Question 412·Hard·Math

Let $x=AC$, the length you need. The words “twice as long” give $x=2BC$. Divide by $2$ to write the other leg in terms of $x$:

Since $C$ is the right angle, $AB$ is the hypotenuse, the side across from it. Apply the Pythagorean theorem to the two legs: $AB^2=x^2+\left(\frac{x}{2}\right)^2$. Square the fraction: $AB^2=x^2+\frac{x^2}{4}$. Combine like terms: $AB^2=\frac54x^2$. Take the positive square root because $AB$ is a length:

Call the point where the altitude meets $AB$ point $D$, so $CD=12$. The altitude is the height when the hypotenuse is the base; it is not a leg of the original triangle. The two legs are perpendicular too, so both base-height pairs give the same area: $\frac12(AC)(BC)=\frac12(AB)(CD)$. Cancel $\frac12$ from both sides:

Substitute the lengths into the area equation: $x\left(\frac{x}{2}\right)=\left(\frac{\sqrt5}{2}x\right)(12)$. Divide both sides by $\frac{\sqrt5}{2}x$, which is nonzero because $x$ is a side length:

Type $x_1/\sqrt5\sim12$ in Desmos. Using $x_1$ and $\sim$ tells Desmos to solve for the unknown; under PARAMETERS, it reports $x_1\approx26.8328$. To give the exact radical instead of a rounded decimal, multiply both sides of $x/\sqrt5=12$ by $\sqrt5$: $x=12\sqrt5$. So $AC$ is $12\sqrt5$ units. Choice C.