333. Problems with the common tangent for two circles.

by Virgil Nicula, Jan 17, 2012, 10:58 AM

PP1. Let $w=C(O,R)$ , $w_1=C(I_1,r_1)$ and $w_2=C(I_2,r_2)$ be three circles so that $w_1$ and $w_2$ are externally tangent and both are internally tangent to $w$ .

$1\blacktriangleright$ Find the length $l$ of the chord from the large circle which is the interior tangent to the small circles (at its common tangent point).

$2\blacktriangleright$ Find the length $L$ of the chord from the larger circle which is the exterior tangent to the small circles (at two points) - two cases.


Proof. Denote $:\ T\in w_1\cap w_2\ ;\  \{A,B\}\subset w$ so that $T\in (AB)$ and $AB\perp I_1I_2\ ;\  TA=x\ ,\ TB=y\ ;\  AB=x+y=l\ ;\ U\in $ $w_1\ ,$

$V\in w_2$ so that $UV$ is common tangent of $w_1$ and $w_2\ ;\ \{C,D\}=UV\cap w$ so that $U\in (CV)\ ;\ CU=u\ ,\ DV=v\ ;\ u+v+UV=L\ (*)$ .

$1\blacktriangleright$ Apply the power of $T$ w.r.t. $w\ :\ TA\cdot TB=R^2-OT^2\iff$ $\boxed{OT^2=R^2-xy}\ (1)$ . Observe that $OI_1=R-r_1$ and $OI_2=R-r_2$ .

Apply the Stewart's theorem to the cevian $OT$ in $\triangle I_1OI_2\ :\ (r_1+r_2)\cdot\left(OT^2+r_1r_2\right)=r_1(R-r_2)^2+r_2(R-r_1)^2\iff$

$OT^2=\frac {R^2(r_1+r_2)-4Rr_1r_2}{r_1+r_2}$ . Using the relation $(1)$ obtain that $\boxed{xy=\frac {4Rr_1r_2}{r_1+r_2}}\ (2)$ . Observe that $UV=2\cdot \sqrt {r_1r_2}$ . Apply the Casey's theorem

to the circles $w_1$ , $w_2$ and the degenerated circles $A$ , $B$ . Obtain that $2\cdot TA\cdot TB=AB\cdot UV\iff$ $l=\frac {xy}{\sqrt {r_1r_2}}\stackrel{(2)}{\iff}$ $\boxed{l=\frac {4R\sqrt {r_1r_2}}{r_1+r_2}}$ .

$2\blacktriangleright$ Appear two cases : the first case when $CD$ separates $A\ ,\ T$ and the second case when $CD$ doesn't separate $A\ ,\ T$ . Denote $T_1\in w\cap w_1$ , $T_2\in w\cap w_2$ .

From an well-known property the rays $(T_1U$ , $(T_2V$ are the bisectors of the angles $\widehat {CT_1D}$ , $\widehat{CT_2D}$ respectively $\implies W\in T_1U\cap T_2V$ belongs to $w$ and it is the

midpoint of the arc $\stackrel{\frown}{CD}$ . Apply the Casey's theorem to the circles $w_1$ , $w_2$ and the degenerated circles $C$ , $D$ . Obtain that $CU\cdot DV+CD\cdot UV=CV\cdot DU\iff$

$uv+2\sqrt{r_1r_2}\cdot L=\left(u+2\sqrt {r_1r_2}\right)\cdot\left(v+2\sqrt{r_1r_2}\right)\iff$ $L=u+v+2\sqrt {r_1r_2}$ what is the evident relation $(*)$ . Soon !!!



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 $M$ and $N$ be the tangency points of $\omega_a$ with $AC$ and $AB$ respectively. By Casey's theorem for the circle $\omega_a$ and

the degenerated circles $B$ , $C$ , $S$ all tangent to the circumcircle $\omega$ we get $ST \cdot BC+SC\cdot BN=CM \cdot SB \Longrightarrow \ ST \cdot BC=SC(CM-BN)\Longrightarrow$

$ST=SC\cdot\frac {b-c}{a}\ (*)$ . By Ptolemy's theorem for $ABSC\implies a\cdot SA=$ $(b+c)\cdot SC\iff$ $SC=SA\cdot\frac {a}{b+c}\stackrel{(*)}{\implies}\ \frac{ST}{SA}=\frac{b-c}{b+c}$ .



PP3 (Balkan MO 2012). Let $ABC$ be an acute-angled triangle with orthocentre $H$ and orthic triangle $DEF$ , where $D\in BC$ , $E\in CA$ , $F\in AB$ . Suppose that

$AH$ meets the circle with diameter $[BC]$ at $P\in (AH)$ . Show that the circumcircles of $\triangle APE$ and $\triangle HPF$ are tangent at $P$ (this problem is modified equivalent).


Proof 1 (metric). In the $P$ right-angled $\triangle BPC$ from the theorem of cathet $[PC]$ obtain that $PC^2=\underline{CD\cdot CB}$ . Since $BDHF$ , $BDEA$ are cyclical quadrilaterals obtain

that $CH\cdot CF=\underline{CD\cdot CB}=CE\cdot CA$ . In conclusion, $PC^2=CH\cdot CF=CE\cdot CA$ , i.e. the circumcircles of $\triangle APE$ and $\triangle HPF$ are tangent at $P$ .

Proof 2 (synthetic). Prove easily that $\left\{\begin{array}{ccc}
\underline{\widehat {CPE}}\equiv\widehat {CBE}\equiv\underline{\widehat {CAP}} & \implies & CP\ \mathrm{is\ tangent\ at\ P\ to\ circumcircle\ of\ \triangle APE}\ .\\\\
\underline {\widehat {CPD}}\equiv\widehat {CBP}\equiv\underline{\widehat {CFP}} & \implies & CP\ \mathrm{is\ tangent\ at\ P\ to\ circumcircle\ of\ \triangle HPF}\ .\end{array}\right\|$



PP4. Let an acute $\triangle ABC$ with $AB \not= AC$ . Let $H$ be the orthocenter of triangle $ABC$ and let $M$ be the midpoint of the side $BC$ .

Let $D$ be a point on the side $AB$ and $E$ a point on the side $AC$ such that $AE=AD$ and the points $D$ , $H$ , $E$ are on the same line.

Prove that the line $HM$ is perpendicular to the common chord of the circumscribed circles of triangle $\triangle ABC$ and triangle $\triangle ADE$ .

Lemma. Let $ABC$ be an acute triangle. Define: the circumcircle $c=C(O)$ and the orthocentre $H$ of the triangle $ABC$; the middlepoint $M$ of the side $[BC]$; the intersection $N$ between

$MH$ and the bisector of the angle $\widehat {BAC}$; $D\in AB$ and $E\in AC$ so that $H\in DE$ and $AD=AE$. Then the point $N$ belongs to the circumcircle $w=C(O_a)$ of the triangle $ADE$.


Proof of the lemma. $AH=2R\cos A$ and $P\in c\cap (AO_a\Longrightarrow$ $AP=2R\cos \frac{B-C}{2}\ ,$ $MP=R(1-\cos A)$ . Thus, $\frac{AN}{AP}=$ $\frac{AH}{AH+MP}=$ $\frac{2\cos A}{1+\cos A}$ $\Longrightarrow $

$\boxed {AN=\frac{2\cos A}{1+\cos A}\cdot AP}\ .$ Denote the point $N'\in AP\cap w$, i.e. $DN'\perp AB$. Thus, $\frac{AD}{\sin \left(B+\frac A2\right)}=\frac{AH}{\cos \frac A2}$ $\Longrightarrow$ $AD=\frac{\cos\frac{B-C}{2}}{\cos \frac A2}\cdot AH$ and $AN'=\frac{AD}{\cos \frac A2}$ $\Longrightarrow$

$AN'=\frac{\cos \frac{B-C}{2}}{\cos^2\frac A2}\cdot AH=$ $\frac{2\cos \frac{B-C}{2}}{1+\cos A}\cdot AH=$ $\frac{AP}{R}\cdot\frac{2R\cos A}{1+\cos A}=$ $\frac{2\cos A}{1+\cos A}\cdot AP$ $\Longrightarrow AN'=AN$ $\Longrightarrow$ $N\equiv N'$, i.e. $N$ belongs to the circumcircle of $\triangle ADE\ .$

Proof of the proposed problem. Denote: $A'\in c\cap (AO$ ; the middlepoint $A_1$ $[AH]$ ; $N\in w\cap (AO_a$. From above lemma results $N\in MH$.

But $AA_1=HA_1=OM$ ($M$ is the middlepoint of the segment $[HA']$), $OA=OA'$ and $O_aA=O_aN$. Thus $A_1$, $O$, $O_a$ are the middlepoints of

$AH$, $AA'$, $AN$ respectively and $N\in HM\equiv HA'$ $\Longrightarrow$ $O_a\in OA_1$ and $OA_1\parallel MH$. Thus, $MH\parallel OO_a$. Remark. I denote ray $(XY$ without $X$.
Attachments:
Casey's theorem and its applications.pdf (252kb)
This post has been edited 77 times. Last edited by Virgil Nicula, Nov 19, 2015, 1:20 PM

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