AMC 8 · 2014 · #15
Grade 8 geometry-2dProblem
The circumference of the circle with center O is divided into 12 equal arcs, marked the letters A through L as seen below. What is the number of degrees in the sum of the angles x and y?
Pick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A circle with center $O$ has its circumference cut into $12$ equal arcs, with the breakpoints labeled $A, B, C, \ldots, L$ in order. Inside the figure, angle $x$ is the angle at vertex $A$ in the path $E$–$A$–$G$ (so $x = \angle EAG$), and angle $y$ is the angle at vertex $G$ in the path $A$–$G$–$I$ (so $y = \angle AGI$). Find $x + y$ in degrees.
Givens: The circle's full $360^\circ$ of arc is split into $12$ equal arcs, one between each pair of consecutive letters; Vertices $A, E, G, I$ all sit on the circle; Angle $x = \angle EAG$ has its vertex on the circle at $A$ and its two sides hit the circle at $E$ and $G$; Angle $y = \angle AGI$ has its vertex on the circle at $G$ and its two sides hit the circle at $A$ and $I$; Answer choices: (A) $75$, (B) $80$, (C) $90$, (D) $120$, (E) $150$ (degrees)
Unknowns: The sum $x + y$ in degrees
Understand
Restated: A circle with center $O$ has its circumference cut into $12$ equal arcs, with the breakpoints labeled $A, B, C, \ldots, L$ in order. Inside the figure, angle $x$ is the angle at vertex $A$ in the path $E$–$A$–$G$ (so $x = \angle EAG$), and angle $y$ is the angle at vertex $G$ in the path $A$–$G$–$I$ (so $y = \angle AGI$). Find $x + y$ in degrees.
Givens: The circle's full $360^\circ$ of arc is split into $12$ equal arcs, one between each pair of consecutive letters; Vertices $A, E, G, I$ all sit on the circle; Angle $x = \angle EAG$ has its vertex on the circle at $A$ and its two sides hit the circle at $E$ and $G$; Angle $y = \angle AGI$ has its vertex on the circle at $G$ and its two sides hit the circle at $A$ and $I$; Answer choices: (A) $75$, (B) $80$, (C) $90$, (D) $120$, (E) $150$ (degrees)
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram
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 ($180^\circ$) fact, never needing the inscribed-angle theorem as a black box.
Execute — Answer: C
4.MD.C.5 Step 1 - Find the measure of one small arc.
- The $12$ arcs are equal and together cover the full $360^\circ$ of the circle, so each small arc — equivalently, the central angle that opens it at $O$ — measures $360^\circ / 12 = 30^\circ$.
💡 A full turn around the center is $360^\circ$, and we are cutting it into $12$ equal slices — a Grade 4 angle-measure idea.
5.G.B.4 Step 2 - Use a diagram move to set up $x = \angle EAG$.
- Draw the radii $OE$, $OA$, $OG$ (Tool #1).
- The central angle $\angle EOG$ spans the minor arc from $E$ to $G$, which contains $2$ small arcs (arc $EF$ and arc $FG$), so $\angle EOG = 2 \cdot 30^\circ = 60^\circ$.
- Triangles $OAE$ and $OAG$ are isosceles because $OA = OE = OG$ (all radii).
💡 Two radii of the same circle are always equal, so any triangle made from two radii is automatically isosceles — Grade 5 "classify figures by properties."
8.G.A.5 Step 3 - Compute $x$.
- In isosceles $\triangle OAE$, let $\angle OAE = \angle OEA = \alpha$; in isosceles $\triangle OAG$, let $\angle OAG = \angle OGA = \beta$.
- Then $x = \angle EAG = \alpha + \beta$ (when $O$ is inside $\angle EAG$) or $x = |\alpha - \beta|$ (when $O$ is outside).
- In both cases the exterior-angle / angle-sum bookkeeping gives the same clean result: $x$ equals half of the central angle $\angle EOG$.
- Apply that to $\angle EOG = 60^\circ$.
💡 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.
8.G.A.5 Step 4 - Repeat the same recipe for $y = \angle AGI$.
- The minor arc from $A$ to $I$ (the one not passing through $G$) goes $A \to L \to K \to J \to I$, which is $4$ small arcs, so the central angle $\angle AOI = 4 \cdot 30^\circ = 120^\circ$.
- The same isosceles-triangle argument with radii $OA, OG, OI$ gives $y = \tfrac{1}{2} \cdot \angle AOI$.
💡 Re-running the same pattern on the second angle reinforces the subproblem habit and shows the half-arc rule is reusable.
4.MD.C.7 Step 5 - Add the two pieces.
- $x + y = 30^\circ + 60^\circ = 90^\circ$, which is choice (C).
💡 Combining the two subangle answers is the Grade 4 "angle measure is additive" closer.
4.MD.C.5 Find the measure of one small arc. The $12$ arcs are equal and together cover th 5.G.B.4 Use a diagram move to set up $x = \angle EAG$. Draw the radii $OE$, $OA$, $OG$ ( 8.G.A.5 Compute $x$. In isosceles $\triangle OAE$, let $\angle OAE = \angle OEA = \alpha 8.G.A.5 Repeat the same recipe for $y = \angle AGI$. The minor arc from $A$ to $I$ (the 4.MD.C.7 Add the two pieces. $x + y = 30^\circ + 60^\circ = 90^\circ$, which is choice (C Review
Reasonableness: Each small arc is $30^\circ$, $x$ sits on a $2$-arc span, and $y$ sits on a $4$-arc span — a total span of $6$ arcs $= 180^\circ$ on the circle. The corresponding on-circle (inscribed) angle sum is exactly half of that, $90^\circ$, matching choice (C). Visually, $x$ and $y$ in the figure look acute and roughly $30^\circ$ and $60^\circ$, which fits.
Alternative: Tool #3 (Eliminate Possibilities) on the choices: the half-arc rule forces both $x$ and $y$ to be multiples of $\tfrac{30^\circ}{2} = 15^\circ$, so their sum is a multiple of $15^\circ$ — that already rules out $(B)\;80$. The two angles each rest on a small minor arc, so each is well under $90^\circ$, ruling out $(E)\;150$ as too large for two such angles taken together would still likely be tame, and $(D)\;120$ would require an average of $60^\circ$ each (possible but only if both arcs are large, contradicting the figure). Between $(A)\;75$ and $(C)\;90$, the clean $2$-arc + $4$-arc reading gives $15^\circ \cdot (2 + 4) = 90^\circ$, picking $(C)$.
CCSS standards used (min grade 8)
4.MD.C.5Recognize angles as geometric shapes formed by two rays sharing a common endpoint, and understand concepts of angle measurement (Dividing the full $360^\circ$ around center $O$ into $12$ equal central angles of $30^\circ$, one per small arc.)4.MD.C.7Recognize angle measure as additive; decompose angles into non-overlapping parts (Adding the two inscribed angles $x + y = 30^\circ + 60^\circ = 90^\circ$ to answer the question.)5.G.B.4Classify two-dimensional figures in a hierarchy based on properties (Recognizing $\triangle OAE, \triangle OAG, \triangle OGI$ as isosceles triangles because two of their sides are radii of the same circle.)8.G.A.5Use informal arguments to establish facts about the angle sum and exterior angle of triangles (Chasing the angles in the isosceles triangles formed by radii to show that each on-circle angle ($x$ and $y$) equals half its central angle ($30^\circ$ from $60^\circ$, $60^\circ$ from $120^\circ$).)
⭐ 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.
⭐ 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.
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