AMC 10 · 2003 · #17

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
coordinate-geometrycircle-equationsystems-of-equations convert-to-algebraidentify-subproblems ↑ Prerequisites: coordinate-geometry
📏 Long solution 💡 3 insights 📊 Diagram
Problem
Square ABCD has side 4, and M is the midpoint of side CD. Draw the circle of radius 2 centered at M and the circle of radius 4 centered at A. Both pass through D, and they cross again at a second point P. Find the perpendicular distance from P to the side AD.

Pick an answer.

(A)
3
(B)
$\frac {16}{5}$
(C)
$\frac {13}{4}$
(D)
$2\sqrt {3}$
(E)
$\frac {7}{2}$
How to solve
Strategy Convert to Algebra

A circle is nothing but a distance rule: "stay 4 from A", "stay 2 from M". Tool #13 (Convert to Algebra) turns those two rules into two equations, and the point we want is the solution we have not already been handed. Tool #1 (Draw a Diagram) chooses the grid so that the answer is easy to read off — put D at the origin with AD along an axis, and the distance to AD becomes just one coordinate. Tool #4 (Introduce a Variable) names P=(x,y). Tool #7 (Identify Subproblems) splits the algebra: both equations contain x²+y², so subtracting one from the other wipes out the squared terms and leaves a straight line through the two crossing points — much easier than solving either circle alone.

1STEP 1

Set the square on a grid

Placing the square on a grid puts AD on an axis, so the answer is just P's x-coordinate.

D=(0,0), C=(4,0), B=(4,4), A=(0,4), M=(2,0)
2STEP 2

Write each circle as an equation

Each circle becomes a distance equation, both satisfied by D and P.

x²+(y-4)²=16 and (x-2)²+y²=4
3STEP 3

Subtract to kill the squared terms

Subtracting kills the squared terms, leaving the line x = 2y through both crossings.

x²+y²=8y, x²+y²=4x → 8y=4x → x=2y
4STEP 4

Solve for the point that is not D

Substituting gives 5y² = 8y; discarding the root that is D leaves y = 8/5.

5y²=8y, y ≠ 0 → y=8/5, x=16/5
5STEP 5

Read the distance off the grid

So the distance is the x-coordinate 16/5, choice (B).

d(P,AD)=|x_P-0|=16/5 → (B)
Answer
16/5
Check that P=(16/5,8/5) really sits on both circles. Distance to A=(0,4): √((16/5)²+(8/5-4)²)=√(256/25+144/25)=√(400/25)=4. Distance to M=(2,0): √((6/5)²+(8/5)²)=√((36+64)/25)=2. Both radii check exactly. The value 16/5=3.2 is also less than the side 4, so P lies inside the square as the figure shows. This is a problem where estimating is not enough: choice (A) is 3 and choice (C) is 3.25, both within a rounding error of 3.2, so only the exact computation separates them.
💡Key takeaway

A circle is just a distance rule, so write both circles as equations; they share the same x²+y² part, and subtracting leaves the line x=2y through both crossing points — throw away the one that is D and the other is P.

  • Set the square on a grid
  • Write each circle as an equation
  • Subtract to kill the squared terms
  • Solve for the point that is not D
  • Read the distance off the grid

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