AMC 10 · 2006 · #21

Grade 11 algebrageometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

logarithm-propertiescompleting-the-squarecircle-equationarea-circles convert-to-algebra ↑ Prerequisites: logarithm-propertiescompleting-the-square

📏 Long solution 💡 3 insights

Problem

Two regions of the plane are described by logarithm inequalities. Find how many times larger the second region's area is.

Pick an answer.

(A)

(B)

(C)

100

(D)

101

(E)

102

How to solve

Strategy Convert to Algebra

A set defined by logarithms is impossible to picture directly, so first strip the logarithms off (Tool #13). That is legal in both directions because log₁₀ is strictly increasing: comparing two logarithms is exactly the same as comparing the two positive numbers inside them. What is left is a quadratic inequality in x and y, and completing the square turns it into a statement about distance from a point — a disk, whose area is π r² (Tool #1 for seeing it). One trap has to be defused before the area is trustworthy. The definition also silently demands x+y ＞ 0, and if part of the disk fell on the wrong side of the line x+y=0, the region would be a cut-off piece of a disk, not a whole one. Check the extreme case (Tool #14): find how small x+y can get for a point that satisfies the quadratic inequality. Finally, keeping the two problems side by side in the same general form (Tool #15) shows exactly where the ratio comes from and why it is not the value a careless scaling argument predicts.

1STEP 1

Strip the logarithms off

Writing the constant as a logarithm makes both sides comparable.

log₁₀(1+x²+y²) ≤ log₁₀(10(x+y)) ⇔ 1+x²+y² ≤ 10(x+y)

2STEP 2

Complete the square to get a disk

Completing the square turns the condition into a disk.

x²-10x+y²-10y ≤ -1 ⟺ (x-5)²+(y-5)² ≤ 49=7²

3STEP 3

Check that the domain cuts nothing off

The logarithm's domain turns out to be automatic, so nothing is trimmed.

10(x+y) ≥ 1+x²+y² ≥ 1 ⟹ x+y ≥ 1/10 ＞ 0

4STEP 4

Repeat the whole argument for the second set

The same argument makes the second region a full disk too.

2+x²+y² ≤ 100(x+y) ⇔ (x-50)²+(y-50)² ≤ 4998

5STEP 5

Turn radii into areas

Each squared radius is already on the right, so no roots are needed.

[S₁]=49π, [S₂]=4998π

6STEP 6

Compare the two areas

Dividing gives 102, just above a round hundred, choice (E).

[S₂]/[S₁]=4998π/49π=(100²/2-2)/(10²/2-1)=4998/49=102 → (E)

Answer

102

The arithmetic is exact, not approximate: 49 × 100=4900 and 49 × 2=98, so 49 × 102=4998, and the ratio is a whole number as the answer choices promise. The scale is also believable — the radius grows from 7 to √(4998)≈ 70.7, about ten times bigger, so the area should be roughly a hundred times bigger, and 102 sits right there. The domain check deserves a second look because it is genuinely tight rather than a formality. The lowest value of x+y on the first disk is 10-7√(2)=10-√(98)≈ 0.1005, and on the second disk it is 100-√(2 · 4998)=100-√(9996)≈ 0.0200; both are positive, but only just, and they match the bounds x+y ≥ 1/10 and x+y ≥ 1/50 proved in Step 3 almost exactly. The condition is not vacuous either: had the problem used 60+x²+y² ≤ 10(x+y), the squared radius would be 50-60 ＜ 0 and the region would be empty. So checking that the disk clears the line x+y=0 is real work, and the answer 100 is the trap for anyone who scales radii by 10 and stops.

💡Key takeaway

Since log₁₀ is increasing, the logarithms come off cleanly and each set turns into a disk; check that the hidden rule x+y ＞ 0 cuts nothing off, then compare π r² to get 4998π/49π=102.

Strip the logarithms off
Complete the square to get a disk
Check that the domain cuts nothing off
Repeat the whole argument for the second set
Turn radii into areas
Compare the two areas