2000 AIME I Problems/Problem 14
In triangle it is given that angles and are congruent. Points and lie on and respectively, so that Angle is times as large as angle where is a positive real number. Find the greatest integer that does not exceed .
Let point be in such that . Then is a rhombus, so and is an isosceles trapezoid. Since bisects , it follows by symmetry in trapezoid that bisects . Thus lies on the perpendicular bisector of , and . Hence is an equilateral triangle.
Now , and the sum of the angles in is . Then and , so the answer is .
Again, construct as above.
Let and , which means . is isosceles with , so . Let be the intersection of and . Since , is cyclic, which means . Since is an isosceles trapezoid, , but since bisects , .
Therefore we have that . We solve the simultaneous equations and to get and . , , so . .
Let the measure of be and . Because is isosceles, . So, . is isosceles too, so . Simplifying, . By double angle formula, we know that . Applying, and . The expression in the parentheses though is triple angle formula! Hence, , . It follows now that , . Giving . .
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