AMC 10 · 2022 · #5

Grade 8 geometry-2d

area-rectanglessimilar-trianglespythagorean-theorem convert-to-algebraidentify-subproblems ↑ Prerequisites: pythagorean-theorem

📏 Medium solution 💡 2 insights

Problem

Square $ABCD$ has side length $1$ . Points $P$ , $Q$ , $R$ , and $S$ each lie on a side of $ABCD$ such that $APQCRS$ is an equilateral convex hexagon with side length $s$ . What is $s$ ?

Pick an answer.

(A)

$\frac{\sqrt{2}}{3}$

(B)

$\frac{1}{2}$

(C)

$2 - \sqrt{2}$

(D)

$1 - \frac{\sqrt{2}}{4}$

(E)

$\frac{2}{3}$

AMC 10 2022 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: Inside a unit square $ABCD$, place points $P, Q, R, S$ on the four sides so that $APQCRS$ is an equilateral convex hexagon — every side the same length $s$. Find $s$.

Givens: $ABCD$ is a square with side length $1$; $P$ on $AB$, $Q$ on $BC$, $R$ on $CD$, $S$ on $DA$; $APQCRS$ traces a closed hexagon with vertices in that order; All six sides equal: $AP = PQ = QC = CR = RS = SA = s$; Answer choices: (A) $\dfrac{\sqrt{2}}{3}$, (B) $\dfrac{1}{2}$, (C) $2 - \sqrt{2}$, (D) $1 - \dfrac{\sqrt{2}}{4}$, (E) $\dfrac{2}{3}$

Unknowns: The common side length $s$ of the hexagon

Understand

Restated: Inside a unit square $ABCD$, place points $P, Q, R, S$ on the four sides so that $APQCRS$ is an equilateral convex hexagon — every side the same length $s$. Find $s$.

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #13 Convert to Algebra

Without a picture, six points and six equal sides are a tangle. Tool #1 (Draw a Diagram) places $A, P, Q, C, R, S$ on the square and immediately reveals the structure: two right triangles cut off at corners $B$ and $D$, each isosceles with legs $1-s$. Tool #7 (Identify Subproblems) names the real question — find $s$ given that one of these triangles has hypotenuse $s$. Tool #13 (Convert to Algebra) writes that Pythagorean condition as $s^2 = 2(1-s)^2$ and isolates $s$. A pure guess-and-check (#6) on the choices would also work but the algebra is short and clean here.

Execute — Answer: C

#1 Draw a Diagram 5.G.A.2 Step 1

Draw the square with $A$ bottom-left, $B$ bottom-right, $C$ top-right, $D$ top-left.
Place $P$ on $AB$ (so $AP = s$, hence $PB = 1 - s$).
The hexagon side $PQ$ runs from $P$ to a point $Q$ on $BC$.
For the next hexagon side $QC$ to also have length $s$, we need $QC = s$, hence $BQ = 1 - s$.
So the corner triangle $\triangle BPQ$ has two legs of length $1 - s$ along the square's sides, meeting at the right angle at $B$.

$$AP = QC = s \;\Rightarrow\; PB = BQ = 1 - s$$

💡 The picture makes the corner-cut visible — Grade 5 "plot points on a grid" labeling is enough to see the small right triangle at $B$.

#7 Identify Subproblems 7.G.B.5 Step 2

$\triangle BPQ$ has a right angle at $B$ (corner of the square) with both legs equal to $1-s$, so it is an isosceles right triangle.
Its hypotenuse $PQ$ is also a side of the hexagon, so $PQ = s$.
The same picture mirrors at corner $D$: $\triangle DRS$ is congruent to $\triangle BPQ$.

$$\triangle BPQ: \text{ legs} = 1-s, \; \text{ hypotenuse} = PQ = s$$

💡 Equal pieces force equal corner cut-offs — Grade 7 angle / triangle facts about right corners do the work.

#13 Convert to Algebra 8.G.B.7 Step 3

Apply the Pythagorean theorem to $\triangle BPQ$: the hypotenuse squared equals the sum of the legs squared.

$$PQ^2 = BP^2 + BQ^2 \;\Rightarrow\; s^2 = (1-s)^2 + (1-s)^2 = 2(1-s)^2$$

💡 The right triangle invites Grade 8 Pythagorean theorem — one line turns geometry into a clean equation in $s$.

#13 Convert to Algebra 8.EE.C.7 Step 4

Solve.
Both sides are positive (since $0 < s < 1$ forces $1 - s > 0$), so take the positive square root: $s = \sqrt{2}\,(1 - s)$.
Expand and gather $s$ terms on one side.
Then divide.

$$s = \sqrt{2}\,(1-s) = \sqrt{2} - \sqrt{2}\,s \;\Rightarrow\; s(1 + \sqrt{2}) = \sqrt{2} \;\Rightarrow\; s = \dfrac{\sqrt{2}}{1 + \sqrt{2}}$$

💡 Standard Grade 8 "solve a linear equation in $s$" — except the constants carry a $\sqrt{2}$, which is the only Grade 8 wrinkle.

#13 Convert to Algebra 8.NS.A.1 Step 5

Rationalize the denominator by multiplying top and bottom by $\sqrt{2} - 1$ (the conjugate).
The denominator becomes $(\sqrt{2})^2 - 1^2 = 1$.

$$s = \dfrac{\sqrt{2}}{1+\sqrt{2}} \cdot \dfrac{\sqrt{2}-1}{\sqrt{2}-1} = \dfrac{\sqrt{2}(\sqrt{2}-1)}{2-1} = 2 - \sqrt{2} \;\Rightarrow\; \textbf{(C)}$$

💡 Conjugate-multiplying clears $\sqrt{2}$ from the denominator — Grade 8 "irrational numbers" technique.

[1] #1 5.G.A.2 Draw the square with $A$ bottom-left, $B$ bottom-right, $C$ top-right, $D$ top-l

[2] #7 7.G.B.5 $\triangle BPQ$ has a right angle at $B$ (corner of the square) with both legs e

[3] #13 8.G.B.7 Apply the Pythagorean theorem to $\triangle BPQ$: the hypotenuse squared equals

[4] #13 8.EE.C.7 Solve. Both sides are positive (since $0 < s < 1$ forces $1 - s > 0$), so take t

[5] #13 8.NS.A.1 Rationalize the denominator by multiplying top and bottom by $\sqrt{2} - 1$ (the

Review

Reasonableness: Numerical sanity. $s = 2 - \sqrt{2} \approx 2 - 1.414 = 0.586$, which sits between $0$ and $1$ — required since $P$ lies strictly inside $AB$. Cross-check the corner triangle: legs $1 - s \approx 0.414$, so hypotenuse $\sqrt{2}(1-s) \approx 1.414 \cdot 0.414 \approx 0.586$, matching $s$. The competing root $2 + \sqrt{2} \approx 3.414$ is rejected because it exceeds $1$.

Alternative: Tool #3 (Eliminate Possibilities) with quick decimals: (A) $\sqrt{2}/3 \approx 0.471$, (B) $0.5$, (C) $0.586$, (D) $1 - \sqrt{2}/4 \approx 0.646$, (E) $0.667$. For each candidate $s$, check whether $s = \sqrt{2}(1-s)$. The simplest test: $s + s\sqrt{2} = \sqrt{2}$, i.e. $s(1 + \sqrt{2}) = \sqrt{2}$. Plug (C): $0.586 \cdot 2.414 \approx 1.414 = \sqrt{2}$ — a match. Other choices fail this test.

CCSS standards used (min grade 8)

5.G.A.2 Represent real-world and mathematical problems by graphing points (Drawing the unit square with vertices labeled and placing the four hexagon points on its sides to expose the corner triangles.)
7.G.B.5 Use facts about supplementary, complementary, vertical, and adjacent angles (Recognizing that the corner of the square gives a right angle at $B$, making $\triangle BPQ$ an isosceles right triangle when its two legs are forced equal.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Translating $\triangle BPQ$ with legs $1-s$ and hypotenuse $s$ into the equation $s^2 = 2(1-s)^2$.)
8.EE.C.7 Solve linear equations in one variable (Manipulating $s = \sqrt{2}(1-s)$ into $s(1+\sqrt{2}) = \sqrt{2}$ and isolating $s$.)
8.NS.A.1 Know that numbers that are not rational are called irrational numbers (Rationalizing $\dfrac{\sqrt{2}}{1+\sqrt{2}}$ via the conjugate $\sqrt{2} - 1$ to land on the closed form $2 - \sqrt{2}$.)

⭐ This AMC 10 problem only needs Grade 8 Pythagorean theorem you already know — the cut-off corner triangle has equal legs $1-s$ and hypotenuse $s$, which gives $s = 2 - \sqrt{2}$.