AMC 10 Step by Step
Printable problems

Angles & Polygons

Angle chasing, polygon angle sums, regular polygons, parallel lines

What you need to know

  • Interior angles. The angles of a polygon with nn sides add up to (n−2)⋅180∘(n-2)\cdot180^\circ, because the diagonals from one corner cut it into n−2n-2 triangles. Example: a hexagon's angles add up to 4⋅180∘=720∘4\cdot180^\circ = 720^\circ.
  • Exterior angles. The exterior angles of any convex polygon (one at each corner) add up to 360∘360^\circ. So a regular nn-gon has exterior angles of 360∘n\frac{360^\circ}{n} and interior angles of 180∘−360∘n180^\circ - \frac{360^\circ}{n}. Example: a regular octagon has 45∘45^\circ exterior and 135∘135^\circ interior angles.
  • Parallel lines. When a line crosses two parallel lines, alternate interior angles (a "Z" shape) are equal, and co-interior angles (a "C" shape) add up to 180∘180^\circ. Example: a 70∘70^\circ angle has a 70∘70^\circ Z-partner and a 110∘110^\circ C-partner.
  • Exterior angle of a triangle. It equals the sum of the two inside angles at the other two corners. Example: angles of 50∘50^\circ and 65∘65^\circ give a 115∘115^\circ exterior angle at the third corner.
  • Isosceles triangles. Their base angles are equal, and a regular polygon's equal sides hide many of them. Example: in regular pentagon ABCDEABCDE, AB=BCAB = BC and ∠B=108∘\angle B = 108^\circ, so ∠BAC=180∘−108∘2=36∘\angle BAC = \frac{180^\circ-108^\circ}{2} = 36^\circ.
  • Angles in a circle. A central angle (corner at the center) equals its arc. An inscribed angle (corner on the circle) is half its arc. Example: a 100∘100^\circ arc has a 100∘100^\circ central angle and a 50∘50^\circ inscribed angle.
  • Diagonals. A convex polygon with nn sides has n(n−3)2\frac{n(n-3)}{2} diagonals: each corner joins n−3n-3 others (not itself or its neighbors), counting each diagonal twice. Example: an octagon has 8⋅52=20\frac{8\cdot5}{2} = 20.

How AMC 10 tests it

  • Problems 1–8: one missing angle in a figure of parallel lines, a triangle, or a kite.
  • "A regular polygon's interior angle is xx (or kk times its exterior angle); how many sides?" Solve 360∘n=\frac{360^\circ}{n} = (exterior angle) for nn. Example: 140∘140^\circ inside means 40∘40^\circ outside, so n=9n = 9.
  • Problems 10–16: a regular pentagon, hexagon, or octagon with diagonals, or with an equilateral triangle or square attached. The trick is spotting the isosceles triangles.

Standard approaches

  1. Name unknown angles with letters, and write equations from triangle and polygon angle sums.
  2. For regular polygons, use exterior angles (360∘n\frac{360^\circ}{n}). The arithmetic is cleaner.
  3. Find isosceles triangles made by equal sides or radii, and copy their base angles.
  4. With a circle, turn angles into arcs, add the arcs, and turn back.

Worked example

Regular pentagon ABCDEABCDE has an equilateral triangle ABFABF drawn inside it, with FF interior to the pentagon. What is the measure, in degrees, of ∠EFC\angle EFC?

(A) 150150 (B) 156156 (C) 162162 (D) 168168 (E) 174174

Solution. Step 1. Each angle of a regular pentagon is 3⋅180∘5=108∘\frac{3\cdot180^\circ}{5} = 108^\circ.

Step 2. Triangle ABFABF is equilateral, so ∠FAB=60∘\angle FAB = 60^\circ, leaving ∠EAF=108∘−60∘=48∘\angle EAF = 108^\circ - 60^\circ = 48^\circ.

Step 3. AE=AB=AFAE = AB = AF, so triangle AEFAEF is isosceles with 48∘48^\circ at AA. Its base angles share 180∘−48∘=132∘180^\circ - 48^\circ = 132^\circ, so ∠AFE=66∘\angle AFE = 66^\circ.

Step 4. The figure is symmetric across line FDFD, which swaps AA with BB and EE with CC. So ∠BFC=66∘\angle BFC = 66^\circ too.

Step 5. The angles EFAEFA, AFBAFB, BFCBFC, CFECFE go once around FF, so they add up to 360∘360^\circ:
∠EFC=360∘−66∘−60∘−66∘=168∘. \angle EFC = 360^\circ - 66^\circ - 60^\circ - 66^\circ = 168^\circ.
The answer is (D) 168\boxed{\textbf{(D)}\ 168}.

Pitfalls

  • Using (n−2)⋅180∘(n-2)\cdot 180^\circ for the exterior angles. Exterior angles always add up to 360∘360^\circ.
  • Deciding a triangle is isosceles, or a quadrilateral is a rectangle, because the drawing looks that way.
  • Giving the arc when the question asks for the inscribed angle. You are off by a factor of two.

Traps that recur

  • Halving the original apothem instead of the circumradius; the crease sits at half the distance from the center to a vertex, not to a side. (2023 AMC 10B #25)
  • Trying to compute the angle by brute-force trigonometry or coordinates, which is slow and error-prone without a calculator. (2022 AMC 10B #20)
  • Trying to solve for the six individual side lengths, which the data does not determine, instead of grouping them into the two triangle sides. (2021 AMC 10A #21)
  • Assuming G is the midpoint of AC, or misreading which points F, G, H, J are, e.g. treating JH as a side of the inner pentagon rather than a diagonal. (2015 AMC 10B #22)

Problems © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. The topic note, tags and key insights are original to this site.