Competition · AMC preparation · step 4 of 4

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

Set coordinates on the square

Put the square on a grid: Z=(0,0), Y=(5,0), X=(5,5), W=(0,5), center O=(52\frac{5}{2},52\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=(5/2,5/2); B=(-2,4), C=(-2,1)
2STEP 2

Find the base BC

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

Turn the fold into a height

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

Compute the height

The base line is x = -2 and O sits at x = 52\frac{5}{2}, so the height is 52\frac{5}{2} - (-2) = 92\frac{9}{2}.

h = dist(O,BC) = 5/2 - (-2) = 5/2 + 2 = 9/2 cm
5STEP 5

Apply the triangle area formula

Plug base 3 and height 92\frac{9}{2} into half-base-times-height: 12\frac{1}{2} · 3 · 92\frac{9}{2} = 274\frac{27}{4} → (C).

[△ ABC] = 1/2 · BC · h = 1/2 · 3 · 9/2 = 27/4 → (C)
Answer
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 = 52\frac{5}{2} and the fold line at x = -2, point A must be at x = -2 - 92\frac{9}{2} = −132\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 92\frac{9}{2} give 12\frac{1}{2} · 3 · 92\frac{9}{2} = 274\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) 154\frac{15}{4} = 3.75 and (B) 214\frac{21}{4} = 5.25 would require a smaller height (under 4), while (D) 212\frac{21}{2} = 10.5 and (E) 272\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 92\frac{9}{2} straight off the picture and the area is 12\frac{1}{2} · 3 · 92\frac{9}{2} = 274\frac{27}{4}.

  • Set coordinates on the square
  • Find the base BC
  • Turn the fold into a height
  • Compute the height
  • Apply the triangle area formula

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