AMC 8 · 1999 · #23
Grade 8 geometry-2dProblem
Square has sides of length 3. Segments and divide the square's area into three equal parts. How long is segment ?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: 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 $CM$.
Givens: $ABCD$ is a square with side length $3$; $M$ is on side $AB$, $N$ is on side $AD$; Segments $CM$ and $CN$ split the square into three equal-area regions: $\triangle CBM$, $\triangle CDN$, and quadrilateral $AMCN$; Answer choices: (A) $\sqrt{10}$, (B) $\sqrt{12}$, (C) $\sqrt{13}$, (D) $\sqrt{14}$, (E) $\sqrt{15}$
Unknowns: The length of segment $CM$
Understand
Restated: 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 $CM$.
Givens: $ABCD$ is a square with side length $3$; $M$ is on side $AB$, $N$ is on side $AD$; Segments $CM$ and $CN$ split the square into three equal-area regions: $\triangle CBM$, $\triangle CDN$, and quadrilateral $AMCN$; Answer choices: (A) $\sqrt{10}$, (B) $\sqrt{12}$, (C) $\sqrt{13}$, (D) $\sqrt{14}$, (E) $\sqrt{15}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
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 $\triangle CBM$, 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.
Execute — Answer: C
3.MD.C.7 Step 1 - Find the area of each of the three equal regions.
- The whole square has area side $\times$ side, and the three regions share that area evenly.
💡 Grade 3 area-of-a-rectangle as side $\times$ side, then a Grade 3 fair-share division by $3$ — this is the entry point that makes the problem accessible long before algebra.
6.G.A.1 Step 2 - Zoom in on $\triangle CBM$.
- From the figure, the angle at $B$ is a corner of the square, so it is a right angle.
- The legs of this right triangle are $BC$ (a full side of the square) and $BM$ (part of side $AB$).
- Use the triangle-area formula to solve for the unknown leg $BM$.
💡 Grade 6 "decompose a polygon and use the triangle-area formula": the figure already cuts the square into pieces; recognizing $\triangle CBM$ as a right triangle lets the standard formula pin down $BM$.
8.G.B.7 Step 3 - Apply the Pythagorean theorem to right triangle $CBM$.
- The hypotenuse is the segment we want, $CM$.
💡 Grade 8 Pythagorean theorem on a clean right triangle with known legs $3$ and $2$. No square root to estimate — $\sqrt{13}$ is already one of the listed choices.
3.MD.C.7 Find the area of each of the three equal regions. The whole square has area side 6.G.A.1 Zoom in on $\triangle CBM$. From the figure, the angle at $B$ is a corner of the 8.G.B.7 Apply the Pythagorean theorem to right triangle $CBM$. The hypotenuse is the seg Review
Reasonableness: 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 $\sqrt{9} = 3$ and $\sqrt{25} = 5$, the answer $\sqrt{13} \approx 3.61$ sits in the right window. The choices $\sqrt{10}, \sqrt{12}, \sqrt{14}, \sqrt{15}$ 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 $\triangle CDN$ should be congruent to $\triangle CBM$. That forces $DN = BM = 2$, which also makes $\triangle CDN$ have area $\tfrac{1}{2} \cdot 3 \cdot 2 = 3$ and the quadrilateral $AMCN$ have area $9 - 3 - 3 = 3$ — all three regions are indeed equal, confirming the setup.
Alternative: Tool #13 (Convert to Algebra) via coordinates. Place $A = (0,0)$, $B = (0,3)$, $C = (3,3)$, $D = (3,0)$ so $M = (0, m)$ on side $AB$ for some $0 < m < 3$. Triangle $CBM$ has vertices $(3,3), (0,3), (0,m)$ and area $\tfrac{1}{2} \cdot 3 \cdot (3 - m) = 3$, giving $m = 1$ and $BM = 3 - m = 2$. Then $CM = \sqrt{(3-0)^2 + (3-1)^2} = \sqrt{9+4} = \sqrt{13}$. Same answer (C), via the distance formula instead of Pythagoras.
CCSS standards used (min grade 8)
3.MD.C.7Relate area to multiplication and find areas of rectangles by multiplying side lengths (Computing the square's area as $3 \times 3 = 9$ and then dividing into three equal regions of area $3$ each.)6.G.A.1Find the area of polygons by composing or decomposing into triangles and other shapes (Recognizing $\triangle CBM$ as a right triangle sliced out of the square and using $\text{area} = \tfrac{1}{2} \cdot \text{base} \cdot \text{height}$ with the known area $3$ and leg $BC = 3$ to solve $BM = 2$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles in real-world and mathematical problems (Computing the hypotenuse $CM = \sqrt{BC^2 + BM^2} = \sqrt{3^2 + 2^2} = \sqrt{13}$ in right triangle $CBM$.)
⭐ 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 $CM = \sqrt{3^2 + 2^2} = \sqrt{13}$ — answer (C).
⭐ 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 $CM = \sqrt{3^2 + 2^2} = \sqrt{13}$ — answer (C).
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