2018 AMC 10A Problems/Problem 13

A paper triangle with sides of lengths 3,4, and 5 inches, as shon, is folded so that point $A$ falls on point $B$ . What is the length in inches of the crease? $[asy] draw((0,0)--(4,0)--(4,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,3), NE); label("$C$", (4,0), SE); label("$4$", (2,0), S); label("$3$", (4,1.5), E); label("$5$", (2,1.5), NW); fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray); [/asy]$ $\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2$

Solution 1

First, we need to realize that the crease line is just the perpendicular bisector of side $AB$ , the hypotenuse of right triangle $\triangle ABC$ . Call the midpoint of $AC$ point $D$ . Draw this line and call the intersection point with $AC$ as $E$ . Now, $\triangle ABC$ is similar to $\triangle ADE$ by $AA$ similarity. Setting up the ratios, we find that $\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{DE}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.$ Thus, our answer is $\boxed{D}$ .

~Nivek

Solution 2 (if you are already out of time)

Simply make a 3x4x5 inch triangle and then cut it out (using fine rips). Then, make the fold and mesure. It will be $\boxed{D} \frac{15}{8}$ inches in length.

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2018 AMC 10A Problems/Problem 13

Solution 1

Solution 2 (if you are already out of time)

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