2025 AMC 10A Problem 15

In the figure $C$ is the point where $\overline{AD}$ crosses $\overline{BE}$, so $\triangle ABC$ has the right angle of the rectangle at $B$ and
$$[ABC] = \tfrac12 \cdot AB \cdot BC = \tfrac12 BC.$$
Everything reduces to finding $BC$.

Step 1: the diagonal $\overline{AE}$. In rectangle $ABEF$, $BE = AF = 7$ and $\angle ABE = 90^\circ$, so
$$AE^2 = AB^2 + BE^2 = 1 + 49 = 50.$$

Step 2: the length $DE$. Since $\angle ADE = 90^\circ$ and $AD = 5$,
$$DE^2 = AE^2 - AD^2 = 50 - 25 = 25, \qquad DE = 5.$$

Step 3: similar triangles. Triangles $ABC$ and $EDC$ have right angles at $B$ and at $D$, and $\angle ACB = \angle ECD$ because $A, C, D$ are collinear and $B, C, E$ are collinear, making these vertical angles. So
$$\triangle ABC \sim \triangle EDC \quad\text{with ratio}\quad \frac{AB}{ED} = \frac{1}{5}.$$
Corresponding sides give
$$DC = 5\,BC \qquad\text{and}\qquad EC = 5\,AC.$$

Step 4: solve. The point $C$ lies on both segments, so $AC + CD = AD = 5$ and $BC + CE = BE = 7$. Substituting,
$$AC + 5\,BC = 5, \qquad BC + 5\,AC = 7.$$
From the first, $AC = 5 - 5\,BC$; putting that into the second,
$$BC + 25 - 25\,BC = 7 \implies 24\,BC = 18 \implies BC = \frac34.$$

Therefore
$$[ABC] = \tfrac12 \cdot 1 \cdot \tfrac34 = \frac{3}{8}.$$

The answer is $\boxed{\textbf{(A)}\ \frac{3}{8}}$.

The data split into two unrelated packets, the rectangle ($AB = 1$, $BE = 7$) and the right triangle ($AD = 5$, $\angle ADE = 90^\circ$), and the only segment they share is $\overline{AE}$. Computing $AE^2 = 50$ two ways is what welds them together and reveals the hidden fact $AD = DE = 5$. After that, two lines crossing at $C$ create a pair of similar triangles whose ratio is known, and the two "the pieces add to the whole" equations turn the geometry into a $2 \times 2$ linear system. The pattern, similar triangles at a crossing plus segment-sum equations, is worth keeping: it avoids the quadratic that appears if one instead writes $AC^2 = 1 + BC^2$.

Coordinates. Put $D = (0,0)$, $A = (0,5)$, and, using $DE = 5$ from Step 2, $E = (5,0)$. The point $B$ satisfies $AB = 1$ and $BE = 7$; subtracting the two circle equations gives the line $y = x + \frac{24}{5}$, and solving yields $B = \left(-\frac35, \frac{21}{5}\right)$, the root on the far side of $\overline{AD}$ from $E$. Line $BE$ meets $x = 0$ at $C = \left(0, \frac{15}{4}\right)$, so $AC = 5 - \frac{15}{4} = \frac54$ and the distance from $B$ to line $AD$ is $\frac35$. Hence $[ABC] = \frac12 \cdot \frac54 \cdot \frac35 = \frac38$.

• Reading the ratio 1 : 5 as a division of BE and concluding BC = 7/6; the similarity pairs BC with DC and AC with EC, never BC with CE.
• Assuming $C$ is the midpoint of $\overline{AD}$ or of $\overline{BE}$, which the figure suggests only because it is not to scale.
• Taking $AB = 1$ and $AD = 5$ as the sides of the rectangle; the rectangle's sides are $AB = 1$ and $AF = BE = 7$, and $\overline{AD}$ is not a side at all.