AMC 10 · 2022 · #12

Grade 11 geometry-3d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length

equilateral-trianglemedian-of-trianglepythagorean-theoremlaw-of-cosinessimilar-figures spatial-visualizationidentify-subproblemsconvert-to-algebra ↑ Prerequisites: pythagorean-theoremlaw-of-cosines

📏 Long solution 💡 3 insights

Problem

A regular tetrahedron has four congruent equilateral faces, so all six edges share one length. Take the midpoint of one edge and draw segments out to the two vertices not on that edge. Find the cosine of the angle they form at the midpoint.

Pick an answer.

(A)

$\frac14$

(B)

$\frac13$

(C)

$\frac25$

(D)

$\frac12$

(E)

$\frac{\sqrt{3}}{2}$

How to solve

Strategy Visualize Spatial Relationships

A tetrahedron is hard to hold in the head, so tool #17 (Visualize Spatial Relationships) does the one move that makes the problem easy: notice that the angle asked about is determined by only three points, and three points always lie in a plane. The solid collapses to the flat triangle CMD, and a three-dimensional problem becomes a two-dimensional one with no loss. From there the shape of the work is fixed — get the three sides of that triangle, then read the angle. Tool #4 (Introduce a Variable) handles the missing size: no edge length is stated, so the answer is scale-free and the edge may simply be set to 1. Tool #1 (Draw a Diagram) supplies the two lengths that are not edges, by drawing each of the faces ABC and ABD flat on the page, where MC and MD are visibly medians of equilateral triangles. Tool #7 (Identify Subproblems) extracts the right triangle hiding inside a face so the Pythagorean Theorem can finish that length. Tool #3 (Eliminate Possibilities) is worth spending one step on before any formula: the side lengths alone rank the angles of the triangle and immediately kill the two choices that assume a nicer angle than the figure allows. Tool #13 (Convert to Algebra) closes the problem, turning three known sides and one unknown cosine into a single linear equation through the Law of Cosines.

1STEP 1

The angle lives in one plane

The three points lie in one plane.

C, M, D determine a plane ⟹ ∠ CMD is an ordinary angle of the flat triangle CMD

2STEP 2

Set the edge length to one

Set the edge length to one.

AB = AC = AD = BC = BD = CD = 1 ⟹ CD = 1, AM = MB = 1/2

3STEP 3

Both faces give the same median

Both faces give the same median.

CA = CB, AM = BM, CM = CM ⟹ △ CAM ≅ △ CBM ⟹ CM ⊥ AB; △ ABD ≅ △ ABC ⟹ DM = CM

4STEP 4

Pythagoras on half a face

Pythagoras on half a face.

CM² = AC² - AM² = 1 - 1/4 = 3/4 ⟹ CM = DM = √(3)/2, CD = 1

5STEP 5

Two choices die immediately

Two choices die immediately.

CD ＞ CM = DM ⟹ ∠ CMD ＞ 60^° ⟹ cos(∠ CMD) ＜ 1/2; CM² + DM² = 3/2 ＞ 1 = CD² ⟹ ∠ CMD ＜ 90^°

6STEP 6

Law of Cosines closes it

The law of cosines gives one third.

1 = 3/4 + 3/4 - 3/2cos(∠ CMD) ⟹ 3/2cos(∠ CMD) = 1/2 ⟹ cos(∠ CMD) = 1/3 (B)

Answer

1/3

Coordinates settle the value independently, with no reuse of the Law of Cosines. Put A = (0,0,0), B = (1,0,0), C = (1/2, √(3)/2, 0), D = (1/2, √(3)/6, √(6)/3). Checking D against A: 1/4 + 1/12 + 2/3 = 3/12 + 1/12 + 8/12 = 1, and by symmetry the same holds for DB and DC, so all six edges really are 1. Then M = (1/2, 0, 0), so MC = (0, √(3)/2, 0) and MD = (0, √(3)/6, √(6)/3). Both have length √(3)/2, since 1/12 + 2/3 = 3/4, confirming Step 4, and their dot product is √(3)/2 · √(3)/6 = 3/12 = 1/4. Dividing by the product of the lengths gives cos(∠ CMD) = 1/4/3/4 = 1/3, confirming (B). The number is also recognisable in its own right: ∠ CMD ≈ 70.53^° is the supplement of the tetrahedral angle 109.47^° that two vertices subtend at the centre of a regular tetrahedron, whose cosine is -1/3 — the same 1/3 with the sign flipped. Finally, price the traps. Choice (D) 1/2 is what a solver gets by treating triangle CMD as equilateral, which would require MC = 1 instead of √(3)/2; the moment Step 4 produces √(3)/2 that reading is dead. Choice (E) √(3)/2 is the length MC itself, copied out of the side-length slot into the cosine slot, and choice (C) 2/5 = 0.4 is close enough to 1/3 ≈ 0.333 that only exact arithmetic separates them — which is why Step 6 cannot be replaced by an estimate.

💡Key takeaway

Three points always lie flat, so shrink the solid down to the one triangle that holds the angle, find its three side lengths, and the Law of Cosines does the rest.

The angle lives in one plane
Set the edge length to one
Both faces give the same median
Pythagoras on half a face
Two choices die immediately
Law of Cosines closes it