AMC 10 · 2015 · #12

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
coordinate-geometryquadratic-equations bound-inequality-then-enumerate ↑ Prerequisites: coordinate-geometry
📏 Medium solution 💡 2 insights
Problem
A circle has center (5,5) and radius 10. Look at every point of the form (x,-x), where x is an integer. Count how many of these points land inside the circle or exactly on it.

Pick an answer.

(A)
11
(B)
12
(C)
13
(D)
14
(E)
15

AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Convert to Algebra

"Inside or on the circle" is a geometric condition, but it has a clean algebraic meaning: the distance from the point to the center is at most the radius. So the main move is tool #13 (Convert to Algebra) — turn "distance ≤ 10" into an inequality in x using the distance formula, then simplify it. Tool #1 (Draw a Diagram) helps picture the circle and the line y=-x that all the candidate points sit on, and tool #4 (Introduce a Variable) is already handed to us since the points are parameterized by x. Once the inequality collapses to a simple bound on x, the count is just an integer count on a number line.

1STEP 1

Turn the circle condition into an inequality

Inside-or-on means the distance from (x,-x) to center (5,5) is at most 10, so the squared distance is at most 100.

(x-5)² + (-x-5)² ≤ 100
2STEP 2

Expand and simplify

Expanding both squares cancels the linear terms (-10x with +10x), collapsing everything to x² ≤ 25.

(x²-10x+25)+(x²+10x+25) ≤ 100 → 2x² + 50 ≤ 100 → x² ≤ 25
3STEP 3

Solve the bound on x

Square-rooting x² ≤ 25 gives |x| ≤ 5, so x lies in the range -5 ≤ x ≤ 5, endpoints included.

x² ≤ 25 → |x| ≤ 5 → -5 ≤ x ≤ 5
4STEP 4

Count the integers

The integers from -5 to 5 number 11 in all — five negatives, five positives, and zero — so the answer is (A).

-5,-4,…,4,5 → 11 integers → (A)
Answer
11
Check the boundary points. At x=5 the point is (5,-5); its squared distance from (5,5) is 0²+(-10)²=100=10², exactly on the circle, so it counts. At x=6 the point is (6,-6), with squared distance 1²+(-11)²=122 ＞ 100, outside, so it correctly does not count. The endpoints ± 5 behave as the inequality predicts, and 11 is one of the answer choices, so (A) is consistent.
💡Key takeaway

"Inside or on a circle" just means "distance from the center is at most the radius" — write that as an inequality, simplify, and count the integers that fit.

  • Turn the circle condition into an inequality
  • Expand and simplify
  • Solve the bound on x
  • Count the integers

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