In the plane of a triangle $ABC,$ in its exterior$,$ we draw the triangles $ABR, BCP, CAQ$ so that $\angle PBC = \angle CAQ = 45^{\circ}$, $\angle BCP = \angle QCA = 30^{\circ}$, $\angle ABR = \angle RAB = 15^{\circ}$. Prove that a.) $\angle QRP = 90\,^{\circ},$ and b.) $QR = RP.$
Problem
Source: IMO ShortList, Netherlands 1, IMO 1975, Day 1, Problem 3
Tags: geometry, trigonometry, Triangle, angles, IMO, IMO 1975
Virgil Nicula
13.11.2005 01:17
In the plane of the triangle $ABC$, in its exterior we draw the triangles $BCP$, $CAQ$, $ABR$ so that ${m(\widehat {BCP})=m(\widehat ACQ})=30^{\circ}$, $m(\widehat {CBP})=m(\widehat {CAQ})=x$ and $m(\widehat {BAR})=m(\widehat {ABR})=60^{\circ}-x$, where $x\in (0,60^{\circ})$. Prove that: $\blacksquare \ 1^{\circ}.\ m(\widehat {PRQ})=2x;$ $\blacksquare \ 2^{\circ}.RP=RQ$ (the triangle $PRQ$ is isosceles in the vertex $R$). Remark. For $x=45^{\circ}$ we obtain the problem from IMO-1975.
LawOfSigns
20.08.2008 00:00
I've worked on this problem some, and it seems to help to trisect the 45 degree angles and bisect the 30 degree angles to have all angles as 15 degree angles or multiples of them, after constructing points exterior to ABC to be the right angled vertices of three isosceles right triangles with hypotenuses as sides of the triangle ABC. It seems that a bit more angle chasing would prove that QRP is right (I haven't been able to do this), and then one could simply use the symmetry of the diagram to finish off the problem. Does anyone have a complete solution?
The QuattoMaster 6000
20.08.2008 01:35
orl wrote: In the plane of a triangle $ ABC,$ in its exterior$ ,$ we draw the triangles $ ABR, BCP, CAQ$ so that $ \angle PBC = \angle CAQ = 45\,^{\circ},$ $ \angle BCP = \angle QCA = 30\,^{\circ},$ $ \angle ABR = \angle RAB = 15\,^{\circ}.$ Prove that a.) $ \angle QRP = 90\,^{\circ},$ and b.) $ QR = RP.$
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$.
sunken rock
06.10.2008 13:00
Construct the equilateral triangle ATB, with T and R on the same side of AB. It is easy to see that the triangles ACQ and ATR are similar and, consequently, a clockwise rotation of 45 deg around A followed by an omothety of ratio AR/AT will map Q to C and R to T. Same way the triangles BCP and BTR are similar and an anticlockwise rotation of 45 deg around B followed by an omothety of ratio BR/BT will map P to C and R to T. Consequently, QR and RP are equal and perpendicular. Note: Above proof (or similar) given at the time in the Romanian "Gazeta Matematica". Best regards, sunken rock
xeroxia
31.07.2009 12:05
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$.
FantasyLover
04.12.2010 06:14
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. $\blacksquare$
v_Enhance
26.04.2023 00:42
Complete the equilateral triangle $ATB$. It is not hard to see that $\triangle TAR \sim \triangle CAQ$, since $\angle ATR = 30^{\circ}$ and $\angle TAR = 45^{\circ}$. Using the fact that spiral similarities come in pairs, we discover that $\triangle TAC \sim \triangle RAQ$. [asy][asy] size(7cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair R = rotate(15)*(Sin(15)/Sin(150))*(A-B)+B; pair P = rotate(-45)*(Sin(30)/Sin(75))*(C-B)+B; pair Q = rotate(45)*(Sin(30)/Sin(75))*(C-A)+A; pair T = rotate(60)*(A-B)+B; draw(T--A--C--cycle, blue); draw(R--A--Q--cycle, red); draw(R--B--T); draw(Q--C--P--B); draw(A--B--C); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$R$", R, dir(110)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$T$", T, dir(T)); /* Source generated by TSQ */ [/asy][/asy] Thus \[ \angle ARQ = \angle ATC \text{ and } RQ = \frac{RA}{TA} \cdot TC. \]In exactly the same fashion we can show that $\angle RBP = \angle BTC$ and $RP = \frac{RB}{TB} \cdot TC$. This implies the result.
shendrew7
02.03.2024 19:25
Rotate $A$ clockwise 90 degrees about $R$ to point $K$. Then $KRB$ is equilateral, inducing \[\triangle BAK \sim \triangle BCP \implies \triangle BKP \sim \triangle BAC.\] It suffices to prove $\triangle RAK \sim \triangle RQP \iff \triangle RAQ \cong \triangle RKP$, which holds as \[\angle RAQ = 60 + \angle B = \angle RKP, \quad QC = \frac{CA}{BC} \cdot PB = PK, \quad AR = KR. \quad \blacksquare\] [asy][asy] size(250); pair A, B, C, P, Q, R, K; A = dir(230); B = dir(310); C = dir(110); P = dir(-45)*.51764*(C-B) + B; Q = dir(45)*.51764*(C-A) + A; R = dir(15)*.51764*(A-B) + B; K = dir(-90)*(A-R) + R; filldraw(R--A--Q--cycle^^R--K--P--cycle, lightgreen); draw(A--B--C--cycle^^A--R--B--P--C--Q--A--K--P--R--Q^^R--K--B); draw(B--K--P--cycle^^B--A--C--cycle, red+linewidth(1.5)); dot("$A$", A, dir(225)); dot("$B$", B, dir(315)); dot("$C$", C, dir(90)); dot("$P$", P, dir(0)); dot("$Q$", Q, dir(180)); dot("$R$", R, dir(270)); dot("$K$", K, dir(135)); [/asy][/asy]
OronSH
28.03.2024 18:59
Construct $X$ on the same side of $BR$ as $A$ satisfying $\triangle BRX$ equilateral, and similarly construct $Y$ on the same side of $AR$ as $B$ satisfying $\triangle ARY$ equilateral. Then from $\angle ARB=150^\circ$ and $AR=BR,$ we see $\triangle ARX,\triangle BRY$ are right isosceles triangles with right angle at $R.$ Now angle chasing gives $\angle ABX=\angle BAY=45^\circ$ and $\angle BAX=\angle ABY=30^\circ,$ so $\triangle BAX\sim\triangle BCP$ and $\triangle ABY\sim\triangle ACQ$ similarly oriented. Then by spiral similarity we get $\triangle BXP\sim\triangle BAC\sim\triangle YAQ,$ similarly oriented. But there is a $90^\circ$ rotation at $R$ sending $B$ to $Y$ and $X$ to $A,$ and this must send $P$ to $Q,$ so we are done.
YaoAOPS
14.08.2024 04:22
Claim: $(ARQ), (PQC), (BPR)$ concur at a point $X$. Proof. Follows since $\measuredangle AQC + \measuredangle CPB + \measuredangle BRA = 0^\circ$. $\blacksquare$ Claim: $X$ lies on $(ABC)$. Proof. Note that \[ 60^\circ = \measuredangle RAB + \measuredangle CAQ = \measuredangle RXB + CXQ = \measuredangle RXQ + \measuredangle CXB = \measuredangle RAQ + \measuredangle CXB \]which implies that $\measuredangle CXB = \measuredangle CAB$. $\blacksquare$ As such, it follows that \[ \measuredangle ARQ + \measuredangle PRB = \measuredangle AXQ + \measuredangle PXB = \measuredangle AXB + \measuredangle PXQ = \measuredangle ACB + \measuredangle PCQ = 60^\circ. \]so $\measuredangle PRQ = 90^\circ$. We can similarly angle chase that \[ \measuredangle BPR + \measuredangle QPC = \measuredangle BXR + \measuredangle QXC = \measuredangle BXC + \measuredangle QXR = \measuredangle BAC + \measuredangle QAR = 60^\circ \]which implies the result.