212. A simple applications of the Casey's theorem.

by Virgil Nicula, Jan 22, 2011, 7:54 AM

Proposed problem 1. Let $ABC$ be an isosceles and $A$-right triangle with the circumcircle $w$ . Consider a circle $w_1=(O_1)$ which is

tangent to $BC$ and interior tangent to $w$ so that $BC$ separates $O_1$ , $A$ . Denote $G\in w_1$ for which $GO_1\perp GA$ . Prove that $AG=AB$ .


Proof. Denote $w=C(O,R)$ and $w_1=C(O_1,r)$ , $F\in BC\cap w_1$ and $E\in w\cap w_1$ . Observe that $OF^2 = $ $OO_1^2 - O_1F^2 = $ $(R - r)^2 - r^2$ $\implies$

$\boxed{OF^2=R(R - 2r)}$ . But we also note that since $OA\perp BC\perp O_1F$, we have $OA\parallel O_1 F$ $\implies$ $O_1A^2= $ $(OA + O_1F)^2 + OF^2=$

$ (R + r)^2 + R(R-2r)$ $\implies$ $\boxed{O_1A^2=2R^2 + r^2}$ $\implies$ $AG^2= O_1A^2 - O_1G^2= 2R^2 + r^2 - r^2$ $\implies$ $AG^2= 2R^2$ $\implies$ $AG = R \sqrt{2} = AB$ .



An easy extension. Let $ABC$ be a triangle with the circumcircle $w=(O)$ . Consider a circle $w_1=(O_1)$ which is tangent to $(BC)$ in $F$

and interior tangent to $w$ so that $BC$ separates $O_1$ , $A$ . Denote $G\in w_1$ for which $GO_1\perp GA$ . Prove that $a\cdot AG=b\cdot FB+c\cdot FC$ .


Proof. I"ll use the Casey's theorem to the circles $A$ , $B$ , $w_1$ and $C$ which are interior tangent to the circle $w$ . Therefore, $a\cdot AG=c\cdot FC+b\cdot FB$ .

Particular case. Let $ABC$ be an isosceles triangle with the circumcircle $w$ . Consider a circle $w_1=(O_1)$ which is tangent to

$BC$ and interior tangent to $w$ so that $BC$ separates $O_1$ , $A$ . Denote $G\in w_1$ for which $GO_1\perp GA$ . Prove that $AG=AB$ .


Proof. Denote $F\in BC\cap w_1$ and $E\in w\cap w_1$ . From the well-known property obtain that $AB=AC\implies A\in EF$ .

Thus, $\widehat {ACF}\equiv\widehat{AEC}\iff$ $\triangle ACF\sim \triangle AEC\iff$ $AC^2=AF\cdot AE=AG^2\implies$ $AG=AC=AB$ .



PP2. Let $ABC$ be a triangle with the incircle $(I)$ and the circumcircle $w$ . Consider the circle $w_a$ which is tangent to $AB$ , $AC$ and which

is internal tangent to $w$ . Denote $\{A,S\}=AI\cap w$ . The tangent from $S$ to $w_a$ touches $w_a$ in $T$ . Prove that $\frac {ST}{SA}=\frac {|b-c|}{b+c}$ .


Proof. Assume w.l.o.g. that $AC > AB$ and let $M,N$ be the tangency points of $\omega_a$ with $AC,AB.$ By Casey's theorem for $\omega_a$ , $B$ , $C$ , $S$

all tangent to $\omega$ , we get $ST \cdot BC+CS\cdot BN=CM \cdot BS \Longrightarrow \ ST \cdot BC=CS(CM-BN)$ If $U$ is the reflection of $B$

across $AS,$ then $CM-BN=UC=AC-AB$ . Hence $ST \cdot BC=CS(AC-AB) \ (*)$ . Using the Ptolemy's theorem

for $ABSC$ we deduce that $SA \cdot BC=CS(AB+AC)$ . Together with the expression $(*)$ we obtain $\frac{ST}{SA}=\frac{AC-AB}{AC+AB}$ .



PP3. Let $w_1$ , $w_2$ be two circles which are interior tangent to the circle $w$ in the points $A$ , $C$ respectively and let $\{B,G\}\subset w$ be two points so that $AC$ is the common interior

tangent in $F$ of $w_1$ , $w_2$ and the line $AC$ separates $B$ , $G$ . Denote the exterior tangent $DE$ of the circles $w_1$ , $w_2$ so that $D\in (AB)$ and $E\in (CB)$ . Prove that $\frac 2{DE}=\frac 1{FB}+\frac 1{FG}$ .


Proof. Apply the Casey's theorem to $w_1\ ,\ w_2$ and the degenerate circles (with the null radii) $(B)\ ,\ (G)\ :\ BF\cdot GF+$ $BF\cdot GF=BG\cdot DE\iff $ $2\cdot FB\cdot FG=DE\cdot (FB+FG)\iff $ $\frac 2{DE}=\frac 1{FB}+\frac 1{FG}$ .


PP4. Let $\triangle ABC$ with the circumcircle $w$. Let the circle $\phi_1$ which is internal tangent to $w$ and is tangent to $AB$ in $M$. Let the circle

$\phi_2$ which is external tangent to $w$ and is tangent to $AB$ in $P$. Prove that $\boxed{s\cdot AM=(s-a)\cdot AP=bc}$ , where $2s=a+b+c$.


Proof. Denote $\left\{\begin{array}{ccc}
N\in AC\cap \phi_1 & ; & AM=AN=x\\\\
Q\in AC\cap\phi_2 & ; & AP=AQ=y\end{array}\right\|$ . Apply the Casey's theorem to the degerate circles $\{A,B,C\}$ and the circles $:$

$\left\{\begin{array}{ccccccc}
\phi_1 & : & ax=b(c-x)+b(b-x) & \implies & 2bc=x(a+b+c) & \implies & AM=x=\frac {bc}s\\\\
 \phi_2 & : & ax=b(x-c)+c(x-b) & \implies & 2bc=x(b+c-a) & \implies & AP=x=\frac {bc}{s-a}\end{array}\right\|$ .
This post has been edited 28 times. Last edited by Virgil Nicula, Nov 26, 2015, 6:19 PM

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