One from Junior Balkan 2004

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

One from Junior Balkan 2004

G H J

Reveal topic tags

perpendicular bisector

geometry proposed

L

G H BBookmark kLocked kLocked NReply

Source: JbMO 2004, PROBLEM 2

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3004 posts

#1 h Jul 20, 2004, 8:04 PM • 3 Y

Y by Relkuyh, Adventure10, Mango247

Let $ABC$ be an isosceles triangle with $AC=BC$ , let $M$ be the midpoint of its side $AC$ , and let $Z$ be the line through $C$ perpendicular to $AB$ . The circle through the points $B$ , $C$ , and $M$ intersects the line $Z$ at the points $C$ and $Q$ . Find the radius of the circumcircle of the triangle $ABC$ in terms of $m = CQ$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1633 posts

liyi

#2 h Jul 21, 2004, 12:38 AM • 3 Y

Y by Relkuyh, Adventure10, Mango247

I think the key to this problem is to prove that $Q$ is on $AM$ , where $M$ is the midpoint of $BC$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

6555 posts

#3 h Jul 21, 2004, 11:46 AM • 3 Y

Y by Relkuyh, Adventure10, Mango247

A full solution:

Indeed the radius of the circumcircle of triangle ABC is $\frac{2}{3}m$ .

Proof. The line Z is a symmetry axis of the isosceles triangle ABC; hence, it is an angle bisector, a median, an altitude and a diameter of the circumcircle simultaneously. Therefore, both the centroid G and the circumcenter O of triangle ABC lie on this line Z. Since the line Z is an angle bisector, actually the angle bisector of the angle BCA, we have < BCQ = < GCM. Also, < BQC = < GMC (since < BQC = < BMC because the point Q lies on the cirle through B, C and M). Therefore, the triangles BCQ and GCM are similar, and $\frac{CQ}{CM}=\frac{BC}{GC}$ . Since CQ = m, we have $\frac{m}{CM}=\frac{BC}{GC}$ .

But the midpoint N of the side AB also lies on the line Z, and the centroid G of triangle ABC satisfies $GC=\frac{2}{3}\cdot NC$ , so we obtain

$\frac{m}{CM}=\frac{BC}{\frac{2}{3}\cdot NC}=\frac{3}{2}\cdot \frac{BC}{NC}$ .

Since the point O is the circumcenter of triangle ABC, it lies on the perpendicular bisector of the side AC, i. e. on the perpendicular to AC at the point M. Therefore, < CMO = 90°. Together with < CNB = 90°, we get < CNB = < CMO. Also, since the line Z bisects the angle BCA, we have < NCB = < MCO. Thus, the triangles CNB and CMO are similar, and $\frac{BC}{NC}=\frac{OC}{MC}$ , so that

$\frac{m}{CM}=\frac{3}{2}\cdot \frac{BC}{NC}=\frac{3}{2}\cdot \frac{OC}{MC}=\frac{3}{2}\cdot \frac{OC}{CM}$ ,

and $m=\frac{3}{2}\cdot OC$ , so $OC=\frac{2}{3}\cdot m$ . But OC is just the radius of the circumcircle of triangle ABC. Hence we are done.

Darij

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

165 posts

#4 h Jul 21, 2004, 11:55 AM • 5 Y

Y by mdelyakova, Relkuyh, Adventure10, Math_legendno12, Mango247

Shortlists solution of Problem 2

Let P be the center of the circle $k_1$ through B, C and M, O be the circumcenter of ABC and K be the midpoint of MC. Since AC=BC, the center O lies on CH. Let KP intersect CH in point L. Since KP and OM are perpendicular to AC, then KP || OM. From MK = KC it follows that OL = CL. On the other hand OP is perpendicular to BC, hence angle LOP = angle COP = 90º - angle BCH. Also we have angle OLP = angle CLK = 90º - angle ACH. Since ABC isosceles and angle BCH = angle ACH, then angle LOP = angle OLP and LP = OP. Since CP = PQ we obtain that angle CLP = angle QOP and CL = OQ. Thus we have CL = LO = OQ, so CO = 2/3CQ. Finally for the radius R of the circumcircle of ABC we obtain R = 2/3m.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

648 posts

Kuba

#5 h Jul 23, 2004, 10:41 PM • 3 Y

Y by Relkuyh, Adventure10, Mango247

this is my solution for problem 2:
the points M,Q,A,C are concycle so if we denote by ${P}=CQ\cap AM$ we get that $PM\cdot PA=PQ\cdot PC$ (*).
We know that $PM=\displaystyle\frac 13 MA$ , $PA=\displaystyle\frac23 MA$ , $\displaystyle CP=\frac23 CC'$ where ${C'}=CQ\cap AB$ .
We know that $MA^2=m_a^2=\displaystyle\frac{2(b^2+c^2)-a^2}4$ (1) and
$CC'^2=m_c^2=\displaystyle\frac{2(b^2+a^2)-c^2}4$ (2)

If we denote by $\alpha=\angle ACQ$ and knowing that $\displaystyle R=\frac{abc}{4S}$ and using (*),(1),(2) and $CP+PQ=m$ with some easy computations we get the desired result.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

6555 posts

#6 h Oct 30, 2005, 11:07 AM • 2 Y

Y by Relkuyh, Adventure10

This is a solution of this problem posted by vedran6 in a different thread:

It is obvious that the BCMQ is cyclic and because of that is and we have that angle <CQM=<CBM (1) let $CZ\cap BM=\left(T\right)$ ; and $CZ\cap AB=\left(D\right)$ it is obvious that is $CT=\frac23\cdot CD$ and because of AC=BC we have <ACQ=<BCT (2) , from (1) and (2) we have that trinagle CQM is similar to triangle CBT. From similarity we have that is $\frac{CQ}{CB}=\frac{CM}{CT}$ which is equal to $CQ=\frac{3\cdot CB\cdot CB}{4\cdot CD}$ (3).
It is well known that in every triangle is $R=\frac{a\cdot b\cdot c}{4P}$ so we use it at this triangle ABC and got $R=\frac{BC\cdot BC}{2\cdot CD}$ (4) . After dividing (3) and (4) we get $CQ=1,5\cdot R$ and the problem is finished

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

7054 posts

#7 h Oct 30, 2005, 5:30 PM • 5 Y

Y by Akababa, Sowmitra, Relkuyh, polrivers, Adventure10

Here is my solution (I think that it differs from the other preceding solutions):

Let $w=C(O,R)$ be a circumcircle of the triangle $ABC$

and let $U$ be the middlepoint of the segment $[AM]$ . From

$\widehat {QCM}\equiv \widehat {QCB}\Longrightarrow QM=QB=QA$

(the point $Q$ is the circumcenter for the triangle $ABM$ ). From

$QU\perp AC,\ MO\perp AC\Longrightarrow QU\parallel MO\Longrightarrow$

$\frac 23 =\frac{CM}{CU}=\frac{CO}{CQ}=\frac{R}{m}\Longrightarrow R=\frac 23 m$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

IstekOlympiadTeam

542 posts

IstekOlympiadTeam

#8 h May 8, 2015, 1:10 PM • 4 Y

Y by Relkuyh, polrivers, Adventure10, Mango247

My Solution
Let $MC=AM=x$ Then $BC=2x$ From sinus we have $\longrightarrow$ $\frac{2x}{cos\alpha}=2R$ Where $R$ be the circumradius of $\triangle{ABC}$ and $\angle{ACQ=BCQ=\alpha}$ . Then from $\triangle{BCQ}$ we have by sinus $\frac{m}{sin(\alpha+\beta)}=\frac{2Rcos\alpha}{sin(2\alpha+\beta)}$ Where $\angle{MBC}=\beta$ . Then we have $m=\frac{2Rcos\alpha\cdot sin(\alpha+\beta)}{sin(2\alpha+\beta)}$ By some trigonometric identies we have $m=\frac{Rsin(2\alpha+\beta)+Rsin\beta}{sin(2\alpha+\beta)}$ $\longrightarrow$ $m=R+\frac{R\cdot sin\beta}{sin(2\alpha+\beta)}$ Loking at $\triangle{MBC}$ by sinus we have $\longrightarrow$ $\frac{x}{sin\beta}=\frac{2x}{sin(2\alpha+\beta)}$ $\longrightarrow$ $\frac{sin\beta}{sin(2\alpha+\beta)}=\frac{1}{2}$ so $m=R+\frac{R\cdot sin\beta}{sin(2\alpha+\beta)}=\frac{3R}{2}$ $\longrightarrow$ $R=\frac{2m}{3}$

Anar Abbas Istek Lyceum

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

80 posts

#9 h Jan 25, 2016, 7:26 PM • 2 Y

Y by Relkuyh, Adventure10

Let $K$ $\in [CA$ , $|MA|=|AK|$ , $M \neq K$ and $O$ is $\triangle ABC$ 's circumcircle center

$|CM|=|MA|=|AK|$ $\Rightarrow$ $|MK|=|AC|=|BC|$ $.$ $\dotsm$ $(i)$

$C,M,Q,B$ is circularly $\Rightarrow$ $\angle QMA = \angle QBC$ $\dotsm$ $(ii)$

$ABC$ is isoscles triangle and $D$ is $BA$ 's midpoint $\Rightarrow$ $\angle ACD = \angle BAD$ $.$ Also $B,Q,M,C$ is circularly $\Rightarrow$ $|MQ|=|BQ|$ $\dotsm$ $(iii)$

Combine $(i),(ii),(iii)$ $\Rightarrow$ $\triangle QMK \equiv \triangle QBC$

İn here $\angle QKM=\angle QCB=\angle OCA=\angle OAC$ $\Rightarrow$ $OA \parallel QK$ $\Rightarrow$ $\frac{AC}{KC} = \frac{OC}{QC}$ $.$ Also $|AC|=2|MC|$ , $|KC|=3|MC|$ $\Rightarrow$ $\frac{OC}{QC} = \frac{2}{3}$ $\Rightarrow$ $|OC|=\frac{2m}{3}$ $Q.E.D$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

134 posts

KRIS17

#10 h Dec 18, 2018, 4:23 AM • 2 Y

Y by Adventure10, Mango247

Let length of side $CB$ = $x$ and length of $QM = a$ . We shall first prove that $QM = QB$ .

Let $O$ be the circumcenter of $\triangle ACB$ which must lie on line $Z$ as $Z$ is a perpendicular bisector of isosceles $\triangle ACB$ .

So, we have $\angle ACO = \angle BCO = \angle C/2$ .

Now $MQBC$ is a cyclic quadrilateral by definition, so we have:
$\angle QMB = \angle QCB = \angle C/2$ and,
$\angle QBM = \angle QCM = \angle C/2$ , thus $\angle QMB = \angle QBM$ , so $QM = QB = a$ .

Therefore in isosceles $\triangle QMB$ we have that $MB = 2 QB \cos C/2 = 2 a \cos C/2$ .

Let $R$ be the circumradius of $\triangle ACB$ .
So we have $CM = x/2 = R \cos C/2$ or $x = 2R \cos C/2$

Now applying Ptolemy's theorem in cyclic quadrilateral $MQBC$ , we get:

$m . MB = x . QM + (x/2) . QB$ or,

$m . (2 a \cos C/2) = (3/2 x) . a = (3/2).(2Ra) \cos C/2 = 3Ra \cos C/2$ or,

$R = (2/3)m$

Attachments:

This post has been edited 1 time. Last edited by KRIS17, Dec 18, 2018, 4:32 AM
Reason: Added pic

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a