AMC 10 · 2023 · #13

Grade 8 geometry-2d

pythagorean-theoremangle-sum-triangleinscribed-anglearc-measure identify-subproblemseasier-related-problemcasework ↑ Prerequisites: pythagorean-theoremangle-sum-triangle

📏 Medium solution 💡 3 insights

Problem

Abdul (A) and Chiang (C) are 48 feet apart. Bharat (B) stands somewhere in the field so that ∠ ABC = 60° (the angle at B in triangle ABC). Among all such positions for Bharat, pick the one that makes AB as large as possible. Report AB².

Pick an answer.

(A)

1728

(B)

2601

(C)

3072

(D)

4608

(E)

6912

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Draw a Diagram): sketch a few possible positions for B keeping ∠ B = 60° and AC fixed. The pictures show that AB stretches longest exactly when the triangle becomes a right triangle with the right angle at C. (Intuition: if C were obtuse, swinging C toward 90° would lengthen AB; past 90° would shrink it.) Tool #9 (Easier Related Problem) replaces "all triangles with ∠ B = 60°" by the specific easier case where ∠ C = 90° — a 30-60-90 right triangle with AC as the side opposite the 60° angle. Tool #7 (Identify Subproblems) then splits the work cleanly: (a) verify ∠ C = 90° maximizes AB, (b) compute AB in the resulting 30-60-90 triangle, (c) square it.

1STEP 1

Place A and C 48 ft apart and sweep B with ∠ABC=60°: B traces a circular arc through A and C, so AB varies along it.

AC = 48, ∠ ABC = 60°

2STEP 2

Tool #9: try the easy case ∠BCA=90°, so ∠BAC=30° (since they sum to 120°): a familiar 30-60-90 triangle.

∠ B = 60°, ∠ C = 90°, ∠ A = 30°

3STEP 3

AB is a chord of the arc, longest when it becomes the diameter — that happens exactly when the inscribed ∠ACB=90°.

∠ ACB = 90° ⇔ AB is a diameter of the arc circle ⇔ AB maximal

4STEP 4

In right triangle ACB, AC=48 is opposite 60°; the 30-60-90 ratio √3:2 for AC:AB gives AB=2/√3·48=32√3.

AB = 2/√(3) · 48 = 96/√(3) = 96√(3)/3 = 32√(3)

5STEP 5

Square it: AB²=(32√3)²=1024·3=3072.

AB² = (32√(3))² = 32² · 3 = 1024 · 3 = 3072

6STEP 6

3072 is exactly choice (C). Check: (E) 6912 needs AB=48√3, impossible since AB can't exceed the arc's diameter.

AB² = 3072 → (C)

Answer

3072

Direction check: as ∠ C slides past 90°, the triangle would force B back toward A (in the picture, the arc curls in), so ∠ C = 90° really is the turning point. Magnitude check: AB = 32√(3) ≈ 55.4 feet — bigger than AC = 48 feet, which makes sense because the 60° angle at B is opposite the shorter side. Squared, 55.4² ≈ 3070, matching 3072. End-to-end Pythagoras on the 30-60-90: BC =

\frac{AB}{2}

= 16√(3), and AC² + BC² = 48² + (16√(3))² = 2304 + 768 = 3072 = AB² — clean.

💡Key takeaway

Holding the angle at B at 60° keeps B on a fixed circular arc through A and C; the chord AB is longest when it becomes the diameter, which happens exactly when the angle at C is 90°. That makes triangle ACB a 30-60-90, so AB = 2/√(3) · 48 = 32√(3) and AB² = 3072.