AMC 10 · 2009 · #19

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Nested Frame / Concentric Difference
area-circlesinradiuspythagorean-theorem area-differenceeasier-related-problem ↑ Prerequisites: area-circles
📏 Long solution 💡 3 insights
Problem
Two regular polygons with different side counts share the same side length. Each has a ring between its circumscribed and inscribed circles. Compare the two ring areas.

Pick an answer.

(A)
$A = \frac {25}{49}B$
(B)
$A = \frac {5}{7}B$
(C)
A = B
(D)
$A= \frac {7}{5}B$
(E)
$A = \frac {49}{25}B$
How to solve
Strategy Draw a Diagram

Tool #1 (Draw a Diagram) does the heavy lifting: one picture — the center, a single side, and the perpendicular from the center to that side — contains both radii at once and shows the two circles are concentric. Tool #13 (Convert to Algebra) turns that picture into R²-r²=1 through the Pythagorean theorem, which is the moment the number of sides disappears. Tool #16 (Change Focus) matters because the natural instinct is to compute R and r for a pentagon and a heptagon separately; the ring area only needs the difference R²-r², so chasing the individual radii is wasted work. Tool #9 (Solve an Easier Related Problem) then pays off: proving the statement for a general regular polygon of side 2 is easier than doing two special polygons, since the side count never enters. Finally Tool #3 (Eliminate Possibilities) reads the choices — four of the five are ratios built from 5 and 7, and the derivation shows the answer cannot depend on the side count at all.

1STEP 1

Both circles share one center

Both circles share one centre.

ring area=π R²-π r²
2STEP 2

The touch point halves the side

The touch point halves a side.

XT=TY=1, OT=r, OX=R
3STEP 3

Pythagoras erases the side count

The Pythagorean theorem then erases the side count.

r²+1²=R² ⟹ R²-r²=1
4STEP 4

Chase the difference, not the radii

Chasing the difference rather than the radii gives a fixed area.

π R²-π r²=π(R²-r²)=π · 1=π
5STEP 5

One calculation covers both polygons

One calculation covers both polygons.

A=π and B=π
6STEP 6

Compare and pick the choice

So the two are equal, choice (C).

A/B=π/π=1 ⟹ A=B (C)
Answer
A = B
Compute the two rings separately and numerically as a check. Each side of the pentagon subtends 360°/5=72° at the center, so the angle at O in triangle OTX is 36°, giving R=1/(sin 36°)≈ 1.70130 and r=1/(tan 36°)≈ 1.37638; then π R²-π r²≈π(2.89443-1.89443)=π. For the heptagon that angle is 180°/7≈ 25.714°, giving R≈ 2.30477 and r≈ 2.07652; then π R²-π r²≈π(5.31193-4.31193)=π. All four radii are different, yet both rings come out to π≈ 3.14159, so A=B and (C) holds. It is worth being precise about what proves this. A square of side 2 gives 2π-π=π and an equilateral triangle of side 2 gives 4π/3-π/3=π, which is suggestive, but a few worked examples would only be a hint — they cannot rule out that some other polygon behaves differently. The proof is the third step, where R²-r²=1 was derived for a general regular n-gon with the side count absent from the equation.
💡Key takeaway

The ring between a regular polygon's two circles has area π times the square of half a side, so with side 2 every such ring is exactly π — the pentagon's ring and the heptagon's ring match, giving A=B, choice (C).

  • Both circles share one center
  • The touch point halves the side
  • Pythagoras erases the side count
  • Chase the difference, not the radii
  • One calculation covers both polygons
  • Compare and pick the choice

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