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

(Created page with "On the sides of an arbitrary triangle ABC; triangles ABR;BCP;CAQ are constructed externally with CBP = CAQ = 45°; BCP = ACQ = 30°; ABR = BAR = 15°: Prove that QRP = 90° and Q...")
 
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On the sides of an arbitrary triangle ABC; triangles ABR;BCP;CAQ are
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==Problems==
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constructed externally with CBP = CAQ = 45°; BCP = ACQ =
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30°; ABR = BAR = 15°: Prove that QRP = 90° and QR = RP:
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On the sides of an arbitrary triangle <math>ABC</math>, triangles <math>ABR, BCP, CAQ</math> are constructed externally with <math>\angle CBP = \angle CAQ = 45^\circ, \angle BCP = \angle ACQ = 30^\circ, \angle ABR = \angle BAR = 15^\circ</math>. Prove that <math>\angle QRP = 90^\circ</math> and <math>QR = RP</math>.
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==Solution==
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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>.
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The above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [https://aops.com/community/p367494]
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== See Also == {{IMO box|year=1975|num-b=2|num-a=4}}

Revision as of 15:09, 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 $QR = RP$.

Solution

If we can find $p\ne q$ such that $(a_p,a_q)=1$, we're done: every sufficiently large positive integer $n$ can be written in the form $xa_p+ya_q,\ x,y\in\mathbb N$. We can thus assume there are no two such $p\ne q$. We now prove the assertion by induction on the first term of the sequence, $a_1$. The base step is basically proven, since if $a_1=1$ we can take $p=1$ and any $q>1$ we want. There must be a prime divisor $u|a_1$ which divides infinitely many terms of the sequence, which form some subsequence $(a_{k_n})_{n\ge 1},\ k_1=1$. Now apply the induction hypothesis to the sequence $\left(\frac{a_{k_n}}u\right)_{n\ge 1}$.

The above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]

See Also

1975 IMO (Problems) • Resources
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 4
All IMO Problems and Solutions
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