AMC 10 A AMC 12 A Problem 21 We divide into cases according to

Amc 10 a amc 12 a problem 21 we divide into cases

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2010 AMC 10 A, Problem #212010 AMC 12 A, Problem #21“We divide into cases according to the three possible values ofn,namelyn= 1,2, or3.”this case. The total number of solutions for all cases is 4.Difficulty:HardNCTM Standard:Number and Operations Standard for Grades 9–12: understand meanings ofoperations and how they relate to one another.
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Three distinct vertices are selected at random from aregular 14-sided polygon (tetradecagon) and used asvertices to form a triangle.What is the probabilitythat the triangle is a right triangle?132010 AMC 10 A, Problem #22“Circumscribe the regular tetradecagon with a circle.The triangle will be a right triangle if and only if twoof the vertices are antipodal points of the circle.”
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TriangleABChas vertices atA(0,0),B(8,14)andC(20,0). The liney=mxdividesABCinto twotriangles of equal area. What ism?2010 AMC 10 A, Problem #23“The liney=mxpasses through vertexA.”
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