2018 AMC 12A Problem 11

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 + \tfrac12 \sqrt{2}

\sqrt{3}

\tfrac{7}{4}

\tfrac{15}{8}

2

Folding

A

onto

B

creases along the perpendicular bisector of

AB;

since

AC \gt BC

it meets

AC

With

D

the midpoint of

AB

and

E

on

AC,

triangle

ADE \sim

triangle

ACB,

so

DE = AD \cdot \tfrac{CB}{AC}

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.

Problem 11 in other years