AMC 10 · 2013 · #13

Grade 8 geometry-2d
sequences-arithmeticangle-sum-polygonangle-sum-trianglesimilar-trianglesoptimization convert-to-algebrasystematic-enumerationextremal-construction ↑ Prerequisites: sequences-arithmeticangle-sum-triangle
📏 Long solution 💡 3 insights
Problem
A diagonal splits a quadrilateral into similar triangles and every angle set steps evenly. Find the largest possible sum of the two biggest angles.

Pick an answer.

(A)
210
(B)
220
(C)
230
(D)
240
(E)
250
How to solve
Strategy Introduce a Variable

Nothing here is numeric. The problem is three overlapping structural conditions — a four-term progression, a fixed similarity, a three-term progression — plus a maximum to find, so the first move is to name things well. Two namings carry the whole solution. First, the similarity collapses every angle in the picture down to three values α, β, γ. Second, the quadrilateral's progression is written around its centre instead of its first term: four terms m-3e, m-e, m+e, m+3e. That second choice is the one that pays, because the sum condition then fixes m = 90 with no fraction in sight, the quantity being maximized becomes a plain multiple of the spread e, and the question turns into "how wide can the spread be". Once both namings are in place the conditions reduce to a short closed list of cases, so a systematic list finishes the search; and because a maximum is claimed, the winning case must be drawn, not just solved for.

1STEP 1

Collapse the figure to three angles

Similarity collapses the figure to three angles.

α = ∠ ADB = ∠ CBD, β = ∠ DBA = ∠ DCB, γ = ∠ BAD = ∠ CDB
2STEP 2

Centre the progression at ninety

The quadrilateral's steps centre on ninety.

90-3e, 90-e, 90+e, 90+3e (0 ≤ e < 30), target = (90+e)+(90+3e) = 180+4e
3STEP 3

Read the corners off the split

The split names each corner directly.

∠ A = γ, ∠ B = α + β, ∠ C = β, ∠ D = α + γ
4STEP 4

Two pairs, one shared gap

Two pairs share the same gap.

{β, β+α}, {γ, γ+α} ⟹ α ∈ {2e, 4e}
5STEP 5

The triangle condition means sixty

The triangle condition forces sixty.

α + β + γ = 180°, (p-k) + p + (p+k) = 3p = 180° ⟹ p = 60°
6STEP 6

Six equations, two survivors

Only two step sizes survive.

α = 2e: {2e, 90-3e, 90+e}; α = 4e: {4e, 90-3e, 90-e}; e ∈ {10, 15}, 180 + 4e ≤ 240
7STEP 7

Build the winner, do not assume it

Building the winner gives 240, choice (D).

∠ A = 75°, ∠ B = 105°, ∠ C = 45°, ∠ D = 135°; 135 + 105 = 240
Answer
240
The formula 180 + 4e lets every answer choice be tested directly, which is a sharper check than confirming one construction. A choice of 210 needs e = 7.5: then α = 2e = 15 gives the triple {15, 67.5, 97.5} and α = 4e = 30 gives {30, 67.5, 82.5} — neither contains 60, so 210 is impossible. A choice of 230 needs e = 12.5: the triples are {25, 52.5, 102.5} and {50, 52.5, 77.5} — no 60, impossible. A choice of 250 needs e = 17.5: the triples are {35, 37.5, 107.5} and {70, 37.5, 72.5} — no 60, impossible. Only 220 (at e = 10, angles 60, 80, 100, 120 with triple {20, 60, 100} or {40, 60, 80}) and 240 (at e = 15) can occur at all, and 240 is the larger. The winning figure also passes every independent sanity test: 45 + 75 + 105 + 135 = 360, all four angles lie strictly between 0° and 180° so the quadrilateral is convex, and 45 + 60 + 75 = 180 for each triangle.
💡Key takeaway

The diagonal adds the same angle to two of the four corners, so the four corner angles must split into two pairs with the same gap — and that one structural fact decides the whole progression.

  • Collapse the figure to three angles
  • Centre the progression at ninety
  • Read the corners off the split
  • Two pairs, one shared gap
  • The triangle condition means sixty
  • Six equations, two survivors
  • Build the winner, do not assume it