Difference between revisions of "1975 IMO Problems/Problem 3"

Revision as of 15:10, 29 January 2021

Problems

On the sides of an arbitrary triangle $ABC$ , triangles $ABR, BCP, CAQ$ are constructed externally with $\angle CBP = \angle CAQ = 45^\circ, \angle BCP = \angle ACQ = 30^\circ, \angle ABR = \angle BAR = 15^\circ$ . Prove that $\angle QRP = 90^\circ$ and $QR = RP$ .

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 ==Solution==
-If we can find <math>p\ne q</math> such that <math>(a_p,a_q)=1</math>, we're done: every sufficiently large positive integer <math>n</math> can be written in the form <math>xa_p+ya_q,\ x,y\in\mathbb N</math>. We can thus assume there are no two such <math>p\ne q</math>. We now prove the assertion by induction on the first term of the sequence, <math>a_1</math>. The base step is basically proven, since if <math>a_1=1</math> we can take <math>p=1</math> and any <math>q>1</math> we want. There must be a prime divisor <math>u|a_1</math> which divides infinitely many terms of the sequence, which form some subsequence <math>(a_{k_n})_{n\ge 1},\ k_1=1</math>. Now apply the induction hypothesis to the sequence <math>\left(\frac{a_{k_n}}u\right)_{n\ge 1}</math>.
-The above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [https://aops.com/community/p367494]
 == See Also == {{IMO box|year=1975|num-b=2|num-a=4}}

1975 IMO (Problems) • Resources
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Difference between revisions of "1975 IMO Problems/Problem 3"

Revision as of 15:10, 29 January 2021

Problems

Solution

See Also