AMC 10 · 2019 · #18

Grade 10 geometry-3d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Axis-Aligned Figure Area from Coordinates

coordinate-geometryarea-trianglesmidpoint-formulaspatial-visualizationpythagorean-theorem spatial-visualizationidentify-subproblemsconvert-to-algebra ↑ Prerequisites: coordinate-geometrypythagorean-theorem

📏 Medium solution 💡 3 insights

Problem

A pyramid has a square base of side 3, and one edge of length 6 rises straight up from a corner, perpendicular to the base, forming the apex. On two of the slanted edges, mark the points one third of the way up; on the remaining slanted edge, mark the point two thirds of the way up. Find the area of the triangle those three points form.

Pick an answer.

(A)

$\frac{3\sqrt2}{2}$

(B)

$\frac{3\sqrt3}{2}$

(C)

$2\sqrt2$

(D)

$2\sqrt3$

(E)

$3\sqrt2$

How to solve

Strategy Visualize Spatial Relationships

Tool #17 (Visualize Spatial Relationships) leads, because the whole problem turns on one spatial fact that is invisible until you look at heights: P and Q end up at the same height while R ends up directly above the midpoint of PQ. That single observation converts a tilted triangle floating in space into an ordinary base-times-height calculation. Tool #1 (Draw a Diagram) is what makes the fact checkable rather than hopeful: the perpendicular edge AE at corner A is already an axis, so the figure hands us a coordinate frame and every point becomes three numbers. Tool #7 (Identify Subproblems) organizes the finish — locate the three points, then measure a base, then measure the height perpendicular to that base, then combine. Trying to attack the area directly with side lengths and Heron's formula also works but does far more arithmetic than the height shortcut needs.

1STEP 1

Put the pyramid in coordinates

Put the pyramid in coordinates.

A=(0,0,0), B=(3,0,0), C=(3,3,0), D=(0,3,0), E=(0,0,6)

2STEP 2

Locate P, Q, and R

Locate all three points.

P=(3,0,0)+1/3(-3,0,6)=(2,0,2) Q=(0,3,0)+1/3(0,-3,6)=(0,2,2) R=(3,3,0)+2/3(-3,-3,6)=(1,1,4)

3STEP 3

Measure PQ as the base

Take one segment as the base.

PQ=√((-2)²+2²+0²)=√(8)=2√2

4STEP 4

R sits directly above PQ's midpoint

The third point sits right above the midpoint.

M=((2+0)/2,(0+2)/2,(2+2)/2)=(1,1,2), and R-M=(0,0,2) (0,0,2)·(-2,2,0)=0 → RM ⊥ PQ, with RM=4-2=2

5STEP 5

Combine base and height

Base times height gives two root two.

[PQR]=1/2 · PQ · RM = 1/2 · 2√2 · 2 = 2√2 → (C)

Answer

2√2

An independent computation avoids the height shortcut entirely. Using the vectors PQ=(-2,2,0) and PR=(-1,1,2), the cross product is (4,4,0) with magnitude 4√2, and half of that is 2√2 — the same value, reached without ever mentioning a midpoint. A third check confirms the shape: PR=√((-1)²+1²+2²)=√6 and QR=√(1²+(-1)²+2²)=√6, so △ PQR is isosceles with base 2√2; the altitude to that base has length √(6-(√2)²)=√4=2, matching the height found from the coordinates. Size-wise, 2√2≈ 2.83 sits comfortably between (B) ≈ 2.60 and (D) ≈ 3.46, and it is small relative to the pyramid's 3 × 3 base, which is right for a triangle whose vertices are pulled well up the slant edges.

💡Key takeaway

When one edge stands straight up from a corner, use it as an axis: "a third of the way along" becomes plain arithmetic on coordinates, and the height of the triangle becomes a subtraction.

Put the pyramid in coordinates
Locate P, Q, and R
Measure PQ as the base
R sits directly above PQ's midpoint
Combine base and height

Put the pyramid in coordinates

Locate P, Q, and R

Measure PQ as the base

R sits directly above PQ's midpoint

Combine base and height

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