Croatian mathematical olympiad, day 2 problem 3

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Croatian mathematical olympiad, day 2 problem 3

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perpendicular bisector

projective geometry

power of a point

geometry unsolved

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#1 h Apr 10, 2011, 4:28 PM • 3 Y

Y by FCBarcelona, Adventure10, Mango247

Let $K$ and $L$ be the points on the semicircle with diameter $AB$ . Denote intersection of $AK$ and $AL$ as $T$ and let $N$ be the point such that $N$ is on segment $AB$ and line $TN$ is perpendicular to $AB$ . If $U$ is the intersection of perpendicular bisector of $AB$ an $KL$ and $V$ is a point on $KL$ such that angles $UAV$ and $UBV$ are equal. Prove that $NV$ is perpendicular to $KL$ .

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#2 h Apr 10, 2011, 10:40 PM • 2 Y

Y by Adventure10, Mango247

Let $KL$ , the perpendicular bisector of $AB$ cut $AB$ , the circumcircle of $~$ $UAB$ at $C$ , $M$
$\angle AVM=\angle BVM$ $~$ and since $~$ $UM$ is its diameter, $~$ $\angle UVM=90^{\circ}$
Thus, $~$ $(A,B/N,C)=1$ $~$ and $N$ is the foot of $T$ -altitude of $~$ $\triangle TAB$
Here $N$ is the intersection of $VM$ and $AB$

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Zhero

#3 h Apr 17, 2011, 10:13 PM • 4 Y

Y by Reynan, Adventure10, Mango247, namanrobin08

Let $M$ be the midpoint of $AB$ and $P$ be the intersection of $KL$ and $AB$ . $K$ and $L$ are the feet of the altitudes $B$ to $AT$ and $A$ to $BT$ , respectively, so $KMNL$ lies on the nine-point circle of $\triangle ABC$ . Hence, $(CU)(CV) = (CB)(CA) = (CL)(CK) = (CN)(CM)$ , so $MNUV$ is cyclic, so $\angle UVN = 180^{\circ} - \angle UMN = 90^{\circ}$ .

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#4 h Apr 26, 2011, 2:12 PM • 2 Y

Y by Adventure10, Mango247

Another solution using harmonics. we use the notations of
Let $O$ be the midpoint of $AB$ , and let $(O)$ be the circle with diameter $AB$ . Now, since $T$ lies on the polar of $C$ wrt $(O)$ , and since $N$ is the point on $OC$ from such that $TN \perp OC$ , therefore, $N$ also lies on the polar of $C$ , leading us to $(AB,NC) = -1$ . Now, since $O$ is the midpoint of $AB$ , therefore, $CN \cdot CO = CA \cdot CB$ . But $CA \cdot CB = CU \cdot CV$ , therefore, $N,O,U,V$ are concylic. Since $UO \perp AB$ , therefore, $UV \perp VN$ , and we're done.

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#5 h Jun 23, 2011, 10:14 AM • 2 Y

Y by Adventure10, Mango247

Zhero wrote:

Let $M$ be the midpoint of $AB$ and $P$ be the intersection of $KL$ and $AB$ . $K$ and $L$ are the feet of the altitudes $B$ to $AT$ and $A$ to $BT$ , respectively, so $KMNL$ lies on the nine-point circle of $\triangle ABC$ . Hence, $(CU)(CV) = (CB)(CA) = (CL)(CK) = (CN)(CM)$ , so $MNUV$ is cyclic, so $\angle UVN = 180^{\circ} - \angle UMN = 90^{\circ}$ .

Do you mean: so $KMNL$ lies on the nine-point circle of $\triangle ABT$ . Hence, $(PU)(PV) = (PB)(PA) = (PL)(PK) = (PN)(PM)$ ?

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jayme

#6 h Jun 24, 2011, 2:04 PM • 1 Y

Y by Adventure10

Dear Mathlinkers,
(O) is the circle with diameter AB
(1) the circle passing through A, U, V, B
(2) the Euler's circle of TAB
(3) the circle with diameter UN
By applying the Monge's theorem (radical axis of three circles are concurrent)
we are done.
Sincerely
Jean-Louis

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RSM

#7 h Jun 27, 2011, 9:21 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Under inversion wrt $C$ with power $CA.CB$ , $A,B$ interchanges and also $U,V$ interchanges. Suppose, $N$ goes to $N'$ . We know inversion preserves cross-ratio. So $(CN,BA)=(\infty N',AB)=-1$
So $N'$ is the mid-point of $AB$ . $N'U\perp AC$ . So $NV\perp KC$ since $N'U$ is anti-parallel to $NV$ wrt $\angle ACK$ . So done.

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#8 h Apr 13, 2020, 12:37 PM • 3 Y

Y by Mango247, Mango247, Mango247

$\angle KVA=180^{\circ}- \angle AVU=\angle UBA=\angle BAU=\angle BVU$ , because quadrilateral $BAVU$ is cyclic. From sine law in triangles $ANC$ and $BNC$ we get: $AN=b\cos \alpha$ and $BN=a\cos \beta$ and $\frac{AN}{BN}=\frac{b \cos \alpha}{a \cos \beta}=\frac{\sin \beta \cos \alpha}{\cos \beta \sin \alpha}$ We also can say that $\frac{AV}{AK}=\frac{\sin \angle AKV}{\sin \angle KVA}$ and $\frac{BV}{BL}=\frac{\sin \angle BLV}{\sin \angle BVL}$ , so $\frac{AV}{BV}=\frac{AK\frac{\sin \angle AKV}{\sin \angle KVA}}{BL\frac{\sin \angle BLV}{\sin \angle BVL}}=\frac{AK\sin \beta}{BL\sin \alpha}$ . Now, $AK=c \cdot \cos \alpha$ and $BL=c \cdot \cos \beta,$ so $\frac{AK\sin \beta}{BL\sin \alpha}=\frac{\sin \beta \cos \alpha}{\cos \beta \sin \alpha}=\frac{AN}{BN}$ . So, $VN$ is bisector of $\angle AVB \implies \angle AVN=\angle BVN$ . Since we now have two pairs of equal angles, that means that $\angle NVL=90^{\circ}$

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#9 h Oct 26, 2024, 5:40 PM

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This problem also appears in January, 2014 Crux Mathematicorum OC103.

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