AMC 10 · 2018 · #13

Grade 10 geometry-2d

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

centroid-2-to-1coordinate-geometrymidpoint-formuladilationarea-rectangles coordinate-geometryconvert-to-algebraidentify-subproblems ↑ Prerequisites: centroid-2-to-1coordinate-geometry

📏 Long solution 💡 3 insights 📊 Diagram

Problem

A square has side 30, and a point P sits inside it at distance 12 from one corner and 26 from another. The square's diagonals are not drawn; instead P is joined to all four corners, cutting the square into four triangles. Mark the centroid — the balance point where the three medians meet — of each of those four triangles. Those four marks form a convex quadrilateral. Find its area.

Pick an answer.

(A)

$100\sqrt{2}$

(B)

$100\sqrt{3}$

(C)

200

(D)

$200\sqrt{2}$

(E)

$200\sqrt{3}$

How to solve

Strategy Convert to Algebra

The picture hides the key fact: a centroid is a pure average, and averages behave beautifully in coordinates. Tool #1 (Draw a Diagram) puts the square on a grid with a corner at the origin so every vertex is a round number. Tool #4 (Introduce a Variable) names the unknown point as P = (3x, 3y) — writing it with a factor of 3 up front so that dividing by 3 for the centroid leaves no fractions. Tool #13 (Convert to Algebra) is the engine: turn 'centroid' into 'average the three coordinates' and just compute all four. Tool #16 (Change Focus) is where the problem cracks open — once the four centroids are written down, the variables x and y shift all four points together and never change the shape, which is why AP = 12 and BP = 26 turn out to be scenery. Tool #7 (Identify Subproblems) reduces the area to the two diagonals, and tool #3 (Eliminate Possibilities) checks the result against the answer list.

1STEP 1

Put the square on a grid

Put the square on coordinates.

A = (0,30), B = (0,0), C = (30,0), D = (30,30)

2STEP 2

Name P as (3x, 3y)

Writing P in thirds kills the fractions.

P = (3x, 3y), where 0 ＜ 3x ＜ 30 and 0 ＜ 3y ＜ 30

3STEP 3

A centroid is an average

A centroid is an average.

G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)

4STEP 4

Compute all four centroids

Compute all four centroids.

G₁ = (x, y+10), G₂ = (x+10, y), G₃ = (x+20, y+10), G₄ = (x+10, y+20)

5STEP 5

The unknowns only slide the shape

The unknowns only slide the shape, never reshape it.

G_i = O + (± 10, 0) or O + (0, ± 10), O = (x+10, y+10)

6STEP 6

Cross the diagonals to get the area

The two diagonals meet at right angles.

G₁G₃ = 20, G₂G₄ = 20, G₁G₃ ⊥ G₂G₄, Area = 1/2 · 20 · 20 = 200

7STEP 7

Match the answer choice

The area is 200.

(10√(2))² = 200 → (C)

Answer

200

Two independent checks agree. First, side lengths: G₁G₂ = √((10)² + (-10)²) = 10√(2), and by symmetry all four sides equal 10√(2); with equal diagonals of 20 that bisect each other at right angles, G₁G₂G₃G₄ is a square, and (10√(2))² = 200 — matching the diagonal computation. Second, scale: the answer should be a modest fraction of the square's area 30² = 900, and 200/900 = 2/9, a clean ratio, with the small quadrilateral comfortably fitting inside — sensible. Finally, the givens are consistent even though they go unused: solving p² + (q-30)² = 144 together with p² + q² = 676 gives q = 1432/60 ≈ 23.9 and p ≈ 10.3, so such a P really does exist strictly inside the square, and the answer 200 is correct for it — and for every other interior P too.

💡Key takeaway

A centroid is just the average of a triangle's three corners, so in coordinates the four centroids always sit 10 left, right, up, and down from one common centre — a square with diagonals 20 and 20, area 200, no matter where P hides.

Put the square on a grid
Name P as (3x, 3y)
A centroid is an average
Compute all four centroids
The unknowns only slide the shape
Cross the diagonals to get the area
Match the answer choice