Firstly, we notice that $\bullet \ TP \ \text{bisects} \ \angle ATB \iff \frac{AT}{BT} = \frac{AP}{PB}$ and let $A'$ and $B'$ be the intersections of the $AT$ and $BT$ with the smaller circle, and let $P'$ be the intersection of $TP$ with the larger circle. Now, by Power of a Point on $A$ and $B$ with respect to the smaller circle, we have that $AP^2 = AA' \cdot AT$, $(*)$ and $BP^2 = BB' \cdot BT$. $(**)$ Also, notice that there is a homothety centered at $T$ which takes the smaller circle to the larger one, so it takes $A'$ to $A$, $B'$ to $B$, and $P$ to $P'$. Hence, $\frac{AT}{BT} = \frac{A'T}{B'T} = \frac{AT - A'T}{BT - B'T} = \frac{AA'}{BB'}$. $(***)$ Now, divide $(*)$ by $(**)$ and we have that $ \left(\frac{AP}{BP} \right)^2 = \frac{AA'}{BB'} \cdot \frac{AT}{BT}$ but if we substitute $(***)$, we have that $\left( \frac{AP}{BP} \right)^2 = \frac{AT}{BT} \cdot \frac{AT}{BT} = \left( \frac{AT}{BT} \right)^2$ $\iff \frac{AP}{BP} = \frac{AT}{BT}$ . QED

The problem has been posted before, including the variant for the circles being externally tangent to each other, which I have no time to search for. http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=345105&hilit=tangent+circles Best regards, sunken rock

Denote $(\omega)$ the smaller circle, and $(\Omega)$ the bigger one. let $(\omega)$ meet $TA$ and $TB$ at $M$ and $N$ respectively. then $\angle{TPM}=\angle{PAT}$ and $\angle{PMT}=\angle{APT}$ then $\angle{PTM}={\angle{PTN}}$ which end the proof.

Dear Mathlinkers, and what can we say when "Two circles are tangent to each other externally at a point T..." Sincerely jean-Louis

jayme wrote: Dear Mathlinkers, and what can we say when "Two circles are tangent to each other externally at a point T..." Sincerely jean-Louis in that case, we choose $B$ nearer to $P$ than $A$, and we call $M$ the intersection of $AT$ and the circle containg $P$. then $PT$ will be the internal bissector of the angle $\angle{BTM}$.

Dear Mathlinkers, my proposition was to precise the nature of the bissector (internal or external)in fonction of the relative position of the two tangent circles. Sincerely Jean-Louis

From an inversion with centre $T$ and radius $\lambda>1$, the larger circle maps to a line $n$ and the smaller circle maps to a line $m$ parallel to it. $A$ and $B$ maps to $A'$ and $B';$ and the line $AB$ maps to the circumcircle of $TA'B'$ which tangents $m$ at $P'$. Let $X$ be a point on $m$ such that $\angle XP'A'<\angle XP'B'$. From alternate segments theorem, $\angle XP'A'=\angle A'TP'$ and also since $m\parallel n,$ we have $\angle XP'A'=\angle P'A'B'$. From the above two equations, we have $\angle A'TP'=\angle P'A'B'=\angle P'TB'$ and when we invert the figure back, the lines $A'T, P'T$ and $B'T$ are mapped to itself and hence, by conformality, $\angle A'TP'=\angle P'TB'\Longrightarrow \angle ATP=\angle PTB\Longrightarrow PT$ bisects $\angle ATB$ as required.

Attachments:

gouthamphilomath wrote: From an inversion with centre $T$ and radius $\lambda>1$, the larger circle maps to a line $n$ and the smaller circle maps to a line $m$ parallel to it. $A$ and $B$ maps to $A'$ and $B';$ and the line $AB$ maps to the circumcircle of $TA'B'$ which tangents $m$ at $P'$. Let $X$ be a point on $m$ such that $\angle XP'A'<\angle XP'B'$. From alternate segments theorem, $\angle XP'A'=\angle A'TP'$ and also since $m\parallel n,$ we have $\angle XP'A'=\angle P'A'B'$. From the above two equations, we have $\angle A'TP'=\angle P'A'B'=\angle P'TB'$ and when we invert the figure back, the lines $A'T, P'T$ and $B'T$ are mapped to itself and hence, by conformality, $\angle A'TP'=\angle P'TB'\Longrightarrow \angle ATP=\angle PTB\Longrightarrow PT$ bisects $\angle ATB$ as required. $\angle A'TP'=\angle P'A'B'=\angle P'TB'$ how is this equal when u did'nt prove it in a procedure ,,may be the $\angle P'B'A'=\angle P'A'B'$ as of same chord intersection at a given point by angular theorem but the next part I can't figure it out cud u explain da

siddharthanand wrote: gouthamphilomath wrote: From an inversion with centre $T$ and radius $\lambda>1$, the larger circle maps to a line $n$ and the smaller circle maps to a line $m$ parallel to it. $A$ and $B$ maps to $A'$ and $B';$ and the line $AB$ maps to the circumcircle of $TA'B'$ which tangents $m$ at $P'$. Let $X$ be a point on $m$ such that $\angle XP'A'<\angle XP'B'$. From alternate segments theorem, $\angle XP'A'=\angle A'TP'$ and also since $m\parallel n,$ we have $\angle XP'A'=\angle P'A'B'$. From the above two equations, we have $\angle A'TP'=\angle P'A'B'=\angle P'TB'$ and when we invert the figure back, the lines $A'T, P'T$ and $B'T$ are mapped to itself and hence, by conformality, $\angle A'TP'=\angle P'TB'\Longrightarrow \angle ATP=\angle PTB\Longrightarrow PT$ bisects $\angle ATB$ as required. $\angle A'TP'=\angle P'A'B'=\angle P'TB'$ how is this equal when u did'nt prove it in a procedure ,,may be the $\angle P'B'A'=\angle P'A'B'$ as of same chord intersection at a given point by angular theorem but the next part I can't figure it out cud u explain da What can't you understand? We have $\angle A'TP'=\angle P'TB'$ and when we invert back, since the angles are preserved, we get $\angle ATP=\angle PTB$ as required.

Please my dear Jayme see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=50&t=155800

This problem completely falls apart by considering a homothety that takes the smaller circle to the larger one about $T$. Then AB gets transformed into $A'B'$ which is tangent to the larger circle at $P'$ and is parallel to $AB$. Therefore, $P'$ is the midpoint of the arc of $AB$ not containing $T$, and $P'PT$ are collinear, so we therefore get that $TP$ is the angle bisector of $\angle ATB$. Cheers, Rofler

Rofler wrote: ... Therefore, $P'$ is the midpoint of the arc of $AB$ not containing $T$, ... Why is this true?

Rofler wrote: This problem completely falls apart by considering a homothety that takes the smaller circle to the larger one about $T$. Then AB gets transformed into $A'B'$ which is tangent to the larger circle at $P'$ and is parallel to $AB$. Therefore, $P'$ is the midpoint of the arc of $AB$ not containing $T$, and $P'PT$ are collinear, so we therefore get that $TP$ is the angle bisector of $\angle ATB$. Cheers, Rofler Just as modularmarc101, I don't understand how you got $P'$ to be midpoint of $AB$. But I have found another solution using your route. Consider a homothety centred at $T$ that maps the smaller circle to the larger one. Let the images be marked with primes. Then, $AB$ is mapped to $A'B'$ tangent to the bigger circle at $P'$ and is parallel to $AB$ by converse of basic proportionality theorem. We have $\angle A'P'A=\angle P'AB=\angle P'TB$ as $A'B'\parallel AB$. Also, by alternate segments theorem, $\angle A'P'A=\angle ATP'$. Hence, $\angle P'TB=\angle ATP'$ Thus proved.

Because $A'B'$ is parallel to the chord $AB$ and tangent to the circle itself, it is tangent at the midpoint by symmetry. Cheers, Rofler

Dear Mathlinkers, because this problem is coming so much on this site, I ask me the following question: can we imagine an similar result when the two circles are secant or not? Sincerely Jean-Louis

Let TP meet large circle at S. by simple angle chasing we have tangent to large circle at S and AB are parallel. the rest is simple angle chasing.

Agung wrote: Two circles are tangent to each other internally at a point $\ T $. Let the chord $\ AB $ of the larger circle be tangent to the smaller circle at a point $\ P $. Prove that the line $\ TP $ bisects $\ \angle ATB $. I just can't believe it's TST... Let $X$ and $Y$ be the points of intersection of the smallest circle with $TB$ and $TA$ respectively. Let $\ell$ be the tangent line to the circles by $T$. Let $\alpha$ be the angle formed by $BX$ and $\ell$ $\Rightarrow \angle YAP=\alpha$ Let $\angle PTX=\theta...(\beta)$ $\Rightarrow \angle XPB=\theta$ Since the angle formed by $TP$ an $\ell$ is $\alpha + \theta$ $\Rightarrow \angle TYP=\alpha + \theta$ In $\triangle AYP:$ $\angle YPA=\theta$ $\Rightarrow \angle YTP=\theta=\angle PTX(\text{By } (\beta)) _\blacksquare$