Competition · AMC preparation · step 4 of 4

AMC 8 · 2014 · #15

Grade 8 geometry-2d

Archetype Arithmetic Expression Decomposition └ Sub-archetype Multi-Segment Path or Count Sum

inscribed-anglearc-measureequal-spacing identify-subproblems ↑ Prerequisites: angle-sum-triangleequal-spacing

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A circle with center O has its circumference cut into 12 equal arcs, with the breakpoints labeled A, B, C, …, L in order. Inside the figure, angle x is the angle at vertex A in the path E–A–G (so x = ∠ EAG), and angle y is the angle at vertex G in the path A–G–I (so y = ∠ AGI). Find x + y in degrees.

Pick an answer.

(A)

(B)

(C)

(D)

120

(E)

150

AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The figure is busy, but the question is just x + y, so Tool #7 (Identify Subproblems) splits the work into three clean sub-questions: (a) what is the measure of one small arc, (b) what is the value of x, (c) what is the value of y. Tool #1 (Draw a Diagram) is the supporting move — adding the radii OE, OG, OA, OI to the figure turns each unknown angle on the circle into two isosceles triangles whose equal sides are radii. With those isosceles triangles in view, we can find x and y from the central angles using only the triangle-angle-sum and the straight-line (180°) fact, never needing the inscribed-angle theorem as a black box.

1STEP 1

Find one small arc

The 12 equal arcs fill 360°, so one small arc — the central angle at O — measures $\frac{360°}{12}$ = 30°.

one small arc = 360°/12 = 30°

2STEP 2

Set up the first angle

Draw radii OE, OA, OG; arc E→G spans 2 small arcs, so ∠ EOG = 60°, and OA = OE = OG makes the radius triangles isosceles.

∠ EOG = 2 · 30° = 60°; OA = OE = OG

3STEP 3

Compute the first angle

The radius triangles are isosceles, and the angle-sum bookkeeping shows x is half of ∠ EOG: x = 30°.

x = 1/2 · ∠ EOG = 1/2 · 60° = 30°

Using triangle-angle sums on two isosceles triangles made of radii is an "informal argument" Grade 8 move — and the punchline is the half-the-central-angle pattern.

The angle x=∠ EAG, whose vertex A lies on the circle and whose two sides pass through the on-circle points E and G, is exactly half of the central angle ∠ EOG that opens the same arc EG.

▸ Why?

Drawing the radii OE, OA, and OG builds two triangles, △ OAE and △ OAG, that are each isosceles, and the two base angles of an isosceles triangle are equal.

▸ Why?

The segments OA, OE, and OG each run from the center to a point on the circle, so they are all one radius long and therefore equal, which is what makes each triangle isosceles.

A circle's radius is constantEvery point on a circle sits exactly one radius from the center, so all radii of a circle are equal.

▸ Why?

A triangle with two equal sides can be folded across the line halfway between them so the equal sides trade places; the fold carries one base angle onto the other, proving those two angles equal.

Rigid motion preserves length and angleSliding, turning, or flipping a shape lays it exactly onto a copy, so every length and angle stays the same (folding is the flip case).

▸ Why?

In each of these triangles the central angle at O works out to twice the base angle at A, so the part of x inside each triangle is half of its central-angle slice, and the two slices joined together are the whole angle ∠ EOG.

▸ Why?

The three angles of the triangle add to 180°, which fixes the angle at O once the two equal base angles are known.

A triangle's angles sum to 180 degreesThe three corners of any flat triangle always add up to one straight angle, 180 degrees.

▸ Why?

The angle at O inside the triangle and the neighbouring central-angle slice lie side by side on a straight radius line and fill 180°, so that slice equals the two base angles added — twice the base angle at A.

A straight angle is 180 degreesAngles sitting side by side along a straight line always fill up 180 degrees.

▸ Why?

The full central angle ∠ EOG is just its two side-by-side slices joined with no gap or overlap, so halving each slice and adding leaves x equal to half of the whole.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

8.G.A.5Identify Subproblems

4STEP 4

Repeat for the second angle

Repeat for y = ∠ AGI: the minor arc A→I is 4 small arcs, so ∠ AOI = 120° and y = 60°.

∠ AOI = 4 · 30° = 120°; y = 1/2 · 120° = 60°

5STEP 5

Add the two angles

Add the parts: x + y = 30° + 60° = 90° — choice (C).

x + y = 30° + 60° = 90° → (C)

Answer

Each small arc is 30°, x sits on a 2-arc span, and y sits on a 4-arc span — a total span of 6 arcs = 180° on the circle. The corresponding on-circle (inscribed) angle sum is exactly half of that, 90°, matching choice (C). Visually, x and y in the figure look acute and roughly 30° and 60°, which fits.

💡Key takeaway

When an angle's vertex sits on a circle, the radii drawn to its sides make isosceles triangles — and the triangle-angle-sum trick turns the angle into half of the arc it spans.

Find one small arc
Set up the first angle
Compute the first angle
Repeat for the second angle
Add the two angles

A parent dashboard for the family lives at sensimlab.com.

Find one small arc

Set up the first angle

Compute the first angle

Repeat for the second angle

Add the two angles

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