2025 AMC 10A Problem 15

In the figure, $C$ is the point where segment $AD$ crosses segment $BE$.

Triangle $ABC$ has a right angle at $B$, because $B$ is a corner of the rectangle. So its area is
$$\tfrac12 \cdot AB \cdot BC = \tfrac12 \cdot 1 \cdot BC = \tfrac12\, BC.$$
Everything comes down to finding $BC$.

Step 1: connect the two parts of the figure. The facts come in two separate groups: the rectangle ($AB = 1$, $AF = 7$) and the right angle at $D$ ($AD = 5$). The segment $AE$ belongs to both groups, so draw it.

1. In the rectangle, opposite sides are equal, so $BE = AF = 7$. Triangle $ABE$ has a right angle at $B$, so by the Pythagorean theorem
$$AE^2 = 1^2 + 7^2 = 50.$$
2. Triangle $ADE$ has a right angle at $D$, and $AE$ is its hypotenuse too. So
$$DE^2 = AE^2 - AD^2 = 50 - 25 = 25, \qquad DE = 5.$$

Step 2: find similar triangles. Look at the two triangles that meet at $C$: triangle $ABC$ (top) and triangle $EDC$ (bottom).

3. Both have a right angle: at $B$ and at $D$.
4. The angles at $C$ are equal, because two straight lines cross there (vertical angles).
5. So the triangles are similar. Match the corners: $A \leftrightarrow E$, $B \leftrightarrow D$, $C \leftrightarrow C$. The side $AB = 1$ matches $ED = 5$, so triangle $EDC$ is $5$ times as big as triangle $ABC$.
6. Matching the other sides with the same corner order:
- $BC$ matches $DC$, so $DC = 5\,BC$.
- $AC$ matches $EC$, so $EC = 5\,AC$.

Step 3: use the pieces of the two segments. $C$ splits each segment into two pieces:
$$AC + CD = AD = 5, \qquad BC + CE = BE = 7.$$
Replace $CD$ and $CE$ using Step 2:
$$AC + 5\,BC = 5, \qquad BC + 5\,AC = 7.$$

1. From the first equation, $AC = 5 - 5\,BC$.
2. Put that into the second: $BC + 5(5 - 5\,BC) = 7$, so $BC + 25 - 25\,BC = 7$, which gives $24\,BC = 18$ and $BC = \frac34$.
3. Quick check: $AC = 5 - \frac{15}{4} = \frac54$, and $1^2 + \left(\frac34\right)^2 = \frac{25}{16} = \left(\frac54\right)^2$. Triangle $ABC$ really is a right triangle with these sides.

So the area is $\tfrac12 \cdot \tfrac34 = \tfrac38$.

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

When a figure gives you two groups of facts that seem unrelated, look for a segment they share; here $AE$ is the hypotenuse of two right triangles, and computing it both ways uncovers $DE = 5$. After that, two lines crossing at a point almost always give a pair of similar triangles. Writing "piece + piece = whole" for each crossing segment turns the picture into two simple equations.

Once you know $DC = 5\,BC$ (Step 2), you can use the Pythagorean theorem in triangle $ABC$ instead of the second equation.

Let $BC = b$. Then $CD = 5b$ and $AC = 5 - 5b$. Triangle $ABC$ has legs $1$ and $b$ and hypotenuse $AC$:
$$(5 - 5b)^2 = 1 + b^2.$$
Expand: $25 - 50b + 25b^2 = 1 + b^2$, so $24b^2 - 50b + 24 = 0$, or $12b^2 - 25b + 12 = 0$. This factors as $(4b - 3)(3b - 4) = 0$, so $b = \frac34$ or $b = \frac43$.

If $b = \frac43$, then $AC = 5 - \frac{20}{3}$ is negative, which is impossible. So $b = \frac34$ and the area is $\frac12 \cdot \frac34 = \frac38$. This way needs a quadratic, which is why the main solution uses the second segment instead.

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