AMC 10 · 2023 · #17

Grade 8 geometry-2d

pythagorean-theoreminteger-pythagorean-triplesarea-rectangles caseworksystematic-enumerationidentify-subproblems ↑ Prerequisites: pythagorean-theoremarea-rectangles

📏 Medium solution 💡 3 insights

Problem

Let $ABCD$ be a rectangle with $AB = 30$ and $BC = 28$ . Point $P$ and $Q$ lie on $\overline{BC}$ and $\overline{CD}$ respectively so that all sides of $\triangle{ABP}, \triangle{PCQ},$ and $\triangle{QDA}$ have integer lengths. What is the perimeter of $\triangle{APQ}$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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: $ABCD$ is a rectangle with $AB = 30$ and $BC = 28$. Point $P$ lies on $\overline{BC}$ and point $Q$ lies on $\overline{CD}$ so that all three right triangles $\triangle ABP$, $\triangle PCQ$, and $\triangle QDA$ have integer side lengths. Find the perimeter of $\triangle APQ$.

Givens: Rectangle $ABCD$ with $AB = 30$ and $BC = 28$, so $CD = 30$ and $AD = 28$; $P$ on $\overline{BC}$, so $\triangle ABP$ is right-angled at $B$ with legs $AB = 30$ and $BP$; $Q$ on $\overline{CD}$, so $\triangle QDA$ is right-angled at $D$ with legs $AD = 28$ and $DQ$; $\triangle PCQ$ is right-angled at $C$ with legs $PC$ and $CQ$; All sides of all three triangles are integers; Answer choices: (A) $84$, (B) $86$, (C) $88$, (D) $90$, (E) $92$

Unknowns: $AP$, $AQ$, $PQ$; Perimeter of $\triangle APQ$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #2 Make a Systematic List, #7 Identify Subproblems

Three right triangles glued inside a rectangle is a classic call for Tool #1 (Draw a Diagram) — labelling sides $BP, PC, DQ, CQ$ on a sketch makes the constraints visible at a glance. Tool #7 (Identify Subproblems) splits the puzzle into the three triangles: each one is an independent "find an integer hypotenuse" problem. Tool #2 (Systematic List) lists all Pythagorean triples with a leg of $30$ and a leg of $28$ — short lists once you remember the $(8,15,17)$ and $(3,4,5)$ families. The third triangle $\triangle PCQ$ is then forced, and we only need its sides to be integers too.

Execute — Answer: A

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

Draw rectangle $ABCD$ with $A$ top-left, $B$ top-right, $C$ bottom-right, $D$ bottom-left.
Mark $P$ on $\overline{BC}$ and $Q$ on $\overline{CD}$.
The three right triangles meet at the corners $B$, $C$, $D$.
The fourth (interior) triangle is $\triangle APQ$, whose perimeter we want.

$$AB = CD = 30,\quad BC = AD = 28$$

💡 Once the picture is on paper, the three right triangles around the rectangle's corners are obvious — and so is the interior triangle whose perimeter we need.

#2 Make a Systematic List 8.G.B.7 Step 2

Subproblem 1: list integer right triangles with a leg of $30$.
Using $a^2 + b^2 = c^2$ and scanning known Pythagorean families: $(8, 15, 17) \times 2 = (16, 30, 34)$ has leg $30$.
Also $(7, 24, 25), (3, 4, 5) \times 6 = (18, 24, 30)$ has $30$ as hypotenuse (not a leg) — discard.
Other small triples scaled give $(30, 40, 50)$ — but the other leg $40$ exceeds $BC = 28$, so $BP = 40$ is impossible.
The only triple with leg $30$ that also has the other leg $\le 28$ is $(16, 30, 34)$.
So $BP = 16$ and $AP = 34$.

$$30^2 + 16^2 = 900 + 256 = 1156 = 34^2 \;\Rightarrow\; BP = 16,\ AP = 34$$

💡 Pythagorean triples are a short, memorable list. Scaling $(8,15,17)$ by $2$ is the unique way to make $30$ a leg while keeping the other leg $\le 28$.

#2 Make a Systematic List 8.G.B.7 Step 3

Subproblem 2: list integer right triangles with a leg of $28$.
Scanning families: $(3, 4, 5) \times 7 = (21, 28, 35)$ has leg $28$.
Also $(7, 24, 25)$ has no $28$.
Could $(28, 45, 53)$ or $(28, 96, 100)$ work?
Those would force $DQ > 30$, which violates $0 < DQ < 30$.
So $DQ = 21$ and $AQ = 35$.

$$28^2 + 21^2 = 784 + 441 = 1225 = 35^2 \;\Rightarrow\; DQ = 21,\ AQ = 35$$

💡 Scaling the most famous triple $(3,4,5)$ by $7$ lands exactly on $28$ as a leg, and that scaling keeps the other leg $\le 30$.

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

Subproblem 3: read off the third triangle.
$PC = BC - BP = 28 - 16 = 12$ and $CQ = CD - DQ = 30 - 21 = 9$.
Check that the integer-side condition holds for $\triangle PCQ$: $PQ^2 = 12^2 + 9^2 = 144 + 81 = 225 = 15^2$, so $PQ = 15$.
The triple $(9, 12, 15)$ is just $(3, 4, 5) \times 3$ — a clean confirmation that our choice of corner triples was right.

$$PC = 12,\ CQ = 9,\ PQ = \sqrt{12^2 + 9^2} = 15$$

💡 Forcing the first two triangles forces the third. The fact that $PQ$ also lands on an integer is the puzzle's signature — only the right choice of corner triples makes it happen.

#7 Identify Subproblems 4.NBT.B.4 Step 5

Add the three sides of $\triangle APQ$.
The perimeter is $AP + AQ + PQ = 34 + 35 + 15$.

$$34 + 35 + 15 = 84 \;\Rightarrow\; \textbf{(A)}$$

💡 Once the three hypotenuses are in hand, the perimeter is just a one-line sum.

[1] #1 4.G.A.2 Draw rectangle $ABCD$ with $A$ top-left, $B$ top-right, $C$ bottom-right, $D$ bo

[2] #2 8.G.B.7 Subproblem 1: list integer right triangles with a leg of $30$. Using $a^2 + b^2

[3] #2 8.G.B.7 Subproblem 2: list integer right triangles with a leg of $28$. Scanning families

[4] #7 8.G.B.7 Subproblem 3: read off the third triangle. $PC = BC - BP = 28 - 16 = 12$ and $CQ

[5] #7 4.NBT.B.4 Add the three sides of $\triangle APQ$. The perimeter is $AP + AQ + PQ = 34 + 35

Review

Reasonableness: Diagram sanity: $BP = 16 < 28 = BC$ and $DQ = 21 < 30 = CD$, so $P$ and $Q$ sit on their intended segments. Triangle-inequality on $\triangle APQ$ with sides $(15, 34, 35)$: $15 + 34 = 49 > 35$, $15 + 35 = 50 > 34$, $34 + 35 = 69 > 15$ — a valid triangle. All three corner triangles are integer Pythagorean triples: $(16, 30, 34)$, $(21, 28, 35)$, $(9, 12, 15)$. The perimeter $84$ matches answer (A), and the other choices $86, 88, 90, 92$ correspond to no consistent assignment of triples on this rectangle.

Alternative: Tool #6 (Guess and Check) directly on $BP$: loop $BP = 1, 2, \dots, 27$ and keep only those that make both $\sqrt{900 + BP^2}$ and the chain $PQ$ integer. Only $BP = 16$ produces integer $AP$ at all (since the only triple with leg $30$ and other leg $\le 28$ is $(16, 30, 34)$). Then $DQ$ is similarly forced. Same conclusion (A) $84$.

CCSS standards used (min grade 8)

4.G.A.2 Classify two-dimensional figures based on presence or absence of parallel or perpendicular lines (Reading the rectangle's right angles at $B$, $C$, $D$ to identify three right triangles around its corners.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Computing the hypotenuses of $\triangle ABP$, $\triangle QDA$, and $\triangle PCQ$ and enforcing that they are integers (Pythagorean triples).)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers using the standard algorithm (Summing $34 + 35 + 15 = 84$ to get the final perimeter.)

⭐ Three right triangles around the rectangle's corners are just three Pythagorean triples in disguise. Scan the short list of triples with a leg matching $30$ and a leg matching $28$: only $(16, 30, 34)$ and $(21, 28, 35)$ fit, and the third triangle $(9, 12, 15)$ falls out for free. Sum the three hypotenuses to get $\textbf{(A) }84$.