1975 IMO Problems/Problem 3

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]

1975 IMO (Problems) • Resources
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1975 IMO Problems/Problem 3

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