2025 AMC 10B Problem 13

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

A cevian-ratio question in a triangle whose shape is completely pinned down is a coordinate problem: two lines, one intersection, and no trigonometry beyond placing the vertices. The final shortcut matters more than it looks. Once two points are known to lie on a common ray from the origin, comparing a single coordinate replaces two distance computations and removes every radical from the arithmetic. Note also that the answer is a pure ratio, so normalising $BC = 1$ costs nothing.

Mass points. The altitude foot divides the hypotenuse as $AF : FB = AC^2 : BC^2 = 3 : 1$, the standard relation in a right triangle. Give $A$ mass $1$ and $B$ mass $3$ so that $F$ balances, and give $C$ mass $3$ so that the midpoint $M$ of $CB$ balances. Then $F$ carries mass $1 + 3 = 4$, and the point $P$ where $AM$ meets $CF$ satisfies
$$CP : PF = \text{mass}(F) : \text{mass}(C) = 4 : 3,$$
giving $\frac{x}{x+y} = \frac{3}{7}$ again.

• Reporting the piece next to the right angle, which is the larger of the two at 4/7 of the altitude, instead of the smaller piece x.
• Drawing the median to the longer leg (the side opposite $60^\circ$) instead of to the shortest side, which splits the altitude $4:1$ and gives $\frac15$.
• Computing $\frac{x}{y} = \frac34$ rather than $\frac{x}{x+y}$; the denominator is the whole altitude.