AMC 10 · 2020 · #14

Grade 10 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

regular-octagonsine-area-formulaarea-trianglesisosceles-right-trianglesymmetry-argument identify-subproblemssymmetry-argumentphysical-representation ↑ Prerequisites: regular-octagonarea-triangles

📏 Long solution 💡 3 insights

Problem

A regular octagon has a certain area. Joining every other vertex gives a quadrilateral. Find the ratio of the quadrilateral's area to the octagon's area.

Pick an answer.

(A)

$\frac{\sqrt{2}}{4}$

(B)

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

(C)

$\frac{3}{4}$

(D)

$\frac{3\sqrt{2}}{5}$

(E)

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

How to solve

Strategy Draw a Diagram

Tool #1 (Diagram): draw the octagon's centre O and the eight spokes out to the vertices. That one extra mark turns two unrelated-looking shapes into two fans of triangles sharing the same hub. Tool #4 (Variable): name the spoke length R instead of the side length, because both areas then come out as a plain number times R² and R cancels. Tool #15 (Reorganize): stop reading ACEG as "a square sitting inside a polygon" and read both figures as sums of triangles measured from O — same hub, same spoke length, only the apex angle differs. Tool #7 (Subproblems): that leaves one small question to answer once — the area of a triangle with two sides R and a known angle between them. Tool #3 (Eliminate): the five choices are far apart numerically, so a decimal estimate confirms the match at the end.

1STEP 1

Mark the centre and spokes

All the spokes are equal.

OA = OB = … = OH, ∠ AOB = ∠ BOC = … = 360^°/8 = 45^°

2STEP 2

See the quadrilateral is a square

Alternate vertices sit at right angles.

∠ AOC = ∠ COE = ∠ EOG = ∠ GOA = 90^°

3STEP 3

Name the spoke

Name the spoke, not the side.

R = OA = OB = … = OH

4STEP 4

Cut both into triangles

Cut them like fans from the centre.

n = 8 · [triangle (R, R, 45^°)], m = 4 · [triangle (R, R, 90^°)]

5STEP 5

Area of one triangle

Two sides and the included angle give the area.

[ R, R, θ ] = 1/2R²sinθ; θ = 90^° → R²/2, θ = 45^° → 1/2 · R·R√(2)/2 = R²√(2)/4

6STEP 6

Add up both fans

Add eight of one and four of the other.

n = 8 · R²√(2)/4 = 2√(2) R², m = 4 · R²/2 = 2R²

7STEP 7

Divide and simplify

Dividing gives root two over two.

m/n = 2R²/2√(2) R² = 1/√(2) = √(2)/2 ≈ 0.707 → (B)

Answer

√(2)/2

Check the split a second way, with a side length instead of a spoke. Set the octagon's side to 1. The square ACEG plus the four corner triangles ABC, CDE, EFG, GHA makes up the whole octagon. Each corner triangle has two sides of length 1 with the octagon's interior angle 135^° between them, so its area is 1/2sin 135^° = √(2)/4 ≈ 0.354; four of them total about 1.414. By the Law of Cosines AC² = 1 + 1 - 2cos 135^° = 2 + √(2), so the square has area 2 + √(2) ≈ 3.414. Adding gives about 4.828 for the octagon, which matches the standard regular-octagon area 2(1+√(2)) ≈ 4.828. The ratio is 3.414/4.828 ≈ 0.707, agreeing with √(2)/2. The size of the answer is also sensible: the four cut-off corners are thin slivers, so the square should keep most of the octagon — clearly more than the 0.354 of choice (A), and clearly not the 0.849 or 0.943 of (D) and (E), which would leave the corners almost no room at all.

💡Key takeaway

Draw the centre and the spokes. Both shapes then become fans of triangles with the same two spoke lengths, and only the angle at the hub differs — 45^° for the octagon's eight pieces, 90^° for the square's four.

Mark the centre and the spokes
See that ACEG is a square
Name the spoke, not the side
Cut both shapes into triangles at O
Area of one spoke triangle
Add up both fans
Divide and clear the radical