AMC 10 · 2024 · #13

Grade 9 algebra

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

completing-the-squareoptimizationsystems-of-equations extreme-principleconvert-to-algebraidentify-subproblems ↑ Prerequisites: completing-the-squarequadratic-equations

📏 Long solution 💡 3 insights

Problem

Four real numbers x, y, h, k make both equations x² + y² - 6x - 8y = h and x² + y² - 10x + 4y = k true at the same time. The same pair (x,y) goes into both equations, so once you pick (x,y) the values of h and k are forced. As (x,y) ranges over all real pairs, find the smallest value h+k can reach.

Pick an answer.

(A)

-54

(B)

-46

(C)

-34

(D)

-16

(E)

How to solve

Strategy Extreme Principle

The question asks for a minimum, so the job is to find a floor that h+k can never go below and then show that floor is actually reached. The trap is to hunt for the floor of h and the floor of k one at a time. That fails, because h bottoms out at one point and k bottoms out at a different point, and a single (x,y) cannot sit in two places. So I change focus: stop tracking h and k separately and track the one quantity the question actually asks about, h+k. Adding the two equations turns h+k into a single expression in x and y, and rewriting that expression as a pile of squares plus a constant makes the floor visible on sight, since squares can never be negative.

1STEP 1

One point feeds both equations

One (x,y) fixes both h and k.

h = x² + y² - 6x - 8y, k = x² + y² - 10x + 4y for the same (x,y)

2STEP 2

Add the equations, not the minimums

Adding leaves two one-variable quadratics.

h + k = (x² + y² - 6x - 8y) + (x² + y² - 10x + 4y) = 2x² + 2y² - 16x - 4y

3STEP 3

Rewrite it as squares

Completing the square exposes negative 34.

h + k = 2(x² - 8x) + 2(y² - 2y) = 2[(x-4)² - 16] + 2[(y-1)² - 1] = 2(x-4)² + 2(y-1)² - 34

4STEP 4

Squares cannot help you go lower

Both squares vanish at the floor.

2(x-4)² ≥ 0 and 2(y-1)² ≥ 0 ⟹ h + k ≥ -34, equality only at (x,y) = (4,1)

5STEP 5

Check the floor is actually reached

At (4,1) it is actually attained.

x=4, y=1: h = 16 + 1 - 24 - 8 = -15, k = 16 + 1 - 40 + 4 = -19, h+k = -34

6STEP 6

Rule out the separate-minimum trap

The two minima differ, so the answer is -34.

h = (x-3)² + (y-4)² - 25 at (3,4); k = (x-5)² + (y+2)² - 29 at (5,-2); (3,4) ≠ (5,-2)

Answer

-34

Test the claimed minimum by nudging away from (4,1). At (4.1, 1) the values are h = -14.79 and k = -19.19, summing to -33.98. At (4, 1.1) they are h = -15.59 and k = -18.39, summing to -33.98 again. At (5,2) the sum is -30, and at (3,0) it is -30. Every nearby point comes out above -34, and the sum climbs symmetrically in all four directions, which is exactly what 2(x-4)² + 2(y-1)² - 34 predicts. The answer also passes a size check against the list: it must be greater than -54, since -54 needs two incompatible points, and it must be well below -16 and 16, since a single point already delivers -34. The geometry agrees too. The first equation says (x,y) lies on a circle centered at (3,4) with radius √(h+25), and the second says it lies on a circle centered at (5,-2) with radius √(k+29). At (4,1) both radii equal √(10), their sum is 2√(10), and the distance between the two centers is √(2² + 6²) = √(40) = 2√(10) as well. The circles are exactly tangent, touching at (4,1), which is the midpoint of the segment joining the centers. That tangency is the picture of a minimum: any smaller pair of radii and the circles would pull apart with no shared point left.

💡Key takeaway

When a question asks for the smallest value of a sum, add the pieces together first and then complete the square, because the two pieces may not be allowed to hit their own smallest values at the same time.

One point feeds both equations
Add the equations, not the minimums
Rewrite it as squares
Squares cannot help you go lower
Check the floor is actually reached
Rule out the separate-minimum trap