2006 Romanian NMO Problems/Grade 9/Problem 3
(a) Prove that there are points which fulfill the above conditions.
(b) Prove that .
(a) Note that, given the positions of and , can be in exactly two places. However, is on segment , and and are on the same side of line , so must be on the same side of as . This shows that, given the positions of , , and , we can determine the position of . Also note that must be in the circumcircle of , which means that must be outside triangle . Without loss of generality, assume that and are on opposite sides of . Now it suffices to show that for ever positioning of , , , and such that , there exists a on such that is cyclic. This is simple; merely extend so that it intersects the circumcircle of again at . An example of such an arrangement of points is shown below; , , , , and .
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Solution 2 (suli) Given the circle , construct equilateral inside with , on its interior such that . Then let the circle centered at with radius intersect circle at . Finally, extend to meet at and we are done!
(b) Let and . I shall now find all of the angles of . We know that , since is equilateral. Therefore . Triangle is isosceles, so we have that , and . Since and are inscribed angles that intercept minor arc , . Now note that . Since is isosceles, . We know that and , so . We are now able to use the Law of Sines on . However, we would only get an equation involving , so if we were to use it to find , we must find the length of . This can easily be done using the Law of Sines on :
And now we use the Law of Sines! On !
Using the Law of Sines on gives that , so it suffices to show that
Canceling out 's and rearranging shows that this statement is equivalent to
Using the sine and cosine addition formulae gives that this statement is equivalent to
Cross-multiplying and using double-angle formulae gives that this statement is equivalent to
Canceling out like terms and dividing both sides by gives that this statement is equivalent to
This is a simple rearrangement of the Pythagorean Identity, which is true. We can work backwards to get that .
Soln. 2 (fmasroor) Let <CBP=2x, so that <BPC=<BCP=90-x (no fractions). Then <ADC=180-<ABC=120-2x, <APD=180-<APB-<BPC=x+30, so then <PAD=x+30. Then PD=PA so se need to prove that ODA is equilateral where O is the center of ABCD. However, since <DAP=<DPA=x+30 DP=AP and so ABPD is a kite; <ABD=60/2=30 and then <DOA=2*30=60, so then ODA is equilateral. Done.
Solution 3 (suli) Notice that the circumcenter of is , so . (Note that is equilateral.) Hence , so is equilateral and . Now if we let then it is easily seen via angle chasing that , so as desired.