2018 AMC 10A Problem 13

Below is the professionally curated solution for Problem 13 of the 2018 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 10A solutions, or check the answer key.

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Concepts:paper folding perpendicular bisector similarity

Difficulty rating: 1420

13.

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?

$1+\dfrac{1}{2} \sqrt{2}$

$\sqrt{3}$

$\dfrac{7}{4}$

$\dfrac{15}{8}$

$2$

Solution:

Note that the crease will the perpendicular bisector of $\overline{AB}.$ Let $\overline{DE}$ be the crease.

By $AA$ similarity, we know that $\triangle ADE \sim \triangle ACB.$ Therefore, $\dfrac{BC}{AC} = \dfrac{DE}{AD}$ Plugging in values: $\dfrac{3}{4} = \dfrac{DE}{\frac{5}{2}}.$

Simplifying gets us that $DE = \dfrac{15}{8}.$

Thus, D is the correct answer.

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