2018 AMC 12A Problem 11

Below is the professionally curated solution for Problem 11 of the 2018 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12A solutions, or check the answer key.

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Concepts:paper folding perpendicular bisector similarity

Difficulty rating: 1570

11.

A paper triangle with sides of lengths $3,$ $4,$ and $5$ inches, as shown, is folded so that point $A$ falls on point $B.$ What is the length in inches of the crease?

$1 + \tfrac12 \sqrt{2}$

$\sqrt{3}$

$\tfrac{7}{4}$

$\tfrac{15}{8}$

$2$

Solution:

The crease lies along the perpendicular bisector of $AB,$ meeting $AC$ at $E$ because $AC \gt BC.$ Let $D$ be the midpoint of $AB,$ so $AD = \tfrac52$ and $\triangle ADE$ is right-angled at $D.$ Since $\triangle ADE \sim \triangle ACB,$ we have $\tfrac{DE}{AD} = \tfrac{CB}{AC} = \tfrac34,$ so $DE = \frac52 \cdot \frac34 = \frac{15}{8}.$

Thus, the correct answer is D.

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