Geometry: Coordinates & Complex

Why it works. The term $\tfrac12(x_iy_{i+1} - x_{i+1}y_i)$ is the signed area of the triangle formed by the origin and the edge from vertex $i$ to vertex $i + 1$: positive when the edge runs counterclockwise around the origin, negative otherwise. Going once around the polygon, the triangles outside the polygon are added once and subtracted once, and those inside are added once, so the total is the polygon's area. That is also why the vertices must be listed in order around the boundary.

Spotting it. Coordinates of every vertex, in order: shoelace, no decomposition needed.

Step 1. The cross terms $x_iy_{i+1} - x_{i+1}y_i$ are $1 \cdot 2 - 8 \cdot 0 = 2$, $8 \cdot 9 - 6 \cdot 2 = 60$, $6 \cdot 6 - 0 \cdot 9 = 36$ and $0 \cdot 0 - 1 \cdot 6 = -6$.

Step 2. They add to $92$, so the area is $46$.

Spotting it. A rectangle with points given by lengths: put $A$ at the origin and every point is known. The only work is intersecting two lines.

Step 1. $A = (0, 0)$, $B = (12, 0)$, $D = (0, 9)$, $E = (4, 9)$. Line $AE$: $y = \tfrac94x$. Line $BD$: $y = 9 - \tfrac34x$.

Step 2. $\tfrac94x = 9 - \tfrac34x$ gives $x = 3$, so $P = \left(3, \tfrac{27}{4}\right)$.

Step 3. $[ABP] = \tfrac12 \cdot 12 \cdot \tfrac{27}{4} = \tfrac{81}{2}$, so $m + n = 83$.

Check by similar triangles: $PED \sim PAB$ with ratio $4 : 12$, so $P$ is $\tfrac34$ of the way up from $AB$ to $CD$, at height $\tfrac{27}{4}$ ✓

Why it works. Multiplying by $e^{i\theta}$ multiplies lengths by $1$ and adds $\theta$ to every argument, which is exactly a rotation about $0$. To rotate about another centre $c$, translate $c$ to the origin ($z - c$), rotate, and translate back. In a square or an equilateral triangle, one side is another side rotated by $90^\circ$ or $60^\circ$, so an unknown vertex is one multiplication away.

Spotting it. The other two vertices are $A$ rotated by $\pm90^\circ$ about the centre of the square, and a quarter turn is multiplication by $i$.

Step 1. The centre is the midpoint $M = 4 + 4i$, and $A - M = -3 - 2i$.

Step 2. $i(-3 - 2i) = 2 - 3i$, so the other vertices are $M \pm (2 - 3i)$: $6 + i$ and $2 + 7i$, i.e. $(6, 1)$ and $(2, 7)$.

Step 3. $6 + 1 + 2 + 7 = 16$. Check: from $A$ to $(6, 1)$ is $(5, -1)$ and from $(6, 1)$ to $C$ is $(1, 5)$: equal lengths, perpendicular ✓

Spotting it. The three cube roots of any nonzero number are equally spaced on a circle: an equilateral triangle, whose area depends only on the radius.

Step 1. $|8i| = 8$, so the radius is $2$. The argument of $8i$ is $90^\circ$, so the roots are at $30^\circ$, $150^\circ$, $270^\circ$: $\sqrt3 + i$, $-\sqrt3 + i$ and $-2i$.

Step 2. The side is $2\sqrt3$ (the distance between $\pm\sqrt3 + i$), so the area is $\tfrac{\sqrt3}{4}(2\sqrt3)^2 = 3\sqrt3$. So $m + n = 6$.

Why it works. Triangles $ABD$ and $ACD$ have the same height from $A$, so their areas are in the ratio $BD : DC$; the same holds for $PBD$ and $PCD$. Subtracting, $\tfrac{BD}{DC} = \tfrac{[ABP]}{[ACP]}$. Likewise $\tfrac{CE}{EA} = \tfrac{[BCP]}{[BAP]}$ and $\tfrac{AF}{FB} = \tfrac{[CAP]}{[CBP]}$. Multiply the three: every area cancels and the product is $1$.

Spotting it. Three concurrent cevians with two ratios known: Ceva gives the third.

Step 1. $\dfrac{6}{4} \cdot \dfrac{5}{3} \cdot \dfrac{AF}{FB} = 1$, so $\dfrac{AF}{FB} = \dfrac{12}{30} = \dfrac25$.

Step 2. $AF = \tfrac27 \cdot 21 = 6$.

Spotting it. One straight line through points on two sides, crossing the third side's extension: Menelaus.

Step 1. $\dfrac{BD}{DC} \cdot \dfrac{CX}{XA} \cdot \dfrac{AM}{MB} = 1$ gives $\dfrac32 \cdot \dfrac{CX}{XA} \cdot 1 = 1$, so $CX : XA = 2 : 3$.

Step 2. Two of the points ($D$, $M$) are inside their sides, so $X$ is outside segment $CA$, and since $CX < XA$ it lies beyond $C$. Then $XA = XC + CA$, so $3k = 2k + 7$, $k = 7$, and $CX = 14$.

Power of a point says that for any line through $P$ meeting a circle at $A$ and $B$, the product $PA\cdot PB$ is the same for every such line, equal to $|d^2 - r^2|$. With a tangent it reads $PT^2 = PA\cdot PB$, which converts a tangency condition into an equation in lengths — exactly what an AIME answer needs.

In a triangle with $a = 7$, $b = 8$, $C = 60^\circ$: $c^2 = 49 + 64 - 2\cdot7\cdot8\cdot\tfrac12 = 57$, so $c = \sqrt{57}$. Stewart's theorem and the median formula are the same identity applied twice, so a configuration with a cevian usually yields to one law-of-cosines pair rather than a construction.

Step 1 — power of the point: $PT^2 = PA\cdot PB$, so $36 = 4\cdot PB$ and $PB = 9$.

Step 2 — $A$ is between $P$ and $B$, so $AB = PB - PA = 9 - 4 = 5$.

Spotting it. The median sits in two triangles that share the angle at $B$ (or $C$); get $\cos B$ from the whole triangle, then use it in the half.

Step 1. In $ABC$: $\cos B = \dfrac{13^2 + 14^2 - 15^2}{2 \cdot 13 \cdot 14} = \dfrac{140}{364} = \dfrac{5}{13}$.

Step 2. In $ABM$ with $BM = 7$: $AM^2 = 169 + 49 - 2 \cdot 13 \cdot 7 \cdot \tfrac{5}{13} = 218 - 70 = 148$.

Check with the median formula $AM^2 = \tfrac{2b^2 + 2c^2 - a^2}{4} = \tfrac{450 + 338 - 196}{4} = 148$ ✓

Spotting it. Power of a point fixes the PRODUCT $CP \cdot PD$; the ratio then fixes the pieces.

Step 1. $CP \cdot PD = 5 \cdot 12 = 60$. With $CP = 3k$, $PD = 5k$: $15k^2 = 60$, so $k = 2$.

Step 2. $CD = 8k = 16$.