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 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]

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

Contents

Problems

Solution 1

Solution 2

Solution 3

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