2017 AMC 8 Problem 25

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To help you master 2017 AMC 8 Problem 25, this page includes:

• The original AMC problem to try, plus the correct answer

• Clear, step-by-step solutions using different methods

• Key ideas, useful formulas, and takeaways you can apply to similar problems

• Related problems from past AMC exams, plus new practice problems that build the same skills

Want more practice? Scroll down to download Think Academy’s AMC 8 practice packs and keep building your AMC 8 skills.

2017 AMC8 Problem 25

Solution

Solution 1

Complementary Area Method

By connecting the vertices with straight lines, we can model the shape as a standard rhombus with two circular segments shaved off its lower edges.

The area of a circular segment is the area of its sector minus the area of the triangle formed by the circle’s center and the chord: $A_{\text{segment}}=A_{\text{sector}}-A_{\text{triangle}}$.

Step 1: Find the dimensions of the upper triangle

We are given that $US=2$, $UT=2$, and $m\angle TUS=60^{\circ }$. Because the triangle has two equal sides with a $60^{\circ }$ angle between them, $△UST$ is an equilateral triangle. Therefore, the base $ST$ is also $2$. We can calculate its area.

$$ST=2$$

$$\text{Area}(△UST)=\frac{\sqrt{3}}{4}{s}^2=\frac{\sqrt{3}}{4}(2{)}^2=\sqrt{3}$$

Step 2: Determine the straight-line distances in the lower region

The problem states that arcs $\widehat{SR}$ and $\widehat{TR}$ are each $1/6$ of a circle of radius $2$. A full circle is $360^{\circ }$, so a $1/6$ arc corresponds to a central angle of $60^{\circ }$. The triangle formed by the circle’s center and the two endpoints of the arc is isosceles with two radii of length $2$ and an included angle of $60^{\circ }$. This makes it an equilateral triangle, meaning the chord length is also $2$. Thus, the straight-line distances are $SR=2$ and $TR=2$.

$$SR=2$$

$$TR=2$$

Step 3: Identify the bounding polygon

If we connect $S$ to $R$ and $T$ to $R$ with straight line segments, we form $△SRT$. Since $ST=2$, $SR=2$, and $TR=2$, $△SRT$ is also an equilateral triangle. Together, the upper $△UST$ and the lower $△SRT$ form a rhombus $USTR$ with side length $2$.

$$\text{Area}(△SRT)=\frac{\sqrt{3}}{4}(2{)}^2=\sqrt{3}$$

$$\text{Area}(USTR)=\text{Area}(△UST)+\text{Area}(△SRT)=2\sqrt{3}$$

Step 4: Calculate the area of one removed circular segment

The visual diagram shows that the arcs $\widehat{SR}$ and $\widehat{TR}$ curve inward, meaning the true region is smaller than the rhombus. Specifically, the region is missing two circular segments (the space between the straight chords and the inward-curving arcs). To find the area of one such segment, we subtract the area of the equilateral triangle from the area of the $60^{\circ }$ sector.

$$\text{Area}(\text{Sector})=\frac{1}{6}\pi r^2=\frac{1}{6}\pi (2{)}^2=\frac{2\pi }{3}$$

$$\text{Area}(\text{Segment})=\frac{2\pi }{3}-\sqrt{3}$$

Step 5: Compute the final total area

The total area of the shape is the area of the rhombus minus the two circular segments. We subtract twice the segment area from the rhombus area.

$$\text{Total Area}=2\sqrt{3}-2\left(\frac{2\pi }{3}-\sqrt{3}\right)$$

$$\text{Total Area}=2\sqrt{3}-\frac{4\pi }{3}+2\sqrt{3}$$

$$\text{Total Area}=4\sqrt{3}-\frac{4\pi }{3}$$

Correct answer: B. $4\sqrt{3}-\frac{4\pi }{3}$
The area is $4\sqrt{3}-\frac{4\pi }{3}$, which corresponds to answer choice B. A quick magnitude check: $4\sqrt{3}-4\pi /3\approx 6.93-4.19=2.74$. The bounding rhombus area is $2\sqrt{3}\approx 3.46$. The calculated area is slightly smaller than the rhombus, which perfectly matches the visual of thin inward-curving edges.

Solution 2

Bounding and Estimation

By establishing the area of the straight-line rhombus that bounds the shape, we can use estimations to eliminate multiple-choice options without exact segment calculations.

In a multiple-choice geometry problem, calculating a strict upper or lower bound is often enough to uniquely identify the correct answer.

Step 1: Determine the bounding upper and lower limits

From the given dimensions, the upper half of the shape is exactly equilateral $△UST$ with area $\sqrt{3}\approx 1.732$. If the lower arcs were straight lines, the total shape would be a rhombus with area $2\sqrt{3}\approx 3.464$. Because the arcs curve inward, the true area must be strictly less than $3.464$, but visually it is still significantly larger than just the top triangle.

$$1.732<\text{True Area}<3.464$$

Step 2: Estimate the answer choices

Substitute $\sqrt{3}\approx 1.73$ and $\pi \approx 3.14$ into the given options to test their magnitudes. – Option C ($2\sqrt{3}\approx 3.46$) is the full rhombus, so it is too large. – Option D ($\approx 6.93-2.09=4.84$) is too large. – Option E ($4+4.19=8.19$) is far too large. This leaves only Options A and B.

$$A\approx 3(1.73)-3.14=5.19-3.14=2.05$$

$$B\approx 4(1.73)-\frac{4(3.14)}{3}=6.92-4.19=2.73$$

Step 3: Select the most reasonable magnitude

The rhombus area is $\approx 3.46$. Option A leaves an area of $2.05$, which means $3.46-2.05=1.41$ was removed. Since the lower triangle alone has an area of $1.73$, Option A would imply almost the entire lower triangle is gone. Option B leaves an area of $2.73$, meaning only $0.73$ was removed. Looking at the diagram, the inward slices are thin and remove significantly less than half of the lower region. Thus, Option B is the only reasonable choice.

Correct answer: B. $4\sqrt{3}-\frac{4\pi }{3}$
The correct answer is B. Estimation successfully bypassed complex fraction and radical arithmetic while maintaining logical certainty.

Similar Problems in Real Exams

The following five AMC 8 problems are closely related to AMC 8 2017 Question 25. These real AMC 8 problems test the same key concept—Composite Area & Geometric Counting. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.

2026 AMC 8 Problem 6

2025 AMC 8 Problem 1

2025 AMC 8 Problem 12

2024 AMC 8 Problem 3

2020 AMC 8 Problem 24

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

• Area of an Equilateral Triangle:

$s$ is the side length of the equilateral triangle. Used to find the area of the upper $△UST$ and the bounding lower $△SRT$.

• Area of a Circular Sector:

$r$ is the radius of the circle, and $\theta$ is the central angle in degrees. Used to calculate the area of the pie slice shape representing $1/6$ of the circle.

• Area of a Circular Segment:

A circular segment is the region bounded by a chord and an arc. The inward-curving regions cut out of the rhombus are exactly circular segments, meaning their areas must be subtracted from the total.

• Chord Length of a 60-Degree Arc:

Applies to any $1/6$ fraction of a circle. This allowed us to immediately deduce that straight-line segments $SR$ and $TR$ are both $2$, without needing to find the exact coordinates of the circle centers.

Key Takeaway: Remember This Strategy for Next Time

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