AMC 10 · 2023 · #13

Grade 8 geometry-2d

pythagorean-theoremangle-sum-triangleinscribed-anglearc-measure identify-subproblemseasier-related-problemcasework ↑ Prerequisites: pythagorean-theoremangle-sum-triangle

📏 Medium solution 💡 3 insights

Problem

Abdul and Chiang are standing $48$ feet apart in a field. Bharat is standing in the same field as far from Abdul as possible so that the angle formed by his lines of sight to Abdul and Chiang measures $60^\circ$ . What is the square of the distance (in feet) between Abdul and Bharat?

Pick an answer.

(A)

1728

(B)

2601

(C)

3072

(D)

4608

(E)

6912

AMC 10 2023 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.

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Toolkit + CCSS Solution

Understand

Restated: Abdul (A) and Chiang (C) are $48$ feet apart. Bharat (B) stands somewhere in the field so that $\angle ABC = 60^\circ$ (the angle at B in triangle ABC). Among all such positions for Bharat, pick the one that makes $AB$ as large as possible. Report $AB^2$.

Givens: $AC = 48$ feet (Abdul-to-Chiang distance); $\angle ABC = 60^\circ$ (the angle at Bharat between his sight lines to Abdul and Chiang); $B$ ranges over all field positions consistent with the angle condition; We want to maximize $AB$ (Abdul-Bharat distance), then report its square; Answer choices: (A) $1728$, (B) $2601$, (C) $3072$, (D) $4608$, (E) $6912$

Unknowns: $AB^2$ at the maximizing position of $B$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #9 Easier Related Problem, #7 Identify Subproblems, #3 Eliminate Possibilities

Tool #1 (Draw a Diagram): sketch a few possible positions for B keeping $\angle B = 60^\circ$ and $AC$ fixed. The pictures show that $AB$ stretches longest exactly when the triangle becomes a right triangle with the right angle at $C$. (Intuition: if $C$ were obtuse, swinging $C$ toward $90^\circ$ would lengthen $AB$; past $90^\circ$ would shrink it.) Tool #9 (Easier Related Problem) replaces "all triangles with $\angle B = 60^\circ$" by the specific easier case where $\angle C = 90^\circ$ — a $30$-$60$-$90$ right triangle with $AC$ as the side opposite the $60^\circ$ angle. Tool #7 (Identify Subproblems) then splits the work cleanly: (a) verify $\angle C = 90^\circ$ maximizes $AB$, (b) compute $AB$ in the resulting $30$-$60$-$90$ triangle, (c) square it.

Execute — Answer: C

#1 Draw a Diagram 8.G.A.5 Step 1

Draw the field.
Place A and C, $48$ feet apart.
Sweep B around so that $\angle ABC = 60^\circ$.
The locus of such B is a circular arc through A and C (an inscribed angle of $60^\circ$ corresponds to all points on one such arc).
Different B's on the arc give different $AB$ lengths.

$$AC = 48, \quad \angle ABC = 60^\circ$$

💡 Grade 8 angle facts: a fixed angle at B with a fixed opposite side $AC$ pins B to an arc — the picture lets you watch $AB$ vary as B slides along that arc.

#9 Easier Related Problem 8.G.A.5 Step 2

Tool #9 (Easier Related Problem): instead of all positions of B, try the specific easier case where the triangle is right-angled at C.
Then $\angle BAC + \angle BCA = 180^\circ - 60^\circ = 120^\circ$, and setting $\angle BCA = 90^\circ$ forces $\angle BAC = 30^\circ$.
The triangle is a familiar $30$-$60$-$90$.

$$\angle B = 60^\circ, \; \angle C = 90^\circ, \; \angle A = 30^\circ$$

💡 A nameable $30$-$60$-$90$ is much easier to reason about than a generic $60^\circ$ triangle — that is exactly Tool #9's "smaller / simpler" move.

#1 Draw a Diagram 8.G.A.5 Step 3

Why is this the maximum?
In the picture, AB is the chord of the arc from A to B; AB is longest when B is as far from A along the arc as possible — which happens when AB is a diameter of the arc's circle, equivalently when the inscribed angle ACB is $90^\circ$ (angles in a semicircle).
So $\angle C = 90^\circ$ really does give the maximizing B.

$$\angle ACB = 90^\circ \iff AB \text{ is a diameter of the arc circle} \iff AB \text{ maximal}$$

💡 Inscribed-angle thinking (Grade 8 angle facts extended): the chord AB grows with the inscribed angle at C — pushed to $90^\circ$, it becomes the diameter.

#7 Identify Subproblems 8.G.B.7 Step 4

In the $30$-$60$-$90$ right triangle ACB (right angle at C), AC is opposite the $60^\circ$ angle (at B).
The standard $30$-$60$-$90$ side ratios are $1 : \sqrt{3} : 2$ for the sides opposite $30^\circ, 60^\circ, 90^\circ$.
So $AC : AB = \sqrt{3} : 2$, giving $AB = \dfrac{2}{\sqrt{3}} \cdot AC$.

$$AB = \dfrac{2}{\sqrt{3}} \cdot 48 = \dfrac{96}{\sqrt{3}} = \dfrac{96\sqrt{3}}{3} = 32\sqrt{3}$$

💡 $30$-$60$-$90$ side ratios follow directly from Pythagoras applied to a half-equilateral triangle — the Grade 8 special right triangle.

#7 Identify Subproblems 8.EE.A.2 Step 5

Square AB to get the requested $AB^2$.

$$AB^2 = (32\sqrt{3})^2 = 32^2 \cdot 3 = 1024 \cdot 3 = 3072$$

💡 Squaring kills the $\sqrt{3}$ cleanly — Grade 8 square-root manipulation gives an integer.

#3 Eliminate Possibilities 8.NS.A.2 Step 6

Match against the choices: $3072$ is exactly answer (C).
Sanity-check the other choices: (A) $1728 = AC^2 \cdot \tfrac{3}{4}$ is far too small, (E) $6912 = 2 \cdot 3072$ would correspond to $AB = 48\sqrt{3}$, which is impossible since $AB$ cannot exceed the arc's diameter.

$$AB^2 = 3072 \;\Rightarrow\; \textbf{(C)}$$

💡 Comparing the closed form $3072$ to the five answer choices is the standard multiple-choice closeout.

[1] #1 8.G.A.5 Draw the field. Place A and C, $48$ feet apart. Sweep B around so that $\angle A

[2] #9 8.G.A.5 Tool #9 (Easier Related Problem): instead of all positions of B, try the specifi

[3] #1 8.G.A.5 Why is this the maximum? In the picture, AB is the chord of the arc from A to B;

[4] #7 8.G.B.7 In the $30$-$60$-$90$ right triangle ACB (right angle at C), AC is opposite the

[5] #7 8.EE.A.2 Square AB to get the requested $AB^2$.

[6] #3 8.NS.A.2 Match against the choices: $3072$ is exactly answer (C). Sanity-check the other

Review

Reasonableness: Direction check: as $\angle C$ slides past $90^\circ$, the triangle would force B back toward A (in the picture, the arc curls in), so $\angle C = 90^\circ$ really is the turning point. Magnitude check: $AB = 32\sqrt{3} \approx 55.4$ feet — bigger than $AC = 48$ feet, which makes sense because the $60^\circ$ angle at B is opposite the shorter side. Squared, $55.4^2 \approx 3070$, matching $3072$. End-to-end Pythagoras on the $30$-$60$-$90$: $BC = AB / 2 = 16\sqrt{3}$, and $AC^2 + BC^2 = 48^2 + (16\sqrt{3})^2 = 2304 + 768 = 3072 = AB^2$ — clean.

Alternative: Tool #6 (Guess and Check) on the choices: each candidate $AB^2$ pairs with a triangle. From the law of cosines on $\angle B = 60^\circ$ (or just the $30$-$60$-$90$ identification), $AC^2 = AB^2 + BC^2 - AB \cdot BC$ (since $\cos 60^\circ = \tfrac{1}{2}$). Plug $AB^2 = 3072$ and $BC^2 = 768$, $BC = 16\sqrt{3}$, $AB = 32\sqrt{3}$: $3072 + 768 - 32\sqrt{3} \cdot 16\sqrt{3} = 3840 - 1536 = 2304 = 48^2$. Confirms (C).

CCSS standards used (min grade 8)

8.G.A.5 Use informal arguments to establish facts about angle sum and exterior angles (Using the triangle-angle-sum to convert $\angle B = 60^\circ$ and $\angle C = 90^\circ$ into $\angle A = 30^\circ$, and locating B on the constant-angle arc.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Recovering the $1 : \sqrt{3} : 2$ side ratios of the $30$-$60$-$90$ triangle (a half-equilateral) to get $AB = \tfrac{2}{\sqrt{3}} \cdot AC$.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Rationalizing $\tfrac{96}{\sqrt{3}} = 32\sqrt{3}$ and squaring $(32\sqrt{3})^2 = 3072$.)
8.NS.A.2 Use rational approximations of irrational numbers to compare their size (Checking $32\sqrt{3} \approx 55.4$ against the magnitude expected from the picture before matching $AB^2 = 3072$ to choice (C).)

⭐ Holding the angle at B at $60^\circ$ keeps B on a fixed circular arc through A and C; the chord AB is longest when it becomes the diameter, which happens exactly when the angle at C is $90^\circ$. That makes triangle ACB a $30$-$60$-$90$, so $AB = \tfrac{2}{\sqrt{3}} \cdot 48 = 32\sqrt{3}$ and $AB^2 = 3072$.