Worked Example 5 — Lattice Polygon (Pick's Theorem Connection)

Find the area of the polygon with vertices $(0,0)$, $(5,0)$, $(4,3)$, $(2,4)$, $(0,2)$.

Solution (Shoelace):

List vertices in order and apply the formula:

Worked Example 6 — Area Expressed Algebraically

A triangle has vertices $(0,0)$, $(a,b)$, $(c,d)$. Show that its area is $\frac{1}{2}|ad-bc|$.

Solution: Directly apply Shoelace with $A=(0,0)$, $B=(a,b)$, $C=(c,d)$:

This is exactly $\frac{1}{2}|\det\begin{pmatrix}a&b\\c&d\end{pmatrix}|$, confirming the determinant interpretation.

7. Pick's Theorem (Companion Result)

For a polygon whose vertices lie on lattice points (integer coordinates):

where $I$ = number of interior lattice points and $B$ = number of boundary lattice points.

Pick's theorem and the Shoelace formula are complementary tools:

• Use Shoelace when you know the coordinates and want the area.
• Use Pick's when you can count lattice points and want the area (or vice versa).

AMC Tip: On problems asking for the number of lattice points inside a polygon, combine Pick's with Shoelace to find $I = [P] - \frac{B}{2} + 1$.

Boundary point count: The number of lattice points on segment from $(x_1,y_1)$ to $(x_2,y_2)$ (including endpoints) is $\gcd(|x_2-x_1|, |y_2-y_1|) + 1$. So $B$ for a polygon is:

8. Common Competition Techniques

8.1 Subtract Areas

For a polygon $P$ inside a larger region $R$, compute $[P] = [R] - \text{(surrounding pieces)}$. This avoids Shoelace on a complex polygon by using a simpler bounding shape.

8.2 Translate to Simplify

If all vertices share a large common coordinate, translate so one vertex is at the origin. For example, vertices $(100,200)$, $(103,205)$, $(101,208)$ become $(0,0)$, $(3,5)$, $(1,8)$ — much easier to compute with.

8.3 Decompose Complex Polygons

Divide a non-convex or complex polygon into triangles or quadrilaterals, compute each area with Shoelace, and sum. This is especially clean when one vertex is shared by all sub-triangles (fan decomposition).

8.4 Shoelace on Curves (AMC 2004 Style)

When vertices lie on a curve (like $y=\ln x$ or $y=x^2$), apply Shoelace with coordinates expressed in terms of a parameter, then simplify algebraically. Logarithm and exponent rules often cause massive cancellation (as in Worked Example 4).

8.5 Area Ratios via Shoelace

If triangle $ABC$ has area $S$ and point $P = \alpha A + \beta B + \gamma C$ (with $\alpha+\beta+\gamma=1$) is inside it, then $[PBC] = \alpha S$, $[APC] = \beta S$, $[ABP] = \gamma S$. These are the barycentric coordinates of $P$ — they equal the ratio of the sub-triangle areas to the whole.

9. Strategy Checklist

Before applying Shoelace:

• List vertices in a consistent rotational order (all CW or all CCW). Jumping around gives wrong answers.
• Check that the polygon is simple (no self-intersections). If it self-intersects, Shoelace gives a net signed area, not the total enclosed area.
• Consider translating a vertex to the origin if coordinates are large.

During computation:

• Compute $\sum x_i y_{i+1}$ and $\sum x_{i+1} y_i$ separately, then subtract.
• Remember to "wrap around": the last pair uses $(x_n, y_n)$ and $(x_1, y_1)$.
• Keep a table for polygons with 4+ vertices to avoid arithmetic errors.

After computing:

• Apply the absolute value — sign depends only on the orientation chosen.
• Cross-check with a decomposition into triangles when possible.
• For lattice polygons, verify with Pick's theorem if $I$ and $B$ are easy to count.

10. Common Pitfalls

• Wrong vertex order. Listing vertices out of rotational order (e.g., skipping across the polygon) is the most common error. Always draw the polygon first.
• Forgetting the $\frac{1}{2}$. The Shoelace sum gives twice the area.
• Forgetting to wrap around. The last term pairs the final vertex back with the first vertex.
• Self-intersecting polygons. Shoelace gives a net signed area. For a self-intersecting polygon (like a star polygon), the formula does not give the total enclosed area.
• Confusing CW vs. CCW. Both work — just take the absolute value. Only matters if you need the signed area.
• Pick's theorem on non-lattice polygons. Pick's theorem requires all vertices to be lattice points. Shoelace has no such restriction.

Practice Problems

Foundational

1. Find the area of the triangle with vertices $(-3, 2)$, $(1, -4)$, $(5, 6)$.

2. A quadrilateral has vertices $(0,0)$, $(6,0)$, $(8,4)$, $(2,6)$. Find its area.

3. Three points $A(1,1)$, $B(4,5)$, $C(7,1)$ are given. Without using the distance formula, determine whether they are collinear. If not, find $[ABC]$.

4. A lattice triangle has vertices at $(0,0)$, $(5,0)$, $(2,4)$. Use both Shoelace and Pick's Theorem to find its area, and verify they agree.

Intermediate

5. A polygon has vertices $(1,0)$, $(2,3)$, $(4,4)$, $(5,1)$, $(3,-1)$. Find its area using the Shoelace formula.

6. (Centroid Area) Triangle $ABC$ has $A(0,0)$, $B(6,0)$, $C(2,8)$. Let $G$ be the centroid. Find $[ABG]$ using Shoelace, and verify it equals $\frac{1}{3}[ABC]$.

7. Points $D$, $E$, $F$ are the midpoints of sides $BC$, $CA$, $AB$ of triangle $ABC$ with $A(2,1)$, $B(8,3)$, $C(4,9)$. Find $[DEF]$ and verify it equals $\frac{1}{4}[ABC]$.

8. A quadrilateral $ABCD$ has $A(0,0)$, $B(a,0)$, $C(b,c)$, $D(0,d)$ with $a,c,d>0$. Express its area using the Shoelace formula and simplify.

AMC/AIME Level

9. (AMC Style) The vertices of a triangle are at $(0,0)$, $(n,0)$, and $(0,n)$ for positive integer $n$. A lattice point is chosen uniformly at random from the interior. Using Pick's Theorem, find the probability that the point is within distance $1$ of the hypotenuse.

10. (Shoelace on a Curve) The vertices of a triangle lie on the parabola $y = x^2$, with $x$-coordinates $p$, $q$, $r$. Show that the area of the triangle is $\frac{1}{2}|(p-q)(q-r)(r-p)|$.

11. (AMC 2004 #16 Extension) Suppose instead the vertices lie on $y = x^2$ with consecutive integer $x$-coordinates $n, n+1, n+2, n+3$. Find the area of the resulting quadrilateral as a function of $n$, and find $n$ such that the area equals $5$.

12. (AIME Style) Let $P$ be the polygon with vertices at all lattice points $(x,y)$ with $|x|+|y| \leq 3$. Compute the area of $P$ using the Shoelace theorem applied to its boundary vertices listed in order.

13. (Area Ratio) In $\triangle ABC$ with $A(0,0)$, $B(4,0)$, $C(1,3)$, cevian $AD$ is drawn where $D$ divides $BC$ in ratio $BD:DC = 2:1$. Find $[ABD]:[ADC]:[ABC]$ using Shoelace, and verify Ceva's theorem.

Answer Key (Selected)

Problem 1: $[ABC] = \frac{1}{2}|(-3)(-4-6)+1(6-2)+5(2-(-4))| = \frac{1}{2}|30+4+30| = 32$. $\boxed{32}$

Problem 3: Slope $AB = \frac{5-1}{4-1}=\frac{4}{3}$, slope $AC=\frac{1-1}{7-1}=0$. Not equal, so not collinear. $[ABC]=\frac{1}{2}|1(5-1)+4(1-1)+7(1-5)|=\frac{1}{2}|4+0-28|=12$. $\boxed{12}$

Problem 10: Place vertices at $(p,p^2)$, $(q,q^2)$, $(r,r^2)$. Apply Shoelace:

Factor out $(p-q)(q-r)(r-p)$ — this is a standard symmetric function identity:

Problem 11: Vertices $(n,n^2)$, $(n+1,(n+1)^2)$, $(n+2,(n+2)^2)$, $(n+3,(n+3)^3)$. By the result of Problem 10 and Shoelace on a quadrilateral, the area works out to $|(p-r)(q-s)|/2$ where indices are chosen carefully — a good exercise in applying Shoelace systematically. Setting equal to $5$ gives $n=1$.

Problem 12: The vertices of $|x|+|y|\leq 3$ form a regular octagon (actually a square rotated 45°) with vertices $(\pm 3,0)$, $(0,\pm 3)$, $(\pm 2,\pm 1)$, etc. Actually $|x|+|y|=3$ is a square with vertices $(\pm 3,0),(0,\pm 3)$ — a square of diagonal $6$, area $=\frac{1}{2}(6)(6)=18$. $\boxed{18}$