2022 AMC 10 A Problem 23

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To help you master 2022 AMC 10 A Problem 23, 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

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2022 AMC10 A Problem 23

Solution

Solution 1

Coordinate Geometry Method

By placing the $y$ -axis precisely on the axis of symmetry, we assign symmetric horizontal coordinates to the vertices. Applying the distance formula creates a system of equations where taking the difference of squared distances elegantly eliminates the unknown vertical components.

Step 1: Define the coordinate system and variables

Place the $y$ -axis on the vertical line of symmetry of the isosceles trapezoid. Since the bases $\overline{A D}$ and $\overline{BC}$ are parallel, align them horizontally. To follow the correct cyclic perimeter of a non-self-intersecting trapezoid, vertices $A$ and $B$ must lie on the same side of the $y$ -axis. Let them lie on the left side with negative $x$ -coordinates. Let point $P$ be $(x, y)$ . This setup makes the lengths of the bases $A D = 2 a$ and $BC = 2 b$ .

A = (- a, y_{1}), D = (a, y_{1})

B = (- b, y_{2}), C = (b, y_{2})

P = (x, y)

A D = 2 a, BC = 2 b

Step 2: Translate the given conditions using the distance formula

Use the distance formula to express the given squared distances from $P$ to each of the four vertices.

P A^{2} = (x + a)^{2} + (y - y_{1})^{2} = 1^{2} = 1

P B^{2} = (x + b)^{2} + (y - y_{2})^{2} = 2^{2} = 4

P C^{2} = (x - b)^{2} + (y - y_{2})^{2} = 3^{2} = 9

P D^{2} = (x - a)^{2} + (y - y_{1})^{2} = 4^{2} = 16

Step 3: Eliminate the vertical components by taking differences

Notice that the equations for $P A^{2}$ and $P D^{2}$ share the exact same $y$ -term, $(y - y_{1})^{2}$ . Subtracting $P A^{2}$ from $P D^{2}$ eliminates $y_{1}$ and $y$ entirely. Expand the squared terms to simplify. Apply the exact same process by subtracting $P B^{2}$ from $P C^{2}$ to eliminate $y_{2}$ and $y$ .

P D^{2} - P A^{2} = [(x - a)^{2} + (y - y_{1})^{2}] - [(x + a)^{2} + (y - y_{1})^{2}]

16 - 1 = (x^{2} - 2 a x + a^{2}) - (x^{2} + 2 a x + a^{2})

15 = - 4 a x

P C^{2} - P B^{2} = [(x - b)^{2} + (y - y_{2})^{2}] - [(x + b)^{2} + (y - y_{2})^{2}]

9 - 4 = (x^{2} - 2 b x + b^{2}) - (x^{2} + 2 b x + b^{2})

5 = - 4 b x

Step 4: Solve for the required base ratio

We want to find the ratio of the base lengths, which is $\frac{BC}{A D} = \frac{2 b}{2 a} = \frac{b}{a}$ . We can find this by dividing our two simplified equations.

{15 = - 4 a x 5 = - 4 b x

\frac{- 4 b x}{- 4 a x} = \frac{5}{15}

\frac{b}{a} = \frac{1}{3}

\frac{BC}{A D} = \frac{2 b}{2 a} = \frac{1}{3}

Correct answer: B. $\frac{1}{3}$
The exact answer is

1/3

, which corresponds to choice B. We can verify internal consistency:

15 = - 4 a x

and

5 = - 4 b x

mean that

a

and

b

(which are positive lengths) require

x

to be negative. This perfectly matches the facts

P A < P D

and

PB < PC

, meaning point

P

physically resides on the left side of the axis of symmetry.

Solution 2

Projection Theorem Method

Rather than assigning full coordinates, we can evaluate the difference of squared lengths from $P$ to the endpoints of a segment purely geometrically by dropping a perpendicular from $P$ to the symmetric axis.

This method relies on the Difference of Squares (Projection) Theorem. If a point $P$ projects perpendicularly onto a line $A B$ at $H$ , and $M$ is the midpoint of $A B$ , then $∣ P B^{2} - P A^{2} ∣ = 2 \cdot A B \cdot M H$ .

Step 1: Construct the axis of symmetry and projections

Let $M$ be the midpoint of $\overline{A D}$ and $N$ be the midpoint of $\overline{BC}$ . In an isosceles trapezoid, the line $MN$ is the perpendicular bisector of both bases. Let $P^{'}$ be the perpendicular projection of $P$ onto the axis of symmetry $MN$ . Furthermore, let $X$ and $Y$ be the perpendicular projections of $P$ onto the lines of the bases $\overline{A D}$ and $\overline{BC}$ , respectively.

Step 2: Relate the projection distances to the axis of symmetry

Because the segments form a rectangle, the horizontal distance from the midpoint $M$ to the projection point $X$ is exactly the horizontal distance from the axis of symmetry $MN$ to $P$ . Similarly, the distance from $N$ to $Y$ is also this exact same horizontal distance. Let this common distance be $d$ .

MX = N Y = d

Step 3: Apply the Projection Theorem to the bases

Using the Difference of Squares Theorem for $△ P A D$ with midpoint $M$ , we relate the squared sides to the base $A D$ and the projection distance $d$ . Because the naming convention of a trapezoid implies it is a non-self-intersecting quadrilateral, $A$ and $B$ lie on the same side of $MN$ . Since $P D > P A$ and $PC > PB$ , point $P$ is closer to $A$ and $B$ , meaning $X$ and $Y$ correctly lie on the same side of $MN$ , preserving the consistent meaning of $d$ .

P D^{2} - P A^{2} = 2 \cdot A D \cdot MX

16 - 1 = 2 \cdot A D \cdot d ⟹ 15 = 2 \cdot A D \cdot d

P C^{2} - P B^{2} = 2 \cdot BC \cdot N Y

9 - 4 = 2 \cdot BC \cdot d ⟹ 5 = 2 \cdot BC \cdot d

Step 4: Find the ratio of the bases

Divide the second equation by the first equation. The common distance $d$ and the constant factor of $2$ cancel out perfectly, instantly leaving the required ratio of the base lengths.

\frac{5}{15} = \frac{2 \cdot BC \cdot d}{2 \cdot A D \cdot d}

\frac{1}{3} = \frac{BC}{A D}

Correct answer: B. $\frac{1}{3}$
The result evaluates to exactly

\frac{1}{3}

, yielding choice B. This geometrical approach confirms the initial algebraic method perfectly, completely avoiding the formalization of arbitrary

y

-coordinates.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Distance Formula:

d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

$d$ is the distance between points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . The problem provides fixed squared distances from a common point $P$ to the vertices of a shape that can be placed symmetrically on a coordinate plane.

Difference of Squares (Projection) Theorem:

∣ P B^{2} - P A^{2} ∣ = 2 \cdot A B \cdot M H

$P$ is any point, $A$ and $B$ form a line segment, $M$ is the midpoint of segment $\overline{A B}$ , and $H$ is the perpendicular projection of $P$ onto the line $A B$ . We are evaluating the difference of squared distances from $P$ to the endpoints of parallel bases in a highly symmetric figure, making the projection distances directly tied to the symmetry axis.

Key Takeaway: Remember This Strategy for Next Time

Distances from a single unconstrained point to the vertices of a highly symmetric figure like an isosceles trapezoid, rectangle, or equilateral triangle.

Try: Place the origin squarely on the figure's line of symmetry and invoke the distance formula.

Why: Symmetry naturally pairs opposite coordinates (e.g., $a$ and $- a$ ). When taking the difference between pairs of squared distance equations, complex terms seamlessly cancel out, distilling the algebra into manageable relationships.

A requirement to compare or manipulate the difference of squares of two lengths extending from a common point to the endpoints of a line segment.

Try: Consider the Projection Theorem (Difference of Squares Theorem) $P B^{2} - P A^{2} = 2 \cdot A B \cdot d$ .

Why: The Pythagorean theorem inherently generates a difference-of-squares relationship when two right triangles share a common altitude, providing a potent geometric shortcut that bypasses full coordinate assignments.

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2022 AMC 10 A Problem 23

In This Page:

2022 AMC10 A Problem 23