AMC 10 · 2016 · #13

Grade 10 geometry-2d
trigonometric-ratiospythagorean-theoremspatial-visualizationestimation spatial-visualizationconvert-to-algebra ↑ Prerequisites: trigonometric-ratiospythagorean-theorem
📏 Medium solution 💡 3 insights
Problem
Two observers a known distance apart look in perpendicular directions at the same object. Find its height.

Pick an answer.

(A)
3.5
(B)
4
(C)
4.5
(D)
5
(E)
5.5
How to solve
Strategy Visualize Spatial Relationships

The plane is above the ground, so this is a three-dimensional picture and the two sightings live in two different vertical planes. The move that flattens it is dropping the plane straight down to a single ground point X: every angle of elevation then becomes a right triangle standing on the ground, and the two houses plus X form one ordinary triangle I can work with on paper. The compass words are the hidden gift here, because 'due north' and 'due west' force a right angle at X, which is exactly what makes the Pythagorean Theorem available. After that I name the altitude h, write both ground distances in terms of h, and the geometry collapses into one equation in one unknown.

1STEP 1

Drop the plane onto the ground

Dropping it down makes a ground point.

PX = h, PX ⊥ ground, ∠ AXP = ∠ BXP = 90°
2STEP 2

North and west meet at a right angle

The two directions meet at a right angle.

∠ AXB = 90° ⟹ AX² + BX² = AB² = 10² = 100
3STEP 3

Turn each elevation into a ground distance

Each elevation gives a ground distance.

tan 30° = h/AX = 1/√(3) → AX = h√(3); tan 60° = h/BX = √(3) → BX = h/√(3)
4STEP 4

One equation in one unknown

That leaves one equation in one unknown.

(h√(3))² + (h/√(3))² = 3h² + h²/3 = 10h²/3 = 100 ⟹ h² = 30 ⟹ h = √(30)
5STEP 5

Pin down √(30) between choices

Bounding the root gives 5.5, choice (D).

5² = 25 < 30 < 30.25 = 5.5², √(30)≈ 5.477, |5.477-5.5| ≈ 0.023 ≪ 0.477 ≈ |5.477-5|
Answer
5.5
Plug the height back in and check the ground triangle closes. With h = √(30), Alice's ground distance is AX = √(30)·√(3) = √(90) ≈ 9.49 miles and Bob's is BX = √(30)/√(3) = √(10) ≈ 3.16 miles. Then AX² + BX² = 90 + 10 = 100 = 10², so the two houses really are 10 miles apart. The split also matches intuition: Bob's steeper 60° sighting should put him much nearer the plane's ground point than Alice's shallow 30° sighting, and 3.16 against 9.49 is exactly that. Finally the altitude 5.48 miles is under both ground distances but the same order of size, which is what a 30°-to-60° pair of viewing angles should produce.
💡Key takeaway

Drop the flying object straight down to the ground first: the shadow point turns a 3D sighting problem into flat right triangles, and compass words like north and west quietly hand you a right angle.

  • Drop the plane onto the ground
  • North and west meet at a right angle
  • Turn each elevation into a ground distance
  • One equation in one unknown
  • Pin down √(30) between choices