Geometry Essentials

Why the exterior angle is the sum of the two far angles: the exterior angle and the interior angle beside it make a straight line, $180^\circ$; the three interior angles also make $180^\circ$. Both totals contain the same interior angle, so what is left over must match — the exterior angle equals the other two interior angles together. Angle chasing is nothing more than using these totals, plus equal angles at parallel lines, one angle at a time.

Step 1 — the exterior angle at a vertex equals the sum of the two interior angles at the OTHER vertices.

Step 2 — $45^\circ + 70^\circ = 115^\circ$. Check the long way: the third angle is $180 - 115 = 65^\circ$, and its exterior is $180 - 65 = 115^\circ$.

Step 1 — $AB$ crosses the parallel lines, so the alternate interior angles are equal: $\angle ABC = 55^\circ$.

Step 2 — the triangle's angles add to $180^\circ$: $\angle BAC = 180 - 55 - 65 = 60^\circ$. Each step names its reason — parallel lines, then the angle sum — which is how you avoid "it looks like $60$".

Step 1 — work with the exterior angle instead: $180 - 140 = 40^\circ$.

Step 2 — the exterior angles of any polygon total $360^\circ$, and in a regular polygon they are equal, so there are $360/40 = 9$ of them.

Step 3 — $9$ sides. Check: $(9 - 2)\cdot180 = 1260$ and $1260/9 = 140$.

Why two equal angles are enough: the third angle is then equal too (the sum is $180^\circ$), and a triangle's shape is fixed by its angles — scaling it changes the size, never the angles. So equal angles mean one triangle is a scaled copy of the other, every length multiplied by the same $k$. Areas multiply by $k^2$ because area is length times length, each scaled by $k$.

Step 1 — the sun's rays are parallel, so the two right triangles (height, shadow, ray) have equal angles and are similar.

Step 2 — height : shadow is the same in both: $\tfrac{1.5}{2} = \tfrac{h}{12}$.

Step 3 — $h = 12\cdot\tfrac{1.5}{2} = 9$ m.

Step 1 — $DE \parallel BC$ makes triangle $ADE$ similar to $ABC$ (shared angle at $A$, equal corresponding angles).

Step 2 — the scale factor is $\tfrac{AD}{AB} = \tfrac25$, so the area ratio is $\left(\tfrac25\right)^2 = \tfrac{4}{25}$.

Step 3 — the trapezoid is the rest: $1 - \tfrac{4}{25} = \tfrac{21}{25}$.

Step 1 — $120^\circ$ is $\tfrac{120}{360} = \tfrac13$ of the circle.

Step 2 — arc: $\tfrac13\cdot2\pi\cdot6 = 4\pi$; area: $\tfrac13\cdot\pi\cdot36 = 12\pi$.

Step 3 — the perimeter is the arc plus two radii: $12 + 4\pi$. Forgetting the two radii is the usual slip.

A quarter circle of radius $4$ inside a $4\times4$ square leaves $16 - 4\pi \approx 3.43$. The technique matters because the shaded piece itself is usually not a shape with a formula, while what surrounds it and what is cut out both are. When several pieces overlap, subtract them one at a time and add back what you removed twice — inclusion–exclusion, in geometry.

This converts "find the area" into "find a ratio". If $D$ is on $BC$ with $BD:DC = 2:3$, then $[ABD]:[ADC] = 2:3$ because the two triangles share the height from $A$ — no lengths needed. Chaining such ratios handles most of the shaded-area problems that look impossible at first glance.

Step 1 — square: $36$.

Step 2 — quarter circle: $\tfrac14\pi(6)^2 = 9\pi$.

Step 3 — subtract: $36 - 9\pi \approx 36 - 28.27 = 7.73$. Leaving the answer as $36 - 9\pi$ is usually what a contest wants; decimals only when asked.

Step 1 — the diagonal cuts the lens into two equal halves.

Step 2 — each half is a quarter circle ($\tfrac14\pi\cdot16 = 4\pi$) minus the right triangle with legs $4$ and $4$ (area $8$): $4\pi - 8$.

Step 3 — the lens is $2(4\pi - 8) = 8\pi - 16 \approx 9.13$. The pieces with formulas were hiding inside the shape without one.

The polygon's own sides are slanted, so its heights are awkward to find. The rectangle's sides are not, and every leftover piece has a horizontal leg and a vertical leg you can count on the grid. It is the subtraction principle again, with the grid supplying every length. For a quick check, the shoelace formula computes the same area straight from the coordinates.

Step 1 — the horizontal gap is $7 - 1 = 6$ and the vertical gap is $10 - 2 = 8$.

Step 2 — they are the legs of a right triangle whose hypotenuse is the distance: $\sqrt{6^2 + 8^2} = \sqrt{100} = 10$. Spotting the $6$-$8$-$10$ triple skips the square root.

Step 1 — the enclosing rectangle runs from $x = 1$ to $7$ and $y = 1$ to $6$: area $6\cdot5 = 30$.

Step 2 — the three corner triangles: below $AB$, legs $6$ and $2$, area $6$; right of $BC$, legs $3$ and $4$, area $6$; left of $CA$, legs $2$ and $5$, area $5$.

Step 3 — $30 - 6 - 6 - 5 = 13$.

Step 1 — boundary points: a side from $(x_1, y_1)$ to $(x_2, y_2)$ passes through $\gcd(|\Delta x|, |\Delta y|)$ grid steps; $\gcd(6, 2) + \gcd(4, 3) + \gcd(2, 5) = 2 + 1 + 1 = 4$, so $B = 4$.

Step 2 — counting the inside points gives $I = 12$.

Step 3 — $12 + \tfrac42 - 1 = 13$, matching the rectangle method. Two independent methods agreeing is worth more than one method done twice.

Step 1 — the hypotenuse: $9$-$12$-$15$ is $3$-$4$-$5$ scaled by $3$, so it is $15$.

Step 2 — area with the legs as base and height: $\tfrac12\cdot9\cdot12 = 54$.

Step 3 — the same area with the hypotenuse as base: $\tfrac12\cdot15\cdot h = 54$, so $h = \tfrac{108}{15} = 7.2$. Computing one area two ways is among the most useful moves in contest geometry.