2018 AMC 10A Problem 13

Problem 13 of 25IntermediateGeometry

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?

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Solution

The crease is the perpendicular bisector of

\overline{AB}.

Let

\overline{DE}

be the crease. By

AA

similarity,

\triangle ADE\sim\triangle ACB.

Therefore,

\dfrac{BC}{AC}=\dfrac{DE}{AD}.

Plugging in the side lengths gives

\dfrac34=\dfrac{DE}{\frac{5}{2}},

so

DE=\dfrac{15}{8}.

Thus, D is the correct answer.

More practice

Concepts: paper folding · perpendicular bisector · similarity

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.