AMC 10 · 2018 · #13

Grade 8 geometry-2d
Archetype Paper Fold-and-Cut Symmetry └ Sub-archetype Fold Maps One Named Point Onto Another
reflection-symmetryperpendicular-bisectorsimilar-trianglespythagorean-theorempaper-folding reflection-unfoldingidentify-subproblems ↑ Prerequisites: reflection-symmetrysimilar-triangles
📏 Medium solution 💡 2 insights 📊 Diagram
Problem
A paper triangle has side lengths 3, 4, and 5 inches, so it is a right triangle. It is folded once so that vertex A lands exactly on vertex B. Find the length of the crease made by the fold.

Pick an answer.

(A)
$1+\frac12 \sqrt2$
(B)
$\sqrt3$
(C)
$\frac74$
(D)
$\frac{15}{8}$
(E)
2

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Visualize Spatial Relationships

Tool #17 (Visualize Spatial Relationships): picture the fold as a mirror. Folding A onto B means the crease is the mirror line that sends A to B, which forces the crease to be the perpendicular bisector of segment AB. Tool #1 (Draw a Diagram): once that line is drawn, a small right triangle appears tucked against vertex A. Tool #7 (Identify Subproblems): recognize that small triangle is similar to the whole 3-4-5 triangle. Tool #4 (Introduce a Variable): call the crease length and read it off a single proportion.

1STEP 1

Read the fold as a mirror

Folding A onto B acts like a mirror, so the crease is the perpendicular bisector of AB — through its midpoint at a right angle.

crease = perpendicular bisector of AB
2STEP 2

Locate the crease

With the right angle at C, let M be the midpoint of AB, so AM=52\frac{5}{2}; the crease runs from M perpendicular to AB, meeting leg AC at P.

AM=AB2\frac{AB}{2}=52\frac{5}{2}
3STEP 3

Find similar triangles

Triangles AMP and ACB share angle A and each has a right angle (at M and at C), so they are similar by angle-angle.

△ AMP ∼ △ ACB
4STEP 4

Set up a proportion and solve

Matching sides give MPCB=AMAC\frac{MP}{CB}=\frac{AM}{AC}, so MP3=524\frac{MP}{3}=\frac{\frac{5}{2}}{4} and the crease MP=158\frac{15}{8} inches — choice (D).

MP3\frac{MP}{3}=524\frac{\frac{5}{2}}{4} → MP=158\frac{15}{8} → (D)
Answer
158\frac{15}{8}
The crease is 158\frac{15}{8}=1.875 inches. It sits inside a triangle whose shortest side is 3, so a crease under 2 inches is sensible, and it is longer than the shorter leg's half — believable. A coordinate check agrees: with A=(0,0), B=(4,3), C=(4,0), the midpoint is M=(2,32\frac{3}{2}), and the perpendicular bisector (slope -43\frac{4}{3}) meets leg AC at (258\frac{25}{8},0). The distance from M to that point is √((98\frac{9}{8})²+(32\frac{3}{2})²)=√(22564\frac{225}{64})=158\frac{15}{8}, matching exactly.
💡Key takeaway

A fold is just a mirror, so the crease is the perpendicular bisector of the segment joining the two matched points; then a pair of similar triangles turns one proportion into the answer 158\frac{15}{8}, choice (D).

  • Read the fold as a mirror
  • Locate the crease
  • Find similar triangles
  • Set up a proportion and solve

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