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 $MN$.
Problem
Source: CSEMO 2005-2
Tags: search, geometry, analytic geometry, geometry solved
shobber
19.07.2005 14:17
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......
nttu
19.07.2005 14:40
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 .]
mecrazywong
21.07.2005 04:09
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.
Ashegh
30.08.2005 19:23
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?
shobber
31.08.2005 10:24
Yes, I love to solve a problem with pure geometry. But the perpendicular lines in this problem really reminded of the coordinate system.
trainer14
31.08.2005 17:25
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
Ashegh
11.09.2005 15:52
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 : :
darij grinberg
11.09.2005 16:14
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
Ashegh
13.09.2005 17:50
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 .
AlastorMoody
11.07.2019 20:48
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