2025 AMC 10B Problem 13

Only a ratio is asked, so pick an easy size. The letters $x$ and $y$ are already used by the problem, so write points as $(u, v)$: $u$ is how far across, $v$ is how far up.

1. Place the triangle. Put the right angle at $C = (0,0)$, with $B = (2, 0)$ and $A = (0, 2\sqrt3)$. The legs are $2$ and $2\sqrt3$ and the hypotenuse is $4$, so this is a $30$-$60$-$90$ triangle. The shortest side is $CB$, the one of length $2$.

2. The altitude from $C$. The hypotenuse $AB$ drops $2\sqrt3$ while moving $2$ across, so its steepness (slope) is $-\sqrt3$, and its line is $v = \sqrt3\,(2 - u)$. The altitude is perpendicular to it, so its slope is $\frac{1}{\sqrt3}$ and, starting at $C$, its line is $v = \frac{u}{\sqrt3}$. Where they meet:
$$\frac{u}{\sqrt3} = \sqrt3(2 - u) \;\Longrightarrow\; u = 3(2-u) \;\Longrightarrow\; u = \frac32.$$
So the foot of the altitude is $F$, with $u = \frac32$.

1. The median to the shortest side. It goes from $A = (0, 2\sqrt3)$ to the midpoint of $CB$, which is $M = (1, 0)$. That line is $v = 2\sqrt3\,(1 - u)$.

4. Where the median crosses the altitude. Set the two heights equal:
$$\frac{u}{\sqrt3} = 2\sqrt3(1-u) \;\Longrightarrow\; u = 6(1-u) \;\Longrightarrow\; u = \frac67.$$
Call this point $P$.

5. Compare lengths along the altitude. $C$, $P$ and $F$ are all on the same straight line from $C$. Along a straight line from $C$, the distance from $C$ grows in step with how far across you go. So
$$\frac{CP}{CF} = \frac{6/7}{3/2} = \frac{4}{7}.$$
The altitude is split into $CP$, which is $\frac47$ of it, and $PF$, which is $\frac37$ of it.

6. Pick the smaller piece. $x$ is the smaller piece, so $x = \frac37 \cdot CF$, and $x + y$ is the whole altitude $CF$. Therefore
$$\frac{x}{x+y} = \frac37.$$

The answer is $\boxed{\textbf{(A)}\ \frac 37}$.

When a triangle's shape is completely fixed, coordinates turn "where do these two segments cross?" into solving two line equations. The last shortcut saves real work: two points on the same line through the origin can be compared by one coordinate, with no distance formula and no square roots. Choosing side lengths that keep the numbers whole also helps.

Use balancing weights (mass points). In a right triangle, the altitude splits the hypotenuse so that $BF : FA = CB^2 : CA^2 = 4 : 12 = 1 : 3$. Put weight $1$ at $A$ and $3$ at $B$ so that $F$ balances $AB$, and weight $3$ at $C$ so that $M$ balances $CB$. Then $F$ carries $1 + 3 = 4$, and $P$ balances $C$ (weight $3$) against $F$ (weight $4$), so $CP : PF = 4 : 3$. The smaller piece is $3$ parts out of $7$.

The answer is a ratio, so fix a convenient size. Using $(u, v)$ for coordinates, put the right angle at $C = (0,0)$ with
$$B = (1, 0), \qquad A = (0, \sqrt3).$$
Then $AB = 2$, and $\tan\angle A = \frac{BC}{AC} = \frac{1}{\sqrt3}$, so $\angle A = 30^\circ$ and $\angle B = 60^\circ$. The shortest side is the one opposite the $30^\circ$ angle, namely $BC$.

The altitude. The hypotenuse lies on $\sqrt3\,u + v = \sqrt3$. The altitude from $C$ is perpendicular to it, so it runs along the direction $(\sqrt3, 1)$, that is, along the line $v = \frac{u}{\sqrt3}$. Substituting into the hypotenuse gives $\sqrt3\,u + \frac{u}{\sqrt3} = \sqrt3$, so $u = \frac34$ and the foot is
$$F = \left(\tfrac34, \tfrac{\sqrt3}{4}\right).$$

The median. The median to $BC$ joins $A = (0,\sqrt3)$ to the midpoint $M = \left(\tfrac12, 0\right)$, so it lies on $v = \sqrt3(1 - 2u)$.

The intersection. Solving $\frac{u}{\sqrt3} = \sqrt3(1 - 2u)$ gives $u = 3(1-2u)$, so $u = \frac37$ and
$$P = \left(\tfrac37, \tfrac{\sqrt3}{7}\right).$$

Both $P$ and $F$ lie on the ray from $C$ through $F$, so lengths measured from $C$ are proportional to first coordinates:
$$\frac{CP}{CF} = \frac{3/7}{3/4} = \frac47 .$$
Therefore $P$ cuts the altitude in the ratio $CP : PF = 4 : 3$. The smaller piece is $x = \frac37 \cdot CF$ while the whole altitude is $x + y = CF$, so
$$\frac{x}{x+y} = \frac37.$$

The answer is $\boxed{\textbf{(A)}\ \frac 37}$.