423. Geometry problems.

by Virgil Nicula, Mar 28, 2015, 11:49 PM

Lemma 1 (well known). Let the triangle $ABC$ . Prove that for any point $D\in (BC)$ there is the relation $\frac {DB}{DC}=\frac {AB}{AC}\cdot\frac {\sin\widehat{DAB}}{\sin\widehat{DAC}}\ .$

Proof. $\frac {DB}{DC}=\frac {[ABD]}{[ACD]}=\frac {AB\cdot AD\cdot\sin\widehat{DAB}}{AC\cdot AD\cdot\sin\widehat{DAC}}\implies$ $\frac {DB}{DC}=\frac {AB}{AC}\cdot\frac {\sin\widehat{DAB}}{\sin\widehat{DAC}}\ .$

Lemma 2 (well known). Let $ABC$ be a triangle and the points $D\in (BC)$ , $E\in (CA)$ , $F\in (AB)$ where $X\in AD\cap EF$ . Prove that $\frac {XF}{XE}=\frac {DB}{DC}\cdot\frac {AF}{AE}\cdot \cdot \frac {AC}{AB}$ .

Proof. Denote $\left\{\begin{array}{ccc}
m\left(\widehat{XAF}\right)=m\left(\widehat{DAB}\right)=x\\\\
m\left(\widehat{XAE}\right)=m\left(\widehat{DAC}\right)=y\end{array}\right\|\ (1)$ . Therefore, $\left\{\begin{array}{ccc}
\frac {XF}{XE} & = & \frac {AF}{AE}\cdot\frac {\sin x}{\sin y}\\\\
\frac {DC}{DB} & = & \frac {AC}{AB}\cdot\frac {\sin y}{\sin x}\end{array}\right\|$ $\bigodot\implies$ $\frac {XF}{XE}=\frac {DB}{DC}\cdot\frac {AF}{AE}\cdot \frac {AC}{AB}$ , i.e. the required relation.

Lemma 3. Let a convex $ABCD$ with $I\in AC\cap BD\ .$ Prove that for any $X\in (AD)$ and $Y\in (BC)$ so that $I\in (XY)$ there is the relation $\frac {XA}{XD}\cdot\frac {YB}{YC}=\frac {IA\cdot IB}{IC\cdot ID}\ .$

Proof. Denote $\left\{\begin{array}{ccccc}
m\left(\widehat{XIA}\right) & = & m\left(\widehat{CIY}\right) & = & x\\\\
m\left(\widehat{XID}\right) & = & m\left(\widehat{YIB}\right) & = & y\end{array}\right\|\ .$ Apply the lemma $(1)\ :\ \left\{\begin{array}{ccc}
X\in (AD)\ ,\ \triangle AID & : & \frac {XA}{XD}=\frac {IA}{ID}\cdot\frac {\sin x}{\sin y}\\\\
Y\in (BC)\ ,\ \triangle BIC & : & \frac {YB}{YC}=\frac {IB}{IC}\cdot\frac {\sin y}{\sin x}\end{array}\right\|$ $\bigodot$ $\implies$ $\frac {XA}{XD}\cdot\frac {YB}{YC}=\frac {IA\cdot IB}{IC\cdot ID}\ .$


PP3. Let a convex pentagon $ABCDE$ where $AB\parallel DE$ , $AB\perp AE$ , $\left\{\begin{array}{c}
BA=BC\\\\
DC=DE\end{array}\right\|$ and $F\in AD\cap BE$ . Prove that $CF\perp BD$ .

Proof. Let $:\ G\in DE\ ,\ BG\perp GE$ $;\ \left\{\begin{array}{c}
BA=BC=a\\\\
DC=DE=b\\\\
AE=BG=m\end{array}\right\|$ and $BD=n$. Thus, $\left\{\begin{array}{ccc}
BE^2=m^2+a^2 & ; & BD^2=CB^2+CD^2\\\\
AD^2=EA^2+ED^2 & ; & BD^2=GB^2+GD^2\end{array}\right\|$ $\implies$ $a^2+b^2=n^2=m^2+(b-a)^2$

Therefore, $m^2=2ab\ (*)\ ;\ \left\{\begin{array}{ccc}
(a+b)\cdot FB & = & a\cdot BE\\\\
(a+b)\cdot FD & = & b\cdot AD\end{array}\right\|\ ;\  CB^2-CD^2=$ $ a^2-b^2\ (1)\ ;\ (a+b)^2\cdot (FB^2-FD^2)= $ $a^2\cdot BE^2-b^2\cdot AD^2= $

$a^2(a^2+m^2)-b^2\cdot (b^2+m^2)=$ $ (a^2-b^2)\cdot (a^2+b^2+m^2)=$ $(a^2-b^2)\cdot (a^2+b^2+2ab)$ from the relation (*). In conclusion, $(a+b)^2\cdot (FB^2-FD^2)= $

$(a^2-b^2)\cdot (a+b)^2\ ,$ i.e. $FB^2-FD^2= a^2-b^2\ (2)\ .$ From the relation $(1)\ \wedge\ (2)$ get $CB^2-CD^2= FB^2-FD^2\ ,$ i.e. $CF\perp BD\ .$



PP4 (Cr.Tello). Let $C\in (AQ)$ and equilateral $\triangle ABC\ ,\ \triangle CPQ$ so that $AQ\cap (BP)=\emptyset$. Let: $AC=a$ , $CQ=b$ ; $M\in AP\cap QB$ , $MC=x$. Prove that $x=\frac {ab}{\sqrt{a^2+ab+b^2}}$ .

Proof. Prove easily that $\triangle ACP\equiv\triangle BCQ\implies$ $\left\{\begin{array}{ccccccc}
m\left(\widehat{QAP}\right)=m\left(\widehat{CBQ}\right)=m\left(\widehat{PCX}\right)=m\left(\widehat{BQP}\right)=\alpha\\\\
m\left(\widehat{AQB}\right)=m\left(\widehat{CPA}\right)=m\left(\widehat{BCX}\right)=m\left(\widehat{PAB}\right)=\beta\end{array}\right\|$ , where $\alpha +\beta =60^{\circ}$ . Let $\left\{\begin{array}{ccc}
MB & = & u\\\\
MP & = & v\end{array}\right\|$ .

In conclusion, $\left\{\begin{array}{ccc}
\triangle BCP & \implies & BP^2=a^2-ab+b^2\\\\
\triangle BMP & \implies & BP^2=u^2+uv+v^2\\\\
\triangle MBC\sim \triangle MCP & \implies & \frac ux=\frac xv=\frac ab\end{array}\right\|$ $\implies$ $\left\{\begin{array}{ccc}
u^2+uv+v^2 & = & a^2-ab+b^2\\\\
\frac uv=\frac {a^2}{b^2}\ ,\ uv=x^2 & ; & \frac u{a^2}=\frac x{ab}=\frac v{b^2}\end{array}\right\|$ $\implies$ $x=\frac {ab}{\sqrt{a^2+ab+b^2}}$ .



PP5 (R.A). Let $\triangle ABC$ with $w=\mathbb C(I,r)$ , $\left\{\begin{array}{c}
D\in BC\cap w\\\\
E\in CA\cap w\\\\
F\in AB\cap w\end{array}\right\|$ , $\left\{\begin{array}{ccccc}
P\in EF & ; & DP\perp EF & ; & DP=n\\\\
D\in BC & ; & AD\perp BC & ; & AD=h\end{array}\right\|$ and $\left\{\begin{array}{c}
\delta_{BC}(F)=x\\\\
\delta_{BC}(E)=y\end{array}\right\|$ . Prove that $\boxed{\sqrt{xy}=n=h\sin\frac A2}$ .

Proof. I"ll use the well known identities $\boxed{\prod\sin\frac A2=\frac r{4R}}\ (*)$ and $\boxed{2Rh_a=bc}\ (1)$ in any $\triangle ABC$ . Thus, $DE\cdot DF=2(s-c)\sin\frac C2\cdot 2(s-b)$ $\sin\frac B2\ \stackrel{(*)}{\implies}$

$\boxed{DE\cdot DF=\frac {r(s-b)(s-c)}{R\sin\frac A2}}\ (2)\ .$ Thus, $DE\cdot DF=$ $2rn\ \stackrel{(2)}{\iff}\ n=\frac {(s-b)(s-c)}{2R\sin\frac A2}\ \stackrel{(2)}{=}$ $\frac {h(s-b)(s-c)}{bc\sin\frac A2}\implies\frac nh=$ $\frac {(s-b)(s-c)}{bc\sin\frac A2}
=$ $\frac {\sin^2\frac A2}{\sin\frac A2}\implies$

$\frac nh=\sin\frac A2\ .$ I"ll prove the first required relation $n^2=xy$ . Indeed, $\left\{\begin{array}{ccc}
\frac hx=\frac c{s-b}\\\\
\frac hy=\frac b{s-c}\end{array}\right\|$ $\bigodot\implies \frac {h^2}{xy}=$ $\frac {bc}{(s-b)(s-c)}\implies$ $\sqrt {xy}=h\sin\frac A2=n\implies n^2=xy$ .



PP6 (R.A). Let $\triangle ABC$ with $w_a=\mathbb C\left(I_a,r_a\right)\ ,\ \left\{\begin{array}{c}
X\in BC\cap w_a\\\\
Z\in CA\cap w_a\\\\
Y\in AB\cap w_a\end{array}\right\|\ ,\ \left\{\begin{array}{ccc}
P\in YZ\ ,\ XP\perp YZ & : & XP=m\\\\
D\in BC\ ,\ AD\perp BC & : & AD=h\end{array}\right\|$ and $\left\{\begin{array}{c}
\delta_{BC}(Y)=x\\\\
\delta_{BC}(Z)=y\end{array}\right\|$. Prove that $\boxed{\sqrt{xy}=m=h\sin\frac A2}$

Proof. Denote the incircle $w=\mathbb C(I,r)$ and observe that $\left\{\begin{array}{ccc}
\widehat{XZY}\equiv\widehat{IBC} & ; & XZ\parallel IC\\\\
\widehat{XYZ}\equiv\widehat{ICB} & ; & XY\parallel IB\end{array}\right\|\ (*)\ .$ Remark that the circumradius of $\triangle BIC$ has

the length $2R\sin\frac A2$ and $w_a=\mathbb C\left(I_a,r_a\right)$ is the circumcircle of $\triangle XYZ$. In conclusion, from the relations $(*)$ obtain that $\triangle XYZ\sim\triangle ICB\implies$

$\frac mr=\frac {r_a}{2R\sin \frac A2}$ and $\frac mh=\frac mr\cdot\frac rh=$ $\frac {r_a}{2R\sin \frac A2}\cdot\frac rh=$ $\frac {rr_a}{2Rh\sin\frac A2}=$ $\frac {(s-b)(s-c)}{bc}\cdot\frac 1{\sin\frac A2}=$ $\sin^2\frac A2\cdot \frac 1{\sin\frac A2}=$ $\sin\frac A2\implies$ $\frac mh=\sin\frac A2$ .

I"ll prove the first relation $m^2=xy$ . Indeed, $\left\{\begin{array}{ccc}
\frac hx=\frac c{s-c}\\\\
\frac hy=\frac b{s-b}\end{array}\right\|$ $\bigodot\implies \frac {h^2}{xy}=$ $\frac {bc}{(s-b)(s-c)}\implies$ $\sqrt {xy}=h\sin\frac A2=m\implies m^2=xy$ .



PP7 (Miguel Ochoa). Let $\triangle ABC$ with $\left\{\begin{array}{c}
E\in (AC)\\\
F\in (AB)\\\
P\in (EF)\end{array}\right\|$ and $\left\{\begin{array}{c}
I\in BE\cap CF\\\
D\in PI\cap BC\\\
M\in AD\cap EF\end{array}\right\|\ .$ Prove that $\frac {MF}{ME}=\frac{PE}{PF}\cdot \frac {IB\cdot IF}{IC\cdot IE}\cdot\frac {AF}{AB}\cdot\frac {AC}{AE}\ .$

Proof. Apply $\left\{\begin{array}{ccc}
\mathrm{\underline{lemma}\ 2}/\triangle ABC & \implies & \frac {MF}{ME}=\frac {DB}{DC}\cdot\frac {AF}{AB}\cdot\frac {AC}{AE}\\\\
\mathrm{\underline{lemma}\ 3}/BCEF & \implies & \frac {DB}{DC}\cdot \frac{PF}{PE}=\frac {IB\cdot IF}{IC\cdot IE}\end{array}\right\|\bigodot$ $\implies$ $\frac {MF}{ME}=\frac{PE}{PF}\cdot \frac {IB\cdot IF}{IC\cdot IE}\cdot\frac {AF}{AB}\cdot\frac {AC}{AE}\ .$

Particular case. If the point $I$ is the incenter of $\triangle ABC$ and $N\in AP\cap BC\ ,$ then $MF=ME\iff \frac {PF}{PE}=\left(\frac {a+c}{a+b}\right)^2\iff \frac {NB}{NC}=\frac {c(a+c)}{b(a+b)}\ .$



PP8 (Ruben Dario). For an interior point $P$ of $\triangle ABC$ let its cevian $\triangle DEF$ , where $D\in (BC)$ , $E\in (CA)$ and $F\in (AB)$ . For an

interior $R$ of $\triangle DEF$ let its cevian $\triangle XYZ$ , where $X\in (EF)$ , $Y\in (DF)$ and $Z\in (DE)$ . Prove that $AX\cap BY\cap CZ\ne\emptyset$ .


Proof. Denote $\left\{\begin{array}{c}
U\in AX\cap BC\\\
V\in BY\cap CA\\\
W\in CZ\cap AB\end{array}\right\|$ and apply the upper lemma $(2)$ in $\triangle ABC$ for the points $\left\{\begin{array}{cc}
X\in (EF)\ ,\ U\in AX\cap BC\ : & \frac {XF}{XE}=\frac {UB}{UC}\cdot \frac {AF}{AE}\cdot \frac {AC}{AB}\\\\
Y\in (FD)\ ,\ V\in BY\cap CA\ : & \frac {YD}{YF}=\frac {VC}{VA}\cdot\frac {BD}{BF}\cdot \frac {BA}{BC}\\\\
Z\in (DE)\ ,\ W\in CZ\cap AB\ : & \frac {ZE}{ZD}=\frac {WA}{WB}\cdot\frac {CE}{CD}\cdot \frac {CB}{CA}\end{array}\right\|$ $\bigodot$ $\implies$

$\frac {XF}{XE}\cdot \frac {YD}{YF}\cdot\frac {ZE}{ZD}=\left(\frac {UB}{UC}\cdot \frac {AF}{AE}\cdot \frac {AC}{AB}\right)\cdot\left(\frac {VC}{VA}\cdot\frac {BD}{BF}\cdot \frac {BA}{BC}\right)\cdot\left(\frac {WA}{WB}\cdot\frac {CE}{CD}\cdot \frac {CB}{CA}\right)=$ $\left(\frac {UB}{UC}\cdot \frac {AF}{AE}\right)\cdot\left(\frac {VC}{VA}\cdot\frac {BD}{BF}\right)\cdot\left(\frac {WA}{WB}\cdot\frac {CE}{CD}\right)$ $\implies$

$\frac {XF}{XE}\cdot \frac {YD}{YF}\cdot\frac {ZE}{ZD}=\left(\frac {UB}{UC}\cdot\frac {VC}{VA}\cdot\frac {WA}{WB}\right)\cdot\left(\frac {AF}{AE}\cdot\frac {BD}{BF}\cdot\frac {CE}{CD}\right)\ (1)$ . Apply the Ceva's theorem to the mentioned points and suitable triangles $:$

$\left\{\begin{array}{cc}
P/\triangle ABC\ : & \frac {DB}{DC}\cdot\frac {EC}{EA}\cdot\frac {FA}{FB}=1\\\\
R/\triangle DEF\ : & \frac {XF}{XE}\cdot\frac {ZE}{ZD}\cdot\frac {YD}{YF}=1\end{array}\right\|\ \stackrel{(1)}{\implies}\ \frac {UB}{UC}$ $\cdot\frac {VC}{VA}\cdot\frac {WA}{WB}=1\implies$ $AU\cap BV\cap CW\ne\emptyset\implies$ $AX\cap BY\cap CZ\ne\emptyset\ .$



PP9 (Ruben Dario). Let $ABCD$ be a rectangle where $m\left\{\widehat{CAD}\right)=x$ and $\left\{\begin{array}{ccc}
E\in AC & ; &BE\perp AC\\\
F\in BC & ; & EF\perp BC\\\
G\in AC & ; & FG\in AC\end{array}\right\|$ .

Prove that $D\in FG\iff DF=DA\iff AG=AB\iff AB\cdot AC=AD^2\iff \tan x=\cos x .$


Proof. Denote $\left\{\begin{array}{ccc}
AB=CD & = & b\\\
AD=BC & = & a\\\
AC=BD & = & d\end{array}\right\|$ and apply the theorem of Cathetus $:\ \left\{\begin{array}{cccc}
\triangle ACD\ : & CD^2=CG\cdot CA & \iff & CG=\frac {b^2}d\\\\
\triangle ABC\ : & BC^2=CE\cdot CA & \iff & CE=\frac {a^2}d\end{array}\right\|\ (1)$ . Thus, $ABFG$ is cyclically $\iff$

$\boxed{CF\cdot CB=CG\cdot CA}\ (2)$ and from the theorem of Cathetus in $\triangle BEC$ obtain that $\boxed{CE^2=CF\cdot CB}\ (3)$ . Therefore, $(2)\ \wedge\ (3)\ \implies\ CE^2=$ $CG\cdot CA\ \stackrel{(1)}{\implies}$

$\left(\frac{a^2}d\right)^2=$ $\frac {b^2}d\cdot d\implies \boxed{bd=a^2}\ (4)$ . In conclusion, $\triangle ACD\sim\triangle DFC$ $\implies$ $\frac {AC}{DF}=\frac {AD}{DC}$ $\implies $ $DF=\frac {bd}a\ \stackrel{(4)}{\implies}\ DF=a\implies \boxed{DF=DA}\ (*)\ .$

Apply the theorem of Cathetus in $\triangle ACD\ :\ AG\cdot AC=$ $AD^2\implies $ $AG=\frac {a^2}d\ \stackrel{(4)}{\implies}\ AG=b$ , i.e. $\boxed{AG=AB}$ and the equivalence $\frac ba=\frac ad\iff$ $\boxed{\tan x=\cos x}$ .


Extension. Let a parallelogram $ABCD$ with $A<90^{\circ}$ and the points $\left\{\begin{array}{ccc}
\{E,G\} & \subset & (AC)\\\\
F & \in & (BC)\end{array}\right\|$ so that $\left\{\begin{array}{ccc}
BE & \parallel & FG\\\\
EF & \parallel & AB\\\\
D & \in & FG\end{array}\right\| .$ Prove that $\boxed{DE=DA\iff \frac {AB}{AC}=\frac {-1+\sqrt 5}2}\ (*).$

Proof (analytic). Let $\left\{\begin{array}{ccc}
A(0.0) & ; & B(b,c)\\\\
C(a+b,c) & ; & D(a,0)\end{array}\right\|$ . There is $\lambda\in (0,1)$ so that $\left\{\begin{array}{c}
F(a\lambda +b,c)\\\\
E[(a+b)\lambda ,c\lambda]\\\\
G\left(\frac {a+b}{2-\lambda},\frac c{2-\lambda}\right)\end{array}\right\|$ . Thus, $D\in FG\iff$ $s_{BE}=s_{DF}\iff$ $\frac {c(\lambda -1)}{(a+b)\lambda -b}=\frac c{a\lambda +b-a}\iff$ $a\lambda^2-3a\lambda +a=0\iff$ $(\lambda -1)(a\lambda +b-a)=(a+b)\lambda -b\iff$ $\boxed{\lambda^2-3\lambda +1=0}\ (1)\ .$ Therefore, $DF=DA\iff$

$DF^2=DA^2\iff$ $(a\lambda +b-a)^2+c^2=a^2\iff$ $\boxed{a^2\lambda^2+2a(b-a)\lambda +\left(b^2+c^2-2ab\right)=0}\ (2)\ .$ Hence the equations $(1)$ and $(2)$ have at least a common root, i.e.

$\left(\begin{array}{cc}
a^2 & b^2+c^2-2ab\\\\
1 & 1\end{array}\right)^2=\left(\begin{array}{cc}
a^2 & 2a(b-a)\\\\
1 & -3\end{array}\right)\cdot\left(\begin{array}{cc}
2a(b-a) & b^2+c^2-2ab\\\\
-3 & 1\end{array}\right)$ $\iff$ $\left[\left(a^2+2ab\right)-\left(b^2+c^2\right)\right]^2=-\left(a^2+2ab\right)\cdot\left[3\left(b^2+c^2\right)-2\left(a^2+2ab\right)\right]$ .

Denote $\left\{\begin{array}{ccc}
b^2+c^2 & = & m\\\\
a^2+2ab & = & n\end{array}\right\|$ . Thus, the previous relation becomes $(n-m)^2+n(3m-2n)=0\iff $ $n^2-mn-m^2=0\iff $ $\boxed{\frac nm=\frac {1+\sqrt 5}2}\ (*)\ .$

Observe that $\left(\frac {AC}{AB}\right)^2=\frac {AC^2}{AB^2}=$ $\frac{(a+b)^2+c^2}{b^2+c^2}=\frac {m+n}m=$ $1+\frac nm=1+\frac {1+\sqrt 5}2=\frac {3+\sqrt 5}2\implies$ $\left(\frac {AB}{AC}\right)^2=\frac {3-\sqrt 5}2\implies$ $\frac {AB}{AC}=\frac {-1+\sqrt 5}2$ .



PP10 (Miguel Ochoa Sanchez). Let $ABCD$ be a square with the points $\left\{\begin{array}{ccc}
G\in AB & ; & B\in (AG)\\\
E\in AD & ; & A\in (DE)\end{array}\right\|$ and $F\in DB\cap CG$ . Prove that $FE\perp FC\iff$ $\frac 1{AB}=\frac 1{BG}-\frac 2{AE}\ .$

Proof. Denote $AD=a$ and $\left\{\begin{array}{ccccccc}
AE=m & ; & m\left(\widehat{BCF}\right) & = & x & ; & \tan x=\frac na\\\
BG=n & ; & m\left(\widehat{DCE}\right) & = & y & ; & \tan y=\frac {a+m}a\end{array}\right\|$ , where $m\left(\widehat{ECF}\right)=90^{\circ}-y+x\ .$ Thus, $FE\perp FC\iff$

$CFED$ is cyclically $\iff$ $\widehat{ECF}\equiv\widehat{EDF}\iff$ $m\left(\widehat{ECF}\right)=45^{\circ}\iff$ $90^{\circ}-y+x=45^{\circ}$ $\iff$ $y=x+45^{\circ}\iff$ $\tan y=\frac {1+\tan x}{1-\tan x}\iff$

$\frac {a+m}a=\frac {a+n}{a-n}\iff$ $(a-n)(a+m)=a(a+n)\iff$ $a^2+a(m-n)-mn=a^2+an\iff$ $\boxed{a(m-2n)=mn}\iff$ $\frac 1a=\frac 1n-\frac 2m\ .$

Remark. $FE\perp FC\iff CFED$ is cyclic $\iff \widehat{CEF}\equiv\widehat{CDF}\iff$ $m\left(\widehat{ECF}\right)=m\left(\widehat{FBG}\right)=45^{\circ}\iff $ $FE=FC$ and $CHBG$ is cyclic, where $H\in BD\cap CE\ .$


An easy extension. Let $ABCD$ be a rectangle with $\left\{\begin{array}{ccc}
AD=a & ; & AB=b\\\
G\in AB & ; & B\in (AG)\\\
E\in AD & ; & A\in (DE)\\\
AE=m & ; & BG=n<b\end{array}\right\|$ and $F\in DB\cap CG\ . $ Prove that $FE\perp FC\iff$ $ab(m-2n)=amn+n(a-b)^2\ .$

Proof. Denote $\left\{\begin{array}{ccccccc}
m\left(\widehat{BCF}\right) & = & x & ; & \tan x=\frac na\\\
m\left(\widehat{DCE}\right) & = & y & ; & \tan y=\frac {a+m}b\\\
m\left(\widehat{EDF}\right) & = & \phi & ; & \tan \phi=\frac ba\end{array}\right\|$ , where $m\left(\widehat{ECF}\right)=90^{\circ}-y+x\ .$ Thus, $FE\perp FC\iff$ $CFED$ is cyclically $\iff$

$\widehat{ECF}\equiv\widehat{EDF}\iff$ $90^{\circ}-y+x=\phi\iff$ $y-x=90^{\circ}-\phi\iff$ $\tan (y-x)=\frac ab\iff$ $\frac {\frac {a+m}b-\frac na}{1+\frac {a+m}b\cdot \frac na}=\frac ab\iff$ $\boxed{ab(m-2n)=amn+n(a-b)^2}\ .$



PP11 (Ruben Dario). Let the circle $w$ with $\{A,D,M\}\subset w$ so that $V\in AA\cap DD$ and $AD$ doesn't separate $M$ and $V$ .

Denote $B\in MM\cap VB$ , $C\in MM\cap VD$ and $N\in AD$ so that $MN\perp AD$ . Prove that $\frac {NA}{ND}=\frac {MB}{MC}\ .$


Proof 1 (Juan Daniel Valdivia Fuentes). Denote $L\in BC\cap AD$ and observe that $\left\{\begin{array}{ccc}
VA & = & VD\\\
BA & = & BM\\\
CD & = & CM\end{array}\right\|\ .$ Apply the Menelaus' theorem to the transversal $\overline{LAD}$ over

$\triangle BVC\ :\ \frac{LB}{LC}\cdot\frac {DC}{DV}\cdot\frac {AV}{AB}=1$ $\implies$ $\frac{LB}{LC}=\frac {AB}{DC}\implies$ $\frac{LB}{LC}=\frac {MB}{MC}\implies$ the division $(B,C;L,M)$ is harmonically. Hence $NM\perp NL\implies$ $\widehat{MNB}\equiv\widehat{MNC}\ .$

Therefore, $\frac {NB}{NC}=\frac {MB}{MC}$ and $\widehat{BNA}\equiv\widehat{CND}$ $\implies$ $\triangle ABN\sim\triangle DCN$ $\implies$ $\frac {NA}{ND}=\frac {NB}{NC}$ $\implies$ $\frac {NA}{ND}=\frac {MB}{MC}\ .$ More exactly, $\frac {NA}{ND}=\frac {MB}{MC}=\frac {NB}{NC}=\frac {AB}{DC}\ .$

Proof 2 (Ruben Dario).
Nice proof ! Denote the midpoints $U$ , $V$ of $[MA]$ , $[MD]$ respectively, i.e. $MA=2\cdot AU$ and $MD=2\cdot DV$ . Therefore,

$\left\{\begin{array}{ccccc}
\triangle ABU\sim\triangle DMN & \implies & \frac {AB}{DM}=\frac {AU}{DN} & \implies & 2\cdot AB\cdot DN=MD\cdot MA\\\\
\triangle DCV\sim\triangle AMN & \implies & \frac {DC}{AM}=\frac {DV}{AN} & \implies & 2\cdot DC\cdot AN=MD\cdot AM\end{array}\right\|$ $\implies$ $AB\cdot DN=DC\cdot AN\implies$ $\frac {NA}{ND}=\frac {AB}{DC}\implies$ $\frac {NA}{ND}=\frac {MB}{MC}\ .$

With other words. Let $\left\{\begin{array}{ccccccc}
m\left(\widehat{BAM}\right) & = & m\left(\widehat{BMA}\right) & = & m\left(\widehat{MDA}\right) & = & x\\\\ 
m\left(\widehat{CDM}\right) & = & m\left(\widehat{CMD}\right) & = & m\left(\widehat{MAD}\right) & = & y\end{array}\right\|\ .$ Thus, $\frac {NA}{ND}=\frac {MA\cos y}{MD\cos x}=\frac {\frac {MA}{2\cos x}}{\frac {MD}{2\cos y}}=$ $\frac {AB}{DC}=\frac {MB}{MC}\implies$ $\frac {NA}{ND}=\frac {MB}{MC}\ .$



PP12 (Ruben Dario). Let $\triangle ABC$ with incircle $w=\mathbb C(I,r)\ .$ Denote $:\ \left\{\begin{array}{ccc}
M\in AB\cap w & ; & N\in AC\cap w\\\\
\{G,H\}\subset AI & ; & \odot \begin{array}{cc}
\nearrow & BG\perp AI\\\\
\searrow & CH\perp AI\end{array}\end{array}\right\|$ . Prove that $\widehat{BMG}\equiv\widehat{CHN}\ .$

Proof 1. $\left\{\begin{array}{cccccc}
BMIG & \mathrm{is\ cyclic} & \implies & \widehat{BMG}\equiv \widehat{BIG} & \implies & m\left(\widehat{BMG}\right)=90^{\circ}-\frac C2\\\\
CNIH & \mathrm{is\ cyclic} & \implies & \widehat{CHN}\equiv \widehat{CIN} & \implies & m\left(\widehat{CHN}\right)=90^{\circ}-\frac C2\end{array}\right\|$ $\implies$ $\widehat{BMG}\equiv\widehat{CHN}\ .$

Proof 2. Denote $\left\{\begin{array}{ccc}
m\left(\widehat{BMG}\right) & = & x\\\\
m\left(\widehat{CHN}\right) & = & y\end{array}\right\|$ and apply the theorem of Sines in the mentioned triangles $:$

$\blacktriangleright\ \triangle BMG\ :\ \frac {\sin\widehat{BMG}}{\sin\widehat{BGM}}=\frac {BG}{BM}\implies$ $\frac {\sin x}{\cos \left(x-\frac A2\right)}=$ $\frac {c\sin\frac A2}{s-b}\implies$ $\frac {\tan x}{\cos\frac A2+\sin\frac A2\tan x}=$ $\frac {c\sin\frac A2}{s-b}\implies$ $\tan x=\frac {c\sin\frac A2\cos\frac A2}{(s-b)-c\sin^2\frac A2}=$

$\frac {c\sin A}{(a+c-b)-c(1-\cos A)}=$ $\frac {a\sin C}{a-b+c\cos A}\implies$ $\tan x=\frac {a\sin C}{a-(b-c\cos A)}=\frac {a\sin C}{a-a\cos C}=$ $\frac {\sin C}{1-\cos C}=\cot \frac C2\implies$ $\tan x=\cot\frac C2\implies x+\frac C2=90^{\circ}\ .$

$\blacktriangleright\ \triangle CNH\ :\ \frac {\sin\widehat{CHN}}{\sin \widehat{CNH}}=\frac {CN}{CH}\implies$ $\frac {\sin y}{\cos \left(y-\frac A2\right)}=$ $\frac {s-c}{b\sin\frac A2}\implies$ $\frac {\tan y}{\cos\frac A2+\sin\frac A2\tan y}=$ $\frac {s-c}{b\sin\frac A2}\implies$ $\tan y=\frac {(s-c)\cos\frac A2}{b\sin\frac A2-(s-c)\sin\frac A2}=$

$\frac {s-c}{b-(s-c)}\cot\frac A2=$ $\frac {s-c}{s-a}\cdot\cot\frac A2=\cot\frac C2\implies$ $\tan y=\cot\frac C2\implies$ $y+\frac C2=90^{\circ}\ .$ In conclusion, $\boxed{x=y=90^{\circ}-\frac C2}\ .$



PP13 (Miguel Ochoa Sanchez). Let $\triangle ABC$ with the incircle $w=\mathbb C(I,r)\ .$ Let $\left\{\begin{array}{c}
M\in BC\cap w\\\\
E\in CA\cap w\\\\
F\in AB\cap w\end{array}\right\|$ and $\left\{\begin{array}{c}
AM\cap w =\{M,N\}\\\\
BE\cap w=\{E,G\}\\\\
CF\cap w=\{F,D\}\end{array}\right\|\ .$ Prove that $DG\cdot EF=3\cdot DE\cdot FG\ .$

Proof. Let $\left\{\begin{array}{c}
AF=AE=x\\\\
BM=BF=y\\\\
CE=CM=z\end{array}\right\|$ . Thus, $\boxed{EF^2=\frac {4x^2yz}{(x+y)(x+z)}}\ (1)$ and $\left\{\begin{array}{c}
\triangle BFG\sim\triangle BEF\implies\frac {BF}{BE}=\frac {FG}{EF}\implies FG=EF\cdot\frac {y}{BE}\\\\
\triangle CED\sim\triangle CFE\implies\frac {CE}{CF}=\frac {ED}{FE}\implies ED=FE\cdot\frac {z}{CF}\end{array}\right\|$ $\bigodot\implies$

$DE\cdot FG=EF^2\cdot \frac {yz}{BE\cdot CF}\implies$ $\boxed{DE\cdot FG=\frac {4x^2y^2z^2}{(x+y)(x+z)}\cdot\frac 1{BE\cdot CF}}\ (2)\ .$ Apply the Pyhagoras' theorem to $:$

$\blacktriangleright\ BE/\triangle BEF\implies$ $BE\cdot GE=BE\cdot (BE-BG)=BE^2-BE\cdot BG=$ $BE^2-BF^2=FE^2+2\cdot BF\cdot FE\cdot\sin\frac A2=$

$FE^2+2y\cdot 2x\sin^2\frac A2=$ $\frac {4x^2yz}{(x+y)(x+z)}+4xy\cdot \frac {yz}{(x+y)(x+z)}=\frac {4xyz}{x+z}\implies$ $\boxed{EG=\frac {4xyz}{x+z}\cdot\frac 1{BE}}\ (3)\ .$

$\blacktriangleright\ CF/\triangle CEF\implies$ $CF\cdot DF=CF\cdot (CF-CD)=CF^2-CF\cdot CD=$ $CF^2-CE^2=$ $FE^2+2\cdot CE\cdot FE\cdot \sin\frac A2=$

$FE^2+2z\cdot 2x\sin^2\frac A2=$ $\frac {4x^2yz}{(x+y)(x+z)}+4xz\cdot \frac {yz}{(x+y)(x+z)}=\frac {4xyz}{x+y}\implies$ $\boxed{FD=\frac {4xyz}{x+y}\cdot\frac 1{CF}}\ (4)\ .$

Hence the relations $(3)\ \wedge\ (4)\implies$ $EG\cdot FD=\frac {4xyz}{x+z}\cdot\frac 1{BE}\cdot \frac {4xyz}{x+y}\cdot\frac 1{CF}\implies$ $\boxed{EG\cdot FD=\frac {16x^2y^2z^2}{(x+y)(x+z)}\cdot\frac 1{BE\cdot CF}}\ \stackrel{(2)}{=}\ 4\cdot DE\cdot FG\implies$

$EG\cdot FD=4\cdot DE\cdot FG\ .$ The Ptolemy's theorem for $DEFG$ , i.e. $4\cdot DE\cdot FG=EG\cdot FD=DE\cdot FG+DG\cdot EF\implies$ $\boxed{DG\cdot EF=3\cdot DE\cdot FG}\ .$

Remark. Denote $:\ \left\{\begin{array}{c}
X\in FM\cap DG\\\\
Y\in EM\cap DG\end{array}\right\|\ ;\ XG=u\ ,\ YD=v\ ,\ XY=w\ .$ Define $\left\{\begin{array}{ccccc}
m\left(\widehat{GFM}\right) & = & m\left(\widehat{GEM}\right) & = & \alpha\\\\
m\left(\widehat{DFM}\right) & = & m\left(\widehat{DEM}\right) & = & \beta\end{array}\right\|\ .$ Thus, $\left\{\begin{array}{ccc}
\frac {XG}{XD} & = & \frac {FG}{FD}\cdot\frac {\sin \alpha}{\sin \beta}\\\\
\frac {YG}{YD} & = & \frac {EG}{ED}\cdot\frac {\sin \alpha}{\sin \beta}\end{array}\right\|$ $\implies$

$\frac {YG}{YD}\cdot \frac {XD}{XG}=\frac {EG}{ED}\cdot\frac {FD}{FG}=4\implies$ $YG\cdot XD=4\cdot YD\cdot XG\implies$ $(u+w)(v+w)=4uv\implies$ $\frac {uv}w=\frac {u+v+w}3\ge \sqrt[3]{uvw}$ $\implies$ $ uv\ge w^2\implies \boxed{XG\cdot YD\ge XY^2}\ .$

See
here the Leo Giugiuc's proof.


PP14 (Ruben Auqui). Let an equilateral $\triangle ABC\ ,\ \left\{\begin{array}{ccccc}
M\in (BC) & ; & E\in (AM) & ; & D\in (BM)\\\\
BD=4 & ; & DM=ME=1 & ; & EA=6\end{array}\right\|$ and $w=C(I,r)$ so that $\{D,E\}\subset w$ . Prove that $r=\frac {\sqrt 3}2$ .

Proof 1. $BM=5$ , $AM=7$ , $AB=a\implies$ the generalized Pythagoras' theorem $:\ AM^2=BA*2+BM^2-BA\cdot BM\iff$ $49=a^2+25-5a\iff a=8$ ,

i.e. $AB=8$ . Let $m\left(\widehat{AMB}\right)=2\phi$ and apply $\tan \frac A2=\sqrt{\frac {(s-b)(s-c)}{s(s-a)}}$ for any $\triangle ABC$ to $\widehat{AMB}$ of $\triangle AMB\ : \tan\phi =\sqrt{\frac {3\cdot 5}{10\cdot 2}}=\frac {\sqrt 3}2\implies$ $r=\tan\phi\implies \frac {\sqrt 3}2$ .

Proof 2. Let $AB=a$ . So $AM=7$ and $MB=5$ . I"ll use the well-known relation (get it easily with Stewart's) $BC^2=AM^2+MB\cdot MC$ , i.e. $a^2=49+5(a-5)\iff$

$a^2-5a-24=0\implies\odot \begin{array}{ccc}
\nearrow & 8 & \searrow\\\\
\searrow & -3 & \nearrow\end{array}\odot\implies$ $\boxed{\ a=8\ }\implies MC=3$ . Let $x=m\left(\widehat{IMB}\right)$ . Thus, $\tan x=\frac {DI}{DM}\implies$ $\boxed{\tan x=r}$ . So $m\left(\widehat{AMC}\right)=180^{\circ}-2x$

and $\cos \widehat{AMC}=-\cos 2x=\frac {MA^2+MC^2-AC^2}{2\cdot MA\cdot MC}=$ $\frac {49+9-64}{2\cdot 7\cdot 3}=-\frac 6{6\cdot 7}\implies$ $\cos 2x=\frac 17\implies$ $\frac {1-\tan^2x}{1+\tan^2x}=\frac 17\implies$ $7(1-r^2)=1+r^2\implies$ $\boxed{r=\frac {\sqrt 3}2}$ .

Proof 3. Let $AB=a$ . I"ll use the well-known relation $BC^2=AM^2+MB\cdot MC$ , i.e. $a^2=49+5(a-5)\iff$ $a^2-5a-24=0\implies\odot \begin{array}{ccc}
\nearrow & 8 & \searrow\\\\
\searrow & -3 & \nearrow\end{array}\odot\implies$

$a=8\implies MC=3\implies DB=DC$ , i.e. $D$ is the midpoint of $[BC]$ . So $\triangle AIE\sim AMD\implies$ $\frac {AE}{AD}=\frac {IE}{MD}\implies$ $\frac 6{4\sqrt 3}=\frac r1\implies$ $\boxed{\ r=\frac {\sqrt 3}2\ }$ .



PP15 (Miguel Ochoa Sanchez). Let $ABCD$ be a square what is inscribed in the circle $w=\mathbb C(O,R)$ . Prove that $(\forall ) P\in w\ ,\ PA^4+PB^4+PC^4+PD^4=6AB^4$.

Proof 1. Suppose w.l.o.g. that $P$ belongs to the small arc $\overarc[]{BC}$ . Let $AB=l=R\sqrt 2$ and $\left\{\begin{array}{ccccc}
PA=a & ; & PB=b & ; & \delta_{AC}(P)=u\\\\
PC=c & ; & PD=d & ; & \delta_{BD}(P)=v\end{array}\right\|$ . Thus, $u^2+v^2=R^2$ and from the right

$\triangle BPD$ and $APC$ obtain that $b^2+d^2=a^2+c^2=2l^2$ , i.e. $\sum a^2=4l^2$ and $\left(a^2+c^2\right)\left(b^2+d^2\right)=4l^4$ . Thus, $\left\{\begin{array}{ccc}
ac & = & 2Ru\\\\
bd & = & 2Rv\end{array}\right|\implies$ $a^2c^2+b^2d^2=$

$4R^2\left(u^2+v^2\right)=4R^4\implies$ $a^2c^2+b^2d^2=l^4$ . So $\sum a^4=\left(\sum a^2\right)^2-2\left[a^2c^2+b^2d^2+\left(a^2+c^2\right)\left(b^2+d^2\right)\right]=$ $\left(4l^2\right)^2-2\cdot\left(l^4+4l^4\right)\implies$ $a^4+b^4+c^4+d^4=6l^4$ .

Proof 2. Let $\left\{\begin{array}{c}
m\left(\widehat{PBC}\right)=x\\\\
m\left(\widehat{PCB}\right)=y\end{array}\right\|$ , where $\boxed{x+y=45^{\circ}}\ (*)$ . Prove easily $\frac {PA}{\sin \left(90^{\circ}+x\right)}=$ $\frac {PB}{\sin y}=\frac {PC}{\sin x}=\frac {PD}{\sin\left(90^{\circ}+y\right)}=2R$ , where $l^2=2R^2$ . So, $\frac 1{16R^4}\cdot\sum PA^4=$

$\sin^4 \left(90^{\circ}+x\right)+$ $\sin^4y+\sin^4x+\sin^4\left(90^{\circ}+y\right)=$ $\cos^4x+$ $\sin^4y+\sin^4x+\cos^4y=$ $\left(1-2\cos^2x\sin^2x\right)+\left(1-2\sin^2y\cos^2y\right)=$ $2-2\cos^2x\sin^2x-$

$2\sin^2y\cos^2y=$ $\frac 12\cdot \left(4-\sin^22x-\sin^22y\right)\ \stackrel{(2x+2y=90^{\circ})}{=}\ \frac 12$ $\cdot \left[4-\left(\sin^22x+\cos^22x\right)\right]=\frac 32\implies$ $\sum PA^4=24R^4=6\left(R\sqrt 2\right)^4=6l^4\implies$ $\boxed{\sum PA^4=6l^4}$ .



PP16 (Miguel Ochoa Sanchez). Let $\triangle ABC$ with incircle $w=\mathbb C(I,r)$ such that $AI^2=10\ ,\ BI^2=5$ and $CI^2=2$ . Prove that $2R=5r$ , where $R$ is length of circumradius

Proof. I"ll use the well-known identity (can prove it easily !) $\boxed{AI^2=bc-x}\ (*)$ , where $x=4Rr$ . Thus, $\left\{\begin{array}{ccccc}
bc & = & 10 & + & x\\\
ca & = & 5 & + & x\\\
ab & = & 2 & + & x\end{array}\right\|$ and $abc=4Rrs\iff$

$\frac 2{4Rr}=\frac 1{bc}+\frac 1{ca}+\frac 1{ab}\iff$ $\frac 2x=\frac 1{x+2}+\frac 1{x+5}+\frac 1{x+10}\iff$ $x^3-80x-200=0\iff$ $(x-10)(x^2+10x+20)=0\iff$ $\boxed{\ x=10\ }$ . In conclusion,

$\left\{\begin{array}{ccccc}
bc &  = & 20 & = & 4\cdot 5\\\
ca & = & 15 & = & 3\cdot 5\\\
ab & = & 12 & = & 3\cdot 4\end{array}\right\|$ $\implies$ $abc=3\cdot 4\cdot 5\implies$ $\boxed{\begin{array}{ccc}
a& = & 3\\\
b & = & 4\\\
c & = & 5\end{array}}\implies$ $ABC$ is a $C$-right triangle $\implies$ $\left\{\begin{array}{ccccccc}
2R & = & c & \implies & R & = & \frac 52\\\\
2r & = & a+b-c & \implies & r & = & 1\end{array}\right\|\implies$ $\boxed{\ \frac R5=\frac r2\ }$ .

$\blacktriangleright\ (*)$ Remark. Let $\triangle ABC$ with the incircle $w=\mathbb C(I,r)$ , the $A$-excircle $w_a=\mathbb C\left(I_a,r_a\right)$ and $\{A,S\}=AI\cap w$ . Is well-known that $SI_a=SI$ , i.e. $II_a=2\cdot IS$ and

$IA\cdot IS=2Rr$ (from the power of $I$ w.r.t. the circumcircle). Thus, $\triangle AIC\sim\triangle ABI_a\iff$ $\frac {AI}{AB}=\frac {AC}{AI_a}\iff$ $AI\cdot AI_a=bc\iff$

$AI(AI+II_a)=bc\iff$ $AI^2=bc-IA\cdot II_a=bc-2\cdot IA\cdot IS\iff$ $\boxed{AI^2=bc-4Rr}$ .



PP17 (Miguel Ochoa Sanchez). Let $\triangle ABC$ and a point $P$ which belongs to the small arc $\overarc[]{AC}$ . Prove that $x^2\sin 2A+y^2\sin 2B+z^2\sin 2C=4S$ , where $S$ is the area of $\triangle ABC$ .

Proof. Apply the Ptolemy's theorem $\boxed{ax+cz=by}\ (*)$ and the generalized Pytagoras' theorem $\boxed{b^2=x^2+z^2+2xz\cos B}\ (1)$ . Prove easily that the relations $:$

$\left\{\left|\begin{array}{c}
a^2\sin 2B+b^2\sin 2A=4S\\\\
b^2\sin 2C+c^2\sin 2B=4S\end{array}\right|\begin{array}{cc}
\odot & x^2\\\\
\odot & z^2\end{array}\right\|\bigoplus$ $\implies$ $4S\left(x^2+z^2\right)=b^2\left(x^2\sin 2A+z^2\sin 2C\right)+\sin 2B\left(a^2x^2+c^2z^2\right)=$ $b^2\left(x^2\sin 2A+z^2\sin 2C\right)+$

$\sin 2B\left[(ax+cz)^2-2acxz\right]\ \stackrel{(*)}{=}$ $b^2\sum x^2\sin 2A-4acxz\sin B\cos B=$ $b^2\sum x^2\sin 2A-8xzS\cos B\implies$

$4S\left(x^2+z^2+2xz\cos B\right)=b^2\sum x^2\sin 2A \stackrel{(1)}{\implies}\ 4b^2S=b^2\sum x^2\sin 2A\implies$ $\sum x^2\sin 2A=4S$ .

Remark. $a^2\sin 2B+b^2\sin 2A=$ $2\cdot \underline{a\sin B}\cdot a\cos B+$ $2\cdot \underline{b\sin A}\cdot b\cos A=$ $2b\sin A(a\cos B+b\cos A)=$ $2bc\sin A=4S\implies$ $\boxed{a^2\sin 2B+b^2\sin 2A=4S}$ .



PP18 (Miguel Ochoa Sanchez). Let $\triangle ABC$ with the $A$-excircle $w_a=\mathbb C\left(I_a,r_a\right)$ and $\left\{\begin{array}{c}
D\in BC\cap w_a\\\\
E\in CA\cap w_a\\\\
F\in AB\cap w_a\end{array}\right\|$ . Prove that $\left\{\begin{array}{ccc}
AD\cap BE\cap CF & \ne & \emptyset\\\\
\frac {PD}{PA}\cdot \frac {PE}{PB}\cdot\frac {PF}{PC} & = & \frac {r_a}{4R}\end{array}\right\|$ .

Proof.Is well-known that $\left\{\begin{array}{ccccc}
BF & = & BD & = & s-c\\\
CE & = & CD & = & s-b\\\
AF & = & AE & = & s\end{array}\right\|$ . Apply the Ceva's theorem to the point $P$ and $\triangle ABC\ :\ \frac {DB}{DC}\cdot\frac {EC}{EA}\cdot \frac {FA}{FB}=\frac {s-c}{s-b}\cdot \frac {s-b}{s}\cdot\frac {s}{s-c}=1\implies$

$AD\cap BE\cap CF\ne \emptyset$ . Apply the Menelaus' theorem to the transversals $:\ \left\{\begin{array}{cccc}
\overline{ADP}/\triangle BCE\ : & \frac {AC}{AE}\cdot \frac {PE}{PB}\cdot\frac {DB}{DC}=1 & \implies & \frac {PE}{PB}=\frac {s(s-b)}{b(s-c)}\\\\
\overline{ADP}/\triangle BCF\ : & \frac {AB}{AF}\cdot\frac {PF}{PC}\cdot\frac {DC}{DB}=1 & \implies & \frac {PF}{PC}=\frac {s(s-c)}{c(s-b)}\\\\
\overline{CPF}/\triangle ABD\ : & \frac {PD}{PA}\cdot\frac {FA}{FB}\cdot\frac {CB}{CD}=1 & \implies & \frac {PD}{PA}=\frac {(s-b)(s-c)}{as}\end{array}\right\|\ \bigodot$ $\implies$

$\frac {PE}{PB}\cdot\frac {PF}{PC}\cdot\frac {PD}{PA}=\frac {s(s-b)}{b(s-c)}\cdot\frac {s(s-c)}{c(s-b)}\cdot\frac {(s-b)(s-c)}{as}=$ $\frac {s(s-b)(s-c)}{abc}=\frac {(s-b)(s-c)}{4Rr}=\frac {rr_a}{4Rr}\implies$ $\frac {PE}{PB}\cdot\frac {PF}{PC}\cdot\frac {PD}{PA}=\frac {r_a}{4R}$ .



PP19 (Gustavo Jimmy Garcia Paytan). Let $\triangle ABC$ with the incircle $w=\mathbb C\left(I,r\right)$ and the points $\left\{\begin{array}{ccccc}
\{R,S\}\subset (AB)\\\\
\{M,N\}\subset (BC)\\\\
\{P,Q\}\subset (CA)\end{array}\right\|$ so that the lines $\left\{\begin{array}{c}
RQ\parallel BC\\\\
MS\parallel CA\\\\
PN\parallel AB\end{array}\right\|$ are tangent to $w$ . Denote

the lengths $\left(r_1,r_2,r_3\right)$ of the inradii for the triangles $ARQ$ , $BMS$ and $CPN$ respectively. Prove that $r^2+r_1^2+r_2^2+r_3^2=\frac {(b+c-a)(c+a-b)(a+b-c)\left(a^2+b^2+c^2\right)}{(a+b+c)^3}$ .


Proof. Observe that $\triangle ABC\sim\triangle ARQ\sim\triangle BMS\sim\triangle CPN\implies$ $\frac {r_1}{s-a}=\frac {r_2}{s-b}=\frac {r_3}{s-c}=\boxed{\frac rs}=$ $\frac {r_1+r_2+r_3}{(s-a)+(s-b)+(s-c)}=\frac {r_1+r_2+r_3}s=$

$\frac {r_1+r_2}c=\frac {r_2+r_3}a=\frac {r_3+r_1}b\implies$ $\boxed{r_1+r_2+r_3=r}\ (*)$ and $\boxed{\frac {r_1+r_2}c=\frac {r_2+r_3}a=\frac {r_3+r_1}b=\frac rs}\ (1)$ $\implies r^2+r_1^2+r_2^2+$ $r_3^2\ \stackrel{(*)}{=}$

$\left(r_1+r_2+r_3\right)^2$ $+r_1^2+r_2^2+r_3^2=$ $\left(r_1+r_2\right)^2+\left(r_2+r_3\right)^2+$ $\left(r_3+r_1\right)^2\ \stackrel{(1)}{=}$ $\frac {r^2}{s^2}$ $\cdot\left(a^2+b^2+c^2\right)$ $\implies$ $r^2+r_1^2+r_2^2+r_3^2=\frac {r^2}{s^2}\cdot\left(a^2+b^2+c^2\right)$ .

Observe that $\frac {(b+c-a)(c+a-b)(a+b-c)}{(a+b+c)^3}=\frac {8(s-a)(s-b)(s-c)}{8s^3}=$ $\frac {sr^2}{s^3}=\frac {r^2}{s^2}$ . In conclusion, the required relation is truly.



PP20 (Miguel O. Sanchez). Let an $A$-isosceles $\triangle ABC$ and its interior $P$ so that $\left\{\begin{array}{c}
\widehat{PBA}\equiv\widehat{PCB}\\\\
\widehat{PCA}\equiv\widehat{PBC}\end{array}\right\|$ . Let $S\in AP\cap BC$ . Prove that the ray $[PS$ is the $P$-symmedian in $\triangle BPC$ .

Proof 1 (trigonometric). Denote $S\in QAP\cap BC$ and $\left\{\begin{array}{c}
m\left(\widehat{PAB}\right)=u\ ;\ m\left(\widehat{PAC}\right)=v\\\
m\left(\widehat{PBA}\right)=\left(\widehat{PCB}\right)=x\\\
m\left(\widehat{PCA}\right)=m\left(\widehat{PBC}\right)=y\end{array}\right\|$ . Apply the trigonometric form

of the Ceva's theorem
$:\ \frac {\sin\widehat{PAB}}{\sin
\widehat{PAC}}\cdot\frac {\sin \widehat{PBC}}{\sin\widehat{PBA}}\cdot\frac {\sin\widehat{PCA}}{\sin\widehat{PCB}}=1\iff$ $\frac {\sin u}{\sin v}\cdot\frac {\sin y}{\sin x}\cdot\frac {\sin y}{\sin x}=1\iff$ $\frac {\sin u}{\sin v}=\left(\frac {\sin x}{\sin y}\right)^2=$ $\left(\frac {PB}{PC}\right)^2\implies$

$\boxed{\frac {\sin u}{\sin v}=\left(\frac {PB}{PC}\right)^2}\ (*)$ . Thus, $\frac {SB}{SC}=\frac {AB}{AC}\cdot\frac {\sin\widehat{SAB}}{\sin\widehat{SAC}}=$ $\frac {\sin u}{\sin v}\ \stackrel{(*)}{\implies}\ \frac {SB}{SC}=$ $\left(\frac {PB}{PC}\right)^2$ .

Proof 2 (trigonometric). Denote $\left\{\begin{array}{c}
m\left(\widehat{PBA}\right)=m\left(\widehat{PCB}\right)=x\\\\
m\left(\widehat{PCA}\right)=m\left(\widehat{PBC}\right)=y\end{array}\right\|$ and $\left\{\begin{array}{ccc}
AB=b & ; & E\in BP\cap AC\\\\
BC=a & ; & F\in CP\cap AB\end{array}\right\|$ . Apply the Ceva's theorem

to the point $P$ and $\triangle ABC\ :\ \boxed{\frac {FB}{FA}\cdot \frac {EA}{EC}\cdot\frac {SC}{SB}=1}\ (*)$ . Observe that $\boxed{\frac {PB}{PC}=\frac {\sin x}{\sin y}}\ (1)$ . I"ll use an well-known the relations

$\left\{\begin{array}{ccc}
\frac {FB}{FA}=\frac {CB}{CA}\cdot\frac {\sin\widehat{FCB}}{\sin\widehat{FCA}}=\frac ab\cdot\frac {\sin x}{\sin y}\\\\
\frac {EA}{EC}=\frac {BA}{BC}\cdot\frac {\sin\widehat{EBA}}{\sin\widehat{EBC}}=\frac ba\cdot\frac {\sin x}{\sin y}\end{array}\right\|\bigodot\ \stackrel{(*)}{\implies}$ $\frac {SB}{SC}=$ $\frac {FB}{FA}\cdot\frac {EA}{EC}=$ $\left(\frac {\sin x}{\sin y}\right)^2=$ $\left(\frac {PB}{PC}\right)^2\ \stackrel{(1)}{\implies}\ \frac {SB}{SC}=$ $\left(\frac {PB}{PC}\right)^2$ .

Proof 3 (metric). Prove easily that $AB$ and $AC$ are tangent to the circumcircle $w$ of $\triangle BPC$ . Denote $\{P,R\}=PS\cap w$ . Hence $\boxed{\frac {SB}{SC}=\frac {BP\cdot BR}{CP\cdot CR}}\ (*)$ .

Therefore, $\left\{\begin{array}{ccc}
\triangle ABP\sim\triangle ARB & \implies & \frac {PB}{BR}=\frac {AB}{AR}\\\\
\triangle ACP\sim \triangle ARC & \implies & \frac {PC}{CR}=\frac {AC}{AR}\end{array}\right\|\implies$ $\frac {PB}{BR}=\frac {PC}{CR}\implies$ $\frac {BR}{CR}=$ $\frac {PB}{PC}\ \stackrel{(*)}{\implies}\ \frac {SB}{SC}=\left(\frac {PB}{PC}\right)^2$ .



PP21 (Ruben Dario). Let an $A$-isosceles $\triangle ABC\ (\ B=C=2\phi\ )$ with the incircle $w=\mathbb C(I,r)$ which touches given triangle at $D\in (BC)$ , $E\in (CA)$ and

$F\in (AB)$ . Prove that for any $M\in (AF)$ and $N\in (AE)$ so that $MN$ is tangent to $w$ there are the relations $:\ \left\{\begin{array}{cccc}
\frac 1{BM}+\frac 1{CN} & = & \frac {BC}{IB^2}=\sin 2\phi & (1)\\\\
\frac {MA}{MB}+\frac{NA}{NC} & = & \tan 2\phi\tan\phi & (2)\end{array}\right\|$ .


Proof. Suppose w.l.o.g. $r=1$ and denote $:\ T\in MN\cap w\ ;\ \left\{\begin{array}{ccc}
MF=MT=x & ; & m\left(\widehat{MIF}\right)=m\left(\widehat{MIT}\right)=u\\\
NE=NT=y & ; & m\left(\widehat{NIE}\right)=m\left(\widehat{NIT}\right)=v\end{array}\right\|$ , where

$\left\{\begin{array}{c}
B=u+v=2\phi\\\\
\phi -u=v-\phi =\frac {v-u}2\end{array}\right\|\ ;$ $DB=DC=a\ ;\ AE=AF=b$ . Prove easily that $\left\{\begin{array}{ccc}
a=\cot\phi & ; & IB=IC=\frac 1{\sin\phi}\\\\
x=\tan u & ; & y=\tan v\end{array}\right\|$ . Thus $:$

$(1)\blacktriangleright\ \frac 1{BM}+\frac 1{CN}=\frac {BC}{IB^2}\iff$ $\frac 1{a+x}+\frac 1{a+y}=2a\sin^2\phi\iff$ $\frac 1{\cot\phi +\tan u}+\frac 1{\cot\phi +\tan v}=2\cot\phi\sin^2\phi =\sin 2\phi\iff$

$\frac {\sin\phi\cos u}{\cos (\phi -u)}+\frac {\sin\phi \cos v}{\cos (\phi -v)}=2\sin \phi\cos\phi\iff$ $\frac {\cos u}{\cos (\phi -u)}+\frac {\cos v}{\cos (\phi -v)}=2\cos\phi\ \stackrel{\phi -u=v-\phi }{\iff}$ $\frac {\cos u+\cos v}{\cos (\phi -u)}=2\cos\phi\iff$

$\cos u+\cos v=2\cos \phi\cos (\phi -u)\iff$ $\cos u+\cos v=\cos u+\cos (2\phi -u)\ \stackrel{u+v=2\phi}{\iff}\ \cos u+\cos v=\cos u+\cos v$ , what is truly.

$(2)\blacktriangleright\ \frac {MA}{MB}+\frac {NA}{NC}=\frac {b-x}{a+x}+\frac {b-y}{a+y}=$ $\frac {\tan 2\phi -\tan u}{\tan \phi +\tan u}+\frac {\tan 2\phi -\tan v}{\tan \phi +\tan v}=\iff$ $\frac {\sin (2\phi -u)}{\cos 2\phi\cos u}\cdot\frac {\sin \phi\cos u}{\cos (\phi -u)}+\frac {\sin (2\phi -v)}{\cos 2\phi\cos v}\cdot\frac {\sin \phi\cos v}{\cos (\phi -v)}=$

$\frac {\sin \phi}{\cos 2\phi}\cdot\left[\frac {\sin(2\phi -u)}{\cos (\phi -u)}+\frac{\sin(2\phi -v)}{\cos (\phi -v)}\right]=$ $\frac {\sin \phi}{\cos 2\phi}\cdot\frac {\sin (2\phi -u)\cos (\phi -v)+\sin (2\phi -v)\cos (\phi -u)}{\cos (\phi -u)\cos (\phi -v)}=$ $\frac {\sin \phi}{\cos 2\phi}\cdot\frac {2\sin\phi +\sin (\phi +v-u) +\sin (\phi +u-v)}{1+\cos (u-v)}=$

$\frac {\sin \phi}{\cos 2\phi}\cdot\frac {2\sin\phi +2\sin\phi\cos (u-v)}{1+\cos (u-v)}=$ $\frac {\sin \phi}{\cos 2\phi}\cdot\frac {2\sin\phi [1+\cos (u-v)]}{1+\cos (u-v)}=$ $\frac {2\sin^2\phi}{\cos 2\phi}=$ $\frac {2\sin^2 \phi\cos\phi}{\cos 2\phi\cos\phi}=$ $\frac {\sin 2\phi\sin\phi}{\cos 2\phi\cos\phi}\implies$ $\frac {MA}{MB}+\frac {NA}{NC}=\tan 2\phi\tan\phi\ .$
This post has been edited 437 times. Last edited by Virgil Nicula, Feb 11, 2018, 3:01 PM

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