2003 AIME I Problems/Problem 10
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 .
and multiplying through by 2 and applying the double angle formulas gives
and so ; since , we must have , so the answer is .
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
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