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| − | ==Problem==
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| − | In rectangle <math>ABCD</math>, <math>AD=1</math>, <math>P</math> is on <math>\overline{AB}</math>, and <math>\overline{DB}</math> and <math>\overline{DP}</math> trisect <math>\angle ADC</math>. What is the perimeter of <math>\triangle BDP</math>?
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| − |
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| − | <asy>
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| − | draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle);
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| − | draw((0,0)--(1.3,2));
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| − | draw((0,0)--(3.4,2));
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| − | dot((0,0));
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| − | dot((0,2));
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| − | dot((3.4,2));
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| − | dot((3.4,0));
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| − | dot((1.3,2));
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| − | label("$A$",(0,2),NW);
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| − | label("$B$",(3.4,2),NE);
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| − | label("$C$",(3.4,0),SE);
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| − | label("$D$",(0,0),SW);
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| − | label("$P$",(1.3,2),N);
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| − | </asy>
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| − | <math>\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)}\ \frac{3+3\sqrt{5}}{2} \qquad\mathrm{(E)}\ 2+\frac{5\sqrt{3}}{3}</math>
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| − | <asy>
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| − | draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle);
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| − | draw((0,0)--(1.3,2));
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| − | draw((0,0)--(3.4,2));
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| − | dot((0,0));
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| − | dot((0,2));
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| − | dot((3.4,2));
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| − | dot((3.4,0));
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| − | dot((1.3,2));
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| − | label("$A$",(0,2),NW);
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| − | label("$B$",(3.4,2),NE);
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| − | label("$C$",(3.4,0),SE);
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| − | label("$D$",(0,0),SW);
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| − | label("$P$",(1.3,2),N);
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| − | label("$1$",(0,1),W);
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| − | label("$2$",(1.7,1),SE);
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| − | label("$\frac{\sqrt{3}}{3}$",(0.65,2),N);
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| − | label("$\frac{2\sqrt{3}}{3}$",(0.85,1),NW);
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| − | label("$\frac{2\sqrt{3}}{3}$",(2.35,2),N);
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| − | </asy>
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| − | <math>AD=1</math>.
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| − | Since <math>\angle ADC</math> is trisected, <math>\angle ADP= \angle PDB= \angle BDC=30^\circ</math>.
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| − | Thus, <math>PD=\frac{2\sqrt{3}}{3}</math>
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| − |
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| − | <math>DB=2</math>
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| − | <math>BP=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}</math>.
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| − | Adding, we get <math>2+\frac{4\sqrt{3}}{3}</math>
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| − |
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| − | <math>\boxed{\text{B}}</math>
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