# 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]