AMC 10 · 2018 · #23

Grade 11 geometry-3d

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

great-circle-arcspatial-visualizationlaw-of-cosinesisosceles-right-trianglepythagorean-theorem spatial-visualizationidentify-subproblems ↑ Prerequisites: law-of-cosinespythagorean-theorem

📏 Medium solution 💡 3 insights

Problem

Two people stand on a perfectly spherical Earth. One is on the equator at 110 degrees east longitude. The other is at 45 degrees north latitude and 115 degrees west longitude. Find the size of the angle formed at the centre of the sphere by the two radii pointing at them.

Pick an answer.

(A)

105

(B)

$112\frac{1}{2}$

(C)

120

(D)

135

(E)

150

How to solve

Strategy Visualize Spatial Relationships

Tool #17 (Visualize Spatial Relationships): latitude and longitude sound like map words, but on a sphere they are two ordinary angles measured at the center, so the first job is to see the two radii CA and CB as sticks pointing out of C and read off the angles between them. Tool #1 (Draw a Diagram): drop B straight down onto the plane of the equator, landing at a point D. That single extra point turns an invisible 3D angle into a picture made of flat triangles. Tool #7 (Identify Subproblems): the tetrahedron ABCD splits into four ordinary triangles, and each one is a familiar exercise — a 45°-45°-90° triangle, a Law of Cosines triangle, a Pythagorean triangle, and finally the triangle holding the angle we want. Tool #4 (Introduce a Variable): the radius is never given, so set CA = CB = 1; the angle cannot depend on the size of the sphere.

1STEP 1

Read latitude and longitude as angles

Read them as angles.

tilt of CB = 45°, turn between meridians = 245° - 110° = 135°

2STEP 2

Drop B onto the equator plane

Drop the northern point onto the equator plane.

∠ BDC = ∠ BDA = 90°, ∠ BCD = 45°, ∠ ACD = 135°, CA = CB = 1

3STEP 3

Solve the 45-45-90 triangle BCD

The dropped triangle is right isosceles.

BD = CD = CB · sin 45° = √(2)/2

4STEP 4

Law of Cosines in triangle ACD

Use the law of cosines in the plane.

AD² = AC² + CD² - 2 · AC · CD · cos 135° = 1 + 1/2 + 1 = 5/2

5STEP 5

Pythagoras in right triangle ABD

Pythagoras gives the distance between them.

AB² = AD² + BD² = 5/2 + 1/2 = 3 → AB = √(3)

6STEP 6

Law of Cosines gives the central angle

The law of cosines again gives 120 degrees.

cos∠ ACB = (1 + 1 - 3)/(2 · 1 · 1) = -1/2 → ∠ ACB = 120° → (C)

Answer

120

Two independent checks agree. First, a chord of a unit sphere cutting off a central angle θ has length 2sinθ/2; here AB = √(3) gives sinθ/2 = √(3)/2, so θ/2 = 60° and θ = 120°. Second, slide B along its meridian and watch the angle. If B sat on the equator at 245° E, the angle would be the full longitude gap, 135°. If B sat at the North Pole, the angle would be 90°, since CA lies in the equatorial plane and the polar axis is perpendicular to it. Billy is between those two spots, so the answer must be strictly between 90° and 135°: that immediately kills (D) 135 and (E) 150, and 120° sits comfortably inside. It is not the halfway value 112 1/2°, because the angle does not shrink at a steady rate as you march north — that is exactly the trap choice (B).

💡Key takeaway

Latitude and longitude are just two angles measured at the Earth's center, so drop the far point straight down to the equator plane and the sphere problem becomes a chain of flat triangles.

Read latitude and longitude as angles
Drop B onto the equator plane
Solve the 45-45-90 triangle BCD
Law of Cosines in triangle ACD
Pythagoras in right triangle ABD
Law of Cosines gives the central angle