2018 AMC 10A Problem 13

Question

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

The crease is the perpendicular bisector of

AB

Use similarity with the original

3

-

4

-

5

triangle

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.

Problem 13 in other years