2003 AIME I Problems/Problem 10
Triangle is isosceles with and Point is in the interior of the triangle so that and Find the number of degrees in
Take point inside such that and .
. Also, since and are congruent (by ASA), . Hence is an equilateral triangle, so .
Then . We now see that and are congruent. Therefore, , so .
From the givens, we have the following angle measures: , . If we define then we also have . Then apply the Law of Sines to triangles and to get
Clearing denominators, evaluating and applying one of our trigonometric identities to the result gives
and multiplying through by 2 and applying the double angle formulas gives
and so ; since , we must have , so the answer is .
Without loss of generality, let . Then, using the Law of Sines in triangle , we get , and using the sine addition formula to evaluate , we get .
Then, using the Law of Cosines in triangle , we get , since . So triangle is isosceles, and .
Note: A diagram would be much appreciated; I cannot make one since I'm bad at asymptote. Also, please make this less cluttered :) ~tauros
First, take point outside of so that is equilateral. Then, connect , , and to . Also, let intersect at . , , and (trivially) , so by SAS congruence. Also, , so , and , making also equilateral. (it is isosceles with a angle) by SAS (, , and ), and by SAS (, , and ). Thus, is isosceles, with . Also, , so .
Solution 5 (Ceva)
Noticing that we have three concurrent cevians, we apply Ceva's theorem:
using the fact that and we have:
By inspection, works, so the answer is
Let Using sine rule on , letting we get : Simplifying, we get that from where Simplifying more, we get that , so NOTE: The simplifications were carried out by the product-to-sum and sum-to-product identities ~Prabh1512
|2003 AIME I (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.