Difference between revisions of "Talk:2021 USAMO Problems/Problem 1"

Revision as of 16:08, 10 September 2021

I spent some time poking around at this with the help of Geogebra, and after a few failed attempts (trying to figure out something from the relationship $\cot \alpha + \cot \beta + \cot \gamma = \cot \alpha \cot \beta \cot \gamma$ when $\alpha + \beta + \gamma = \pi$ and using applications of the Cosine rule to try to get something in terms of the triangle's angles) I realized that the point of concurrency is the intersection of the circumcircles of the three rectangles. The main difficulty after that is proving that this point is also the intersection point of the diagonals.

Clearly, if we construct two rectangles on $AB$ and $BC$ and then circumscribe them, the circles meet in two points, $B$ and $O$ . Equally clearly, the angles $|A_1OB_2| = |A_1AB_2| = |A_1BB_2| = \alpha$ and $|B_1OC_2| = |B_1BC_2| = |B_1CC_2| = \beta$ .

Now we can construct a circumcircle for the triangle $AOC$ which will intersect the perpendiculars from $AC$ at $A, C$ in $A_2, C_1$ , and $|A_2AC_1| = |A_2OC_1| = |A_2CC_1| = \gamma$ . Now *if* the triples $(A_1,O,B_2), (C_1, O, B_2), (B_1, O, C_2)$ are collinear, then the angle $|B_20C_1| = |A_2OC_1| = \gamma$ and by construction $\alpha + \beta + \gamma = \pi$ and we're done. But I can't figure out how to prove the collinearity (that is, to prove that the intersection of the circumcircles is on the diagonal).

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Difference between revisions of "Talk:2021 USAMO Problems/Problem 1"

Revision as of 16:08, 10 September 2021