2000 AMC 10 Problems/Problem 7

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[asy] 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((3.4,0)); dot((1.3,2)); label("$A$",(0,2),NW); label("$B$",(3.4,2),NE); label("$C$",(3.4,0),SE); label("$D$",(0,0),SW); label("$P$",(1.3,2),N); [/asy]

$AD=1$.

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

Thus, $PD=\frac{2\sqrt{3}}{3}$. $DB=2$ $BP=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}$.

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

B.

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