Posted on May 1, 2026
This week's puzzle was a collaboration with high-schooler Connor Hill, the First Place Winner from this year's Regeneron Science Talent Search. Connor's project involved an original, computer-assisted proof to find the complete list of noble polyhedra, in which all the faces are indistinguishable and all the vertices are indistinguishable. The five Platonic solids are perhaps the best known examples of noble polyhedra, but Connor documented all of them, including two infinite sets and 146 exceptional cases, among which were 85 new examples.
Here's the puzzle from Connor, unrelated to his work, which is about a frog that makes very particular hops:
A frog is hopping around a chessboard, always from the center of one square to the center of another square. Each square has side length 1, but the board itself is not necessarily 8-by-8. Instead, it's \(N\)-by-\(N\), where \(N\) is some large whole number.
Every jump the frog makes must be the same distance, which we'll call \(L\). The frog wants to make four jumps such that:
What is the smallest jumping distance \(L\) for which this is possible?
Put the starting square at \((0,0)\), where all square centers have integer coordinates. The frog's jump length is the distance between two lattice points.
If \(L<\sqrt{65}\), then the only possible lattice distances are
\[ 1,\quad \sqrt{2},\quad 2,\quad \sqrt{5},\quad 2\sqrt{2},\quad 3,\quad \sqrt{10},\quad \sqrt{13},\quad 4,\quad \sqrt{17},\quad 3\sqrt{2},\quad 2\sqrt{5},\quad 5,\quad \sqrt{26},\quad \sqrt{29},\quad 4\sqrt{2},\quad \sqrt{34},\quad 6,\quad \sqrt{37},\quad 2\sqrt{10},\quad \sqrt{41},\quad 3\sqrt{5},\quad 7,\quad 5\sqrt{2},\quad 2\sqrt{13},\quad \sqrt{53},\quad \sqrt{58},\quad \sqrt{61},\quad 8. \]
Rather than ruling these out one at a time, it is better to use the geometry of a four-step loop. If the loop is
\[ (0,0)\to u \to u+v \to v \to (0,0), \]
then \(u\) and \(v\) must be lattice vectors of the same length \(L\). The loop is a rhombus, and it is not a square exactly when \(u\cdot v\neq 0\).
The new rule says that the three non-start vertices \(u\), \(v\), and \(u+v\) must all avoid both kinds of forbidden lines from the origin:
For every lattice length with \(L^2<65\), there is only one nontrivial way to write \(L^2\) as a sum of two squares. So, up to signs and swapping coordinates, the only possible edge vectors are the same pattern \((a,b)\) and \((\pm a,\pm b)\) or \((\pm b,\pm a)\).
Those possibilities all fail:
If \(L=1\), then every move is horizontal or vertical by one square. Any closed four-step walk through four distinct squares is necessarily the boundary of a unit square, which is forbidden.
If \(L=\sqrt{2}\), then the very first hop from \((0,0)\) lands on \((\pm 1,\pm 1)\), a square diagonal from the start, which is also forbidden.
For every larger case below \(65\), choosing the second edge by sign changes or coordinate swaps either makes the loop a square, or puts one of \(u\), \(v\), or \(u+v\) on a rook line or bishop line from the start.
So \(L\) must be at least \(\sqrt{65}\).
Now take the four squares
\[ (0,0)\to (8,1)\to (15,5)\to (7,4)\to (0,0). \]
Each jump has length
\[ \sqrt{8^2+1^2}=\sqrt{7^2+4^2}=\sqrt{65}. \]
The frog visits exactly four distinct squares, returns to the start after the fourth jump, and the loop is a rhombus rather than a square. Also, the intermediate squares \((8,1)\), \((15,5)\), and \((7,4)\) are neither diagonal from \((0,0)\) nor horizontal or vertical from it.
Answer: \( \boxed{\sqrt{65}} \).
The frog is jumping around the board with the same minimum distance \(L\) just found.
But this time, the frog also wants to be able to hop to every location on the chessboard. What is the minimum value of \(N\) for which this is possible?
With \(L=\sqrt{65}\), the legal moves are
\[ (\pm 8,\pm 1),\quad (\pm 1,\pm 8),\quad (\pm 7,\pm 4),\quad (\pm 4,\pm 7). \]
If \(N\le 13\), then the central square (or central block of squares) is isolated: every legal move changes one coordinate by at least \(7\), so from the middle of such a board there is nowhere to go.
On a \(14\times 14\) board, however, the move graph is connected, so every square is reachable from every other square. Thus \(14\) is the first nontrivial board on which the frog can eventually hop to every location.
Extra credit answer: \( \boxed{14} \).
Degenerate note: if one allows the trivial \(1\times 1\) board, then there is only one location to reach. The first board on which the frog can actually move and still reach every square is \(14\times 14\).
The main demo focuses on the updated 4-jump loop with both bishop and rook restrictions. The extra-credit demo focuses on reachability and a full-board walk. Click any square to change the starting square.
Try different jump lengths for the updated problem. The board size is set automatically to the smallest \(N\) that supports a valid 4-jump loop for that value of \(L\).
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This one is about visiting every square. The starting square counts as already visited, and when the board is fully reachable the demo shows an explicit covering walk.
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