AMC 10 · 2011 · #11

Grade 8 number-theorygeometry-2d

Archetype Branching State Enumeration └ Sub-archetype Numeric Operation Tree

parityinteger-pythagorean-triplesinvariant-monovariant systematic-enumeration ↑ Prerequisites: parity

📏 Long solution 💡 3 insights

Problem

Every hop covers the same distance and must land on a whole-number point. Find the fewest hops to a nearby target.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Organize Information in More Ways

Chasing actual positions is hopeless — after three jumps there are already thousands of reachable points. The winning move is Tool #15: stop tracking the pair (x,y) and track a single bit of it, whether x+y is even or odd. That one number changes in a completely predictable way, which turns "can the frog get there?" into "does the count of jumps have the right parity?". Tool #4 turns a jump into the equation a²+b²=25 and Tool #2 lists all of its integer solutions, which is what makes the parity claim checkable. Tool #3 then deletes the impossible answer choices. Finally, because a smallest-number question is only half answered by a lower bound, Tool #11 (Work Backwards from the finish) builds an explicit path that meets the bound.

1STEP 1

Write one jump as an equation

One hop is a single whole-number equation.

√(a²+b²)=5 ⟺ a²+b²=25, a,b integers

2STEP 2

List every legal jump

There are only 12 legal hops.

(± 5, 0), (0, ± 5), (± 3, ± 4), (± 4, ± 3) — 12 jumps in all

3STEP 3

Every jump flips odd and even

Every hop flips the coordinate sum's parity.

a²+b² = 25 is odd → exactly one of a,b is odd → a+b is odd

4STEP 4

Count the flips

So the number of hops must be odd.

x+y ≡ n (mod 2); target sum = 1 is odd → n is odd

5STEP 5

Kill two jumps a second way

Two hops also fail for a second reason.

p²+q² = (p-1)²+q² = 25 → 2p-1 = 0 → p = 1/2 ∉ Z

6STEP 6

Rule out one jump

One hop is plainly the wrong distance.

√((1-0)²+(0-0)²) = 1 ≠ 5 → n ≠ 1, so n ≥ 3

7STEP 7

Build a three-jump path

A three-hop path exists, so the answer is 3.

(0,0) → (3,4) → (6,0) → (1,0)

Answer

Both halves of the claim are closed, which is what a "smallest" question demands. Lower bound: parity rules out every even count, and the length mismatch rules out one jump, so n ≥ 3. Upper bound: an explicit legal path does it in 3. The two attacks on the two-jump case share nothing — one counts odd-even flips of x+y, the other subtracts the two circle equations and gets p = 1/2 — yet they agree, which is strong evidence the reasoning is sound. Checking every one of the 12 jump vectors confirms both facts directly: none of them is distance 5 from (1,0) as well as from (0,0), and the only way to write (1,0) as a sum of three jump vectors is (3,4)+(3,-4)+(-5,0) in some order, so the path found is essentially the only one that works. Choice (D) 5 is the designed trap: it has the correct parity and paths of length 5 do exist, but it is not the smallest. Choices (A), (C), (E) are even and fail parity.

💡Key takeaway

Every length-5 jump on the grid changes x+y by an odd number, so the frog's even-odd colour flips every time; reaching (1,0) needs an odd count of jumps, one jump is too short, and three jumps really do it.

Write one jump as an equation
List every legal jump
Every jump flips odd and even
Count the flips
Kill two jumps a second way
Rule out one jump
Build a three-jump path