2018 AMC 10A Problem 13

In the figure the right angle is at $C$, with $AC = 4$, $BC = 3$, and hypotenuse $AB = 5$.

Folding $A$ onto $B$ reflects $A$ across the crease, so the crease lies on the perpendicular bisector of $\overline{AB}$. Let $M$ be the midpoint of $AB$, so $AM = \tfrac52$, and let the crease meet leg $\overline{AC}$ at $P$. The crease inside the paper is segment $MP$.

Triangle $AMP$ has a right angle at $M$ and shares angle $A$ with the big triangle. In $\triangle ABC$, $\tan A = \dfrac{BC}{AC} = \dfrac34$, so in $\triangle AMP$
$$MP = AM \cdot \tan A = \frac52 \cdot \frac34 = \frac{15}{8}.$$

(Also $AP = AM / \cos A = \tfrac52 \cdot \tfrac54 = \tfrac{25}{8} < 4$, confirming $P$ really lies on $AC$ rather than beyond $C$.)

The answer is $\boxed{\textbf{(D)}\ \frac{15}{8}}$.

Every fold is a reflection, and the crease is the mirror line: the perpendicular bisector of the segment joining a point to its image. Once that line is identified, the crease is trapped inside a small right triangle similar to the original one, and a single ratio finishes it. Ask "which side does the crease exit through?" before computing; here it exits through the longer leg.

Coordinates: $A = (0,0)$, $C = (4,0)$, $B = (4,3)$. The midpoint of $AB$ is $(2, \tfrac32)$ and the perpendicular slope is $-\tfrac43$, so the bisector is $y - \tfrac32 = -\tfrac43(x - 2)$. It hits $y = 0$ at $x = \tfrac{25}{8}$; the distance from $(2, \tfrac32)$ to $(\tfrac{25}{8}, 0)$ is $\sqrt{(\tfrac98)^2 + (\tfrac{12}{8})^2} = \tfrac{15}{8}$.

• Assuming the crease runs to the opposite vertex or to leg BC, or measuring the crease as the whole perpendicular bisector rather than the part inside the triangle.
• Using $\tan A = \tfrac43$ (the ratio for the other acute angle), which gives $\tfrac{10}{3}$, not among the choices, or using $\sin$ instead of $\tan$ and getting $\tfrac32$.