2000 AMC 10 Problem 7

$\angle ADC = 90^\circ$ is cut into three equal parts, so $\angle ADP = \angle PDB = \angle BDC = 30^\circ$, and therefore $\angle ADB = 60^\circ$.

Triangle $ADP$ has a right angle at $A$ and a $30^\circ$ angle at $D$, with the leg $AD = 1$ adjacent to the $30^\circ$ angle. In a 30-60-90 triangle the short leg is the long leg divided by $\sqrt3$ and the hypotenuse is twice the short leg:
$$AP = \frac{1}{\sqrt3} = \frac{\sqrt3}{3}, \qquad DP = \frac{2}{\sqrt3} = \frac{2\sqrt3}{3}.$$

Triangle $ADB$ has a right angle at $A$ and $\angle ADB = 60^\circ$, so $AD = 1$ is now the short leg:
$$AB = \sqrt3, \qquad DB = 2.$$

Then $BP = AB - AP = \sqrt3 - \frac{\sqrt3}{3} = \frac{2\sqrt3}{3}$, and the perimeter of $\triangle BDP$ is
$$DP + DB + BP = \frac{2\sqrt3}{3} + 2 + \frac{2\sqrt3}{3} = 2 + \frac{4\sqrt3}{3}.$$

The answer is $\boxed{\textbf{(B)}\ 2+\frac{4\sqrt{3}}{3}}$.

A trisected right angle is a signal that every triangle in sight is 30-60-90, so one known length propagates to all the others through fixed ratios. Anchor each ratio on the side you know ($AD = 1$) and be careful about whether that side is the short leg or the long leg in each triangle. Segments on a shared side, like $BP$, come from subtraction.

Since $\angle ABD = 30^\circ$ (complement of $60^\circ$) and $\angle PDB = 30^\circ$, triangle $BDP$ is isosceles with $PB = PD$. After finding $DP = \frac{2\sqrt3}{3}$ and $DB = 2$, the perimeter is $2 + 2 \cdot \frac{2\sqrt3}{3}$ with no need to compute $AB$.

• Taking BP to be all of AB instead of AB minus AP, or swapping which leg is opposite the 30-degree angle.
• Writing $DB = \sqrt3$ by confusing the hypotenuse of $ADB$ with its long leg; the diagonal is $2$.
• Leaving $AP$ as $1/\sqrt3$ and failing to match it to the rationalized forms in the choices.