AMC 10 · 2017 · #23

Grade 8 geometry-2d
combinations-basiccoordinate-geometryslope-intercept complementary-countingcaseworksystematic-enumeration ↑ Prerequisites: combinations-basic
📏 Long solution 💡 3 insights
Problem
Look at the 5× 5 grid of points (i,j) where i and j are whole numbers from 1 to 5. Pick any 3 of these points as the corners of a triangle. How many of those choices give a real triangle — one with actual area, not a flat sliver?

Pick an answer.

(A)
2128
(B)
2148
(C)
2160
(D)
2200
(E)
2300

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

Counting good triangles directly is hard because the bad cases hide everywhere. Tool #16 (Count the Complement) flips it: count every way to pick 3 of the 25 points, then subtract only the bad picks — the ones where the 3 points lie on a single line and the area is zero. Tool #7 (Identify Subproblems) sorts those flat triples by the kind of line they sit on (flat, upright, and slanted), and Tool #2 (Make a Systematic List) counts each kind without missing or double-counting any line.

1STEP 1

Count every triple of points

25 points, and a triangle uses any 3 of them in any order, so C(25, 3) = 2300 possible picks — but some are not real triangles yet.

C(25, 3) = (25· 24· 23)/6 = 2300
2STEP 2

Spot the bad picks: collinear triples

Three points make no triangle exactly when they are collinear — zero area — so subtract every collinear triple from the 2300.

triangles = 2300 - (number of collinear triples)
3STEP 3

Count lines that hold 5 points

5 rows, 5 columns, and 2 main diagonals — 12 lines of 5 points — each give C(5, 3) = 10 collinear triples, for 120.

(5 + 5 + 2)·C(5, 3) = 12· 10 = 120
4STEP 4

Count the shorter diagonals (slope ± 1)

The shorter slope ± 1 diagonals — four with 4 points and four with 3 points — add 4·C(4, 3) + 4·C(3, 3) = 20 more.

4·C(4, 3) + 4·C(3, 3) = 16 + 4 = 20
5STEP 5

Count the steeper and shallower lines

Lines with slope 2, -2, 1/2, or -1/2 fit only 3 points, in 3 spots each — 4× 3 lines give 12·C(3, 3) = 12 more.

4 slopes× 3 lines×C(3, 3) = 12
6STEP 6

Add the bad picks and subtract

Add the flat triples: 120 + 20 + 12 = 152 collinear. Subtract from 2300: 2300 - 152 = 2148, which is answer (B).

2300 - (120 + 20 + 12) = 2300 - 152 = 2148 → (B)
Answer
2148
The collinear count 152 is tiny next to the 2300 total, so almost every triple should be a real triangle — and 2148 is indeed just below 2300, matching that picture. Checking the subtraction: 2300 - 152 = 2148. The flat triples break into 120 + 20 + 12 = 152, and each piece is a clean count of full lines, so nothing is double-counted: rows, columns, and diagonals are all different lines. Answer 2148 is choice (B).
💡Key takeaway

Count every way to pick 3 points, then take away the picks that land on one straight line — what is left are the real triangles.

  • Count every triple of points
  • Spot the bad picks: collinear triples
  • Count lines that hold 5 points
  • Count the shorter diagonals (slope ± 1)
  • Count the steeper and shallower lines
  • Add the bad picks and subtract