bisected segment

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zhaoli

417 posts

zhaoli

#1 h Jul 17, 2005, 3:25 PM • 1 Y

Y by Adventure10

Circle $C$ (with center $O$ ) does not have common point with line $l$ . Draw $OP$ perpendicular to $l$ , $P \in l$ . Let $Q$ be a point on $l$ ( $Q$ is different from $P$ ), $QA$ and $QB$ are tangent to circle $C$ , and intersect the circle at $A$ and $B$ respectively. $AB$ intersects $OP$ at $K$ . $PM$ , $PN$ are perpendicular to $QB$ , $QA$ , respectively, $M \in QB$ , $N \in QA$ . Prove that segment $KP$ is bisected by line $MN$ .

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shobber

3498 posts

shobber

#2 h Jul 19, 2005, 11:17 AM • 2 Y

Y by Adventure10, Mango247

After making a sketch, I have a strong feeling that this problem can be solved by using analytic geometry. But I am having troubles finding the equations of the tangants......

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nttu

486 posts

nttu

#3 h Jul 19, 2005, 11:40 AM • 2 Y

Y by Adventure10, Mango247

Grobber gave a solution long time ago . Try to search with the keyword : " circle_and_tangent "

[Moderator edit: In fact, see http://www.mathlinks.ro/Forum/viewtopic.php?t=5198 .]

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mecrazywong

606 posts

mecrazywong

#4 h Jul 21, 2005, 1:09 AM • 2 Y

Y by Adventure10, Mango247

Let QA and QB intersect OP at X and Y respectively. Then the problem is equivalent to the fact P,X,K,Y are harmonic conjugate(why

), which is easy to justify.

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Ashegh

858 posts

Ashegh

#5 h Aug 30, 2005, 4:23 PM • 1 Y

Y by Adventure10

dear shober i love solving problem by using analytic geometry.but i think :
solving th e problems with out using analytic geometry is more powerfull and nice.
dont u think so?

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shobber

3498 posts

shobber

#6 h Aug 31, 2005, 7:24 AM • 2 Y

Y by Adventure10, Mango247

Yes, I love to solve a problem with pure geometry. But the perpendicular lines in this problem really reminded of the coordinate system.

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trainer14

2 posts

trainer14

#7 h Aug 31, 2005, 2:25 PM • 1 Y

Y by Adventure10

draw PX perpendicular to AB ( X is a point on AB)
We know that O, A, P, Q, B are cyclic.
Then apply Simson's theorem from point P to the triangle ABQ.
K, N, M are on the same line.
angle QOP = angle QAP = angle NXP = angle KPX (because OQ are parallel to XP)
So KP is bisected by line NM

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Ashegh

858 posts

Ashegh

#8 h Sep 11, 2005, 12:52 PM • 2 Y

Y by Adventure10, Mango247

dear zaoli ithink u question is wrang

.
be cause u know by moving Q on line L the locus of point X (inter section ofAQ andKP)is only a point .
and it is stable.but by by changing Qthe locus of Kwont be stable .and it change on the line OP.
then how can line QA bisect KP

:

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darij grinberg

6555 posts

darij grinberg

#9 h Sep 11, 2005, 1:14 PM • 1 Y

Y by Adventure10

ashegh wrote:

dear zaoli ithink u question is wrang

.
be cause u know by moving Q on line L the locus of point X (inter section ofAQ andKP)is only a point .
and it is stable.but by by changing Qthe locus of Kwont be stable .and it change on the line OP.
then how can line QA bisect KP

:

Why on earth is X stable?

And yes, K is stable. This is something most solvers showed in their solutions as an auxiliary fact. Please read the solutions posted by others before posting such comments, since they could clear up your doubts.

And, by the way, it's MN and not QA that bisects KP.

Darij

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Ashegh

858 posts

Ashegh

#10 h Sep 13, 2005, 2:50 PM • 2 Y

Y by Adventure10, Mango247

ya ya i finally found where was the mistake.
but itsreallyanicequestion .isnt it?
because as u know the locus of point X is fixed(X is the inter section ofKP and MN).
and again the lucas of point K,will be stable.
because K is the pole of line L.and for each point like Q on L we have:
the polar of Q will go through the pole of L wich is K.and it means k is stable

.

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AlastorMoody

2125 posts

AlastorMoody

#11 h Jul 11, 2019, 5:48 PM • 2 Y

Y by Adventure10, Mango247

Lemma: Let $\Delta ABC$ be an isosceles triangle with $AB=AC$ , $H$ as its orthocenter and $D$ as $A-$ antipode. Take a point $F$ on $\odot(ABC)$ . Let $FD\cap BC=G$ . Then the simson line of $F$ is parallel to $GH$
Proof: Assume, $F$ on minor arc $AB$ . Let $N$ be foot from $F$ to $AB$ and $DF$ intersects $F-$ simson line at $O$ . $\angle FON=\angle 180^{\circ} -\angle FAN-\angle FAC=180^{\circ}-2\angle FAD=\angle FGH \qquad \blacksquare$

Using the fact, of orthocentric bisection by simson line, the problem is proved

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