# Difference between revisions of "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.