Problem 4, Iberoamerican Olympiad 2011

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Problem 4, Iberoamerican Olympiad 2011

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23 posts

#1 h Oct 2, 2011, 3:49 PM • 4 Y

Y by ImSh95, Adventure10, Mango247, Spiritpalm

Let $ABC$ be an acute-angled triangle, with $AC \neq BC$ and let $O$ be its circumcenter. Let $P$ and $Q$ be points such that $BOAP$ and $COPQ$ are parallelograms. Show that $Q$ is the orthocenter of $ABC$ .

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1347 posts

WakeUp

#2 h Oct 2, 2011, 5:11 PM • 5 Y

Y by Wizard_32, ImSh95, Adventure10, Mango247, Spiritpalm

The problem is just begging for vectors!

We have $\bold{q}=\bold{p}+\overrightarrow{PQ}=\bold{b}+\overrightarrow{BP}+\overrightarrow{PQ}=\bold{b}+\bold{a}+\bold{c}$ so $Q$ is the orthocentre.

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6591 posts

#3 h Oct 2, 2011, 5:17 PM • 2 Y

Y by ImSh95, Adventure10

It has been some time since I last allowed myself to indulge in geometry. Please verify the following solution.

Since $BOAP$ is a parallelogram and $BO=AO=R$ , where $R$ is the circumradius of $\triangle ABC$ , $BOAP$ is a rhombus of side length $R$ . So $BP=AP=R$ . Since $COPQ$ is a parallelogram with $CO=P$ , we know that the opposite side $QP=R$ . Then $QP=BP=AP=R$ . Without loss of generality, suppose that $\angle B>\angle A$ (generality is not lost as the following arguments can be tailored for the case $A>B$ by replacing each $B$ with $A$ , and each $A$ with $B$ ; the problem states that $AC\neq BC\iff \angle A\neq\angle B$ , so that case needs not be considered). Note that $\angle APC,\angle BPC=\frac{\angle APB}{2}=\frac{\angle AOB}{2}=\angle C$ . Also, $\angle BOC = 2\angle A\implies \angle POC = \angle C + 2\angle A\implies \angle OPQ = \angle B - \angle A$ .
We can now say that $\begin{cases}\angle BPQ = \angle C + \angle A - \angle B=\pi - 2\angle B\implies BQ= \sqrt{BP^2 + PQ^2-(BP)(QP)\cos \angle BQP}=2R\cos B\\ \angle APQ = \angle C + \angle B - \angle A=\pi-2\angle A\implies AQ= \sqrt{AP^2 + PQ^2-(AP)(QP)\cos \angle AQP}=2R\cos A\end{cases}$ .
Let $H$ be the orthocenter of $\triangle ABC$ . It is well-known that $\begin{cases}BH = 2R\cos \angle B\\ AH =2R\cos \angle A\end{cases}$ . There are up to two points that satisfy these distances from $A,B$ ; i.e. the interesections of the circle with center $B$ and radius $2R\cos \angle B$ , and the circle with center $A$ and radius $2R\cos \angle A$ . These two points are clearly the same point or on different sides of $AB$ . Since $PB=PQ=PA=R$ , $Q$ is inside $\triangle ABC$ , and since $\triangle ABC$ is acute, $Q$ is the orthocenter.

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#4 h Oct 2, 2011, 7:19 PM • 4 Y

Y by DominicanAOPSer, ImSh95, Adventure10, and 1 other user

Let $M$ be midpoint of $AB$ . We have $CQ \perp AB$ since $CQ \parallel OP$ and clearly $OP$ is the perpendicular bisector of $AB$ .

Also $CQ = OP$ and $OP = 2OM = CH$ and therefore $CQ = CH$ . Hence $Q=H$ .

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#5 h Oct 4, 2011, 7:33 AM • 2 Y

Y by ImSh95, Adventure10

Clearly $OA=OB$ , hence $P$ is on the perpendicular to $AB$ through its midpoint $M$ , and $P$ is the symmetric of $O$ with respect to $AB$ . Let $N$ be the midpoint of $CP$ , clearly inside $ABC$ (otherwise $O$ would be outside $ABC$ ), because $ABC$ is acute. Note now that
$MN^2=\frac{ON^2+PN^2}{2}-\frac{OP^2}{4}=\frac{OC^2+OP^2}{4}-\frac{PC^2}{8}+\frac{PN^2}{2}-\frac{OP^2}{4}=\frac{OC^2}{4},$
hence $N$ is at a distance equal to $\frac{R}{2}$ from $M$ , on the line halfway between the perpendicular bisector of $AB$ and the altitude from $C$ (by Thales' theorem and because $N$ is the midpoint of $PC$ ), and inside $ABC$ . Hence $N$ is the nine-point center, and $Q$ the symmetric of $O$ with respect to it, thus the orthocenter.

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semy

#6 h Dec 12, 2011, 2:57 PM • 4 Y

Y by Vladimir_Djurica, ImSh95, Adventure10, Mango247

This problem is just Servois' theorem itself.

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#7 h Mar 4, 2012, 2:59 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

semy wrote:

This problem is just Servois' theorem itself.

I've looked for this, but couldn't find it, where can I find a source which talks about this theorem?

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#8 h Feb 4, 2018, 1:22 AM • 2 Y

Y by ImSh95, Adventure10

This problem is actually trivial if you know that $2MO=CH$ . Because of this $COPH$ is a parallelogram and so is $COPQ$ , thus $H=Q$

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#9 h Apr 1, 2018, 4:05 AM • 4 Y

Y by ImSh95, Adventure10, Mango247, AlexCenteno2007

Solution

Attachments:

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Arkmmq

#10 h Apr 1, 2018, 9:23 AM • 4 Y

Y by Vladimir_Djurica, ImSh95, Adventure10, Spiritpalm

Easy by complex numbers.....just set ( $ABC$ ) as the unit circle and notice that $p=a+b$ and $q=a+b+c$ ..done.

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9775 posts

jayme

#11 h Jun 27, 2020, 10:34 AM • 5 Y

Y by ImSh95, Mango247, Mango247, Mango247, Spiritpalm

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%207.pdf p. 54...

Sincerely
Jean-Louis

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#12 h Jun 5, 2021, 1:19 AM • 1 Y

Y by ImSh95

Let $H$ be the orthocenter of $ABC$ . Let us prove $COPH$ is a parallelogram.

Claim 1: $HP = OC$

Notice that $BP=OA=OC=OB=PA$ . Thus $O$ and $P$ are reflections of each other over line $AB$ . Realizing the circle centered at $P$ passing through $A$ (say, $\omega)$ has the same radius as the circle centered at $P$ passing through $A$ (say, $\omega')$ . The conclusion follows $\omega$ and $\omega'$ are reflections of each other about $AB$ . Recall, the reflection of $H$ about $AB$ lies on $\omega$ so, $H \in \omega'$ . Hence, $OH = OA = OB$ . Proving the claim.

Claim 2: $CH \parallel OP$

Clearly, $O$ and $P$ lie on the perpendicular bisector of $AB$ so, $OP \perp AB$ . Clearly, $CH \perp AB$ so $CH \parallel OP$ . Proving the claim.

Claim 1 and Claim 2 are sufficient to ensure $COPH$ is a parallelogram. So, we are done.

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primesarespecial

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primesarespecial

#13 h Jun 22, 2021, 12:40 PM • 3 Y

Y by Vladimir_Djurica, ImSh95, Spiritpalm

Well Just use complex numbers
Let $a,b,c$ lie on the unit circle .
I don't like vectors (Alone) , so we use the fact that the diagonals of a parallelogram bisect each other to get

$p=a+b$ and $q=a+b+c$ ,
Thus $q$ is the orthocenter .
Moral of the story , Don't be sticky with Synthetic solutions.
Like complex numbers finish this in a second or two.

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#14 h Dec 5, 2022, 12:23 AM • 3 Y

Y by ImSh95, Mango247, Spiritpalm

We have $OA=OB=OC$ because $O$ is the circumcenter of $ABC$ , so the paralellograms imply that $PA=PB=PQ$ , hence $P$ is the circumcenter of $ABQ$ , this gives $\angle AQB=180^\circ -\frac{\angle BPA}{2}=180^\circ -\frac{\angle AOB}{2}=180^\circ -\angle ACB.$ Observe that $OA=OB$ and $PA=PB$ imply that $OP\perp AB$ , so $CQ\perp AB$ . Hence $Q$ is the orthocenter of $ABC$ .

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#15 h Dec 6, 2022, 4:20 PM • 2 Y

Y by ImSh95, Spiritpalm

βλέπε σχήμα:

Attachments:

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Ianis

#16 h Jun 16, 2023, 6:10 PM • 2 Y

Y by ImSh95, Spiritpalm

$Q=P+C=A+B+C$

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#17 h Aug 17, 2023, 2:57 PM • 2 Y

Y by Spiritpalm, Vahe_Arsenyan

Clarly $P$ is the reflection of $O$ wrt $AB.$ So $Q$ is a point in the $C-altitude$ such that $OP=CQ.$ Clarly the orthocenter satisfies this lenght condition, and thus we are donne.

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#18 h Aug 17, 2023, 4:07 PM

Y by

So we use complex numbers...
Set $(ABC)$ as the unit circle.
We got $p+0=a+b$ so $p=a+b$ .
We got $c+p=0+q$ so $q=c+p=a+b+c$ which means $Q$ is the orthocenter.

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cursed_tangent1434

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cursed_tangent1434

#19 h Feb 25, 2024, 8:47 AM

Y by

Complex is actually quite immediate but we give a synthetic solution anyways. First, we show $Q$ lies on $\overline{CH}$ .

Claim : The line $\overline{CQ}$ is infact the $C$ -altitude of $\triangle ABC$ .
We note that,
$\begin{align*} \measuredangle QCO &= \measuredangle POC \\ &= \measuredangle POB + \measuredangle BOC\\ &= \measuredangle OPA + \measuredangle BOC \\ &= \measuredangle AOP + \measuredangle BOC\\ &= 90 + \measuredangle OAB + \measuredangle BOC \\ &= \measuredangle ACB + 2\measuredangle BAC \end{align*}$ Also, we can verify that
$\begin{align*} \measuredangle HCO & = \measuredangle BCA + 2 \measuredangle ACO\\ &= \measuredangle BCA + \measuredangle AOC\\ &= \measuredangle BCA + 2 \measuredangle ABC \\ &= \measuredangle ACB + 2 \measuredangle BAC \end{align*}$ from which it is clear that $\measuredangle QCO = \measuredangle HCO$ which implies that $Q$ lies on $\overline{CH}$ as claimed.

Now, it is well known that (and easy to prove) if $M_C$ is the midpoint of $AB$ and $N_C$ is the midpoint of $CH$ , then $CN_CM_CO$ is a parallelogram. First note that since in a parallelogram, diagonals bisect each other, $PO$ passes through the midpoint of $AB$ (which is $M_C$ ).

Thus,
$QC = OP = 2OM_C = 2CN_C=CH$ But combining this with the previous fact, concludes that indeed $Q$ is the orthocenter of $\triangle ABC$ and we are done.

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#20 h Feb 25, 2024, 11:21 AM • 1 Y

Y by Spiritpalm

Clearly the point $Q$ is unique. Let $H$ be the orthocentre of $\Delta ABC$ . Our goal is to prove that $COPH$ is a parallelogram.
Observe that if $F$ is the midpoint of $AB$ , then $OF \perp AB \implies P \text{ is the reflection of } O \text{about line} AB$ . Now clearly $OP \parallel CH$ , and also $OP = 2R\cos C = CH$ where $R$ is the circumradius of $\Delta ABC$ , from standard trigonometry results, so we are done.

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