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2000 AMC 10 Problems/Problem 7

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Problem

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?

draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle);draw((0,0)--(1.3,2));draw((0,0)--(3.4,2));dot((0,0));dot((0,2));dot((3.4,2));dot(...

\mathrm{(A)}\ 3+\frac{\sqrt{3}}{3} \qquad\mathrm{(B)}\ 2+\frac{4\sqrt{3}}{3} \qquad\mathrm{(C)}\ 2+2\sqrt{2} \qquad\mathrm{(D...

Solution

draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle);draw((0,0)--(1.3,2));draw((0,0)--(3.4,2));dot((0,0));dot((0,2));dot((3.4,2));dot(...

AD=1.

Since \angle ADC is trisected, \angle ADP= \angle PDB= \angle BDC=30^\circ.

Thus, PD=\frac{2\sqrt{3}}{3}

DB=2

BP=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}.

Adding, 2+\frac{4\sqrt{3}}{3}.

\boxed{\text{B}}

See Also

2000 AMC 10 (Problems • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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