Albanian IMO TST 2010 Question 1

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Albanian IMO TST 2010 Question 1

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#1 h May 22, 2010, 4:42 PM • 2 Y

Y by Adventure10, Mango247

$ABC$ is an acute angle triangle such that $AB>AC$ and $\hat{BAC}=60^{\circ}$ . Let's denote by $O$ the center of the circumscribed circle of the triangle and $H$ the intersection of altitudes of this triangle. Line $OH$ intersects $AB$ in point $P$ and $AC$ in point $Q$ . Find the value of the ration $\frac{PO}{HQ}$ .

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#2 h May 22, 2010, 5:01 PM • 2 Y

Y by Adventure10, Mango247

Is not hard to prove with angle relations that $\Delta APQ$ is equlilateral. Now the nine point centre lies on the bisector of $\widehat{BAC}$ . Because if $B',A'$ are the middle points of $AB,AC$ and $O'$ the nine point centre, the quadrilateral $O'B'A'C'$ is concyclic $( \widehat{B'O'C'}=120^0)$ ,So $H,O$ are symmetric points about the bisector of $\angle BAC$ , So $OP=HQ$

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Bugi

#3 h May 22, 2010, 8:52 PM • 2 Y

Y by Adventure10, Mango247

It's well known that in any triangle $\angle CAH=\angle BAO$ . Also it's well known that $AH=AO$ iff $\angle BAC=60^\circ$ . Also, APQ is equilateral. So triangles $AQH$ and $APO$ are congruent, and so $PO=QH$

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Ahwingsecretagent

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Ahwingsecretagent

#4 h May 22, 2010, 10:48 PM • 1 Y

Y by Adventure10

This is a nice but recycled problem. See here: http://www.bmoc.maths.org/home/bmo2-2007.pdf .

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#5 h May 23, 2010, 7:03 AM • 2 Y

Y by Adventure10, Mango247

Also in our national olympiad there was a problem from British Math Olympiad. We like very much Britain!

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Obel1x

#6 h May 23, 2010, 11:00 AM • 2 Y

Y by Adventure10, Mango247

Bugi wrote:

It's well known that in any triangle $\angle CAH=\angle BAO$ . Also it's well known that $AH=AO$ iff $\angle BAC=60^\circ$ .

Possibly could you explain me why $\angle CAH=\angle BAO$ and if $\angle BAC=60^\circ \implies AH=AO$ .
Where can I find a proof , any appropriate link ?

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#7 h May 23, 2010, 7:52 PM • 2 Y

Y by Adventure10, Mango247

To answer Obel1x's first question:
$\angle CAH$
$= 90 - \angle C$ (since $AH$ is an altitude)
$= 90 - \frac{1}{2} \angle AOB$ (since $\angle ACB$ is inscribed in the circle with center $O$ )
$=90- \frac{1}{2} (180- 2 \angle BAO)$ (since $\triangle AOB$ is isoceles since $OA=OB=R$ )
$=\angle BAO$
This is true in any triangle

For the second question:
Lemma: Let $X$ be the perpendicular from $O$ to $AB$ . In any triangle, $CH=2OX$
Proof: Extend $AO$ to meet the circumcircle at $D$ , then draw $BD$ . $\angle ABD=90$ since it's inscribed in a semicircle, so $\triangle AOX \sim \triangle ADB$ , and since $AB=2AX, BD=2OX$ . Now we have $CH \parallel BD$ , and $\angle DCB= \angle DAB= 90- \angle ADB= 90 - \angle C= \angle CBH$ . Thus, $DC \parallel HB$ , so $DCHB$ is a parallelogram, so $CH=DB=2OX$ .

Now for the question. Let $Y$ be the perpendicular from $H$ to $AC$ . Then $\angle CHY=60$ so $\triangle CYH$ is a $30-60-90$ right triangle, so $HY= \frac{1}{2} CH= \frac{1}{2}(2OX)=OX$ . Thus, since $\angle AYH= \angle AXO=90$ and $\angle HAY= \angle OAX$ , $\triangle HYA \cong \triangle OXA$ so $AH=AO$ .

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Obel1x

#8 h May 23, 2010, 10:35 PM • 2 Y

Y by Adventure10, Mango247

thanks dgreenb801, your explanation was so needed.

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Anni

#9 h May 24, 2010, 12:31 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

ridgerso u kualifikove?

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#10 h May 24, 2010, 1:06 PM • 2 Y

Y by Adventure10, Mango247

Denote the second intersection $S$ of the line $AI$ ( $I$ is incenter) with the circumcircle $C(O,R)$ of $\triangle ABC$ . Since $m(\widehat {BOC})=m(\widehat {BHC})=m(\widehat{BIC})=120^{\circ}$ obtain that the points $O$ , $H$ , $I$ belong to the circle $C(S,R)$ and the quadrilateral $AOSH$ is a rhombus, $OH\perp AS$ a.s.o.

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#11 h May 24, 2010, 8:03 PM • 2 Y

Y by Adventure10, Mango247

Anni wrote:

ridgerso u kualifikove?

Po, kete vit do ikim vetem 4 vete!

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Anni

#12 h May 25, 2010, 12:06 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

bravo vlla!mu be shume qejfi sinqerisht.ma dergo dhe njehere numrin se kam ndryshuar numer qe te pime ndonje kafe...

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#13 h May 26, 2010, 4:03 PM • 2 Y

Y by Adventure10, Mango247

Anni wrote:

bravo vlla!mu be shume qejfi sinqerisht.ma dergo dhe njehere numrin se kam ndryshuar numer qe te pime ndonje kafe...

0672051285 te enjten mbas ores 5 e gjysem jam ne tirane!

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#14 h May 26, 2010, 4:16 PM • 2 Y

Y by Adventure10, Mango247

Please, english language ... There are and national communities on this site. Thank you.

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#15 h May 26, 2010, 4:19 PM • 2 Y

Y by Adventure10, Mango247

Virgil Nicula wrote:

Please, english language ... There are and national communities on this site. Thank you.

He wrote personal things just for me, and nobody opened so far the requested Albanian Community!

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#16 h May 26, 2010, 9:40 PM • 2 Y

Y by Adventure10, Mango247

ridgers wrote:

Virgil Nicula wrote:

Please, english language ... There are and national communities on this site. Thank you.

He wrote personal things just for me, and nobody opened so far the requested Albanian Community!

Exists and private messages, my dear ....

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#17 h Apr 25, 2025, 7:38 PM

Y by

Kinda classic
Claim 1: $AO=AH$
Proof: Notice that $\angle{BOC}=\angle{BHC}$ and so $BXHC$ is cyclic and so by simple angle chasing we get that $\angle{AXH}=\angle{AHX}$
Claim 2: $\angle{AOP}=\angle{AHQ}$
Proof: $\angle{AOP}=180-\angle{AOH}$ and $\angle{AHQ}=180-\angle{AHO}$
Claim 3: $\angle{PAO}=\angle{HAQ}$
Proof: $\angle{PAO}=\angle{HAQ}=90-\angle{C}$
From the $3$ claims we get that $\triangle APO \cong \triangle AQH$
Thus the result :cool:

:cool:

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