2017 AMC 10B Problem 22

Draw the segment $BC$. Since $AB$ is a diameter and $C$ lies on the circle, $\angle ACB = 90^\circ$ (angle in a semicircle).

Now look at $\triangle ADE$: it is right-angled at $D$ with $AD = AB + BD = 4 + 3 = 7$ and $DE = 5$, so
$$AE = \sqrt{7^2 + 5^2} = \sqrt{74}, \qquad [ADE] = \frac12 \cdot 7 \cdot 5 = \frac{35}{2}.$$

Triangles $ACB$ and $ADE$ share the angle at $A$ and each has a right angle ($C$ and $D$ respectively), so $\triangle ACB \sim \triangle ADE$, with hypotenuses $AB = 4$ and $AE = \sqrt{74}$. Areas of similar triangles scale by the square of the ratio of corresponding sides:
$$[ABC] = [ADE] \cdot \left(\frac{AB}{AE}\right)^2 = \frac{35}{2} \cdot \frac{16}{74} = \frac{35 \cdot 8}{74} = \frac{140}{37}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{140}{37}}$.

Whenever a point lies on a circle whose diameter is given, the right angle at that point is the intended tool. The right angle at $C$ and the constructed right angle at $D$ share vertex $A$, making the two triangles similar, and the area ratio is then a single squared ratio of hypotenuses. No coordinates of $C$ are ever needed.

Coordinates: $A = (-2, 0)$, $B = (2, 0)$, $D = (5, 0)$, $E = (5, 5)$. Line $AE$ has slope $\frac57$; $C$ is the foot of the perpendicular from $B$ to this line. The distance from $B$ to the line is $BC = \frac{4 \cdot 5}{\sqrt{74}} = \frac{20}{\sqrt{74}}$, then $AC = \sqrt{16 - \frac{400}{74}} = \frac{28}{\sqrt{74}}$, and $[ABC] = \frac12 \cdot \frac{20 \cdot 28}{74} = \frac{140}{37}$.

• Forgetting that C on the circle with AB a diameter forces a right angle at C, and instead trying to intersect the line with the circle by algebra.
• Scaling the area by the ratio $\frac{4}{\sqrt{74}}$ instead of its square, or using $AD = 5$ (confusing $BD$ with $AD$).