7.2 Grid Perimeter, Shaded Area, Sub-Shape Counting & Shortest Lattice Paths

7.2 Grid Perimeter, Shaded Area, Sub-Shape Counting & Shortest Lattice Paths

Visual-quantitative problems on the TR-YÖS bridge geometry and combinatorics. Rather than asking you to count dozens of overlapping shapes or trace paths by hand—which invites double-counting or omission errors under time pressure—these items reward clean structural formulas for sub-shape enumeration, unit-grid area/perimeter, and lattice path routing.

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The rectangle and square counts below use sides along the drawn grid lines; tilted squares require a different count.

1. Counting Rectangles, Squares, and Triangles in a Grid (Şekil Sayma)

Rectangles and Squares in an $m \times n$ Grid

Consider an $m \times n$ grid made of $m$ rows and $n$ columns of $1 \times 1$ unit squares. Geometrically, this grid is formed by $m + 1$ horizontal lines and $n + 1$ vertical lines:

1. Total Number of Rectangles (Including Squares): Every rectangle is uniquely determined by choosing any $2$ of the $m + 1$ horizontal lines (top and bottom borders) and any $2$ of the $n + 1$ vertical lines (left and right borders):

$$N_{\text{rectangles}} = \binom{m+1}{2} \times \binom{n+1}{2} = \frac{m(m+1)}{2} \cdot \frac{n(n+1)}{2}$$

1. Total Number of Squares: A $k \times k$ square can be placed in $(m - k + 1)$ vertical positions and $(n - k + 1)$ horizontal positions. Summing over all possible side lengths $k = 1, 2, \dots, \min(m, n)$ gives the countdown product sum:

$$N_{\text{squares}} = \sum_{k=1}^{\min(m,n)} (m - k + 1)(n - k + 1) = m n + (m-1)(n-1) + (m-2)(n-2) + \dots$$

In the special case of an $n \times n$ square grid, this simplifies to the sum of squares:

$$N_{\text{squares }(n \times n)} = 1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$$

To find the number of non-square rectangles, compute $N_{\text{rectangles}} - N_{\text{squares}}$.

1. Rectangles and Squares Containing a Specific Marked Cell $(r, c)$: Suppose the unit cell at row $r$ ($1 \le r \le m$ from the top) and column $c$ ($1 \le c \le n$ from the left) contains a star or shaded mark. For a rectangle to enclose cell $(r, c)$:
   • Its top border must be chosen from the $r$ horizontal grid lines on or above row $r$.
   • Its bottom border must be chosen from the $m - r + 1$ horizontal grid lines on or below row $r$.
   • Its left border must be chosen from the $c$ vertical grid lines on or to the left of column $c$.
   • Its right border must be chosen from the $n - c + 1$ vertical grid lines on or to the right of column $c$. By the multiplication principle:

$$N_{\text{containing }(r,c)} = r(m - r + 1) \times c(n - c + 1)$$

If a question asks for squares of side $k$ containing $(r, c)$, count how many valid $k \times k$ windows overlap row $r$ and column $c$ for each $k = 1, 2, \dots, \min(m, n)$ and sum the valid placements.

Counting Triangles in Cevian Fans from Vertex $A$

Suppose $k$ interior line segments (cevians) are drawn from apex $A$ of $\triangle ABC$ down to base $BC$, partitioning $BC$ into $n = k + 1$ elementary base segments:

• Any choice of $2$ rays out of the $n + 1$ rays emanating from $A$ forms a unique triangle with apex $A$: there are $\binom{n+1}{2} = \frac{n(n+1)}{2}$ such triangles.
• If $m$ horizontal lines (including base $BC$ itself) are drawn parallel to $BC$ across the fan, each horizontal line acts as an independent base for those $\frac{n(n+1)}{2}$ apex-$A$ triangles:

$$N_{\text{triangles}} = m \times \frac{n(n+1)}{2}$$

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2. Shaded Grid Area & Polyomino Perimeter (Birim Kareli Zeminde Alan ve Çevre)

Shaded Lattice Polygons: Bounding Box vs. Pick's Theorem

When a tilted triangle or polygon is drawn with vertices on the intersections (lattice points) of a unit-square grid, its side lengths are irrational square roots, so avoid using $\frac{1}{2}bh$ with tilted bases. Instead, apply one of two exact methods:

1. Bounding-Box Subtraction (Çerçeve Alanından Dış Üçgenleri Çıkarma): Enclose the shaded polygon inside the tightest axis-parallel rectangle of area $W \times H$, then subtract the areas of the unshaded right triangles ($\frac{1}{2}ab$) and corner rectangles around the outside.
2. Pick's Theorem (Pick Teoremi): For any simple polygon whose vertices all lie on integer lattice points:

$$\text{Area} = I + \frac{B}{2} - 1$$

where $I$ is the number of lattice points strictly inside the polygon and $B$ is the number of lattice points lying on its boundary perimeter. Along a straight segment from $(x_1, y_1)$ to $(x_2, y_2)$, the number of boundary steps is $\gcd(\lvert x_2 - x_1 \rvert, \lvert y_2 - y_1 \rvert)$.

Polyomino Perimeter Formula

When $N_{\text{squares}}$ unit squares (each having perimeter $4$) are glued together edge-to-edge, every shared internal edge eliminates two unit segments ($1$ from each adjacent square) from the exterior boundary:

$$\text{Perimeter} = 4 N_{\text{squares}} - 2 N_{\text{shared internal edges}}$$

The shared-edge formula counts every boundary, including hole boundaries. To obtain only the outer perimeter of a shape with holes, subtract the boundary lengths of those holes. Use direct edge counting when a bounding-box shortcut is not justified.

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3. Shortest Lattice Paths (En Kısa Yol / Izgara Yolları)

A traveler moving along an $m \times n$ rectangular street grid from bottom-left corner $A(0,0)$ to top-right corner $B(m,n)$ takes a shortest path if and only if every step moves strictly East ($E$, $+x$) or North ($N$, $+y$).

1. Unrestricted Grid ($m$ steps East, $n$ steps North): Every shortest path is a permutation of $m$ identical $E$'s and $n$ identical $N$'s of total length $m + n$:

$$\text{Paths}(A \to B) = \binom{m+n}{m} = \frac{(m+n)!}{m!\,n!}$$

1. Mandatory Checkpoint $C(x_1, y_1)$: By the multiplication rule, multiply the paths from $A$ to $C$ by the paths from $C$ to $B$:

$$\text{Paths}(A \to C \to B) = \binom{x_1 + y_1}{x_1} \times \binom{(m - x_1) + (n - y_1)}{m - x_1}$$

1. Blocked Intersection or Missing Road Segment — Pascal Node-Labeling (Köşe Toplama Yöntemi):
   • If a single intersection $X$ is blocked, subtract the forbidden paths: $\text{Paths}(A \to B) - \text{Paths}(A \to X \to B)$.
   • If multiple intersections/edges are missing or the grid is L-shaped/irregular, label the starting corner $A$ with $1$, propagate $1$'s along the outer bottom and left boundary edges (up to any break), and fill every interior node recursively using:

$$P(x, y) = P(x - 1, y) + P(x, y - 1)$$

 setting $P = 0$ at any blocked intersection and omitting the term from any missing road segment.

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Worked Examples

Worked Example 7.2.1: Sub-Shape Counting & Marked Cell in a $4 \times 5$ Grid

Problem: A $4 \times 5$ rectangular grid consists of $4$ rows and $5$ columns of unit squares, with a star ($\star$) placed inside the unit square at row $r = 2$ (from the top) and column $c = 3$ (from the left). Determine: (a) the total number of rectangles in the grid, (b) the total number of squares in the grid, and (c) how many rectangles contain the starred cell $(2, 3)$.

Solution:

1. Total Rectangles ($m = 4, n = 5$):

$$N_{\text{rectangles}} = \binom{4+1}{2} \times \binom{5+1}{2} = \binom{5}{2} \times \binom{6}{2} = 10 \times 15 = 150$$

1. Total Squares:

$$N_{\text{squares}} = (4 \times 5) + (3 \times 4) + (2 \times 3) + (1 \times 2) = 20 + 12 + 6 + 2 = 40$$

1. Rectangles Containing Cell $(r = 2, c = 3)$:
   • Top border choices: $r = 2$; Bottom border choices: $4 - 2 + 1 = 3$.
   • Left border choices: $c = 3$; Right border choices: $5 - 3 + 1 = 3$.

$$N_{\text{containing }(2,3)} = (2 \times 3) \times (3 \times 3) = 6 \times 9 = 54\text{ rectangles.}$$

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Worked Example 7.2.2: Shortest Lattice Paths with Checkpoint and Blocked Node

Problem: On a $5 \times 4$ city street grid ($5$ blocks East, $4$ blocks North from corner $A(0,0)$ to corner $B(5,4)$), how many shortest paths from $A$ to $B$ pass through checkpoint $C(2,2)$ without passing through a closed construction intersection at $D(4,3)$?

Solution:

1. Count shortest paths from $A(0,0)$ to $C(2,2)$ ($2$ East, $2$ North):

$$\text{Paths}(A \to C) = \binom{2+2}{2} = \binom{4}{2} = 6$$

1. Count all shortest paths from $C(2,2)$ to $B(5,4)$ ($3$ East, $2$ North):

$$\text{Paths}_{\text{total}}(C \to B) = \binom{3+2}{2} = \binom{5}{2} = 10$$

1. Subtract paths from $C(2,2)$ to $B(5,4)$ that hit blocked node $D(4,3)$:
   • From $C(2,2)$ to $D(4,3)$ is $2$ East, $1$ North: $\binom{2+1}{1} = 3$ paths.
   • From $D(4,3)$ to $B(5,4)$ is $1$ East, $1$ North: $\binom{1+1}{1} = 2$ paths.
   • Forbidden paths from $C$ to $B$ via $D$: $3 \times 2 = 6$.
   • Valid paths from $C$ to $B$ avoiding $D$: $10 - 6 = 4$.
2. Multiply stages $A \to C$ and $C \to B$:

$$\text{Valid Paths} = 6 \times 4 = 24$$

Review the Method

Shortest paths across an $m \times n$ street grid equal $\binom{m+n}{m}$, whereas grids with missing road segments or multiple obstacles are solved via Pascal-triangle node addition (Köşe Toplama Yöntemi).