and the same at (1838,-18) and (-1274,-18). Verify one of them a second way, without that formula, by the enclosing-box computation for A=(1874,18): 2[ADE] = |(680-1874)(389-18)-(689-1874)(380-18)| = |{-1194} · 371 + 1185 · 362| = |{-442974}+428970| = 14004, so [ADE]=7002. The symmetric pairing also checks out numerically: 1874+(-1274)=600 and (-1238)+1838=600. Scale is sensible too — DE has length only 9√(2)≈ 12.7 while BC has length 223, yet ADE has more than triple the area of ABC, so A must be enormously far from line DE compared with its distance 18 from line BC; x-values around ± 1500 are exactly that. Finally, stress-test the method on a shrunken copy with the same B,C,D,E but areas 223 and 9: then |y|=2 and |x-y-300|=2, giving x = 304, 300, 300, 296, which still sums to 1200=4 · 300, confirming the areas do not affect the sum. That small copy also exposes why the wording had to be checked: there two different solution points share the x-coordinate 300, so summing distinct values would give 900 instead. In the actual problem the four x-values 1874,-1238,1838,-1274 are all different, so no such ambiguity arises.
💡Key takeaway

Each area condition traps A on a pair of parallel lines, the two pairs cross in exactly four points arranged symmetrically about the spot where line BC meets line DE, so the four x-coordinates add to four times that spot's x-coordinate.

  • Name the unknown point, aim at a set
  • The first area pins y to ± 18
  • Line DE, and the vertical gap to it
  • Split along AA': the second area is 9/2v
  • Two pairs of lines cross in exactly four points
  • Re-read the four points as one symmetry
  • Total, and what each wrong choice is

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