the finest problem of apmc

[the arrivals]

Given a circle $\Gamma$ with center $O$ and radius $r$.let $AB$ be a fixed diameter of $\Gamma$. $K$ be a fixed point of segment $AO$. Denote by $t$ the line tangent to $\Gamma$ at $A$. For any chord $CD$ (other than $AB$) passing through $K$. Let $P$ and $Q$ be the points of intersection of lines $BC$ and $BD$ with $t$.
prove that the product $AP\cdot AQ$ remains costant as the chord $CD$ varies.

( dont give too stress on the line that THE CHORD CD VARIES...just fix AB the diameter and chose any chord CD on your own accord).
very very very very very nice problem.100% pure( ) euclidian geometry

[the arrivals]

i am totally astonish isnt there anybody to answer THIS ques??
i was quite eager when i was posting this prob.cuz i wanna see the ideas,but no response!!!

[Moon]

When posting a geometry problem, you should attach a diagram (or at least write it legibly with LaTeX) to make it more attractive to users.
BTW off topic, why don't you introduce yourself here: viewforum.php?f=10

I have solved your problem. But I'll be back after I add a feature in the forum.

[Moon]

Given a circle $\Gamma$ with center $O$ and radius $r$.let $AB$ be a fixed diameter of $\Gamma$. $K$ be a fixed point of segment $AO$. Denote by $t$ the line tangent to $\Gamma$ at $A$. For any chord $CD$ (other than $AB$) passing through $K$. Let $P$ and $Q$ be the points of intersection of lines $BC$ and $BD$ with $t$.
Prove that the product $AP\cdot AQ$ remains costant as the chord $CD$ varies.

Let $C'D'$ be any other chord through $K$. Define $P',Q'$ the same way we did $P,Q$.

Lemma 1: $PCDQ (=\omega_1), P'C'D'Q'(=\omega_2)$ both are cyclic.
Proof: $\angle QPB=90^o-\angle APB=\angle CAB=\angle CDB$. So $PCDQ$ is cyclic. Analogously $P'C'D'Q'$ is also cyclic.

Lemma 2: $P'PCC'$ cyclic.
Proof: $\angle AP'B=\angle C'D'B=\angle C'CB$.

From lemma 2, we have $BC'\cdot BP'=BC\cdot BP$ (power of point). $B$ is on the radical axis of both $\omega_1,\omega_2$. Again from power of point $C'K\cdot KD'=CK \cdot KD'$. So $BK$ is the radical axis of $\omega_1,\omega_2$.

Let $\omega_1, \omega_2$ intersect at $X,Y$. Then from power of point $PA\cdot AQ=YA \cdot AX=P'A \cdot AQ'$.
Proved!

apmc_02.png (81.32KiB)Viewed 3689 times

[the arrivals]

umm...you think complex.at least your diagram is enough to make any novice afraid of geometry.
well but nice solution indeed.