AMC 10 · 2022 · #23

Grade 8 geometry-2d

coordinate-geometrypythagorean-theoremratio-proportion identify-subproblemsconvert-to-algebra ↑ Prerequisites: coordinate-geometry

📏 Medium solution 💡 3 insights

Problem

Isosceles trapezoid $ABCD$ has parallel sides $\overline{AD}$ and $\overline{BC},$ with $BC < AD$ and $AB = CD.$ There is a point $P$ in the plane such that $PA=1, PB=2, PC=3,$ and $PD=4.$ What is $\tfrac{BC}{AD}?$

$\textbf{(A) }\frac{1}{4}\qquad\textbf{(B) }\frac{1}{3}\qquad\textbf{(C) }\frac{1}{2}\qquad\textbf{(D) }\frac{2}{3}\qquad\textbf{(E) }\frac{3}{4}$

Pick an answer.

(A)

$\frac{1}{4}$

(B)

$\frac{1}{3}$

(C)

$\frac{1}{2}$

(D)

$\frac{2}{3}$

(E)

$\frac{3}{4}$

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Toolkit + CCSS Solution

Understand

Restated: An isosceles trapezoid $ABCD$ has $\overline{AD} \parallel \overline{BC}$ with $BC < AD$ and legs $AB = CD$. There is a point $P$ somewhere in the plane with $PA = 1, PB = 2, PC = 3, PD = 4$. Find the ratio $\frac{BC}{AD}$.

Givens: Isosceles trapezoid $ABCD$ with $\overline{AD} \parallel \overline{BC}$, $BC < AD$, $AB = CD$; Point $P$ in the plane with $PA = 1, PB = 2, PC = 3, PD = 4$; Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{1}{3}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{2}{3}$, (E) $\tfrac{3}{4}$

Unknowns: The ratio $\frac{BC}{AD}$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #13 Convert to Algebra, #3 Eliminate Possibilities

Tool #1 (Diagram) — place the trapezoid in coordinates so its symmetry simplifies the algebra. Let the symmetry axis be the $y$-axis, $AD$ on the $x$-axis: $A = (-a, 0)$, $D = (a, 0)$, $B = (-b, h)$, $C = (b, h)$ with $b < a$. The wanted ratio is $\frac{BC}{AD} = \frac{2b}{2a} = \frac{b}{a}$. Tool #7 (Subproblems) — focus separately on the two pairs of symmetric vertices: $\{A, D\}$ and $\{B, C\}$. Tool #13 (Algebra) — write each distance squared, then subtract the two equations in each pair. The $a$ and $b$ pop out cleanly without ever needing $y$ or $h$. Tool #3 (Eliminate) — sanity check against the listed fractions.

Execute — Answer: B

#1 Draw a Diagram 6.NS.C.8 Step 1

Place the trapezoid using its symmetry axis.
Put the axis of symmetry on the $y$-axis and the long base $AD$ on the $x$-axis.
By isosceles symmetry: $A = (-a, 0)$, $D = (a, 0)$ with $a > 0$, and $B = (-b, h)$, $C = (b, h)$ with $0 < b < a$ and $h \neq 0$.
Let $P = (x, y)$.

$$A = (-a, 0), D = (a, 0), B = (-b, h), C = (b, h)$$

💡 Grade 6 — use the line of symmetry as the $y$-axis, so symmetric vertices have opposite $x$-coordinates.

#13 Convert to Algebra 8.G.B.8 Step 2

Write the four distance-squared equations.
Since the wanted ratio is $\frac{2b}{2a} = \frac{b}{a}$, both $h$ and $y$ are nuisance variables we want to eliminate.

$$\begin{aligned} PA^2 &= (x+a)^2 + y^2 = 1 \\ PD^2 &= (x-a)^2 + y^2 = 16 \\ PB^2 &= (x+b)^2 + (y-h)^2 = 4 \\ PC^2 &= (x-b)^2 + (y-h)^2 = 9 \end{aligned}$$

💡 Grade 8 — distance squared in the coordinate plane is $(\Delta x)^2 + (\Delta y)^2$, no square roots needed.

#7 Identify Subproblems 8.EE.C.8 Step 3

Subtract within each symmetric pair so the $y^2$ and $(y-h)^2$ terms cancel.
From $PA^2$ and $PD^2$: $[(x+a)^2 - (x-a)^2] = 1 - 16$, which simplifies to $4ax = -15$.
From $PB^2$ and $PC^2$: $[(x+b)^2 - (x-b)^2] = 4 - 9$, which simplifies to $4bx = -5$.

$$4ax = -15, \quad 4bx = -5$$

💡 Grade 8 — subtract paired squares so the unknown vertical coordinates cancel; only $a, b, x$ remain.

#13 Convert to Algebra 6.RP.A.3 Step 4

Divide the two equations.
Both $a$ and $x$ are nonzero (else $4ax = -15$ would fail), so we can divide.
$\frac{4bx}{4ax} = \frac{-5}{-15}$ gives $\frac{b}{a} = \frac{1}{3}$.

$$\frac{b}{a} = \frac{-5}{-15} = \frac{1}{3}$$

💡 Grade 6 — divide one equation by another to get a clean ratio.

#13 Convert to Algebra 6.RP.A.3 Step 5

Translate back.
$\frac{BC}{AD} = \frac{2b}{2a} = \frac{b}{a} = \frac{1}{3}$, choice (B).

$$\frac{BC}{AD} = \frac{b}{a} = \frac{1}{3} \;\Rightarrow\; \textbf{(B)}$$

💡 Grade 6 ratio — the answer is exactly the $b/a$ we just found.

[1] #1 6.NS.C.8 Place the trapezoid using its symmetry axis. Put the axis of symmetry on the $y$

[2] #13 8.G.B.8 Write the four distance-squared equations. Since the wanted ratio is $\frac{2b}{

[3] #7 8.EE.C.8 Subtract within each symmetric pair so the $y^2$ and $(y-h)^2$ terms cancel. Fro

[4] #13 6.RP.A.3 Divide the two equations. Both $a$ and $x$ are nonzero (else $4ax = -15$ would f

[5] #13 6.RP.A.3 Translate back. $\frac{BC}{AD} = \frac{2b}{2a} = \frac{b}{a} = \frac{1}{3}$, cho

Review

Reasonableness: Sanity. The distances $1, 2, 3, 4$ are very asymmetric — $P$ is much closer to $A$ than to $D$, so $P$ is way off-center (large $|x|$), and similarly $P$ is closer to $B$ than to $C$. The pair $PA, PD$ feels the long base $AD$; the pair $PB, PC$ feels the short base $BC$. The differences $PA^2 - PD^2 = -15$ and $PB^2 - PC^2 = -5$ scale with the bases, and their ratio $5/15 = 1/3$ directly gives $BC/AD$. Choice (B) $\tfrac{1}{3}$ is well in the listed range ($\tfrac{1}{4}$ to $\tfrac{3}{4}$) and matches the gut feeling that $BC$ is noticeably shorter than $AD$ but not tiny.

Alternative: Tool #3 (Eliminate): only one method exists that uses the distances in the clean $\Delta(\text{squares}) \propto (\text{base length})$ form, and the differences $PD^2 - PA^2 = 15$ and $PC^2 - PB^2 = 5$ are themselves in ratio $3 : 1$. So $\frac{AD}{BC} = 3$, giving $\frac{BC}{AD} = \tfrac{1}{3}$. Any answer not equal to $\tfrac{1}{3}$ would require these specific distance values to be a coincidence — they aren't.

CCSS standards used (min grade 8)

6.NS.C.8 Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane (Placing the trapezoid in coordinates with the symmetry axis on the $y$-axis.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Reading $\frac{b}{a} = \frac{1}{3}$ as the wanted side ratio $\frac{BC}{AD}$.)
8.G.B.8 Apply the Pythagorean theorem to find the distance between two points in a coordinate system (Writing $PA^2, PB^2, PC^2, PD^2$ as squared coordinate distances.)
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations (Subtracting paired equations to eliminate $y^2$ and $(y-h)^2$, leaving $4ax = -15$ and $4bx = -5$.)

⭐ This AMC 10 problem only needs Grade 8 coordinate distance you already know — drop the trapezoid onto axes using its symmetry, square the four distances, subtract within each symmetric pair to kill the $y$'s, divide $4bx = -5$ by $4ax = -15$, and read $\frac{BC}{AD} = \frac{1}{3}$.