In triangle $ABC$ with $AB\neq AC$, let its incircle be tangent to sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. The internal angle bisector of $\angle BAC$ intersects lines $DE$ and $DF$ at $X$ and $Y$, respectively. Let $S$ and $T$ be distinct points on side $BC$ such that $\angle XSY=\angle XTY=90^\circ$. Finally, let $\gamma$ be the circumcircle of $\triangle AST$.
(a) Show that $\gamma$ is tangent to the circumcircle of $\triangle ABC$.
(b) Show that $\gamma$ is tangent to the incircle of $\triangle ABC$.
Circle is tangent to circumcircle and incircle
 Kazi_Zareer
 Posts: 86
 Joined: Thu Aug 20, 2015 7:11 pm
 Location: Malibagh,Dhaka1217
Circle is tangent to circumcircle and incircle
We cannot solve our problems with the same thinking we used when we create them.
Re: Circle is tangent to circumcircle and incircle
Without loss of generality, let $Y$ be closer to $A$ than $X$. The crux move is to show that $AD$ bisects $\angle SAT$. Let $L$ be the midpoint of $XY$. Clearly $X,Y,S,T$ all lie on a circle centered at $L$. Denote this circle by $\ell$.
Lemma : $CY\perp AX$ and $BX\perp AX$
Proof : Well known. Chase angles.
Lemma : The incircle and $\ell$ are orthogonal.
Proof : It suffices to show that $IY\cdot IX=r^2=IE^2$. Let $IC$ intersect $ED$ at $M$. Then since $\angle XYC = \angle XMC =90^{\circ}$, $XYMC$ is a cyclic quadrilateral. Therefore $IY\cdot IX = IM\cdot IC = IE^2$. We are done.
Lemma : $LA$ bisects $\angle SAT$.
Proof : Being radius of the same circle, $LS=LT$. Therefore it suffices to prove that $L$ lies on the circumcircle of $\triangle AST$.Let $AX$ intersect $BC$ at $P$. It therefore suffices to show that $LP\cdot LA=LT^2=LY^2$.
$$XY=AXAY=c\cos\dfrac{A}{2} b\cos\dfrac{A}{2}$$
implying $LY=\dfrac{1}{2}XY = \dfrac12(cb)\cos\dfrac{A}{2}$.
$$LA=LY+YA=\dfrac12(b+c)\cos\dfrac{A}{2}$$
meaning $LP\cdot LA = (LAAP)\cdot LA = LA^2  LA\cdot AP = \dfrac14(b+c)^2\cos\dfrac{A}{2}\dfrac12(b+c)\cos\dfrac{A}{2}\cdot \dfrac{2bc}{b+c}\cos\dfrac{A}{2}$ which calculates to $\dfrac14(\cos\dfrac{A}{2})^2(bc)^2=LY^2$
thus ending the proof.
For part (a), let the circumcircle of $AST$ intersect $AC$ and $AC$ at $W$ and $Z$, It suffices to show that $WZ$ is parallel to $BC$, which is obvious since by our previous lemma, $\angle WAS = \angle TAZ$.
For part (b), invert with respect to $\ell$. This leaves the incircle right where it is, but takes the circumcircle of $AST$ to line $BC$, since $L$ also lies on circle $AST$. Since $BC$ and the incircle are tangent, we are done.
Lemma : $CY\perp AX$ and $BX\perp AX$
Proof : Well known. Chase angles.
Lemma : The incircle and $\ell$ are orthogonal.
Proof : It suffices to show that $IY\cdot IX=r^2=IE^2$. Let $IC$ intersect $ED$ at $M$. Then since $\angle XYC = \angle XMC =90^{\circ}$, $XYMC$ is a cyclic quadrilateral. Therefore $IY\cdot IX = IM\cdot IC = IE^2$. We are done.
Lemma : $LA$ bisects $\angle SAT$.
Proof : Being radius of the same circle, $LS=LT$. Therefore it suffices to prove that $L$ lies on the circumcircle of $\triangle AST$.Let $AX$ intersect $BC$ at $P$. It therefore suffices to show that $LP\cdot LA=LT^2=LY^2$.
$$XY=AXAY=c\cos\dfrac{A}{2} b\cos\dfrac{A}{2}$$
implying $LY=\dfrac{1}{2}XY = \dfrac12(cb)\cos\dfrac{A}{2}$.
$$LA=LY+YA=\dfrac12(b+c)\cos\dfrac{A}{2}$$
meaning $LP\cdot LA = (LAAP)\cdot LA = LA^2  LA\cdot AP = \dfrac14(b+c)^2\cos\dfrac{A}{2}\dfrac12(b+c)\cos\dfrac{A}{2}\cdot \dfrac{2bc}{b+c}\cos\dfrac{A}{2}$ which calculates to $\dfrac14(\cos\dfrac{A}{2})^2(bc)^2=LY^2$
thus ending the proof.
For part (a), let the circumcircle of $AST$ intersect $AC$ and $AC$ at $W$ and $Z$, It suffices to show that $WZ$ is parallel to $BC$, which is obvious since by our previous lemma, $\angle WAS = \angle TAZ$.
For part (b), invert with respect to $\ell$. This leaves the incircle right where it is, but takes the circumcircle of $AST$ to line $BC$, since $L$ also lies on circle $AST$. Since $BC$ and the incircle are tangent, we are done.
Last edited by rah4927 on Sun Feb 05, 2017 12:35 am, edited 3 times in total.

 Posts: 61
 Joined: Tue Dec 08, 2015 4:25 pm
 Location: Bashaboo , Dhaka
Re: Circle is tangent to circumcircle and incircle
Will bisector $\angle BAC$ intersect both $DE$ and $DF$?
"(To Ptolemy I) There is no 'royal road' to geometry."  Euclid
 Kazi_Zareer
 Posts: 86
 Joined: Thu Aug 20, 2015 7:11 pm
 Location: Malibagh,Dhaka1217
Re: Circle is tangent to circumcircle and incircle
Extend $DF$ and it will intersect the bisector of $\angle BAC$ at $Y$.Absur Khan Siam wrote:Will bisector $\angle BAC$ intersect both $DE$ and $DF$?
We cannot solve our problems with the same thinking we used when we create them.