AMC 8 · 2005 · #9

Grade 8 geometry-2d

Archetype Arithmetic Expression Decomposition

isosceles-triangleangle-sum-triangle identify-subproblems ↑ Prerequisites: angle-sum-triangleisosceles-triangle

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

In quadrilateral $ABCD$ , sides $\overline{AB}$ and $\overline{BC}$ both have length 10, sides $\overline{CD}$ and $\overline{DA}$ both have length 17, and the measure of angle $ADC$ is $60^\circ$ . What is the length of diagonal $\overline{AC}$ ?

Pick an answer.

(A)

13.5

(B)

(C)

15.5

(D)

(E)

18.5

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Toolkit + CCSS Solution

Understand

Restated: In quadrilateral $ABCD$, the two sides meeting at $B$ have length $\overline{AB} = \overline{BC} = 10$, and the two sides meeting at $D$ have length $\overline{CD} = \overline{DA} = 17$. The angle at $D$ is $\angle ADC = 60^\circ$. Find the length of diagonal $\overline{AC}$.

Givens: $\overline{AB} = \overline{BC} = 10$ (the two sides at vertex $B$); $\overline{CD} = \overline{DA} = 17$ (the two sides at vertex $D$); $\angle ADC = 60^\circ$; Answer choices: (A) $13.5$, (B) $14$, (C) $15.5$, (D) $17$, (E) $18.5$

Unknowns: The length of diagonal $\overline{AC}$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

Tool #1 (Draw a Diagram) shows that the diagonal $\overline{AC}$ slices the quadrilateral into two triangles, $\triangle ABC$ and $\triangle ADC$. Only one of them carries useful given information: $\triangle ADC$ has two known sides ($DA = CD = 17$) and the angle between them ($60^\circ$). The sides $AB$ and $BC$ at vertex $B$ are a distraction. Tool #7 (Identify Subproblems) reduces the original quadrilateral question to a single subproblem about $\triangle ADC$: find its third side $AC$. The isosceles structure plus the $60^\circ$ apex angle forces all three angles to be $60^\circ$, making the triangle equilateral, so $AC = 17$ with no calculation beyond the angle sum.

Execute — Answer: D

#1 Draw a Diagram 4.G.A.2 Step 1

Draw the quadrilateral and the diagonal $\overline{AC}$.
The diagonal splits $ABCD$ into $\triangle ABC$ (sides $10$, $10$, and $AC$) and $\triangle ADC$ (sides $17$, $17$, and $AC$).
The problem only tells us an angle inside $\triangle ADC$, so that triangle is where the work happens.

$$\triangle ADC: \;\; DA = 17, \;\; CD = 17, \;\; \angle ADC = 60^\circ$$

💡 Picking the triangle that carries the given side lengths and angle is the Grade 4 "classify a figure by its known properties" move.

#7 Identify Subproblems 4.G.A.2 Step 2

Focus on $\triangle ADC$ as a subproblem.
Two of its sides are equal ($DA = CD = 17$), so it is isosceles.
The angles opposite those equal sides — $\angle DCA$ (opposite $DA$) and $\angle DAC$ (opposite $CD$) — must be equal too.

$$DA = CD \;\Rightarrow\; \angle DCA = \angle DAC$$

💡 "Two equal sides force two equal opposite angles" is the Grade 4 isosceles fact in its simplest form.

#7 Identify Subproblems 8.G.A.5 Step 3

Use the triangle angle sum.
Let $x = \angle DAC = \angle DCA$.
The three angles of $\triangle ADC$ add to $180^\circ$.

$$x + x + 60 = 180 \;\Rightarrow\; 2x = 120 \;\Rightarrow\; x = 60$$

💡 The $180^\circ$ triangle-angle-sum fact is the Grade 8 "informal arguments about triangle angles" standard.

#1 Draw a Diagram 4.G.A.2 Step 4

All three angles of $\triangle ADC$ are $60^\circ$, so the triangle is equiangular — and therefore equilateral.
Every side has the same length, so $AC = DA = CD = 17$.

$$\angle ADC = \angle DAC = \angle DCA = 60^\circ \;\Rightarrow\; AC = 17 \;\Rightarrow\; \textbf{(D)}$$

💡 Recognizing the equiangular-equals-equilateral fact is the Grade 4 "classify $2$-D figures" payoff.

[1] #1 4.G.A.2 Draw the quadrilateral and the diagonal $\overline{AC}$. The diagonal splits $AB

[2] #7 4.G.A.2 Focus on $\triangle ADC$ as a subproblem. Two of its sides are equal ($DA = CD =

[3] #7 8.G.A.5 Use the triangle angle sum. Let $x = \angle DAC = \angle DCA$. The three angles

[4] #1 4.G.A.2 All three angles of $\triangle ADC$ are $60^\circ$, so the triangle is equiangul

Review

Reasonableness: The answer $AC = 17$ is one of the listed choices and exactly matches $DA = CD = 17$, which fits the equilateral conclusion. It is also between the longest pair of sides ($17$) and the shortest pair ($10$): the diagonal of a convex quadrilateral has to be shorter than the sum of two adjacent sides ($17 + 17 = 34$ and $10 + 10 = 20$) and longer than their difference ($17 - 10 = 7$), which $17$ comfortably satisfies. The fact that the $10$s at vertex $B$ were never used is a hint that the problem really did want only the $\triangle ADC$ subproblem.

Alternative: Tool #6 (Guess and Check) on the choices: the Law of Cosines gives $AC^2 = 17^2 + 17^2 - 2(17)(17)\cos 60^\circ = 289 + 289 - 289 = 289$, so $AC = 17$. The shortcut for kids is the same observation without the law: with two equal sides and a $60^\circ$ included angle, the triangle must be equilateral, so the third side equals the other two — choice (D).

CCSS standards used (min grade 8)

4.G.A.2 Classify two-dimensional figures based on properties of their lines and angles (Identifying $\triangle ADC$ as isosceles from $DA = CD$, then concluding it is equilateral once all three angles are $60^\circ$ so $AC = 17$.)
8.G.A.5 Use informal arguments to establish facts about the angle sum of triangles (Applying the $180^\circ$ triangle angle sum to $x + x + 60 = 180$ and finding the base angles $x = 60^\circ$.)

⭐ The diagonal $\overline{AC}$ lives inside two triangles, but only one of them carries the given $17$, $17$, and $60^\circ$. That isosceles triangle with a $60^\circ$ apex angle has to be equilateral, so the third side just equals $17$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 8)

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