AMC10 2018 10A, Problem 13

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? [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]

Solution: [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("$D$", (2,1.5), NW); label("$E$", (3.125,0), S); draw ((2,1.5)--(3.125,0),linewidth(1));  draw(rightanglemark((0,0),(2,1.5),(3.125,0))); [/asy] 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 $AB$ point $D$. Draw this line and call the intersection point with $AC$ as $E$. Now, $\triangle ACB$ is similar to $\triangle ADE$ by $AA$ similarity. Setting up the ratios, we find that \[\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{DE}{\frac{5}{2}} E=\frac{15}{8}.\] Thus, our answer is $\fbox{\frac{15}{8}}$.

Answer: \frac{15}{8}