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 .
Official Solution (MAA)
Let . Because is exterior to isosceles triangle its measure is and has the same measure. Because is exterior to its measure is . Let . It follows that and that . Two of the angles of triangle have measure , and thus the measure of is . It follows that . Because and , it also follows that . Now apply the Law of Sines to triangle to find because . Hence . Since , this implies that , i.e. . Thus and which implies that . So the answer is .
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 .
Solution 2 (Law of sines)
Let and be the measure of . Since and are isoceles, and . Because and both have a side length opposite , by the law of sines:
Simplifying, this becomes
From the first two fractions,
Substituting, we have from the first and third fractions,
Because , is acute, so
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 . .
Solution 4 (Trig identities)
Let and . is isosceles, so and . is isosceles too, so . Using the expression for , we get by the triple angle formula! Thus and . It follows now that , , giving , which implies that . So the answer is .
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