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2022 AMC 10A Problem 23

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2022 AMC 10A Problem 23AMC 10 Step by Step
Name:                             Date: October 28, 2026
Quadrilaterals & Polygon Areas · difficulty 4 of 5 · about 6 min · position P21-25

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

  1. A)14\frac{1}{4}
  2. B)13\frac{1}{3}
  3. C)12\frac{1}{2}
  4. D)23\frac{2}{3}
  5. E)34\frac{3}{4}

Answer

B

Key insight

Put the trapezoid's mirror line on the y-axis. Then PD squared minus PA squared, and PC squared minus PB squared, are each a base times the same unknown.

Solution

Step 1: set up coordinates using the symmetry. An isosceles trapezoid has a mirror line through the middles of both bases. Put that line on the yy-axis:
A=(−a,0),D=(a,0),B=(−b,h),C=(b,h).A = (-a, 0), \quad D = (a, 0), \quad B = (-b, h), \quad C = (b, h).
Then AD=2aAD = 2a and BC=2bBC = 2b. Let P=(x,y)P = (x, y).

Step 2: why subtract? Each squared distance has many unknowns, but AA and DD are mirror images, so the difference PD2−PA2PD^2 - PA^2 should lose most of them.

Step 3: the bottom base.
PA2=(x+a)2+y2,PD2=(x−a)2+y2.PA^2 = (x + a)^2 + y^2, \qquad PD^2 = (x - a)^2 + y^2.
The y2y^2 parts cancel when we subtract, and
(x−a)2−(x+a)2=(x2−2ax+a2)−(x2+2ax+a2)=−4ax.(x - a)^2 - (x + a)^2 = (x^2 - 2ax + a^2) - (x^2 + 2ax + a^2) = -4ax.
The given lengths make PD2−PA2=16−1=15PD^2 - PA^2 = 16 - 1 = 15. So −4ax=15-4ax = 15.

Step 4: the top base, the same way. The (y−h)2(y - h)^2 parts cancel, and
PC2−PB2=(x−b)2−(x+b)2=−4bx.PC^2 - PB^2 = (x - b)^2 - (x + b)^2 = -4bx.
The given lengths make this 9−4=59 - 4 = 5. So −4bx=5-4bx = 5.

Step 5: divide. The unknown xx cancels:
BCAD=2b2a=−4bx−4ax=515=13.\frac{BC}{AD} = \frac{2b}{2a} = \frac{-4bx}{-4ax} = \frac{5}{15} = \frac13.

The answer is (B) 13\boxed{\textbf{(B)}\ \frac{1}{3}}.

Why this works

For two points that are mirror images across a line, the difference of the squared distances from any point PP is simple: it only depends on how far apart the two points are and how far PP is from the mirror line. Both bases share the same mirror line, so the same unknown appears twice and cancels. When several distances from one point are given, try subtracting their squares.

The trap

Trying to find the size of the trapezoid or where P is. Neither is fixed by the problem; only the ratio of the bases is.

Concise solution

Use the trapezoid's symmetry: place its axis on the yy-axis with
A=(−a,0),D=(a,0),B=(−b,h),C=(b,h),A = (-a, 0),\quad D = (a, 0),\quad B = (-b, h),\quad C = (b, h),
so AD=2aAD = 2a, BC=2bBC = 2b, and b<ab < a. Let P=(x,y)P = (x, y).

Compare the distances to the two ends of each base. On the lower base,
PD2−PA2=[(x−a)2+y2]−[(x+a)2+y2]=−4ax,PD^2 - PA^2 = \big[(x-a)^2 + y^2\big] - \big[(x+a)^2 + y^2\big] = -4ax,
and the given lengths make this 16−1=1516 - 1 = 15. On the upper base,
PC2−PB2=[(x−b)2+(y−h)2]−[(x+b)2+(y−h)2]=−4bx,PC^2 - PB^2 = \big[(x-b)^2 + (y-h)^2\big] - \big[(x+b)^2 + (y-h)^2\big] = -4bx,
which equals 9−4=59 - 4 = 5.

Dividing the two equations, the unknown xx cancels:
BCAD=2b2a=−4bx−4ax=515=13.\frac{BC}{AD} = \frac{2b}{2a} = \frac{-4bx}{-4ax} = \frac{5}{15} = \frac13.

The answer is (B) 13\boxed{\textbf{(B)}\ \frac{1}{3}}.

Review in the AoPS books

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta · Place the figure on coordinates and compute · Exploit symmetry to reduce work or pair up objects