Competition · AMC preparation · step 4 of 4

AMC 8 · 1999 · #23

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length
area-trianglespythagorean-theoremcoordinate-geometry identify-subproblemscoordinate-geometry ↑ Prerequisites: area-trianglespythagorean-theorem
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
Square ABCD has side length 3. Inside the square, segments CM and CN are drawn so that they split the square into three regions of equal area. Point M lies on side AB and point N lies on side AD. Find the length of segment CM.

Pick an answer.

(A)
$\sqrt{10}$
(B)
$\sqrt{12}$
(C)
$\sqrt{13}$
(D)
$\sqrt{14}$
(E)
$\sqrt{15}$

AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

The figure is already given, but the key step is reading the diagram for what it really shows: a right triangle CBM tucked against the corner B — exactly the Tool #1 (Draw a Diagram) move of spotlighting the load-bearing sub-figure inside a larger picture. The square gets carved into three equal pieces, which is the Tool #7 (Break into Subproblems) move: instead of solving the whole square at once, isolate △ CBM, find its leg BM from its area, then run the Pythagorean theorem on that single triangle. No algebra system, no coordinates — one triangle does all the work.

1STEP 1

Find each region's area

The square's total area is 9; splitting it into three equal parts gives each region an area of 3.

area of square = 3 × 3 = 9, area of each region = 9/3 = 3
2STEP 2

Use the area to find BM

In right triangle CBM the right angle sits at B, so its area 12\frac{1}{2} · 3 · BM = 3 forces BM = 2.

area = 1/2 · BC · BM → 3 = 1/2 · 3 · BM → BM = 2
3STEP 3

Apply the Pythagorean theorem

The Pythagorean theorem on legs 3 and 2 gives CM² = 13, and CM is its square root.

CM² = BC² + BM² = 3² + 2² = 9 + 4 = 13 → CM = √(13) → (C)
Answer
√(13)
Two quick checks. (1) Length sanity: CM is the hypotenuse of a right triangle with legs 3 and 2, so it must be longer than each leg but shorter than 3 + 2 = 5. Since √(9) = 3 and √(25) = 5, the answer √(13) ≈ 3.61 sits in the right window. The choices √(10), √(12), √(14), √(15) are all in the same window too, so the sanity check alone does not pick a single choice — the exact computation 9 + 4 = 13 does. (2) Symmetry check: the figure is symmetric across the diagonal AC, so △ CDN should be congruent to △ CBM. That forces DN = BM = 2, which also makes △ CDN have area 12\frac{1}{2} · 3 · 2 = 3 and the quadrilateral AMCN have area 9 - 3 - 3 = 3 — all three regions are indeed equal, confirming the setup.
💡Key takeaway

Slice the square into three equal areas, focus on the corner right triangle CBM (area 3, one leg 3) to get the other leg BM = 2, then Pythagoras gives CM = √(3² + 2²) = √(13) — answer (C).

  • Find each region's area
  • Use the area to find BM
  • Apply the Pythagorean theorem

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