In rectangle ABCD, AD = 1, P is on AB, and DB and DP trisect angle ADC. Write the perimeter of triangle BDP in simplest form as: w+(x*sqrt(y))/z, where w, x, y, z are positive integers. What is w + x + y + z?
+9466 
DB and DP trisect angle ADC, so m∠ADP = m∠PDB = m∠BDC = 90° / 3 = 30°
∠ABD and ∠BDC are alternate interior angles, so they have the same measure.
If we draw a line from P to BD that is perpendicular to side BD, we will have congruent triangles.
The angle measures of these triangles are 30°, 60°, and 90° .
The side opposite the 60° angle = 1 , so
the side opposite the 90° angle = 2 / √3
perimeter of triangle BDP = 2 + 2/√3 + 2/√3 = 2 + 4/√3 = 2 + 4√3 / 3
2 + 4 + 3 + 3 = 12
+9466
DB and DP trisect angle ADC, so m∠ADP = m∠PDB = m∠BDC = 90° / 3 = 30°
∠ABD and ∠BDC are alternate interior angles, so they have the same measure.
If we draw a line from P to BD that is perpendicular to side BD, we will have congruent triangles.
The angle measures of these triangles are 30°, 60°, and 90° .
The side opposite the 60° angle = 1 , so
the side opposite the 90° angle = 2 / √3
perimeter of triangle BDP = 2 + 2/√3 + 2/√3 = 2 + 4/√3 = 2 + 4√3 / 3
2 + 4 + 3 + 3 = 12
+128407 Angle ADP = 30° and angle APD = 60°
So.....since we have a 30-60-90 triangle......AP = 1/√3 = √3/3
And PD = 2 / √3 = 2√3 / 3
And angle DPB = 120° and angle PDB = 30° ..... so angle PBD = 30°
So...triangle DPB isosceles with PD = PB = 2√3 / 3
And using the Law of Sines
DB/ sin(120) = PB / sin (30)
DB = sin (120) * [ 2√3 / 3 ] / (1/2) = √3 * 2√3 / 3 = 2
So...the perimeter of triangle BDP =
PD + PB + DB = 4√3 / 3 + 2 =
2 + 4√3 / 3
So w = 2, x = 4 , y = 3 and z = 3
And their sum is 12
