Art of Problem Solving
AoPS Online AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy AoPS Academy
Small live classes for advanced math
and language arts learners in grades 1-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Stay ahead of learning milestones! Enroll in a class over the summer!
geometry
B
No tags match your search
M
G
Topic
First Poster
Last Poster
No topics here!
H
J
H
Albanian IMO TST 2010 Question 1
ridgers   16
N Apr 25, 2025 by ali123456
$ABC$ 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 $\frac{PO}{HQ}$.
16 replies
ridgers
May 22, 2010
ali123456
Apr 25, 2025
J
Albanian IMO TST 2010 Question 1
G H J
Reveal topic tags
ratio
geometry
circumcircle
parallelogram
incenter
rhombus
geometry unsolved
L
G H BBookmark kLocked kLocked NReply
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ridgers
713 posts
ridgers
#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 $\frac{PO}{HQ}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Kenny O
116 posts
Kenny O
#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 $OP=HQ$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Bugi
1857 posts
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 $PO=QH$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ahwingsecretagent
151 posts
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 .
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ridgers
713 posts
ridgers
#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!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Obel1x
524 posts
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 ?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
dgreenb801
1896 posts
dgreenb801
#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 $AH=AO$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Obel1x
524 posts
Obel1x
#8 h  May 23, 2010, 10:35 PM • 2 Y
Y by Adventure10, Mango247
thanks dgreenb801, your explanation was so needed.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Anni
74 posts
Anni
#9 h  May 24, 2010, 12:31 PM • 3 Y
Y by Adventure10, Mango247, and 1 other user
ridgerso u kualifikove?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Virgil Nicula
7054 posts
Virgil Nicula
#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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ridgers
713 posts
ridgers
#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!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Anni
74 posts
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...
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ridgers
713 posts
ridgers
#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!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Virgil Nicula
7054 posts
Virgil Nicula
#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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ridgers
713 posts
ridgers
#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!
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Virgil Nicula
7054 posts
Virgil Nicula
#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 ....
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ali123456
52 posts
ali123456
#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:
Z K Y
N Quick Reply
G
H
=
a