2018 AMC 10A Problem 13

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

1. Where is the crease? Folding lays $A$ exactly on $B$. Any point on the crease stays put, so it is the same distance from $A$ as from $B$. The points like that form the line through the middle of $AB$ at a right angle to $AB$.
2. Name the points. Let $M$ be the midpoint of $AB$, so $AM = \tfrac52$. Draw a line through $M$ at a right angle to $AB$; it runs down to the bottom side $AC$ at a point $P$. The crease is the segment $MP$.
3. Spot a smaller copy. Triangle $AMP$ has a right angle at $M$ and shares the angle at $A$ with triangle $ACB$, which has its right angle at $C$. Two angles match, so the two triangles have the same shape. In $AMP$, the right angle is at $M$, so $AM$ matches $AC$ and $MP$ matches $CB$.
4. Use the matching ratio. In the big triangle, the short leg divided by the long leg is $\dfrac{CB}{AC} = \dfrac34$. The same ratio holds in the small triangle:
$$\frac{MP}{AM} = \frac34, \qquad MP = \frac34 \cdot \frac52 = \frac{15}{8}.$$
5. Check that $P$ is really on side $AC$. In triangle $AMP$ the long side is $AP$, which matches $AB$: $\dfrac{AP}{AM} = \dfrac{AB}{AC} = \dfrac54$, so $AP = \dfrac54 \cdot \dfrac52 = \dfrac{25}{8}$. That is less than $4 = AC$, so $P$ lands on the side, not past $C$.

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

A fold is a mirror: the crease is the mirror line, and it always cuts the segment between a point and its image in half at a right angle. After that, the crease sits in a small right triangle that has the same shape as the paper, so one ratio finishes the job. Before computing, check which side the crease runs out through.

Coordinates. Put $A = (0,0)$, $C = (4,0)$, $B = (4,3)$. The midpoint of $AB$ is $M = \left(2, \tfrac32\right)$. The slope of $AB$ is $\tfrac34$, so the crease has slope $-\tfrac43$. Going down $\tfrac32$ from $M$ at that slope moves right by $\tfrac32 \cdot \tfrac34 = \tfrac98$, reaching $P = \left(\tfrac{25}{8}, 0\right)$. Then
$$MP = \sqrt{\left(\tfrac98\right)^2 + \left(\tfrac{12}{8}\right)^2} = \sqrt{\tfrac{81 + 144}{64}} = \tfrac{15}{8}.$$

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