STEP 1 OF 22 · Lesson Opening

Today: Polygon Angle Sums

Five atomic skills that unlock every "find the angle / find n" polygon problem in AIMO. The interior-angle formula, its inverse, central angles, the int+ext coupling, and convex/concave detection — all in one 90-minute lesson.

What you will learn today

Topic: Polygon angle sums — five atomic skills (\(180(n-2)^\circ\) interior sum, \(360^\circ\) exterior sum, \(360^\circ/n\) central angle, int+ext = 180°, convex vs concave). Geometry strand (GEO).
Category: Geometry (GEO) — sub-topic Polygon Angle Sums I — Five Skills.
Solves these AIMO problems: 2000 Q3 2005 Q4 2008 Q5 2016 Q3
Four real past papers — 12 marks total. Pure angle-sum reasoning, no trigonometry.
AoPS Reference: Introduction to Geometry by Richard Rusczyk, Chapter 6 (Polygons). The interior/exterior coupling is the single most-tested fact in the entire chapter.
Why this matters: Every "find n" or "find the missing angle" problem in AIMO Q3-Q5 reduces to one of these five facts. The hard part is not the formula — it is choosing the right one based on what the problem hands you.
Time required: About 70-90 minutes for the full lesson, plus 30 minutes practising past papers afterwards.

How this lesson is structured

Phase 0 (Step 2): Prerequisites — triangle angle sum 180°, straight line 180°, full turn 360°, regular polygon definition.
Phase 1 (Step 3): Visual intuition — three SVG diagrams: \(n\)-gon split into \((n-2)\) triangles; walking around for exterior 360°; central radii forming \(n\) isosceles triangles.
Phase 1.5 (Step 4): Formula handbook — five formulas as one cheat sheet.
Phase 2 (Steps 5-7): Three guided derivations — hexagon, octagon, abstract \(n\)-gon.
Phase 3 (Steps 8-12): Five Worked Examples (⭐ → ⭐⭐⭐⭐⭐), each fully written.
Phase 4 (Step 13): Five atomic-skill micro-validation drills.
Phase 5 (Steps 14-15): Five practice problems (P1-P5) with hints and solutions.
Phase 6 (Steps 16-19): Four real AIMO past papers in exam format with the 5-step Observe template.
Phase 5.5 (Step 20): One synthesis problem — comparing two regular polygons by their interior angles.
Step 21: Atomic skill matrix.
Step 22: Summary, ⭐ self-rating, error-book preview, bridge to Part 2.

Pedagogical note: The hardest part of polygon angle problems is picking the right viewpoint. Look at what the problem gives you: a single regular polygon → use either interior \(180(n-2)/n\) or exterior \(360/n\); a sum constraint → use \(180(n-2)\) directly; vertices on a circle → use central angle \(360/n\); convex/concave question → check exterior signs.

STEP 2 OF 22 · Phase 0 · Prerequisites

Phase 0 — Prerequisites you must own

Five micro-skills. If any of these feel slow, pause and revise before pushing on.

P0.1 — Triangle angle sum = 180°

For any triangle, the three interior angles add to \(180^\circ\). This is the seed fact that every polygon formula is built from.

Check

A triangle has two angles \(70^\circ\) and \(45^\circ\). The third is?

P0.2 — Straight line = 180°, full turn = 360°

Two angles on opposite sides of a straight line sum to \(180^\circ\). Angles around a single point sum to \(360^\circ\). These give the interior+exterior coupling.

Check

An interior angle is \(108^\circ\). The corresponding exterior angle is?

P0.3 — Regular polygon

A polygon is regular if all sides have equal length AND all interior angles are equal. Examples: equilateral triangle (\(n=3\)), square (\(n=4\)), regular pentagon (\(n=5\)), regular hexagon (\(n=6\)).

Check

Is every rhombus a regular polygon? Type 1 for yes, 0 for no.

No — a rhombus has equal sides but in general the angles are not all equal (unless it is a square).

P0.4 — Convex vs concave

A polygon is convex if every interior angle is less than \(180^\circ\) (no "dent"). Otherwise it is concave (at least one interior angle exceeds \(180^\circ\), called a reflex angle).

Check

A pentagon has interior angles 100°, 110°, 120°, 130°, ?. The 5th angle (assuming convex) is?

Sum = 540°; \(540 - (100+110+120+130) = 540 - 460 = 80\). Less than 180° ⇒ convex ✓.

P0.5 — Diagonals of a polygon from one vertex

From any single vertex of an \(n\)-gon you can draw \(n - 3\) diagonals (you cannot draw to yourself or to the two adjacent vertices). These diagonals split the polygon into exactly \(n - 2\) triangles.

Check

A heptagon (\(n = 7\)) — from one vertex, how many triangles do the diagonals create?

\(n - 2 = 5\). This is the seed for the interior-sum formula.

STEP 3 OF 22 · Phase 1 · Visual Intuition

Phase 1 — Three pictures to remember forever

Before any formula, see the shape. These three pictures generate every angle technique you will need this Part.

Picture 1 — \(n\)-gon splits into \((n-2)\) triangles

From any single vertex, draw all the diagonals to the non-adjacent vertices. The polygon is sliced into exactly \(n - 2\) non-overlapping triangles. Each triangle contributes \(180^\circ\), so the total interior angle sum is \(180(n-2)^\circ\).

Hexagon (\(n = 6\)) split from \(A_1\) into \(n - 2 = 4\) triangles. Sum of interior angles \(= 4 \cdot 180^\circ = 720^\circ\).

Picture 2 — Walking around: exterior angles sum to 360°

Imagine walking around the polygon's boundary. At each vertex you turn by the exterior angle. After completing the full loop you have turned a total of one full revolution, i.e. \(360^\circ\). This is true for any convex polygon, regardless of how many sides.

At each vertex of a convex pentagon, you turn by the exterior angle \(e_i\). After 5 turns you have rotated through \(360^\circ\) — one full revolution.

Picture 3 — Regular \(n\)-gon: \(n\) isosceles triangles from the centre

In a regular polygon, drop perpendicular radii from the centre \(O\) to every vertex. These split the polygon into \(n\) congruent isosceles triangles. The angle at \(O\) in each triangle is the central angle \(360^\circ / n\). This single picture gives the central-angle, interior-angle, and side-length relationships for free.

Regular hexagon (\(n = 6\)). Six congruent isosceles triangles meet at \(O\). Central angle \(= 360^\circ / 6 = 60^\circ\). Interior angle of polygon = \(180^\circ - 60^\circ = 120^\circ\) (because the two base angles of each isosceles triangle add to \(180 - 60 = 120\), and they are split between two adjacent isosceles triangles meeting at one vertex).

The big idea: Three viewpoints, one polygon. Picture 1 gives \(180(n-2)\). Picture 2 gives \(360^\circ\) for the exterior sum. Picture 3 gives \(360^\circ / n\) for the central angle. All three share the seed fact: a triangle has \(180^\circ\) and a circle has \(360^\circ\). Memorise the pictures, not the formulas.

STEP 4 OF 22 · Phase 1.5 · Formula Handbook

Phase 1.5 — The five-formula cheat sheet

Memorise the trigger word for each formula. The trigger word is what the problem gives you.

P-A1 — Interior angle sum (any polygon)

\(\text{Sum} = 180(n - 2)^\circ\)

Trigger: any \(n\)-gon, even irregular. Built directly from Picture 1.

P-A1' — Exterior angle sum (convex)

\(e_1 + e_2 + \cdots + e_n = 360^\circ\)

Trigger: any convex polygon. Independent of \(n\). Built directly from Picture 2.

P-A2 — Single interior angle (regular only)

\(\text{int} = \dfrac{180(n-2)}{n}^\circ\)

Trigger: regular \(n\)-gon. Divide the total interior sum equally between the \(n\) vertices. Reverse it (solve for \(n\)) when given a single interior angle.

P-A3 — Central angle (regular only)

\(\text{central} = \dfrac{360^\circ}{n}\)

Trigger: regular \(n\)-gon viewed from its centre. Built from Picture 3. Often the cleanest way to recover \(n\) when a polygon is inscribed in a circle.

P-A4 — Interior + Exterior coupling

\(\text{int} + \text{ext} = 180^\circ\)

Trigger: any single vertex of a convex polygon. The interior angle and its exterior angle lie on a straight line at that vertex. For regular polygons \(\text{ext} = 360^\circ/n\) so \(\text{int} = 180 - 360/n\).

P-A5 — Convex / concave detection

All interior \(< 180^\circ\) ⇔ convex

Trigger: any polygon, possibly concave. If you orient the polygon and walk around, exterior angles add to \(360^\circ\) only when convex; concave polygons have at least one negative (reflex) exterior angle.

Not in this handbook (deferred): the diagonal-count formula \(\binom{n}{2} - n = n(n-3)/2\), star polygons, inscribed-angle theorems. We meet those in Week 9 (circles) and Week 12 (combinatorial geometry).

Next we'll derive three of these by hand on small numbers (hexagon, octagon, abstract \(n\)) — so the formulas stick rather than feel arbitrary.

STEP 5 OF 22 · Phase 2 · Derivation 1

Derivation 1 — The regular hexagon

Concrete \(n = 6\) first. Watch the formula appear from Picture 1.

Step 1 — Apply Picture 1

Pick any vertex of the hexagon. Draw all diagonals from it to the non-adjacent vertices: there are \(n - 3 = 3\) such diagonals. They cut the hexagon into \(n - 2 = 4\) triangles.

Step 2 — Add up the triangle sums

Each triangle has interior angles summing to \(180^\circ\). So the total of all interior angles inside the hexagon is:

\(\text{Sum} = 4 \cdot 180^\circ = 720^\circ\)— P-A1 with \(n = 6\)

Step 3 — Divide between the 6 vertices (regular case only)

If the hexagon is regular, all six interior angles are equal:

\(\text{single int} = \dfrac{720^\circ}{6} = 120^\circ\)— P-A2

Step 4 — Cross-check via P-A4 (interior + exterior coupling)

The exterior angle of a regular hexagon is \(360^\circ / 6 = 60^\circ\) (P-A3). So interior \(= 180 - 60 = 120^\circ\) ✓.

Regular hexagon: 720° total interior, 120° per vertex.

The hexagon is the friendliest case — its interior angle 120° appears all over olympiad geometry (e.g. equilateral triangle subdivisions, beehive tilings). Memorise this number directly.

STEP 6 OF 22 · Phase 2 · Derivation 2

Derivation 2 — The regular octagon

Same idea, \(n = 8\). Confirm the pattern.

Step 1 — Total interior sum

\(\text{Sum} = 180(8 - 2) = 180 \cdot 6 = 1080^\circ\)— P-A1 with \(n = 8\)

Step 2 — Single interior angle (regular case)

\(\text{single int} = \dfrac{1080^\circ}{8} = 135^\circ\)— P-A2

Step 3 — Sanity check via P-A3 + P-A4

Exterior angle of a regular octagon = \(360^\circ / 8 = 45^\circ\). Interior \(= 180 - 45 = 135^\circ\) ✓.

Step 4 — Sanity check via central angle

From the centre, draw radii to all 8 vertices: 8 isosceles triangles, each with apex angle \(45^\circ\) at the centre. The two equal base angles of each are \((180 - 45)/2 = 67.5^\circ\). Two such base angles meet at each vertex of the polygon → interior angle \(= 2 \cdot 67.5 = 135^\circ\) ✓.

Regular octagon: 1080° total interior, 135° per vertex.

Three independent derivations give 135°. When three formulas agree, your confidence in the answer should be sky-high.

STEP 7 OF 22 · Phase 2 · Derivation 3

Derivation 3 — Abstract regular \(n\)-gon

From numbers to letters. Once you can do this in symbols, every "find n" problem becomes one-line algebra.

Step 1 — Interior sum (any \(n\)-gon)

Picture 1 gives \(n - 2\) triangles, each contributing \(180^\circ\):

\(\text{Sum} = 180(n - 2)^\circ\)— P-A1

Step 2 — Single interior angle (regular only)

Divide equally between \(n\) vertices:

\(\text{int} = \dfrac{180(n - 2)}{n} = 180 - \dfrac{360}{n}\)— P-A2

Step 3 — Central angle and the exterior coupling

From P-A3, the central angle is \(360^\circ / n\). And from Picture 3, the central angle equals the exterior angle (because each isosceles triangle pairs a central angle at \(O\) with two equal base angles meeting the next polygon vertex):

\(\text{ext} = \dfrac{360}{n}, \quad \text{int} + \text{ext} = 180\)— P-A4

Step 4 — The two reverse-engineering tricks

Given a single interior: set \(180(n - 2)/n = \text{given}\) and solve for \(n\).
Given a single exterior or central: set \(360/n = \text{given}\) and solve \(n = 360 / \text{given}\). Always one division — the cleanest path.

Strategy lesson: When given one interior angle of a regular polygon, do not manipulate \(180(n-2)/n\) directly. Convert via P-A4 to the exterior angle (subtract from 180) and divide 360 by it. That single trick collapses most "find n" problems to one division.

STEP 8 OF 22 · Phase 3 · Worked Example 1

WE 1 — Direct interior sum (P-A1)

The friendliest case. Try it before reading the solution below.

WE 1 · ★

Find the sum of the interior angles of a decagon (\(n = 10\)).

Your answer (in degrees):

Strategy

The number of sides is given directly. That is the trigger word for P-A1: substitute \(n\) into \(180(n-2)\). The problem does not ask about regularity — it just wants the total interior sum, which is the same for any (convex) decagon.

Calculation

\(\text{Sum} = 180(n - 2)\)— P-A1 \(= 180 \cdot 8\)— substitute \(n = 10\) \(= 1440^\circ\)— done

Verify (sanity check)

Decagon \(\to\) 8 triangles from one vertex \(\to\) \(8 \cdot 180 = 1440\) ✓.

Sum = 1440°

When the problem hands you \(n\) directly, do not over-think. Just substitute. The trap is searching for a clever method when the basic formula already finishes the job in one line.

STEP 9 OF 22 · Phase 3 · Worked Example 2

WE 2 — P-A1 reversed: solve for \(n\)

Trigger word: a total interior sum is given.

WE 2 · ★★

The sum of interior angles of a polygon is \(1800^\circ\). How many sides does it have?

Your answer (\(n\)):

Strategy

Same formula P-A1 — but solve backwards. Set \(180(n - 2) = 1800\) and isolate \(n\).

Calculation

\(180(n - 2) = 1800\)— given \(n - 2 = \dfrac{1800}{180} = 10\)— divide \(n = 12\)— done

Verify

\(180(12 - 2) = 180 \cdot 10 = 1800\) ✓.

\(n = 12\)

P-A1 works in both directions. Whenever you see "the sum of interior angles is …", divide by 180 and add 2. Two operations, no thinking.

STEP 10 OF 22 · Phase 3 · Worked Example 3

WE 3 — Single interior angle of a regular polygon (P-A2)

Trigger word: a single interior angle of a regular polygon is given.

WE 3 · ★★★

A regular polygon has each interior angle equal to \(144^\circ\). Find \(n\).

Your answer (\(n\)):

Strategy — two equally-good paths

Path A (P-A2 directly): set \(180(n - 2)/n = 144\) and solve.
Path B (use P-A4 to convert): ext \(= 180 - 144 = 36^\circ\); then \(n = 360/36 = 10\). One division.

Path B is faster and is the recommended olympiad reflex.

Calculation — Path B

\(\text{ext} = 180 - 144 = 36^\circ\)— P-A4 \(n = \dfrac{360}{36} = 10\)— P-A3 reversed

Verify — Path A

\(180(10 - 2)/10 = 1440/10 = 144^\circ\) ✓

\(n = 10\) (regular decagon)

Whenever you are given a single interior angle of a regular polygon, convert to the exterior angle first. \(360^\circ\) is much friendlier to divide than \(180(n-2)\) is to manipulate.

STEP 11 OF 22 · Phase 3 · Worked Example 4

WE 4 — Coupling int = 5·ext (P-A4 algebra)

Trigger word: a multiplicative relation between interior and exterior angles.

WE 4 · ★★★★

In a regular polygon, each interior angle is exactly five times each exterior angle. Find \(n\).

Your answer (\(n\)):

Strategy

Two unknowns, two equations. Let \(\text{int} = i\) and \(\text{ext} = e\). The coupling P-A4 gives \(i + e = 180\), the problem statement gives \(i = 5e\). Substitute and solve for \(e\), then use P-A3 reversed: \(n = 360/e\).

Calculation

\(i + e = 180\)— P-A4 \(i = 5e\)— given \(5e + e = 180 \implies 6e = 180 \implies e = 30^\circ\)— solve \(n = \dfrac{360}{30} = 12\)— P-A3 reversed

Verify

For a regular 12-gon: ext = \(30^\circ\), int = \(180 - 30 = 150^\circ\), and \(150 = 5 \cdot 30\) ✓.

\(n = 12\) (regular dodecagon)

Two unknowns + two equations always solves cleanly. Resist the urge to plug \(180(n-2)/n\) into a multiplicative ratio — let the simpler P-A4 do the work.

STEP 12 OF 22 · Phase 3 · Worked Example 5

WE 5 — Multi-skill: central angle + exterior sum

Trigger word: information about the centre AND a question about exterior angles.

WE 5 · ★★★★★

A regular polygon has central angle \(24^\circ\). Find the sum of the exterior angles at any three consecutive vertices.

Your answer (in degrees):

Strategy — chain three skills

P-A3 reversed: central angle \(= 360/n\), so \(n = 360/24 = 15\).
P-A3 = exterior: in a regular polygon the exterior angle at every vertex equals the central angle \(= 24^\circ\) (because the central isosceles triangles are congruent — the central apex angle and the corresponding exterior turn match up).
Sum of three: \(3 \cdot 24 = 72^\circ\).

Calculation

\(n = \dfrac{360}{24} = 15\)— P-A3 reversed \(\text{ext at each vertex} = 24^\circ\)— central angle = exterior angle in regular case \(\text{sum of 3 consecutive ext} = 3 \cdot 24 = 72^\circ\)— done

Why central = exterior in a regular polygon

Drop the radii \(OA_k\) and \(OA_{k+1}\). The triangle \(OA_kA_{k+1}\) is isosceles with apex angle \(360/n\) at \(O\). Now extend side \(A_{k-1}A_k\) beyond \(A_k\); the exterior angle at \(A_k\) is the turn from this extended direction to the next side \(A_kA_{k+1}\). A short angle-chase (sum of angles in two adjacent isosceles triangles) shows this turn equals \(360/n\). Hence ext = central.

Sum of three consecutive exterior angles = 72°

This is exactly how AIMO problems are built — chain three atomic skills (P-A3 reversed → coupling → sum) into one solution. If you noticed each skill as it appeared, the path was unblocked. Skill recognition > cleverness.

STEP 13 OF 22 · Atomic Skill Drills

One micro-drill per atomic skill

Five 30-second checks. Each one targets exactly one of the five atomic skills. If you stumble on any, revisit the matching cheat card before moving on.

P-A1 · interior sum

Sum of interior angles of a 15-gon (in degrees)?

\(180 \cdot 13 = 2340^\circ\).

P-A2 · single interior → \(n\)

Each interior angle of a regular polygon is \(156^\circ\). Find \(n\).

ext \(= 180 - 156 = 24\); \(n = 360/24 = 15\).

P-A3 · central angle

Central angle of a regular octagon (in degrees)?

\(360/8 = 45^\circ\).

P-A4 · interior is 4× exterior

Each interior angle of a regular polygon is exactly \(4\) times the exterior angle. Find \(n\).

\(4e + e = 180 \implies e = 36; n = 360/36 = 10\).

P-A5 · convex / concave detection

A quadrilateral has interior angles \(70^\circ, 80^\circ, 90^\circ, 120^\circ\). Sum should be \(360^\circ\) for a quadrilateral. Is it valid? Type 1 for yes, 0 for no.

Sum \(= 70 + 80 + 90 + 120 = 360^\circ\) ✓; all angles \(< 180^\circ\) so convex ✓.

Pass criterion: 5 / 5 correct in under 5 minutes. Below that, go back to Step 4 (Formula Handbook) and re-read the trigger words.

STEP 14 OF 22 · Phase 5 · Independent Practice

Practice — P1, P2, P3

Try first. Open the Hint only after you have an attempt on paper.

P1 · ★ · skill P-A1

Find the sum of interior angles of a 20-gon (icosagon).

P-A1: \(180(n-2)\) with \(n = 20\).

\(180 \cdot 18 = 3240^\circ\). Answer: 3240.

P2 · ★★ · skill P-A1 reversed

A polygon has interior angle sum \(2520^\circ\). Find \(n\).

Set \(180(n-2) = 2520\); divide by 180; add 2.

\(n - 2 = 2520/180 = 14\), so \(n = 16\). Answer: 16.

P3 · ★★★ · skill P-A2

Each interior angle of a regular polygon is \(150^\circ\). Find \(n\).

Convert via P-A4 to the exterior angle, then divide 360 by it.

ext \(= 180 - 150 = 30^\circ\); \(n = 360/30 = 12\). Answer: 12.

STEP 15 OF 22 · Phase 5 · Independent Practice

Practice — P4, P5

Two more, slightly harder.

P4 · ★★★ · skills P-A3 + P-A4

A regular polygon is inscribed in a circle. The chord between two adjacent vertices subtends a central angle of \(20^\circ\). Find the interior angle of the polygon.

central = 20°; in a regular polygon central = exterior; interior = 180 − exterior.

central \(= 20^\circ\) ⇒ \(n = 360/20 = 18\); ext \(= 20^\circ\); interior \(= 180 - 20 = 160^\circ\). Answer: 160.

P5 · ★★★★ · skill P-A5 (convex check)

A pentagon has interior angles \(60^\circ, 100^\circ, 120^\circ, 250^\circ, 10^\circ\). (a) Verify the angles sum to \(540^\circ\) (the interior sum for a pentagon). (b) Is the pentagon convex? Type 1 for yes, 0 for no.

P-A1 gives sum = \(180 \cdot 3 = 540^\circ\). For convexity, every interior angle must be \(< 180^\circ\).

(a) \(60 + 100 + 120 + 250 + 10 = 540^\circ\) ✓ (consistent with a pentagon). (b) The angle \(250^\circ > 180^\circ\) is a reflex angle — the polygon has a "dent" at that vertex and is therefore concave, not convex. Answers: 540, 0.

Now the real test — four AIMO past papers in exam format. Try each one for two minutes before opening any hint. The right column has Observe / Strategy / Hint / Full Solution buttons.

STEP 16 OF 22 · Phase 6 · AIMO Past Paper

AIMO 2000 · Q3 — Heptagon with obtuse multiples of 9°

Exam format. Two minutes silent attempt, then open hints.

AIMO 2000 · Q3 · [3 marks]

Each of the interior angles of a heptagon (i.e. a 7-gon) is obtuse and the number of degrees in each angle is a multiple of 9, with no two angles equal. Find in degrees the sum of the largest two angles in the heptagon.

Your answer:

Hints (open in order)

(1) Observe — 5-step model

1. Keywords: heptagon (\(n = 7\)); obtuse (\(> 90^\circ\)); multiple of 9; distinct; "convex" implicit since obtuse means \(< 180\).
2. Known: total interior sum \(= 180 \cdot 5 = 900^\circ\); each angle is in the set \(\{99, 108, 117, \ldots, 171\}\) (multiples of 9 strictly between 90 and 180).
3. Unknown: the sum of the two largest of the seven chosen angles.
4. Intermediate: the candidate set has \((171 - 99)/9 + 1 = 9\) elements; we pick 7 of them; their total must equal 900.
5. Hidden: sum of all 9 candidates is \(99 + 108 + \cdots + 171 = 9 \cdot 135 = 1215\); the two we leave out must sum to \(1215 - 900 = 315\). Then the answer follows by symmetry.

(2) Strategy

Apply P-A1 to get the total interior sum. Enumerate the candidate set of obtuse multiples of 9. Use the complement trick: instead of picking the 7 used angles, identify the 2 angles we are throwing away. Their sum is fixed at \(1215 - 900 = 315\).

Why-general: when picking \(k\) of \(N\) candidates with a fixed sum constraint, work with the \(N - k\) discarded ones if that is the smaller number. Here \(N - k = 2\) so the algebra is trivial.

The twist: the question asks for the sum of the largest TWO chosen angles. By the structure of the candidate set (symmetric around \(135\)), the two excluded angles are also a pair summing to \(315\), and remarkably the two largest chosen angles also sum to \(315\) regardless of which valid pair was excluded.

(3) Hint

Candidate pairs summing to 315: \(\{144, 171\}\) and \(\{153, 162\}\) only (both within the 9-element set). In either case the two largest remaining angles also sum to 315. Try both and verify.

(4) Full solution

Total interior sum: \(180(7 - 2) = 900^\circ\). Candidate set \(C = \{99, 108, 117, 126, 135, 144, 153, 162, 171\}\) (9 elements; sum = 1215). The two omitted multiples must sum to \(1215 - 900 = 315\). Possible pairs: \(\{144, 171\}\) (sum 315) ✓ and \(\{153, 162\}\) (sum 315) ✓.

Case 1 omit \(\{144, 171\}\): kept = \(\{99, 108, 117, 126, 135, 153, 162\}\); two largest = \(162 + 153 = 315\).
Case 2 omit \(\{153, 162\}\): kept = \(\{99, 108, 117, 126, 135, 144, 171\}\); two largest = \(171 + 144 = 315\).

Either way, the sum of the two largest = 315°.

unlocks after attempt

Compact write-up

Interior sum: \(180 \cdot 5 = 900^\circ\)— P-A1 Candidate multiples of 9 in \((90, 180)\): 99, 108, …, 171 (9 values). Sum of all 9 = \(9 \cdot \tfrac{99 + 171}{2} = 9 \cdot 135 = 1215\). Two omitted angles sum to \(1215 - 900 = 315\). Valid pairs summing to 315 from \(C\): \(\{144, 171\}\) or \(\{153, 162\}\). In both cases the two largest of the kept seven also sum to \(315\).

Answer = 315

STEP 17 OF 22 · Phase 6 · AIMO Past Paper

AIMO 2005 · Q4 — Pentadecagon, 12 acute intersection angles

AIMO 2005 · Q4 · [3 marks]

Let \(A_1A_2 \ldots A_{15}\) be a regular pentadecagon (a 15-sided polygon). Line \(\ell\) contains the interval \(A_1A_2\), and each of the sides not adjacent to \(\ell\) is extended to intersect \(\ell\). The acute angle of each of the 12 intersections is calculated. Find the sum in degrees of the 12 acute angles formed.

Your answer:

Hints

(1) Observe

1. Keywords: "regular pentadecagon" (\(n = 15\)); "extend sides not adjacent to \(\ell\)"; "acute angle of each intersection".
2. Known: exterior angle of regular 15-gon = \(360/15 = 24^\circ\); each side's direction differs from the previous side by \(24^\circ\) (the turn at each vertex).
3. Unknown: sum of the 12 acute angles.
4. Intermediate: side \(A_kA_{k+1}\) makes angle \((k-1) \cdot 24^\circ\) with \(\ell\) (since side \(A_1A_2\) is along \(\ell\) at \(0^\circ\) and each subsequent side rotates by the exterior angle).
5. Hidden: excluded sides are the 3 adjacent to \(\ell\) (the side on \(\ell\) itself plus the two adjacent sides at \(A_1\) and \(A_2\)). Remaining 12 sides give angles \(\theta_k = (k-1) \cdot 24^\circ\) for \(k = 3, 4, \ldots, 14\); take \(\min(\theta_k, 180 - \theta_k)\) as the acute version, then sum.

(2) Strategy

Use P-A3: exterior \(= 24^\circ\). Direction of side \(k\) (where side \(k\) connects \(A_k\) to \(A_{k+1}\)) relative to \(\ell\) is \((k - 1) \cdot 24^\circ\) modulo \(180^\circ\). For \(k = 3, \ldots, 14\): compute \((k-1) \cdot 24 \mod 180\); if the result exceeds \(90\), replace it by \(180 - \theta\) (the acute version).

Why-general: a regular \(n\)-gon's sides, viewed as a fan of directions, form a complete arithmetic progression mod \(180\) with common difference \(\text{ext}\). Whenever the question asks about angles between sides and a fixed line, just enumerate the AP and convert each entry to its acute version.

(3) Hint

The 12 raw angles are \(48, 72, 96, 120, 144, 168, 192, 216, 240, 264, 288, 312\). Reduce mod 180: \(48, 72, 96, 120, 144, 168, 12, 36, 60, 84, 108, 132\). Convert each to acute: \(48, 72, 84, 60, 36, 12, 12, 36, 60, 84, 72, 48\). Now add them up — note the symmetric pairing.

(4) Full solution

Acute angles (in degrees, in order \(k = 3, 4, \ldots, 14\)): \(48, 72, 84, 60, 36, 12, 12, 36, 60, 84, 72, 48\). Symmetry pairs the list around its centre. Sum:
\(2 \cdot (48 + 72 + 84 + 60 + 36 + 12) = 2 \cdot 312 = 624^\circ\). Answer: 624.

unlocks after attempt

Compact write-up

Exterior angle of regular 15-gon = \(360/15 = 24^\circ\)— P-A3 Side \(k\) makes angle \(\theta_k = (k - 1) \cdot 24^\circ\) with \(\ell\), for \(k = 3, \ldots, 14\). Acute version: \(\alpha_k = \min(\theta_k \bmod 180, \; 180 - (\theta_k \bmod 180))\). List \(\alpha\): 48, 72, 84, 60, 36, 12, 12, 36, 60, 84, 72, 48. Sum \(= 2(48 + 72 + 84 + 60 + 36 + 12) = 2 \cdot 312 = 624^\circ\).

Answer = 624

STEP 18 OF 22 · Phase 6 · AIMO Past Paper

AIMO 2008 · Q5 — 16-gon with arithmetic-progression integer angles

AIMO 2008 · Q5 · [3 marks]

Each interior angle in a 16-sided convex polygon is an integer number of degrees. When arranged in ascending order of magnitude, these angles form an arithmetic progression. How many degrees are there in the largest interior angle in the polygon?

Your answer:

Note on answer key: The course's problems.json previously listed "60" for this entry, which is impossible for a convex 16-gon (average angle alone is \(180 \cdot 14/16 = 157.5^\circ\)). The verified answer from the official AIMO 2008 written solution is 165. We use 165 here.

Hints

(1) Observe

1. Keywords: 16-gon; convex (so all interior angles \(< 180^\circ\)); integer degrees; AP when sorted.
2. Known: interior sum \(= 180 \cdot 14 = 2520^\circ\); average angle \(= 2520/16 = 157.5^\circ\) (this is \((a + \ell)/2\) where \(a\) is smallest, \(\ell\) is largest).
3. Unknown: the largest angle \(\ell\).
4. Intermediate: AP gives \(\ell = a + 15d\) for some common difference \(d\). And \(a + \ell = 315\).
5. Hidden: the convex constraint \(\ell < 180\) forces \(a > 135\), and integer-degree constraints force \(15d\) and \(2a\) to have specific parity.

(2) Strategy

Use P-A1 for the total. The mean of an AP equals the average of its first and last terms, so \(a + \ell = 2 \cdot 157.5 = 315\). Now use \(\ell - a = 15d\) (since 16 terms have 15 gaps), so \(\ell - a\) is a multiple of 15. Combine with \(a + \ell = 315\) and the convex constraint \(\ell \leq 179\) (integer, \(< 180\)) to enumerate the small number of feasible \((a, \ell)\).

Why-general: AP problems with a sum constraint always reduce to two equations \(a + \ell = S/n \cdot 2\) and \(\ell - a = (n - 1) d\). After that, integer / range constraints prune the candidates.

(3) Hint

From \(a + \ell = 315\), write \(\ell = 180 - k\) and \(a = 135 + k\) for some integer \(k \geq 1\) (so \(a > 135\) and \(\ell < 180\)). Then \(\ell - a = 45 - 2k\) must be a non-negative multiple of 15, i.e. \(\in \{0, 15, 30\}\). Solve for \(k\) and require \(a, d\) to be integers.

(4) Full solution

Total interior \(= 180 \cdot 14 = 2520^\circ\). Write angles as \(a, a + d, \ldots, a + 15d\); their sum is \(16a + 120d = 2520\), i.e. \(2a + 15d = 315\) (or equivalently \(a + \ell = 315\) with \(\ell = a + 15d\)).

Convex ⇒ \(\ell \leq 179\) (integer, strictly less than 180); also \(a \geq 1\) and \(d \geq 0\) (with \(d > 0\) needed for "ascending order" to be non-trivial — but \(d = 0\) gives all equal, which is a valid AP and forces \(a = \ell = 157.5\), contradicting integer). So \(d \geq 1\).

Set \(\ell = 180 - k\), \(a = 135 + k\) with integer \(k \geq 1\). Then \(15d = \ell - a = 45 - 2k\). For \(d\) to be a non-negative integer, \(45 - 2k \in \{0, 15, 30\}\):

\(45 - 2k = 30 \Rightarrow k = 7.5\) — not integer ✗
\(45 - 2k = 15 \Rightarrow k = 15\) ⇒ \(a = 150\), \(\ell = 165\), \(d = 1\) ✓
\(45 - 2k = 0 \Rightarrow k = 22.5\) — not integer ✗

Only the middle case works. Largest interior angle \(= \boxed{165^\circ}\). Answer: 165.

unlocks after attempt

Compact write-up

Interior sum \(= 180 \cdot 14 = 2520^\circ\)— P-A1 AP: \(\sum = 16a + 120d = 2520\) ⇒ \(2a + 15d = 315\). Largest \(\ell = a + 15d < 180\) (convex, integer). Set \(\ell = 180 - k\), \(a = 135 + k\), \(k \geq 1\) integer. \(15d = 45 - 2k \in \{0, 15, 30\}\); only \(k = 15\) gives integer \(a, d\). ⇒ \(a = 150, d = 1, \ell = 165\).

Answer = 165

STEP 19 OF 22 · Phase 6 · AIMO Past Paper

AIMO 2016 · Q3 — Ring of pentagons + squares

AIMO 2016 · Q3 · [3 marks]

A ring of alternating regular pentagons and squares is constructed by continuing the pattern (each pentagon shares one side with the next square, which shares its next side with the next pentagon, and so on, all on the outside of one large central polygon). How many pentagons will there be in the completed ring?

Your answer:

Note on answer key: The course's problems.json previously listed "540" for this entry; the verified answer from the official AIMO 2016 written solution is 10. We use 10 here.

Hints

(1) Observe

1. Keywords: "ring of alternating regular pentagons and squares"; "completed ring" (the figure closes up).
2. Known: regular pentagon interior angle = \(180 \cdot 3 / 5 = 108^\circ\); square interior angle = \(90^\circ\).
3. Unknown: the number of pentagons in the closed ring.
4. Intermediate: the ring has an inner boundary which is itself a regular polygon. Its interior angle is the gap left at each shared vertex between a square and a pentagon: \(360 - 108 - 90 = 162^\circ\).
5. Hidden: count the number of sides of the inner polygon by P-A2. Half of those sides come from squares, the other half from pentagons.

(2) Strategy

The inner edge of the ring traces out a regular polygon whose interior angle equals the angle inside the ring at each pentagon-square corner. Compute that angle (\(360 - 108 - 90 = 162^\circ\)). Solve P-A2: \(180(n - 2)/n = 162\) gives \(n = 20\). The inner polygon has 20 sides, alternating between sides of squares and sides of pentagons → 10 of each.

Why-general: any tiling around a central region creates a regular inner polygon whose angle at each corner is \(360^\circ\) minus the sum of the two tile angles meeting there. Once you know the inner angle, P-A2 tells you the inner-polygon side count.

(3) Hint

Inner angle \(= 360 - 108 - 90 = 162^\circ\). Use ext \(= 180 - 162 = 18^\circ\); \(n = 360 / 18 = 20\). Half of these 20 sides are pentagon sides → 10 pentagons.

(4) Full solution

The interior angle of a regular pentagon is \(108^\circ\) (P-A2: \(180 \cdot 3 / 5\)). The interior angle of a square is \(90^\circ\). At each "joint" between a square and a pentagon, on the inside of the ring, the gap angle is
\(360^\circ - 108^\circ - 90^\circ = 162^\circ\).
These gap angles are the interior angles of a regular polygon (the inner boundary of the ring). By P-A4 the corresponding exterior angle is \(180 - 162 = 18^\circ\), so the inner polygon has \(n = 360/18 = 20\) sides. Sides alternate (one from a square, next from a pentagon, etc.), so half = 10 are pentagon sides ⇒ 10 pentagons.

unlocks after attempt

Compact write-up

Pentagon interior \(= 180 \cdot 3/5 = 108^\circ\)— P-A2 Square interior \(= 90^\circ\). Inner angle of ring \(= 360 - 108 - 90 = 162^\circ\). Exterior of inner polygon \(= 180 - 162 = 18^\circ\)— P-A4 Side count \(= 360/18 = 20\)— P-A3 reversed Pentagons \(= 20/2 = 10\).

Answer = 10

STEP 20 OF 22 · Phase 5.5 · Synthesis

Synthesis — Two regular polygons compared

Combine three atomic skills (P-A2, P-A4, algebraic manipulation) into one chain.

SYN · ★★★★

A regular \(n\)-gon's interior angle is exactly \(6^\circ\) less than the interior angle of a regular \((n + 3)\)-gon. Find \(n\).

Your answer (\(n\)):

Strategy — three skills in sequence

Skill P-A2 (single interior of regular polygon): the interior angle of a regular \(m\)-gon is \(180(m - 2)/m = 180 - 360/m\).
Skill P-A4 (cleaner via exterior): equivalently, interior \(= 180 - \text{ext}\) where ext \(= 360/m\). The "interior is 6° smaller" translates to "exterior is 6° larger".
Algebra: \(360/n - 360/(n+3) = 6\). Multiply through by \(n(n+3)/6\) to land on a quadratic in \(n\), solve, and pick the positive integer root.

Calculation

int\(_n\) \(= 180 - \dfrac{360}{n}\), int\(_{n+3}\) \(= 180 - \dfrac{360}{n+3}\)— P-A2 int\(_{n+3}\) \(-\) int\(_n\) \(= 6\) ⇒ \(\dfrac{360}{n} - \dfrac{360}{n+3} = 6\)— given \(\dfrac{360 \cdot 3}{n(n+3)} = 6\) ⇒ \(\dfrac{1080}{n(n+3)} = 6\)— common denom \(n(n+3) = 180\)— solve \(n^2 + 3n - 180 = 0 \Rightarrow n = \dfrac{-3 + \sqrt{9 + 720}}{2} = \dfrac{-3 + 27}{2} = 12\)— quadratic formula

Verify

For \(n = 12\): int\(_{12} = 180 - 360/12 = 150^\circ\). For \(n + 3 = 15\): int\(_{15} = 180 - 360/15 = 156^\circ\). Difference = \(156 - 150 = 6^\circ\) ✓.

\(n = 12\)

This is exactly how AIMO problems are built — three or four atomic skills chained together. If you noticed each skill as it appeared, the path was unblocked. The key trick is to write each interior as \(180 - 360/m\) instead of \(180(m-2)/m\) — the constant \(180\) cancels and you are left with a clean equation in just \(360/n\) and \(360/(n+3)\).

STEP 21 OF 22 · Skill Matrix

Atomic skill matrix — what mapped to what

A one-page summary of which atomic skill solved which problem.

Skill	Description	Solved in
P-A1	Sum of interior = \(180(n-2)\). Forward and reversed.	WE1, WE2, P1, P2, AIMO 2000 Q3, AIMO 2008 Q5
P-A2	Single interior of regular polygon = \(180(n-2)/n\). Reversed: solve for \(n\).	WE3, P3, AIMO 2016 Q3, Synthesis
P-A3	Central angle of regular polygon = \(360/n\). Equals exterior angle.	WE5, P4, AIMO 2005 Q4, AIMO 2016 Q3
P-A4	int + ext = 180. Use to convert between the two viewpoints.	WE3, WE4, P3, P4, AIMO 2008 Q5, AIMO 2016 Q3, Synthesis
P-A5	Convex iff every interior \(< 180^\circ\); concave iff at least one reflex.	P5; convex constraint critical in AIMO 2008 Q5

Notice: P-A4 (the interior-exterior coupling) appears in nearly every problem. It is the bridge between the two natural viewpoints — interior-summing and exterior-summing — and the single most useful skill for "find n" questions.

Common pitfalls.

Forgetting that P-A2 (single interior \(= 180(n-2)/n\)) requires the polygon to be regular.
Forgetting that the exterior-angle-sum P-A1' (= 360°) is for convex polygons; concave polygons need signed exterior angles.
In P-A3, mixing up "central angle" (apex of isosceles wedge at the centre) with "interior angle" of the polygon.
In WE/AIMO problems with integer constraints (like AIMO 2008 Q5), forgetting to enforce that \(d\) and \(a\) must be integers separately, not just the total sum.

STEP 22 OF 22 · ⭐ Self-Assessment

Rate yourself on each atomic skill

Be honest. 1★ = struggled today; 2★ = could solve with a hint; 3★ = solo, confident, fast.

P-A1 — Sum of interior \(= 180(n-2)\). Forward and reversed.

P-A2 — Single interior of regular polygon, and reverse-solving for \(n\).

P-A3 — Central angle = exterior angle = \(360/n\) (regular only).

P-A4 — int + ext = 180; the bridge between viewpoints.

P-A5 — Convex / concave detection via interior or exterior signs.

⭐ 0 / 15 — click stars

📒 Your error book (this session)

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Wrap-up. Five atomic angle skills, four AIMO past papers, one synthesis problem — all powered by two seed facts: a triangle has \(180^\circ\) and a circle has \(360^\circ\). Carry forward to Part 2 — Quadrilateral Properties, where you will apply the same angle-sum machinery to special quadrilaterals (parallelograms, trapeziums, kites, cyclic).

Open Part 2 — Parallelogram & Trapezoid Basics →