2000 AMC 10 Problem 7

1. Angle $ADC$ is a right angle, $90^\circ$. Two lines cut it into three equal parts, so each part is $30^\circ$:
$$\angle ADP = \angle PDB = \angle BDC = 30^\circ, \qquad \angle ADB = 60^\circ.$$
2. Recall the 30-60-90 triangle: the sides are in the ratio $1 : \sqrt3 : 2$. The short leg is across from the $30^\circ$ angle, the long leg is across from the $60^\circ$ angle, and the hypotenuse is twice the short leg.
3. Triangle $ADP$. It has a right angle at $A$ and $30^\circ$ at $D$. The side $AD = 1$ is across from the $60^\circ$ angle at $P$, so it is the long leg. Divide by $\sqrt3$ to get the short leg, then double:
$$AP = \frac{1}{\sqrt3} = \frac{\sqrt3}{3}, \qquad DP = \frac{2\sqrt3}{3}.$$
4. Triangle $ADB$. It has a right angle at $A$ and $60^\circ$ at $D$. Now $AD = 1$ is across from the $30^\circ$ angle at $B$, so it is the short leg:
$$AB = \sqrt3, \qquad DB = 2.$$
5. Side $BP$. $P$ lies on $AB$, so $BP$ is what is left of $AB$ after removing $AP$:
$$BP = \sqrt3 - \frac{\sqrt3}{3} = \frac{2\sqrt3}{3}.$$
6. Perimeter. Add the three sides of triangle $BDP$:
$$DP + DB + BP = \frac{2\sqrt3}{3} + 2 + \frac{2\sqrt3}{3} = 2 + \frac{4\sqrt3}{3}.$$

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

Cutting a right angle into three equal parts always makes $30^\circ$ and $60^\circ$ angles, so every right triangle in the picture is a 30-60-90 triangle. Then one known length gives all the others. In each triangle, first decide whether the known side is the short leg or the long leg; that is where most mistakes happen.

Look at triangle $BDP$ itself. The angle at $B$ is $\angle ABD = 90^\circ - 60^\circ = 30^\circ$, and the angle at $D$ is $\angle PDB = 30^\circ$. Two equal angles mean two equal sides, so $PB = PD = \frac{2\sqrt3}{3}$. With $DB = 2$, the perimeter is $2 + 2 \cdot \frac{2\sqrt3}{3}$, and you never need $AB$.