AMC 8 · 2003 · #25

Grade 8 geometry-2d

Archetype Paper Fold-and-Cut Symmetry └ Sub-archetype Fold Maps One Named Point Onto Another

reflection-symmetrypaper-foldingisosceles-trianglearea-trianglesspatial-visualization reflection-unfoldingphysical-representationidentify-subproblems ↑ Prerequisites: area-trianglesreflection-symmetryperfect-squares

📏 Long solution 💡 4 insights 📊 Diagram

Problem

Square WXYZ has area 25 cm², so its side length is 5 cm. Four 1 cm-by-1 cm squares frame the figure (one at each corner of the surrounding rectangle), so the segment BC runs vertically just outside side WZ. Triangle ABC is isosceles with AB = AC, and folding it over BC lands A exactly on O, the center of square WXYZ. Find the area of △ ABC.

Pick an answer.

(A)

$\frac{15}4$

(B)

$\frac{21}4$

(C)

$\frac{27}4$

(D)

$\frac{21}2$

(E)

$\frac{27}2$

AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The figure is the whole problem, so Tool #1 (Draw a Diagram) goes first: place coordinates on the picture so every length we need is a coordinate difference. Once axes are set, the area question splits cleanly into Tool #7 (Identify Subproblems): (a) find the base BC by reading the vertical positions of B and C off the side of the square minus the two 1-cm trims, and (b) find the height by using the fold. The fold is the key Grade 8 reflection idea — line BC is the perpendicular bisector of AO, so the height from A to BC equals the distance from O to BC, which we read off coordinates. Multiply, halve, done.

1STEP 1

Put the square on a grid: Z=(0,0), Y=(5,0), X=(5,5), W=(0,5), center O=( $\frac{5}{2}$ , $\frac{5}{2}$ ); the vertical base sits at B=(-2,4), C=(-2,1).

Z=(0,0), W=(0,5), X=(5,5), Y=(5,0); O=(

\frac{5}{2}

\frac{5}{2}

); B=(-2,4), C=(-2,1)

2STEP 2

Sub-problem A: B and C share an x, so BC is vertical; the 5-cm side loses 1 cm top and 1 cm bottom, giving BC = 3.

BC = y_B - y_C = 4 - 1 = 3 cm

3STEP 3

Sub-problem B: the fold makes BC the perpendicular bisector of AO, so the triangle's height equals the distance from O to line BC.

fold A → O → BC is the perpendicular bisector of AO → h = dist(A,BC) = dist(O,BC)

4STEP 4

The base line is x = -2 and O sits at x = $\frac{5}{2}$ , so the height is $\frac{5}{2}$ - (-2) = $\frac{9}{2}$ .

h = dist(O,BC) =

\frac{5}{2}

- (-2) =

\frac{5}{2}

+ 2 =

\frac{9}{2}

5STEP 5

Plug base 3 and height $\frac{9}{2}$ into half-base-times-height: $\frac{1}{2}$ · 3 · $\frac{9}{2}$ = $\frac{27}{4}$ → (C).

[△ ABC] =

\frac{1}{2}

· BC · h =

\frac{1}{2}

· 3 ·

\frac{9}{2}

\frac{27}{4}

→ (C)

Answer

\frac{27}{4}

Sanity check the size. The fold lands A on O across BC, so A sits on the far side of BC at the same distance as O. With O at x =

\frac{5}{2}

and the fold line at x = -2, point A must be at x = -2 -

\frac{9}{2}

\frac{-13}{2}

, well to the left of the small-square frame — which matches the diagram showing A far outside the squares. A direct area check: base 3 and height

\frac{9}{2}

give

\frac{1}{2}

· 3 ·

\frac{9}{2}

\frac{27}{4}

= 6.75 cm², the same order of magnitude as a triangle with a 3-cm base and a roughly 4.5-cm height. Choices (A)

\frac{15}{4}

= 3.75 and (B)

\frac{21}{4}

= 5.25 would require a smaller height (under 4), while (D)

\frac{21}{2}

= 10.5 and (E)

\frac{27}{2}

= 13.5 would require a height larger than the whole figure. Only (C) fits.

💡Key takeaway

Folding A onto O across BC means A and O are mirror images across the fold line, so the triangle's height equals the distance from O to BC. Read base 3 and height $\frac{9}{2}$ straight off the picture and the area is $\frac{1}{2}$ · 3 · $\frac{9}{2}$ = $\frac{27}{4}$ .

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