AMC 10 Daily Practice Round 4

Complete problem set with solutions and individual problem pages

Problem 19 Easy

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

  • A.

    3+ \frac{ \sqrt{3}}{3}

  • B.

    2+ \frac{ 4\sqrt{3}}{3}

  • C.

    2+ 2\sqrt{2}

  • D.

    \frac{ 3+3\sqrt{5}}{2}

  • E.

    2+ \frac{ 5\sqrt{3}}{3}

Answer:B

Since \angle A D C is trisected, \angle A D P=\angle P D B=\angle B D C=30^{\circ}. Thus, P D=\frac{2 \sqrt{3}}{3}, D B=2, and B P=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2 \sqrt{3}}{3}. Adding them, we get \boxed{\text{(B) }2+\frac{4 \sqrt{3}}{3}}.

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