Prove an eq triangle

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Prove an eq triangle

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Source: Indian RMO 2000 Problem 1

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Rushil

#1 h Oct 26, 2005, 9:44 AM • 2 Y

Y by Adventure10, Mango247

Let $AC$ be a line segment in the plane and $B$ a points between $A$ and $C$ . Construct isosceles triangles $PAB$ and $QAC$ on one side of the segment $AC$ such that $\angle APB = \angle BQC = 120^{\circ}$ and an isosceles triangle $RAC$ on the other side of $AC$ such that $\angle ARC = 120^{\circ}.$ Show that $PQR$ is an equilateral triangle.

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#2 h Jan 22, 2006, 6:01 AM • 2 Y

Y by Adventure10, Mango247

By the law of cosine: $RP^2=AP^2+AR^2-2 \cdot AP \cdot AR \cdot cos{\angle{PAR}}$ . Since $\angle{PAR}=\angle{RAC}+\angle{PAB}=30^o+30^o=60^o$ , so $RP^2=AP^2+AR^2-AP \cdot AR$ . Similarly we have $RQ^2=CQ^2+CR^2-CQ \cdot CR$ . Since $AR=CR$ , so:

$RP^2-RQ^2=AP^2-CQ^2+AR \cdot (CQ-AP)$
$=(AP+CQ)(AP-CQ)+AR \cdot (CQ-AP)$
$=(AP-CQ)(AP+CQ-AR)$ .

Because: $AP \cdot \sin{30^o}+ CQ \cdot \sin{30^o}=\frac{1}{2}AC=AR \sin{30^o}$ , so $AP+CQ=AR$ , thus $AP+CQ-AR=0$ , thus $PR^2=QR^2$ , $PR=QR$ .

$PQ^2=PA^2+QC^2-2 \cdot PA \cdot QC \cdot \cos{120^o}=PA^2+QC^2+PA \cdot QC$ .
Let $RP^2-PQ^2$ , we can get:
$RP^2-PQ^2=(CQ+AR)(CQ-AR+AP)=0$

Therefore $PR=QP$ , thus $PR=RQ=QP$ , so $\triangle{PQR}$ is an equilateral triangle.

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3498 posts

#3 h Jan 22, 2006, 6:03 AM • 2 Y

Y by Adventure10, Mango247

Another interesting thing is that the circumcircles of $\triangle{APB}$ , $\triangle{BQC}$ and $\triangle{ARC}$ meet at one point.

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delta

#4 h Jan 23, 2006, 10:40 AM • 1 Y

Y by Adventure10

let P1,Q1,R1 be the points symmetrical to P,Q,R with respect to AC. Then PAP1, QAQ1 are equilateral , and PP1R,QQ1R,PQr1 are equal. So PR=RQ=QR.

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#5 h Jan 24, 2006, 12:27 AM • 2 Y

Y by Adventure10, Mango247

isn't this just napoleon's theorem, except the triangle is degenerate
(the one relating to the first fermat point?)

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3498 posts

#6 h Jan 24, 2006, 9:58 AM • 3 Y

Y by math_and_me, Adventure10, Mango247

Altheman wrote:

isn't this just napoleon's theorem, except the triangle is degenerate
(the one relating to the first fermat point?)

Yes. If we construct equilateral triangles with side $AB$ , $BC$ , $AC$ , then $P$ , $Q$ , $R$ are the centres of the triangles, thus $\triangle{PQR}$ is equilateral due to Napoleon's Theorem.

But I don't know how it is related the Fermat point...?

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#7 h Jan 25, 2006, 12:32 AM • 2 Y

Y by Adventure10, Mango247

the circumcircles of the equilateral triangles drawn on the sides of the triangle in napolean's theorem concur at the first fermat point, only for acute triangles though, so you are correct that it would not be the case for this triangle, but that is how i remember it

just curious, if you wrote that on the actual exam, would you get full credit, your solution would be like a sentence

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3498 posts

#8 h Jan 25, 2006, 3:35 AM • 1 Y

Y by Adventure10

Altheman wrote:

just curious, if you wrote that on the actual exam, would you get full credit, your solution would be like a sentence

I'd still use the law of cosine. I don't think that Napolean's theorem on a straight line is acceptable...

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7054 posts

#9 h Jan 25, 2006, 2:03 PM • 3 Y

Y by math_and_me, Adventure10, Mango247

Indeed, the Altheman's remark is nicely: this problem is a particular case (just on the limit) of the well-known Napoleon's problem.

A equivalent enunciation. Let $ABCD$ be a rhombus with $m(\widehat {ABC})=60^{\circ}$ . For a point $M\in (BD)$ denote the points $N\in (AB)$ , $P\in (AD)$ so that $MN\parallel AD$ and $MP\parallel AB$ . Then the triangle $CNP$ is equilaterally.

Proof using the complex numbers.

A generalization of the proposed problem.

If $PA=PQ$ , $QB=QC$ , $RA=RC,\ m(\widehat {APB})=m(\widehat {BQC})=m(\widehat {ARC})=\phi$ ,
then $PR=PQ\Longleftrightarrow RP=RQ\Longleftrightarrow QP=QR\Longleftrightarrow \phi=120^{\circ}$
and in this case the triangle $PQR$ is equilaterally.

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#10 h Oct 22, 2011, 4:34 PM • 1 Y

Y by Adventure10

Brute forced with coordinate geometry.
I didn't realise it was a degenerated form of Napoleon's theorem (although in the rmo, I don't think it would be possible to simply state this)

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4380 posts

#11 h Nov 23, 2011, 4:28 PM • 4 Y

Y by SAT1001, math_comb01, Adventure10, Mango247

$\{D\}\in AP\cap CQ$ ; see that $ARCD$ is a rhombus, with $m(\widehat{RAD})=60^\circ$ , $PBQD$ a parallelogram, so $AP=DQ, DR=AR, \angle DAR=\angle CDR$ , hence $\triangle PAR\equiv \triangle QDR$ and, a $60^\circ$ rotation around $R$ will map each other, i.e. $PR=RQ, m(\widehat{PRQ})=60^\circ$ , done.

Best regards,
sunken rock

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#12 h Feb 18, 2013, 12:24 PM • 2 Y

Y by Adventure10, Mango247

Let $AP=PB=a$ and $BQ=QC=b$ and $AR=RC=c$ .

In $\Delta APB$ , by sine rule $AB=\dfrac{AP\sin{120^\circ}}{\sin{30^\circ}}=a\sqrt{3}$
Similarly, $BC=b\sqrt{3}$ and $AC=c\sqrt{3}$ .
But $AB+BC=AC$ which gives $a+b=c...(1)$ or $a^2+c^2-ac=b^2+c^2-bc...(2)$

In $\Delta QRC$ by cosine rule $QR^2=b^2+c^2-bc$ and in $\Delta PRA$ $PR^2=a^2+c^2-ac$ . Hence from (2) $PR=QR$ and $PR^2=QR^2=a^2+b^2+ab$

Now by applying sine rule and cosine rule in $\Delta QRC$ and $\Delta PRA$ for angles $QRC$ and $ARP$ respectively we(after a lot of calculations) we find $\sin{(\angle{QRC}+\angle{ARP})}=\sqrt{3}/2$ which gives $\angle{PRQ}=60^\circ$ and we are done.

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#13 h Feb 18, 2013, 2:47 PM • 2 Y

Y by Adventure10, Mango247

Complex numbers, $p=\frac{a-b\epsilon}{1-\epsilon}, q=\frac{b-c\epsilon}{1-\epsilon}, r=\frac{c-a\epsilon}{1-\epsilon}$ , so $r+p\epsilon+q\epsilon^2=0$ ...

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2502 posts

ATGY

#14 h May 5, 2021, 9:55 AM

Y by

We can start off by labelling the angles. As $\triangle{APB}$ and $\triangle{BQC}$ are isosceles triangles, $\angle{PAB} = \angle{PBA} = 30^{\circ}$ and $\angle{QBC} = \angle{QCB} = 30^{\circ}$ . Similarly, $\triangle{RAC}$ is isosceles. This means that $\angle{RAC} = \angle{RCA} = 30^{\circ}$ .

Now, extend line $\overline{PA}$ to point $D$ where $D$ is in line with point $C$ . Using the fact that angles on a line add up to $180^{\circ}$ , $\angle{BPD} = \angle{BQD} = 60^{\circ}$ and $\angle{PBQ} = 120^{\circ}$ . Since the angles of a quadrilateral add up to $360^{\circ}$ , $\angle{D} = 120^{\circ}$ . This implies that $DPQB$ is a parallelogram.

Let $PD = x$ and $PB = y$ . As $DPQB$ is a parallelogram, $DQ = y$ and $BQ = x$ . Since $\triangle{APB}$ and $\triangle{BQC}$ are isosceles triangles, $PA = y$ and $QC = x$ . We know that $\angle{DAR} = \angle{DCR}$ and $\angle{CDA} = \angle{ARC}$ . This means that $ADCR$ is a parallelogram. As $DA = DC = x + y$ , $CR = AR = x + y$ .

Using Law of Cosines on $\triangle{PDQ}$ (to find side PQ), we see that:
$PQ^2 = x^2 + y^2 - 2xy\cos{120^{\circ}} = x^2 + y^2 + xy$ Therefore, $PQ = \sqrt{x^2 + y^2 + xy}$ .

We can use the Law of Cosines on triangles $PAR$ and $QCR$ as well (to find sides PR and QR).
$PR^2 = y^2 + (x + y)^2 - 2y(x + y)\cos{60^{\circ}} = x^2 + y^2 + xy$ Therefore, $PR = \sqrt{x^2 + y^2 + xy}$ .

$QR^2 = x^2 + (x + y)^2 - 2x(x + y)\cos{60^{\circ}} = x^2 + y^2 + xy$ Therefore, $QR = \sqrt{x^2 + y^2 + xy}$ .

All the sides of triangle $PQR$ are equal, implying that it is equilateral. $\blacksquare$

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#15 h Feb 25, 2024, 6:47 AM

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Consider the composition $\Gamma = \text{Rot}(R, 120^{\circ}) \ \circ \ \text{Rot}(Q,120^{\circ}) \ \circ \ \text{Rot}(P, 120^{\circ})$ $\Gamma$ has to be a translation, since the angles add to $360^\circ$ , and it takes $A \rightarrow B \rightarrow C \rightarrow A$ so it has a fixed point $A$ . Thus $\Gamma$ must be the identity transformation.

Now, note that the inverse function of the rotation $\text{Rot}(A,\alpha)$ is simply the rotation $\text{Rot}(A,-\alpha)$ , therefore, composing $\Gamma$ on the left with inverse of $\text{Rot}(R,120^{\circ})$ , we get that (since $\Gamma$ is just the identity), that
$\text{Rot}(Q,120^{\circ}) \ \circ \text{Rot}(P,120^{\circ}) = \text{Rot}(R,-120^{\circ})$ Therefore, by the composition law of rotations, $R$ must be the point of intersection of the lines which make an angle of $60^{\circ}$ with $PQ$ at $P,Q$ . Hence, $\angle RPQ = \angle RQP = 60^{\circ}$ , and we’re through. $\square$

This post has been edited 3 times. Last edited by Rijul saini, Feb 25, 2024, 7:19 AM

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