2000 AIME I Problems/Problem 14
Problem
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 .
Contents
[hide]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 .
Solution 1
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
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 3 (Trig identities)
Let and . is isosceles, so and . is isosceles too, so . Using the expression for , we get by the triple angle formula! It follows now that , , giving .
The answer is .
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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