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Difference between revisions of "1975 IMO Problems/Problem 3"
 
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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>.
 
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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==Solution 1==
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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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Consider <math> X</math> and <math> Y</math> so that <math> \triangle CQA\sim \triangle CPX</math> and <math> \triangle CPB\sim \triangle CQY</math>. Furthermore, let <math> AX</math> and <math> BY</math> intersect at <math> F</math>. Now, this means that <math> \angle PBC = 45 = \angle QAC = \angle PXC</math> and <math> \angle PCB = 15 = \angle QCA = \angle PCX</math>, so <math> \triangle BPC\cong \triangle XPC</math>. Thus, <math> XC = BC</math> and <math> \angle BCX = 30 + 30 = 60</math>, so <math> \triangle BCX</math> is equilateral. Similarly is <math> \triangle ACY</math>. Yet, it is well-known that the intersection of <math> BY</math> and <math> AX</math>, which is <math> F</math>, must be the Fermat Point of <math> \triangle ABC</math> if <math> \triangle AYC</math> and <math> \triangle BXC</math> are equilateral. Now, <math> \angle PBX = 60 - 45 = 15</math>. Similarly, <math> \angle PXB = 15</math>, so <math> \triangle BRA\sim \triangle BPX</math>, so a spiral similarity maps <math> \triangle BPR\sim \triangle BXA</math>. This implies that <math> \angle BRP = \angle FAB</math>. Similarly, <math> \angle ARQ = \angle FBA</math>, so <math> \angle BRP + \angle ARQ = \angle FBA + \angle FAB = 180 - \angle AFB = 60</math>. Then, <math> \angle PRQ = (180 - 15 - 15) - 60 = 90</math>. We also realize that <math> \frac {AX}{RP} = \frac {AB}{RB} = \frac {AB}{AR} = \frac {BY}{RQ}</math>. Now, <math> Y</math> is a rotation of <math> A</math> <math> 60</math> degrees around <math> C</math> and <math> B</math> is the same rotation of <math> X</math> around <math> C</math>, so <math> AX</math> maps to <math> YB</math> from this rotation, so <math> AX = YB</math>. It follows that <math> RP = RQ</math>.
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The above solution was posted and copyrighted by The QuattoMaster 6000. The original thread for this problem can be found here: [https://aops.com/community/p1229257]
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==Solution 2==
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Let <math> \triangle ABT</math> be the equilateral triangle constructed such that <math> T</math> and <math> R</math> are on the same side. <math> \triangle RTB \sim \triangle PCB \sim \triangle QCA</math>. We have <math> \frac {AT}{AC} = \frac {AR}{AQ}</math> from similarity. Also we have <math> \angle TAC = \angle RAQ</math> . So <math> \triangle ACT \sim \triangle ARQ</math>. Then <math> \angle ATC = \angle ARQ = m</math> and <math> \frac {AR}{AT} = \frac {RQ}{TC}</math>. Similar calculations for <math> B</math>. We will have <math> \angle BTC = \angle BRP = 60-m</math> and <math> \frac {BR}{AT} = \frac {RP}{TC}</math>. Also from the question we have <math> AR = BR</math>. So <math> \angle PRQ = 180 - (60-x) - (60-x) -m -(60-m) = 2x</math> and <math> PR = RQ</math>.
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The above solution was posted and copyrighted by xeroxia. The original thread for this problem can be found here: [https://aops.com/community/p1581352]
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==Solution 3==
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Define <math>X</math> to be a point such that <math>\triangle AQC</math> is directly similar to <math>\triangle AXB</math>.
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Then, it is trivial to show that <math>\angle AXR=60^{\circ}</math> and that <math>\angle RXB=45^{\circ}</math>; that is, <math>AX=AR=XR=RB</math> and <math>\triangle RBX</math> is a right isosceles triangle. If we prove that <math>\triangle RPQ</math> is similar to <math>\triangle RBX</math>, then we will be done.
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According to the property of spiral similarity, it suffices to prove that <math>\triangle RBP\sim\triangle RXQ.</math>
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Since <math>\triangle AQC\sim \triangle AXB</math>, we have <math>\triangle AQX\sim\triangle ACB</math>, and this gives <math>\angle RBP=60^{\circ}+\angle  CBA=60^{\circ}+\angle QXA=\angle RXQ</math>. It remains to prove that <math>\frac{BP}{RB}=\frac{XQ}{RX}</math>.
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As <math>RB=RX</math>, we must prove that <math>BP=XQ</math>. From Law of Sines on <math>\triangle BPC</math>, we have <math>BP=\frac{a}{2\cos 15^{\circ}}</math>. Since <math>\frac{XQ}{AX}=\frac{a}{c}</math>, it remains to prove that <math>AX=AR=\frac{c}{2\cos 15^{\circ}}</math>, which is easily verified. We are done. <math>\blacksquare</math>
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The above solution was posted and copyrighted by fantasylover. The original thread for this problem can be found here: [https://aops.com/community/p2107142]
  
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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]
 
  
 
== See Also == {{IMO box|year=1975|num-b=2|num-a=4}}
 
== See Also == {{IMO box|year=1975|num-b=2|num-a=4}}

Latest revision as of 15:14, 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 1

Consider $X$ and $Y$ so that $\triangle CQA\sim \triangle CPX$ and $\triangle CPB\sim \triangle CQY$. Furthermore, let $AX$ and $BY$ intersect at $F$. Now, this means that $\angle PBC = 45 = \angle QAC = \angle PXC$ and $\angle PCB = 15 = \angle QCA = \angle PCX$, so $\triangle BPC\cong \triangle XPC$. Thus, $XC = BC$ and $\angle BCX = 30 + 30 = 60$, so $\triangle BCX$ is equilateral. Similarly is $\triangle ACY$. Yet, it is well-known that the intersection of $BY$ and $AX$, which is $F$, must be the Fermat Point of $\triangle ABC$ if $\triangle AYC$ and $\triangle BXC$ are equilateral. Now, $\angle PBX = 60 - 45 = 15$. Similarly, $\angle PXB = 15$, so $\triangle BRA\sim \triangle BPX$, so a spiral similarity maps $\triangle BPR\sim \triangle BXA$. This implies that $\angle BRP = \angle FAB$. Similarly, $\angle ARQ = \angle FBA$, so $\angle BRP + \angle ARQ = \angle FBA + \angle FAB = 180 - \angle AFB = 60$. Then, $\angle PRQ = (180 - 15 - 15) - 60 = 90$. We also realize that $\frac {AX}{RP} = \frac {AB}{RB} = \frac {AB}{AR} = \frac {BY}{RQ}$. Now, $Y$ is a rotation of $A$ $60$ degrees around $C$ and $B$ is the same rotation of $X$ around $C$, so $AX$ maps to $YB$ from this rotation, so $AX = YB$. It follows that $RP = RQ$.

The above solution was posted and copyrighted by The QuattoMaster 6000. The original thread for this problem can be found here: [1]

Solution 2

Let $\triangle ABT$ be the equilateral triangle constructed such that $T$ and $R$ are on the same side. $\triangle RTB \sim \triangle PCB \sim \triangle QCA$. We have $\frac {AT}{AC} = \frac {AR}{AQ}$ from similarity. Also we have $\angle TAC = \angle RAQ$ . So $\triangle ACT \sim \triangle ARQ$. Then $\angle ATC = \angle ARQ = m$ and $\frac {AR}{AT} = \frac {RQ}{TC}$. Similar calculations for $B$. We will have $\angle BTC = \angle BRP = 60-m$ and $\frac {BR}{AT} = \frac {RP}{TC}$. Also from the question we have $AR = BR$. So $\angle PRQ = 180 - (60-x) - (60-x) -m -(60-m) = 2x$ and $PR = RQ$.

The above solution was posted and copyrighted by xeroxia. The original thread for this problem can be found here: [2]

Solution 3

Define $X$ to be a point such that $\triangle AQC$ is directly similar to $\triangle AXB$.

Then, it is trivial to show that $\angle AXR=60^{\circ}$ and that $\angle RXB=45^{\circ}$; that is, $AX=AR=XR=RB$ and $\triangle RBX$ is a right isosceles triangle. If we prove that $\triangle RPQ$ is similar to $\triangle RBX$, then we will be done.

According to the property of spiral similarity, it suffices to prove that $\triangle RBP\sim\triangle RXQ.$

Since $\triangle AQC\sim \triangle AXB$, we have $\triangle AQX\sim\triangle ACB$, and this gives $\angle RBP=60^{\circ}+\angle CBA=60^{\circ}+\angle QXA=\angle RXQ$. It remains to prove that $\frac{BP}{RB}=\frac{XQ}{RX}$.

As $RB=RX$, we must prove that $BP=XQ$. From Law of Sines on $\triangle BPC$, we have $BP=\frac{a}{2\cos 15^{\circ}}$. Since $\frac{XQ}{AX}=\frac{a}{c}$, it remains to prove that $AX=AR=\frac{c}{2\cos 15^{\circ}}$, which is easily verified. We are done. $\blacksquare$

The above solution was posted and copyrighted by fantasylover. The original thread for this problem can be found here: [3]


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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