Question

In rectangle $ABCD,$ $AD = 1,$ $P$ is on $\overline{AB},$ and $\overline{DB}$ and $\overline{DP}$ trisect $\angle ADC.$ What is the perimeter of $\triangle BDP?$

$3 + \dfrac{\sqrt3}{3}$

$2 + \dfrac{4\sqrt3}{3}$

$2 + 2\sqrt2$

$\dfrac{3 + 3\sqrt5}{2}$

$2 + \dfrac{5\sqrt3}{3}$

Solution:

The right angle $\angle ADC = 90^\circ$ is trisected into three $30^\circ$ angles, so $\angle ADP = 30^\circ$ and $\angle ADB = 60^\circ.$

In right triangle $ADP,$ with $AD = 1,$ we get $DP = \dfrac{1}{\cos 30^\circ} = \dfrac{2\sqrt3}{3}$ and $AP = \tan 30^\circ = \dfrac{\sqrt3}{3}.$

In right triangle $ADB,$ with $AD = 1,$ we get $DB = \dfrac{1}{\cos 60^\circ} = 2$ and $AB = \tan 60^\circ = \sqrt3.$

Then $PB = AB - AP = \sqrt3 - \dfrac{\sqrt3}{3} = \dfrac{2\sqrt3}{3}.$

The perimeter of $\triangle BDP$ is $DP + PB + DB = \dfrac{2\sqrt3}{3} + \dfrac{2\sqrt3}{3} + 2 = 2 + \dfrac{4\sqrt3}{3}.$

Thus, the correct answer is B.

2000 AMC 10 Problem 7

Solution:

Problem 7 in Other Years