AMC 10 · 2011 · #24

Grade 8 number-theorygeometry-2d

Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Lattice Constraint Analysis

slope-interceptcoordinate-geometryfraction-arithmetic optimization ↑ Prerequisites: slope-intercept

📏 Long solution 💡 4 insights

Problem

A lattice point is a point

(x,y)

whose two coordinates are both integers. The line

y = mx + 2

must pass through no lattice point with

0 \lt x \le 100

, and that has to hold for every slope

m

with

\frac{1}{2} \lt m \lt a

. Find the largest possible value of

a

Pick an answer.

(A)

$\frac{51}{101}$

(B)

$\frac{50}{99}$

(C)

$\frac{51}{100}$

(D)

$\frac{52}{101}$

(E)

$\frac{13}{25}$

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Extreme Principle

The whole question is a boundary hunt, so Tool #14 (Extreme Principle) is primary: the largest safe a is pinned by the single closest "bad" slope sitting just above 1/2. To even see those bad slopes, use Tool #16 (Change Focus) — instead of chasing lattice points, translate "line hits a lattice point" into "the slope equals a fraction k/x with x ≤ 100." Tool #4 (Introduce a Variable) names that fraction so we can measure how far above 1/2 it sits. Finally Tool #3 (Eliminate Possibilities) checks the five answer choices against the boundary we find, catching the even-denominator trap 51/100.

1STEP 1

Reframe as a slope-is-a-fraction question

Since $2$ is whole, $y=mx+2$ is an integer exactly when $mx$ is — so a lattice point appears only if $m=\frac{k}{x}$ for some $x \le 100$ .

y = mx + 2 ∈ Z ⇔ mx ∈ Z ⇔ m = k/x

2STEP 2

Name the target value of a

The interval breaks the moment it swallows a forbidden fraction, so the biggest $a$ is the smallest such fraction above $\frac{1}{2}$ .

a_max = min{ k/x ＞ 1/2 : 1 ≤ x ≤ 100 }

3STEP 3

Gap above one-half for a fixed denominator

Fix $x$ : the first $k$ with $\frac{k}{x} \gt \frac{1}{2}$ sits a gap of $\frac{2k-x}{2x}$ above a half, and $2k-x$ is a positive integer.

k/x - 1/2 = (2k - x)/2x ≥ 1/2x

4STEP 4

Odd denominators get closest

That numerator is $1$ only when $x$ is odd; an even $x$ forces it to $2$ , so an odd denominator hugs a half twice as tightly.

x odd → gap = 1/2x; x even → gap ≥ 1/x

5STEP 5

Push the denominator to its maximum

The gap $\frac{1}{2x}$ shrinks as $x$ grows, so the largest odd $x=99$ wins: $\frac{50}{99}$ , gap $\frac{1}{198}$ , beating $\frac{51}{100}$ .

50/99 - 1/2 = 1/198 ＜ 1/100 = 51/100 - 1/2

6STEP 6

Confirm the boundary and the answer

No fraction with denominator $\le 100$ lies in $(\frac{1}{2},\frac{50}{99})$ , and $m=\frac{50}{99}$ hits $(99,52)$ , so $a=\frac{50}{99}$ — choice (B).

a_max = 50/99 → (B)

Answer

50/99

All five choices cluster just above 1/2, which fits: the answer must be a fraction barely larger than 0.5. Numerically 50/99 ≈ 0.50505, and the only choice below it is 51/101 ≈ 0.50495 — a safe but sub-maximal value, exactly the kind of near-miss the problem plants. The tempting wrong answer 51/100 = 0.51 is the best even-denominator fraction; our parity step is precisely what rules it out, since the odd denominator 99 hugs 1/2 twice as tightly. The winning slope 50/99 does hit the lattice point (99, 99·50/99+2) = (99, 52), confirming it is genuinely the first obstacle.

💡Key takeaway

Turn "the line dodges all lattice points" into "the slope isn't any fraction k/x with x ≤ 100," then find the fraction that hugs 1/2 most tightly from above — an odd denominator wins, and x=99 gives 50/99.

Reframe as a slope-is-a-fraction question
Name the target value of a
Gap above one-half for a fixed denominator
Odd denominators get closest
Push the denominator to its maximum
Confirm the boundary and the answer