AMC 10 · 2022 · #19

Grade 11 geometry-2d
median-of-trianglecentroid-2-to-1law-of-cosinesequilateral-trianglesupplementary-angles identify-subproblemsconvert-to-algebra ↑ Prerequisites: law-of-cosinescentroid-2-to-1
📏 Long solution 💡 3 insights
Problem
Two medians are drawn in a triangle and cross at a point. One vertex, that crossing point, and one midpoint happen to form an equilateral triangle. That single condition fixes the triangle's shape, so one angle's cosine is determined. Write it in the requested form and add the three whole numbers.

Pick an answer.

(A)
44
(B)
48
(C)
52
(D)
56
(E)
60
How to solve
Strategy Introduce a Variable

The problem hands over no numbers at all, only a shape condition, so the first move is to supply the missing number yourself. Call the side of the equilateral triangle s. That one name is enough, because every other length in the picture is chained to it: E is a midpoint, so AC = 2s immediately, and the centroid's 2:1 rule converts AG and GE into the full medians and their remaining pieces. At the end s cancels, since cos C is a ratio, which is exactly what the absence of numbers predicted. The second idea is to stop looking at △ ABC and look instead at the small triangles that share the vertex G. The equilateral condition puts a known 60° at G, and because both medians pass straight through G, the other angles there are forced to 120° and 60° by supplementary and vertical pairs. Each small triangle then has two known sides and the angle between them, which is precisely the Law of Cosines setup. Two applications produce AB and BC; a third application, this time in the big triangle at vertex C, produces cos C.

1STEP 1

Name the equilateral side

Give the side a name.

AG = GE = EA = s ⟹ AC = 2s, EC = s
2STEP 2

Unfold with the two-to-one rule

Each median splits two to one.

BG = 2 GE = 2s, BE = 3s; AG = 2/3AD = s ⟹ AD = 3/2s, GD = 1/3AD = 1/2s
3STEP 3

Read the angles at the crossing

Vertical angles fix the angles.

∠ AGE = 60° ⟹ ∠ AGB = 180° - 60° = 120°, ∠ DGB = ∠ AGE = 60° (vertical angles)
4STEP 4

Find the first side

The law of cosines gives one side.

AB² = s² + (2s)² - 2 · s · 2s·cos 120° = s² + 4s² + 2s² = 7s² ⟹ AB = s√(7)
5STEP 5

Find the second side

It gives another side too.

BD² = (2s)² + (1/2s)² - 2 · 2s·1/2s·1/2 = 4s² + 1/4s² - s² = 13/4s² ⟹ BD = √(13)/2s, BC = 2 BD = s√(13)
6STEP 6

Get the cosine from three sides

Collect all three and take the cosine.

cos C = (a²+b²-c²)/2ab = (13s² + 4s² - 7s²)/(2 · s√(13) · 2s) = 10s²/4s²√(13) = 5/2√(13)
7STEP 7

Rationalize and add

Rationalizing and adding gives 44.

cos C = 5/2√(13)·√(13)/√(13) = 5√(13)/26 ⟹ m=5, n=26, p=13 ⟹ m+n+p = 44
Answer
44
Rebuild the triangle from scratch with coordinates and see whether the hypothesis really comes back out. Put G at the origin and E at (1,0) with s=1. Since B, G, E are collinear with BG = 2, the point B sits on the opposite ray at (-2,0). The equilateral condition puts A at angle 60° and distance 1 from G, so A = (1/2, √(3)/2). Because E is the midpoint of AC, C = 2E - A = (3/2, -√(3)/2). Two checks confirm the construction is legitimate rather than assumed: the midpoint of BC is (-1/4, -√(3)/4), which lies on line AG at distance 1/2 from G, matching GD = 1/2s exactly; and the average of A, B, C is (0,0) = G, confirming G really is the centroid. Measuring, AC = 2, AB = √(7)≈ 2.6458, BC = √(13)≈ 3.6056, all matching the derivation. The dot product gives CA·CB = (-1)(-7/2) + (√(3))(√(3)/2) = 5, so cos C = 5/2√(13) ≈ 0.69338, and 5√(13)/26 ≈ 0.69338 as well. The three angles come out A ≈ 100.89°, B ≈ 33.00°, C ≈ 46.10°, summing to 180°, and cos C lands safely inside [-1,1]. The ordering is also sensible: C faces the middle-length side √(7) and is indeed the middle-sized angle. One more check confirms the angle assignment in Step 3 was forced, not lucky. Swapping the two angles at G — putting 120° in △ BGD and 60° in △ ABG — would give AB² = 3s² and BC² = 21s², hence cos C = (21+4-3)/(2 · 2√(21)) = 11/2√(21) ≈ 1.20, which is impossible for a cosine. The wrong assignment destroys itself.
💡Key takeaway

When a shape problem gives no numbers, name one length yourself and let the fixed rules — a midpoint doubles, a centroid splits 2:1, a straight line makes 180° — carry that name to every other length.

  • Name the equilateral side
  • Unfold both medians with the 2 to 1 rule
  • Read the angles around G
  • Law of Cosines in triangle ABG gives AB
  • Law of Cosines in triangle BGD gives BC
  • All three sides, then Law of Cosines at C
  • Rationalize, then read off m, n, p