AMC 10 · 2023 · #17

Grade 11 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

law-of-cosinessine-area-formulasequences-arithmeticquadratic-equations convert-to-algebraidentify-subproblemsextreme-principle ↑ Prerequisites: law-of-cosinessequences-arithmetic

📏 Long solution 💡 3 insights

Problem

A triangle's three side lengths form an arithmetic progression. The shortest side is 6, and somewhere in the triangle sits an angle of 120 degrees. Find the area of the triangle.

Pick an answer.

(A)

$12\sqrt{3}$

(B)

$8\sqrt{6}$

(C)

$14\sqrt{2}$

(D)

$20\sqrt{2}$

(E)

$15\sqrt{3}$

How to solve

Strategy Introduce a Variable

The phrase "arithmetic progression" is a gift, because it collapses three unknown side lengths down to one unknown. Name the common difference d and the sides are forced to be 6, 6+d, and 6+2d — a whole family of triangles controlled by a single knob. That leaves exactly one condition still unused, the 120° angle, and one knob to turn, so the condition should pin d down to a number. Before that can happen there is a decision to make: the problem hides which of the three angles is the 120° one. The Extreme Principle settles it, because the biggest angle in a triangle always faces the biggest side, and 120° has to be the biggest angle here. Once the angle is parked at the correct vertex, the Law of Cosines converts the geometry into a quadratic in d, and the area formula 1/2absin C finishes the job using the same angle that was given for free.

1STEP 1

All three sides, one letter

One common difference controls all three.

sides = 6, 6+d, 6+2d with d ＞ 0

2STEP 2

Which side the angle faces

Being the only obtuse angle, it faces the longest side.

120° is the unique obtuse angle ⟹ it is the largest angle ⟹ it faces the longest side 6+2d

3STEP 3

Use the law of cosines

The law of cosines gives an equation.

(6+2d)² = 6² + (6+d)² - 2 · 6·(6+d)cos 120° = 6² + (6+d)² + 6(6+d)

4STEP 4

Solve the quadratic

Discard the negative root.

36 + 24d + 4d² = 108 + 18d + d² ⟹ 3d² + 6d - 72 = 0 ⟹ (d+6)(d-4)=0 ⟹ d = 4

5STEP 5

Area from two sides and the angle

Two sides and the angle give the area.

Area = 1/2 · 6 · 10·sin 120° = 1/2 · 6 · 10·√(3)/2 = 15√(3)

6STEP 6

Verify the angle

Verifying gives an area of fifteen root three.

cos A = 13/14, cos B = 11/14, cos C = -1/2 ⟹ A ≈ 21.79°, B ≈ 38.21°, C = 120°, sum = 180°

Answer

15√(3)

Compute the area a completely different way as a cross-check. Heron's formula uses only the side lengths, with no angle and no assumption about where the 120° sits. The semiperimeter is s = (6+10+14)/2 = 15, so the area is √(15(15-6)(15-10)(15-14)) = √(15 · 9 · 5 · 1) = √(675) = √(225 · 3) = 15√(3). Two independent routes land on the same exact value, and numerically both give 25.98. Next, confirm that Step 2's placement of the angle was not just convenient but forced, by testing the two rejected placements algebraically. If the 120° angle faced the shortest side 6, the Law of Cosines would demand 36 = (6+d)² + (6+2d)² + (6+d)(6+2d); the right side already equals 108 at d=0 and only grows as d increases, and its roots are d=-6 and d=-12/7, both negative and both rejected. If the 120° angle faced the middle side 6+d, the requirement would be (6+d)² = 36 + (6+2d)² + 6(6+2d), which simplifies to d² + 8d + 24 = 0 with discriminant 64 - 96 = -32 ＜ 0, so there is no real solution at all. Only the longest-side placement survives, exactly as the largest-angle argument predicted. Finally, a size check against the answer list: the triangle fits inside a 6-by-14 frame, so its area is under 84, and it should comfortably exceed the area of a right triangle on legs close to 6 and 10 scaled down by the obtuse squash. The value 15√(3)≈ 25.98 sits in that range. The nearest decoys are all smaller — 8√(6)≈ 19.60, 14√(2)≈ 19.80, 12√(3)≈ 20.78 — and 12√(3) in particular is exactly what 1/2 · 6 · 8·sin 120° produces if the common difference is misread as d=2, so getting d right is the whole problem.

💡Key takeaway

When a problem says an angle exists but not where, put it where the rules force it — the obtuse angle always faces the longest side — and then one equation is enough to pin down everything else.

One knob controls all three sides
The 120 degree angle faces the longest side
Law of Cosines turns geometry into algebra
Solve the quadratic, discard the impossible root
Area from two sides and the angle between them
Confirm the triangle really has a 120 degree angle