433. Some properties of remarkable lines in a triangle II.

by Virgil Nicula, Nov 16, 2015, 2:27 PM

P1. Let $\triangle ABC$ , the midpoint $M$ of $[BC]$ , $\left\{\begin{array}{ccc}
E\in AB & ;  & ME\perp AB\\\\
F\in AC & ; & MF\perp AC\end{array}\right\|$ and intersection $D$ of tangent lines at $E$ , $F$ to circle with diameter $[AM]$ . Prove that $DM\perp BC$ .

Proof 1 (synthetical). Let $L$ be a point for which $LA\parallel BC$ and $LM\perp BC$ . Denote $AM=m_a$ . Observe that $[MAB]=[MAC]$ $\implies$ $\boxed{\frac c{MF}=\frac b{ME}}\ (*)$

and $\left\{\begin{array}{c}
[ABL]=[AML]=[ACL]\\\\
\widehat{AEL}\equiv\widehat{AML}\equiv\widehat{AFL}\end{array}\right\|$ $\implies$ $c\cdot LE=m_ah_a=b\cdot LF$ $\stackrel{(*)}{\implies}$ $MF\cdot LE=ME\cdot LF$ , i.e. $LEMF$ is an harmonic cyclic quadrilateral.

From an well-known property obtain that $D\in LM$ - the bisector of the side $[BC]$ , i.e. $DB=DC\iff DM\perp BC$ . I was inspired by phuongtheong. Thank you.

Proof 2. Let $\left\{\begin{array}{c}
x=m(\widehat {BAM})\\\\
y=m(\widehat {CAM})\end{array}\right\|$ , the circle $w$ with the diameter $[AM]$ and the intersections $U$ , $V$ of the perpendicular bisector of the side $[BC]$ with the tangent lines to t$w$ at $E$ , $F$

respectively. We observe that $ME=m_a\sin x$ and $MF=m_a\sin y$ , i.e. $\frac{ME}{MF}=\frac{\sin x}{\sin y}\ (1)$ . From the Sinus' theorem applied in $\triangle MEU$ , $\triangle MFV$ results $\frac{MU}{\sin x}=\frac{ME}{\sin (B-x)}$

and $\frac{MV}{\sin y}=\frac{MF}{\sin (C-y)}$ . Therefore, the point $D$ belongs to the perpendicular bisector of the side $[BC]$ $\Longleftrightarrow$ $MU=MV$ $\Longleftrightarrow$ $\frac{ME\sin x}{\sin (B-x)}=\frac{MF\sin y}{\sin (C-y)}$ $\Longleftrightarrow$

$\frac{\sin (B-x)}{\sin (C-y)}=\frac{\sin ^2x}{\sin ^2y}$ , what is truly, from the relation $(1)$ and the identity posted
here

Remark. Prove easily that $ MU=\frac a2\cdot\frac {1}{\cot x-\cot B}$ , $MV=\frac a2\cdot\frac {1}{\cot y-\cot C}$ . Since $MB=MC\iff$ $c\cdot\sin x=b\cdot\sin y\iff$ $\sin C\sin x=\sin B\sin y$ obtain that

$MU=MV=MD$ $\iff$ $\frac{\sin (B-x)}{\sin (C-y)}=\frac{\sin ^2x}{\sin ^2y}\iff$ $\frac{\sin (B-x)}{\sin (C-y)}=\frac {\sin B}{\sin C}\cdot \frac{\sin x}{\sin y}\iff$ $\frac {\sin (B-x)}{\sin B\sin x}=\frac {\sin (C-y)}{\sin C\sin y}\iff$ $\cot x-\cot B=\cot y-\cot C=\frac {a}{2\cdot MD}$ .

Otherwise. $\cot x-\cot B=$ $\frac {c^2+m_a^2-\frac {a^2}{4}}{2S}-$ $\frac {a^2+c^2-b^2}{4S}=$ $\frac {2c^2+b^2+c^2-a^2}{4S}-\frac {a^2+c^2-b^2}{4S}=$ $\frac {b^2+c^2-a^2}{2S}=$ $\frac {2bc\cdot\cos A}{bc\cdot\sin A}=$ $2\cot A\implies$

$\boxed{\cot x-\cot B=\cot y-\cot C=2\cot A\ \ \wedge\ \ MD=\frac {MT}{2}=\frac a4\cdot\tan A}$ , where $T$ is the intersection of the tangent lines at $B$ , $C$ to the circumcirle of $\triangle ABC$ .


P2. Let $ABC$ be an $A$-isosceles triangle. For $Q\in (BC)$ denote second intersection $P$ of the line $AQ$ with the circumcircle of $\triangle ABC$ . Prove that $\frac {1}{PB}+\frac {1}{PC}=\frac {k}{PQ}$ , $k=\frac {BC}{AB}$ .

Proof 1. $\left\{\begin{array}{ccc}
\triangle PBQ\sim \triangle CAQ & \implies & \frac {PQ}{PB}=\frac {CQ}{CA}\\\\
\triangle PCQ\sim \triangle BAQ & \implies & \frac {PQ}{PC}=\frac {BQ}{BA}\end{array}\right\|\ \bigoplus\ \implies$ $PQ\cdot\left(\frac {1}{PB}+\frac {1}{PC}\right)=\frac{BC}{AB}\implies$ $\boxed{\frac {1}{PB}+\frac {1}{PC}=\frac {k}{PQ}}$ , $ k=\frac {BC}{AB}$ .

Proof 2 (trigonometric). Denote $\left\{\begin{array}{c}
m\left(\widehat{QAB}\right)=x\\\
m\left(\widehat{QAC}\right)=y\\\
m\left(\widehat{AQB}\right)=\phi\end{array}\right\|$ . Observe that $B+y=\phi$ , $B+\phi =180^{\circ}-x$ and $BC=2\cdot AB\cos B$ . Thus, $\frac {1}{PB}+\frac {1}{PC}=\frac {k}{PQ}\iff$

$\frac {PQ}{PB}+\frac {PQ}{PC}=\frac {BC}{AB}\iff$ $\frac {\sin y}{\sin \phi}+\frac {\sin x}{\sin\phi}=2\cos B\iff$ $\sin x+\sin y=2\cos B\sin\phi =\sin (B+\phi)+\sin (\phi-B)$ , what is truly.


Extension. Let $\triangle ABC$ and for $Q\in (BC)$ denote the second intersection $P$ of the line $AQ$ with the circumcircle of $\triangle ABC$ . Prove that $\boxed{\frac {AC}{PB}+\frac {AB}{PC}=\frac {BC}{PQ}}$ .

Proof. $\left\{\begin{array}{ccc}
\triangle ABQ\sim\triangle CPQ & \implies & \frac {AB}{CP}=\frac {BQ}{PQ}\\\\
\triangle ACQ\sim\triangle BPQ & \implies & \frac {AC}{BP}=\frac {CQ}{PQ}\end{array}\right\|\ \bigoplus\ \implies\ \frac {AC}{PB}+$ $\frac {AB}{PC}=\frac {BC}{PQ}$ .


P3. Let $\triangle ABC$ and the midpoints $M$ , $N$ of $[AB]$ , $[AC]$ respectively and the intersections $P$ , $Q$ of the $A$-symmedian with $BN$ , $CM$ respectively. Prove that $\widehat{ABQ}\equiv\widehat{ACP}$ .

Proof 1 (metric). Denote $\left\{\begin{array}{c}
S\in \overline{AQP}\cap BC\\\\
R\in AC\cap BQ\\\\
T\in CP\cap AB\end{array}\right\|$ . It's well-known that $\frac {SB}{SC}=\frac {c^2}{b^2}$ . Apply the Menelaus' theorem to the transversals:

$\left\{\begin{array}{cccc}
\overline{APS}/\triangle BNC\ : & \frac {AN}{AC}\cdot\frac {SC}{SB}\cdot\frac {PB}{PN}=1 & \implies & \frac {PB}{PN}=\frac {2c^2}{b^2}\\\\
\overline{AQS}/\triangle CMB\ : & \frac {AM}{AB}\cdot\frac {SB}{SC}\cdot\frac {QC}{QM}=1 & \implies & \frac {QC}{QM}=\frac {2b^2}{c^2}\end{array}\right\|$ $\blacktriangleleft\ \mathrm{and}\ \blacktriangleright$ $\left\{\begin{array}{cccc}
\overline{BQR}/\triangle AMC\ : & \frac {BM}{BA}\cdot\frac {RA}{RC}\cdot\frac {QC}{QM}=1 & \implies & \frac {RA}{c^2}=\frac {RC}{b^2}=\frac {b}{b^2+c^2}\\\\
\overline{CPT}/\triangle ANB\ : & \frac {CN}{CA}\cdot\frac {TA}{TB}\cdot\frac {PB}{PN}=1 & \implies & \frac {TA}{b^2}=\frac {TB}{c^2}=\frac {c}{b^2+c^2}\end{array}\right\|\implies$

$AR\cdot AC=AT\cdot AB=\frac {b^2c^2}{b^2+c^2}\implies$ $BTRC$ is cyclic $\implies$ $\widehat{TBR}\equiv\widehat{TCR}\implies$ $\widehat{ABQ}\equiv\widehat{ACP}$ .

Proof 2 (metric). Denote $\left\{\begin{array}{c}
S\in \overline{AQP}\cap BC\\\\
R\in AC\cap BQ\\\\
T\in CP\cap AB\end{array}\right\|$ . Apply the Ceva's theorem to the points $P$ and $Q\ :\ \left\{\begin{array}{ccc}
\frac {SB}{SC}\cdot\frac {NC}{NA}\cdot\frac {TA}{TB}=1 & \implies & \frac {TA}{b^2}=\frac {TB}{c^2}=\frac {c}{b^2+c^2}\\\\
\frac {SB}{SC}\cdot\frac {RC}{RA}\cdot\frac {MA}{MB}=1 & \implies & \frac {RA}{c^2}=\frac {RC}{b^2}=\frac {b}{b^2+c^2}\end{array}\right\|\implies$

$AR\cdot AC=AT\cdot AB=\frac {b^2c^2}{b^2+c^2}\implies$ $BTRC$ is cyclic $\implies$ $\widehat{TBR}\equiv\widehat{TCR}\implies$ $\widehat{ABQ}\equiv\widehat{ACP}$ .

Remark. Denote $m\left(\widehat{ABQ}\right)=\phi$ . Apply an well-known relation $\frac {c^2}{b^2}=\frac {RA}{RC}=\frac {BA}{BC}\cdot\frac {\sin\widehat{RBA}}{\sin\widehat{RBC}}=$ $\frac ca\cdot\frac {\sin\phi}{\sin(B-\phi )}\implies$

$\frac {\tan\phi}{\sin B-\cos B\tan\phi}=\frac {ac}{b^2}\implies$ $\tan\phi =\frac {ac\cdot\sin B}{b^2+ac\cdot\cos B}=$ $\frac{4S}{2b^2+\left(a^2+c^2-b^2\right)}\implies$ $\boxed{\tan\phi =\frac {4S}{a^2+b^2+c^2}}$ .



P4. Let $\triangle ABC$ , the midpoint $M$ of $[AB]$ , $H\in BC$ which satisfy $\widehat{CAH}\equiv\widehat{CBA}$ . The bisector of $\angle{BAH}$ meet $BC$ at $E$ . Let $F\in ME\cap AH$ . Prove that $CF\parallel AE$ .

Proof 1 (metric). Observe that $\triangle CAH\sim\triangle CBA\implies$ $\frac {CA}{CB}=\frac {AH}{BA}=\frac {HC}{AC}\implies$ $\frac ba=\frac {AH}{c}=\frac {HC}{b}\implies$ $\boxed{HA=\frac {bc}{a}}$ and $\boxed{HC=\frac {b^2}{a}}$ . $m\left(\widehat{CEA}\right)=$

$m\left(\widehat{ABE}\right)+m\left(\widehat{BAE}\right)=$ $m\left(\widehat{HAC}\right)+m\left(\widehat{HAE}\right)=$ $m\left(\widehat{CAE}\right)$ $\implies$ $CE=AC=b$ $\implies$ $\boxed{BE=a-b}$ . Thus, $EH=CE-CH=b-\frac {b^2}{a}\implies$ $\boxed{EH=\frac {b(a-b)}{a}}$ .

Apply the Menelaus' theorem to the transversal $\overline{FEM}$ and $\triangle ABH\ \ :\ \ \frac {FH}{FA}\cdot$ $\frac {MA}{MB}\cdot\frac {EB}{EH}=1\implies$ $\frac {FH}{FA}=\frac {EH}{EB}=\frac ba\implies$ $\frac {FH}{FA}=\frac ba=\frac {CH}{CE}\implies$ $AE\parallel CF$ .

Proof 2 (synthetic). Prove easily that that $CA = CE$ . Denote $D\in BC$ for which $AD\perp AE$ . Since $m\left(\widehat{EAD}\right)=90^{\circ}$ and $CA=CE$ obtain that $C$ is the midpoint of $[ED]$ , i.e.

$CD=CE=CA$ . Notice that the division $(B,H;E,D)$ is harmonically $\implies (B,A;M,X)=F(B,H;E,D)\cap AB$ , where $X\in  FD\cap AB$ is an harmonical division. Since $M$

is the midpoint of $AB$ obtain $X\equiv \infty$ , i.e. $FD\parallel AB$ . Hence $\widehat{CDF}\equiv\widehat{ABD}\equiv\widehat{FAC}$ . Together with $CA = CD$ this means that $AFD$ is an isosceles triangle and

$m(\angle CFD) = m(\angle CFA) = 90^{\circ}- m(\angle FAD) = m(\angle EAF)$ . In conclusion, $AE\parallel FC$ .



Lemma. Let $\triangle ABC$ with incenter $I$ and $X\in AB$ , $Y\in AC$ so that $I\in XY$ . Then $b\cdot \frac {XB}{XA}+c\cdot\frac {YC}{YA}=a$ .

Denote $D\in AI\cap BC$ , the line $d\parallel BC$ so that $A\in d$ and $T\in XY\cap BC$ , $Z\in XT\cap d$ . Suppose w.l.o.g. $c<b$ and $B\in (TC)$ . Thus, $\left\{\begin{array}{cc}
\frac {XB}{XA}=\frac {TB}{AZ} & \bigodot\ b\\\\
\frac {YC}{YA}=\frac {TC}{AZ} & \bigodot\ c\end{array}\right|$ $\implies$

$b\cdot\frac {XB}{XA}+c\cdot\frac {YC}{YA}=$ $\frac {b\cdot TB+c\cdot TC}{AZ}=$ $\frac {(b+c)\cdot TD}{AZ}=$ $(b+c)\cdot \frac {ID}{IA}=$ $(b+c)\cdot\frac {a}{b+c}=a$ $\implies$ $b\cdot \frac {XB}{XA}+c\cdot\frac {YC}{YA}=a$ .
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P5. Let $AD$ be a bisector of $\triangle ABC$ , where $D\in (BC)$ . The line trough the incenters of $\triangle ABD$ and $\triangle ACD$ meets $AB$ in $M$ and $AC$ in $N$ . Prove that $BN\cap CM\cap AD\ne \emptyset$ .

Proof. Is well-known that $\frac {DB}{c}=\frac {DC}{b}=\frac {a}{b+c}$ . Denote $P\in I_1I_2\cap AD$ and apply the upper lemma: $\left\{\begin{array}{cc}
AD\cdot\frac {MB}{MA}+c\cdot\frac {PD}{PA}=BD=\frac {ac}{b+c} & \div c\\\\
AD\cdot\frac {NC}{NA}+b\cdot\frac {PD}{PA}=DC=\frac {ab}{b+c} & \div b\end{array}\right|$ $\implies$

$\left\{\begin{array}{c}
\frac {AD}{c}\cdot\frac {MB}{MA}+\frac {PD}{PA}=\frac {a}{b+c}\\\\
\frac {AD}{b}\cdot\frac {NC}{NA}+\frac {PD}{PA}=\frac {a}{b+c}\end{array}\right|$ $\implies$ $\frac {1}{c}\cdot\frac {MB}{MA}=\frac {1}{b}\cdot\frac {NC}{NA}$ . Since $\frac bc=\frac {DC}{DB}$ obtain that $\frac {MB}{MA}\cdot \frac {NA}{NC}\cdot \frac {DC}{DB}=1\implies$ $BN\cap CM\cap AD\ne \emptyset$ .



P6. Let $\triangle ABC$ and $D$ , $E$ , $F$ be the tangent points of the incircle $w(I)$ with $BC$ , $CA$ , $AB$ respectively. $AD$ meets again $w$ at $Q$ . Show that $EQ$ bisects $AF\iff CA=CB$ .

Proof 1. Let $X\in EQ\cap AB\implies$ $XA=XF\iff$ $XA^2=XQ\cdot XE\iff$ $\widehat{BAD}\equiv\widehat{XEA}\iff$ $\widehat{BAD}\equiv\widehat{ADE}\iff$ $DE\parallel AB\iff$ $CI\perp AB\iff CA=CB$ .

Proof 2 (metric). $\left\{\begin{array}{c}
U\in EF\cap AD\\\
X\in EQ\cap AB\end{array}\right|\implies$ $\left\{\begin{array}{c}
\frac {UE}{UF}=\frac {DC}{DB}\cdot\frac {AE}{AC}\cdot\frac {AB}{AF}=\frac {c(s-c)}{b(s-b)}\\\\
\frac {UE}{UF}=\frac {DE}{DF}\cdot\frac {\sin\widehat{UDE}}{\sin\widehat{UDF}}=\frac {(s-c)\sin \frac C2}{(s-b)\sin\frac B2}\cdot \frac {\sin\widehat{XEA}}{\sin\widehat{XEF}}\end{array}\right|\implies$ $\boxed{\frac {\sin\widehat{XEA}}{\sin\widehat{XEF}}=\frac {c\cdot \sin\frac B2}{b\cdot\sin \frac C2}}\ (*)$ . Prove easily that $\frac {XA}{XF}=$ $\frac {EA}{EF}\cdot\frac {\sin\widehat{XEA}}{\sin\widehat{XEF}}\stackrel{(*)}{=}$

$\frac {1}{2\sin\frac A2}\cdot\frac {c\cdot \sin\frac B2}{b\cdot\sin \frac C2}=$ $\frac{c}{2b}\cdot\sqrt{\frac {(s-a)(s-c)}{ac}\cdot\frac {ab}{(s-a)(s-b)}\cdot\frac {bc}{(s-b)(s-c)}}\implies$ $\boxed{\frac {XA}{XF}=\frac {c}{a+c-b}}$ . In conclusion, $XA=XF$ $\iff$ $ b=a$ $\iff$ $CA=CB$ .



Lemma. Let $\triangle ABC$ and an interior $P$ for which denote $D\in AP\cap BC$ . Prove that for any

$E\in (AC)$ and $F\in (AB)$ so that $P\in EF$ exists the relation $\frac {FB}{FA}\cdot DB+\frac {EC}{EA}\cdot DB=\frac {PD}{PA}\cdot BC$ .


Remark. In the proof of PP apply the particular cases: $\left\|\begin{array}{ccc}
P:=I & \implies & b\cdot\frac {FB}{FA}+c\cdot\frac {EC}{EA}=a\\\\
P:=O & \implies & b\cos B\cdot\frac {FB}{FA}+c\cos C\cdot\frac {EC}{EA}=a\cos A\end{array}\right|$ .

P7. Let an acute $\triangle ABC$ with orthocenter $H$ , incenter $I$ and circumcenter $O$ and $\left\{\begin{array}{cc}
B_1\in BH\cap AC\ ; & C_1\in CH\cap AB\\\\
B_2\in BI\cap AC\ ; & C_2\in CI\cap AB\end{array}\right|$ . Prove that $I\in B_1C_1\iff O\in B_2C_2$

Proof. $\underline{I\in B_1C_1}\iff$ $b\cdot \frac {C_1B}{C_1A}+c\cdot\frac {B_1C}{B_1A}=a\iff$ $b\cdot \frac {a\cos B}{b\cos A}+c\cdot\frac {a\cos C}{c\cos A}=a\iff$ $\underline{\cos B+\cos C=\cos A}$ . $\underline{O\in B_2C_2}$

$\iff$ $b\cos B\cdot \frac {C_2B}{C_2A}+c\cos C\cdot \frac {B_2C}{B_2A}=a\cos A$ $\iff$ $b\cos B\cdot\frac ab+c\cos C\cdot\frac ac=a\cos A$ $\iff$ $\underline{\cos B+\cos C=\cos A}$ . In conclusion,

$\boxed{I\in B_1C_1\iff \cos B+\cos C=\cos A\iff O\in B_2C_2}$ .



P8. Given $\triangle ABC$ . The bisector of $ \angle BAC $ meets $BC$ and circumcircle of $\triangle ABC$ at $D$ , $E$ respectively. Let $M$ , $N$ be midpoints of $BD$ , $CE$ respectively. The circumcircle of

$\triangle ABD$ meets again $AN$ at $Q$ . The circle passing through $A$ and which is tangent to $BC$ at $D$ meets $AM$ , $AC$ respectively at $P$ , $R$ . Show that $B, P, Q, R$ belong to same line.


Proof. Prove easily that $\triangle ABD\sim\triangle AEC$ . Thus, $\left\{\begin{array}{c}
MB=MD\\\
NE=NC\end{array}\right|\implies \triangle ABM\sim\triangle AEN\ (*)$ .

Therefore, $\left\{\begin{array}{ccccc}
MP\cdot MA=MD^2=MB^2 & \implies & \triangle MBP\sim\triangle MAB & \implies & \widehat{MBP}\equiv\widehat{MAB}\\\\
\widehat{QBD}\equiv\widehat{QAD}\equiv\widehat {NAE} & \implies & \widehat{QBD}\stackrel{(*)}{\equiv}\widehat{MAB} & \implies & \widehat{QBD}\equiv\widehat{MAB}\end{array}\right|$ $\implies$ $\widehat{QBD}\equiv\widehat{PBD}$ $\implies$ $B\in PQ$ . Thus, $\left\{\begin{array}{c}
\widehat{PBD}\equiv\widehat{PAB}\\\\
\widehat{PDB}\equiv\widehat{PAD}\end{array}\right|$

$\implies$ $\widehat{QPD}=\widehat{PBD}+\widehat{PDB}=$ $\widehat{PAB}+\widehat{PAD}=$ $\widehat{BAD}=\frac A2$ . Since $APDR$ is cyclically obtain that $\widehat{RPD}\equiv\widehat{RAD}=\frac A2$ . In conclusion, $\widehat{QPD}\equiv\widehat{RPD}\implies$ $P\in QR$ .



P9. Let $[AD$ be the $A$-bisector of $\triangle ABC$ , where $D\in BC$ . The incircles of the triangles $ABD$ , $ACD$

touch $BC$ at the points $M$ , $N$ respectively. Prove that $\boxed{\frac {2}{AD}=\frac 1{DN}-\frac 1{BM}=\frac 1{DM}-\frac 1{CN}}$ .


Proof. I"ll use the relations $\left\{\begin{array}{c}
\frac {DB}{c}=\frac {DC}{b}=\frac {a}{b+c}\\\\
AD^2=\frac {4bcs(s-a)}{(b+c)^2}\end{array}\right|$ and $\left\{\begin{array}{c}
2\cdot BM=c+BD-AD\\\\
2\cdot DN=DA+DC-b\end{array}\right|$. Thus, $\left\{\begin{array}{c}
b-DC=b-\frac {ab}{b+c}=\frac {2b(s-a)}{b+c}\\\\
c+BD=c+\frac {ac}{b+c}=\frac {2cs}{b+c}\end{array}\right|$ and $\frac {2}{AD}+\frac 1{BM}=\frac 1{DN}\iff$

$DN\cdot(2\cdot BM+AD)=AD\cdot BM\iff$ $(DA+DC-b)(c+BD)=AD(c+BD-AD)\iff$ $AD^2=(b-DC)(c+BD)\iff$ $AD^2=\frac {4bcs(s-a)}{(b+c)^2}\ ,$ truly.



P10. Let $\triangle ABC$ with $\{N,B'\}\subset (AC)$ and $\{M,C'\}\subset (AB)$ so that $O\in BB'\cap  CC'\cap MN$ . Prove that $\frac{MB}{MA}\cdot \frac{C'A}{C'B}+\frac{NC}{NA}\cdot \frac{B'A}{B'C}=1$ .

Proof. Denote $O'\in AO\cap BC$ and apply an well-known identity $\boxed{\frac {MB}{MA}\cdot \frac {CA'}{CB}+\frac {NC}{NA}\cdot\frac {BA'}{BC}=\frac {OA'}{OA}}\ (*)$ . Apply the Menelaus' theorem to the transversals:

$\left\{\begin{array}{cccc}
\overline{COC'}/\triangle ABA'\ : & \frac {CA'}{CB}\cdot\frac {C'B}{C'A}\cdot\frac {OA}{OA'}=1 & \implies & \frac {CA'}{CB}=\frac {C'A}{C'B}\cdot\frac {OA'}{OA}\\\\
\overline{BOB'}/\triangle ACA'\ : & \frac {BA'}{BC}\cdot\frac {B'C}{B'A}\cdot\frac {OA}{OA'}=1 & \implies & \frac {BA'}{BC}=\frac {B'A}{B'C}\cdot\frac {OA'}{OA}\end{array}\right\|\ \stackrel{(*)}{\implies}$ $\frac {MB}{MA}\cdot \frac {C'A}{C'B}\cdot\frac {OA'}{OA}+\frac {NC}{NA}\cdot\frac {B'A}{B'C}\cdot\frac {OA'}{OA}=\frac {OA'}{OA}\implies$ $\frac{MB}{MA}\cdot \frac{C'A}{C'B}+\frac{NC}{NA}\cdot \frac{B'A}{B'C}=1$ .



P11 (Samuel Palacios Paulino). Let $\triangle ABC$ with the incircle $w$ which touches $AB$ , $BC$ at $F$ , $D$ respectively. Denote the midpoint $M$ of $[BC]$ , the

projection $P$ of $A$ on $BC$ , the point $N\in AM$ so that $DN\parallel AB$ and the intersections $K\in BN\cap AD$ , $L\in DF\cap AP$ . Prove that $KL\parallel BC$ .


Proof. Suppose w.l.o.g. that $b>c$ . Observe that $DN\parallel AB\implies$ $\frac {KD}{KA}=\frac {DN}{AB}=\frac {MD}{MB}=\frac {\frac {b-c}2}{\frac a2}\implies$ $\boxed{\frac {KD}{KA}=\frac {b-c}a}\ (*)$ . The triangles $AFL$ and $PDL$ have $\widehat{ALF}\equiv\widehat{PLD}$

and $m\left(\widehat{AFL}\right)+m\left(\widehat{PDL}\right)=180^{\circ}$ . From an well-known property obtain that $\frac {LP}{LA}=$ $\frac {DP}{AF}\ \stackrel{(1)}{=}\ \frac {\frac {(b-c)(s-a)}a}{s-a}\implies$ $\boxed{\frac {LP}{LA}=\frac {b-c}a}\ \stackrel{(*)}{\implies}\ \frac {LP}{LA}=$ $\frac {KD}{KA}\implies KL\parallel BC\ .$

Remark 1. $b^2-c^2=PC^2-PB^2=2a\cdot PM\implies$ $DP=PM-DM=\frac {b^2-c^2}{2a}-\frac {b-c}2=$ $\frac {b-c}2\cdot\left(\frac {b+c}a-1\right)\implies$ $\boxed{DP=\frac {b-c)(s-a)}a}\ (1)\ .$

Remark 2. Apply the Menelaus' theorem to $\overline{DLF}/\triangle ABP\ :\ \frac {DP}{DB}\cdot \frac {FB}{FA}\cdot\frac {LA}{LP}=1\implies$ $\frac {LP}{LA}=\frac {DP}{FA}\ \stackrel{(1)}{=}\ \frac {\frac {(b-c)(s-a)}a}{s-a}\implies$ $\boxed{\frac {LP}{LA}=\frac {b-c}a}\ .$



P12 (Samuel Palacios Paulino). Let $\triangle ABC$ with the incircle $w=\mathbb C(I,r)$ which touches $AC$ , $AB$ at $E$ , $F$ respectively

and $P\in (EF)$ , $Q\in (BC)$ so that $I\in PQ$ and $[IQ$ is the bisector of $\widehat{BIC}$ . Prove that $IQ=2\cdot IP\iff A=60^{\circ}$ .


Proof. Prove easily that $\left\{\begin{array}{ccccc}
IB\cdot\sin \frac B2 & = & IE\cdot \cos\frac {A-C}2 & = & r\\\\
IC\cdot\sin \frac C2 & = & IF\cdot \cos\frac {A-B}2 & = & r\end{array}\right\|\ (*)$ . I"ll use the length $l_a$ of the $A$-bisector in the standard $\triangle ABC\ :\ l_a=\frac {2bc\cos\frac A2}{b+c}$ . Therefore,

$\left\{\begin{array}{c}
IQ=\frac {2\cdot IB\cdot IC}{IB+IC}\cdot\cos\widehat{QIC}\\\\
IP=\frac {2\cdot IE\cdot IF}{IE+IF}\cdot\cos\widehat{PIF}\end{array}\right\|\implies$ $2=$ $\frac {IQ}{IP}=\frac {IB}{IE}\cdot\frac {IC}{IF}\cdot\frac {IE+IF}{IB+IC}=$ $\frac {\frac 1{IF}+\frac 1{IE}}{\frac 1{IB}+\frac 1{IC}}\stackrel{(*)}{=}\frac {\cos\frac {A-B}2+\cos\frac {A-C}2}{\sin\frac C2+\sin\frac B2}\iff$ $2\left(\sin\frac C2+\sin\frac B2\right)=$ $\cos\frac {A-B}2+\cos\frac {A-C}2\iff$

$2\sin\frac {B+C}4=\cos\frac {B+C-2A}4\iff$ $\boxed{2\sin\left(45^{\circ}-\frac A4\right)=\cos \left(45^{\circ}-\frac {3A}4\right)}\ (1)$ . With $45^{\circ}-\frac A4=\phi$ , i.e. $\boxed{\frac A4=45^{\circ}-\phi}\ (2)$ , the relation $(1)$ becomes

$2\sin\phi =\cos\left(3\phi-90^{\circ}\right)\iff 2\sin\phi =\sin 3\phi\iff$ $2\sin\phi =\sin\phi\left(3-4\sin^2\phi\right)\iff$ $4\sin^2\phi=1\iff$ $\sin\phi =\frac 12\iff$ $\phi =30^{\circ}\stackrel{(2)}{\iff} A=60^{\circ}\ .$

Remark. Prove similarly that an easy its extension $:\ \boxed{IQ=k\cdot IP\ \iff\ \tan A=\frac {(k-1)\sqrt {(k+1)(3-k)}}{1+2k-k^2}}$ , where $1<k<1+\sqrt 2$ . Very nice problem!



P13. Let $\triangle ABC$ with orthic $\triangle DEF$ where $D\in (BC)$ , $E\in (CA)$ , $F\in (AB)$ , orthocenter $H$ and $P\in (EF)$ ,

$Q\in (BC)$ so that $H\in PQ$ and $[HQ$ is the bisector of $\widehat{BHC}$ . Prove that $HP=HQ\cdot\cos A\iff A=60^{\circ}$ .


Proof. Prove easily that $\left\{\begin{array}{ccccc}
HB=2R\cos B & ; & HE=2R\cos A\cos C\\\\
HC=2R\cos C & ; & HF=2R\cos A\cos B\end{array}\right\|\ (*)$ . I"ll use the length $l_a$ of the $A$-bisector in the standard $\triangle ABC\ :\ l_a=\frac {2bc\cos\frac A2}{b+c}$ . Therefore,

$\left\{\begin{array}{c}
HP=\frac {2\cdot HE\cdot HF}{HE+HF}\cdot\cos\widehat{PHF}\\\\
HQ=\frac {2\cdot HB\cdot HC}{HB+HC}\cdot\cos\widehat{QHC}
\end{array}\right\|\implies$ $\frac {HP}{HQ}=\frac {HE}{HB}\cdot\frac {HF}{HC}\cdot\frac {HB+HC}{HE+HF}\ \stackrel{(*)}{=}\ \frac {\cos A\cos C}{\cos B}\cdot$ $\frac {\cos A\cos B}{\cos C}\cdot\frac {\cos B+\cos C}{\cos A(\cos B+\cos C)}=\cos A\iff$ $HP=HQ\cdot\cos A$ .



P14..Let $ \omega$ be the circumcircle of $\triangle ABC$ and $ AA'$ be the diameter of $ \omega$ . Consider collinear points $ D\in BC$ , $ E\in AC$ , $ F\in AF$ . Prove that $ A'D\perp EF\iff \frac {A'E}{A'F}=\frac {A'C}{A'B}\ .$

Proof. Let $ A'D\perp EF$ we have $ A'B\perp AB$ therefore $ BDA'F$ is cyclic hence $ \angle DFA'=\angle DBA'=\angle A'AC$ hence $ AEA'F$ is cyclic. By similarity

of triangles $ EA'F$ and $ BA'C$ we have : $ \frac{A'F}{A'E}=\frac{A'B}{A'C}$ .Now consider $ \frac{A'F}{A'E}=\frac{A'B}{A'C}$ . Hence $ \frac{A'F}{A'B}=\frac{A'E}{A'C}$ and $ \angle A'CE=\angle A'BF=\frac{\pi}{2}$ .

Hence $ \angle A'FA=\angle A'EC$ , i.e. $ AEA'F$ is cyclic. Therefore $ \angle A'BC=\angle EAA'=\angle EFA'$ . Therefore $ BDA'F$ is cyclic then $ \angle FDA'=\frac{\pi}{2}\ .$



Lemma. For $\triangle ABC$ let incenter $I$, $A$-excenter $I_{a}\ ,$ reflection $A'$ of $A$ w.r.t. the circumcenter and second intersection $M$ of $\overline{AII_{a}}$ with the circumcircle.

Incircle and $A$- excircle touch $[BC]$ in $D$ and $D'$ respectively. Then the intersections $S\in IA'\cap DM$ and $S'\in I_{a}A'\cap D'M$ belong to the circumcircle.


Proof. Denote $p_{w}(X)$- the power of the point $X$ w.r.t. the circumcircle $w=C(O,R)$ of the triangle $ABC\ .$ I recall

the well-known relations : $-p_{w}(I)=IA\cdot IM=R^{2}-OI^{2}=2Rr$ and $p_{w}(I_{a})=I_{a}A\cdot I_{a}M=OI^{2}_{a}-R^{2}=2Rr_{a}\ .$

$\blacktriangleright$ $\widehat{DIM}\equiv\widehat{IAA'}$ and $\frac{ID}{IM}=\frac{AI}{AA'}$, i.e. $IA\cdot IM=2Rr$ $\Longrightarrow$ $\triangle IDM\sim\triangle AIA'$ $\Longrightarrow$ $\widehat{DMI}\equiv\widehat{IA'A}$ $\Longrightarrow$ $S\in w\ .$

$\blacktriangleright$ $\widehat{D'I_{a}M}\equiv\widehat{I_{a}AA'}$ and $\frac{I_{a}D'}{AI_{a}}=\frac{I_{a}M}{AA'}$, i.e. $I_{a}M\cdot I_{a}A=2Rr_{a}$ $\Longrightarrow$ $\triangle I_{a}D'M\equiv\triangle AI_{a}A'$ $\Longrightarrow$ $\widehat{D'MI_{a}}\equiv\widehat{I_{a}A'A}$ $\Longrightarrow$

$\widehat{AMS'}\equiv\widehat{AA'S'}$ $\Longrightarrow$ $S'\in w\ .$ Observe the "identity" between the two above proofs : $I\rightarrow I_{a}\ ;\ r\rightarrow r_{a}\ ;\ D\rightarrow D'\ ;\ \ S\rightarrow S'\ .$


Enunciation. For $\triangle ABC$ denote the incenter $I$, the $A$- exincenter and the reflection $A'$ of the point $A$ w.r.t. the circumcenter.

Define the intersections : $\left\{\begin{array}{c}E_{b}\in IA\cap BA'\ ,\ E_{c}\in IA\cap CA'\mathrm{\ .\ }\\\\F_{b}\in IA'\cap AC\ ,\ F_{c}\in IA'\cap AB\mathrm{\ .\ }\\\\F'_{b}\in I_{a}A'\cap AC\ ,\ F'_{c}\in I_{a}A'\cap AB\ .\end{array}\right\|$ . Prove that : $\left\{\begin{array}{c}1.\blacktriangleright\mathrm{The\ lines\ }BC\ ,\ E_{b}F_{b}\ ,\ E_{c}F_{c}\mathrm{\ are\ concurrently\ .}\\\\ 2.\blacktriangleright\mathrm{The\ lines\ }BC\ ,\ E_{b}F'_{b}\ ,\ E_{c}F'_{c}\mathrm{\ are\ concurrently\ .}\end{array}\right\|\ .$


Proof. I"ll use the notations from the above lemma. Apply the Pascal's theorem in the following two pairs of hexagons :

$\left\{\begin{array}{c}ABCA'SM\Longrightarrow D\in E_{c}F_{c}\\\\ ACBA'SM\Longrightarrow D\in E_{b}F_{b}\end{array}\right\|$ $\Longrightarrow \boxed{\ D\in E_{b}F_{b}\cap E_{c}F_{c}\ }\mathrm{\ \ \ ;\ \ \ }\left\{\begin{array}{c}ABCA'S'M\Longrightarrow D'\in E_{c}F'_{c}\\\\ ACBA'S'M\Longrightarrow D'\in E_{b}F'_{b}\end{array}\right\|$ $\Longrightarrow \boxed{\ D\in E_{b}F'_{b}\cap E_{c}F'_{c}\ }\ .$



P15. Let an acute $\triangle ABC$ with the orthocenter $H$ and the orthic $\triangle DEF$ , where $D\in BC$ , $E\in CA$ and $F\in AB$ . Prove that $:$

$1\blacktriangleright$ Denote $\left\{\begin{array}{ccc}
K\in AC\ ,\ DK\perp AC & ; & L\in AB\ ,\ DL\perp AB\\\\
X\in BK\cap DL & ; & Y\in CL\cap DK\end{array}\right\|$ . Prove $XY\parallel BC$ .

$2\blacktriangleright$ Denote the points $U\in (BF)\ ,\ V\in (CE)$ so that $H\in UV$ and $\left\{\begin{array}{ccc}
M\in BE & ; & UM\perp UV\\\\
N\in CF & ; & VN\perp VU\end{array}\right\|$ . Prove that $MN\parallel BC$ .


Proof.

$1\blacktriangleright$ Let $P$ , $Q$ be second intersections of $BK$ , $CL$ respectively with the circumcircle of $ALDK$ . Thus, $AL\cdot AB=AD^2=$

$AK\cdot AC\implies$ $BCKL$ cyclic, i.e. $\widehat{LKB}\equiv\widehat{LCB}$, i.e. $\overarc {PD}=\overarc{DQ}\implies$ $\widehat {YLX}\equiv \widehat{QLD}\equiv\widehat{PKD}\equiv\widehat{XKY}$ , i.e. $LXYK$ is cyclic.

$2\blacktriangleright$ $\triangle HUM\sim\triangle HEV, \triangle HUF\sim\triangle HNV\implies \frac{HM}{HN}=\frac{FH}{HE}=\frac{HB}{HC}\implies$ $\frac{HM}{HN}=\frac{HB}{HC}\implies$ $MN\parallel BC$ .



P16. Let $\triangle ABC$ and $D\in (BC)$ such that $AD$ bisects angle $A$ . IF $DA = 6$ , $DB = 4$ and $DC = 3$ , THEN find $AB$ and $AC$ .

Proof 1. $\boxed{\frac {DB}c=\frac {DC}b}\iff$ $\frac {4}c=\frac {3}b$ . Let the common second point $S$ of $AD$ with the circumcircle of $\triangle ABC$ . Thus, $\triangle ABS\sim\triangle ADC\iff$

$\frac {AB}{AD}=\frac {AS}{AC}\iff$ $AD\cdot AS=bc\iff$ $AD(AD+DS)=bc\iff$ $AD^2=bc-DA\cdot DS\iff$ $\boxed{AD^2=bc-DB\cdot DC}$

Thus, $36=bc-12$ , i.e. $bc=48$ . Therefore, $\frac {4}c=\frac {3}b=\sqrt{\frac {4\cdot 3}{bc}}=\sqrt {\frac {12}{48}}=\frac 12$ . In conclusion, $\frac 3b=\frac 4c=\frac 12\implies \odot \begin{array}{ccc}
\nearrow & b=6 & \searrow\\\\
\searrow & c=8 & \nearrow\end{array}\odot$ .

Proof 2. Exists $k>0$ so that $\left|\begin{array}{cc}
b=3k\\\
c=4k\end{array}\right|$ . Apply the Stewart's relation $:\ AC^2\cdot DB+AB^2\cdot DC=$ $AD^2\cdot BC+DB\cdot DC\cdot BC\iff$

$4b^2+3c^2=6^2(4+3)+3.4.7\iff$ $4(3k)^2+3(4k)^2=336\iff$ $k=2\implies \odot\begin{array}{ccc}
\nearrow & b=6 & \searrow\\\\
\searrow & c=8 & \nearrow\end{array}\odot$ .

Proof 3. Exists $k>0$ so that $\{\begin{array}{c}
b=3k\\\
c=4k\end{array}\|$ . Apply the generalized theorem of Cosines $:\ \frac {b^6+6^2-3^2}{2\cdot6\cdot b}=\cos\frac A2=\frac {c^2+6^2-4^2}{2\cdot 6\cdot c}\implies$

$\frac {b^2+27}b=\frac {c^2+20}{c}\iff$ $\frac {9k^2+27}{3k}=\frac {16k^2+20}{4k}\iff$ $3k^2+9=4k^2+5\iff$ $k^2=4\iff$ $k=2\iff\odot \begin{array}{ccc}
\nearrow & b=6 & \searrow\\\\
\searrow & c=8 & \nearrow\end{array}\odot$ .

Remark. If $DA=l\ ,\ DB=m$ and $DC=n$ , then $\left\{\begin{array}{c}
c=mk\ ;\ b=nk\\\\
l^2=bc-mn\end{array}\right\|\implies$ $k=\sqrt{1+\frac {l^2}{mn}}\implies$ $\boxed{\frac cm=\frac bn=\sqrt{1+\frac {l^2}{mn}}}$ .



P17. Let $\triangle ABC$ with the incenter $I$ . The circumcircle $w$ of $\triangle BIC$ cut the sidelines $AB$ , $AC$ in the points $E$ , $D$ respectively. Prove that $BE=CD$ .

Proof 1. Suppose w.l.o.g. $b<c$ , i.e. $B\in (AE)$ and $D\in (AC)$ . Thus, the $A$-excenter $I_a\in w$ , $[I_aI]$ is a diameter of $w$ and $\triangle ABI_a\sim\triangle AIC\iff$

$\frac {AB}{AI}=\frac {AI_a}{AC}\iff $ $\boxed{AI_a\cdot AI=bc}\ (*)\ .$ Apply the power of point $A$ w.r.t. $w\ :\ AB\cdot AE=AI\cdot AI_a=AC\cdot AD\ \stackrel{(*)}{\iff}$

$c\cdot AE=bc=b\cdot AD\ \odot\begin{array}{ccccccc}
\nearrow & AE=b & \implies & BE=AE-AB & \implies & BE=b-c & \searrow\\\\
\searrow & AD=c & \implies & CD=AC-AD & \implies & CD=b-c & \nearrow\end{array}\odot $ $\implies \boxed{BE=CD=b-c}\ .$

Proof 2. $A\in I_aI$ , where $[I_aI]$ is diameter of $w$ and $AI_a$ is the bisector of $\widehat {BAC}\implies$ $AB=AD$ , $AE=AC\implies AE-AB=AC-AD\implies BE=CD$ and $BD\parallel CE$ .



P18. Let an acute $\triangle ABC$ with incircle $ \omega = C(I,r)$ and circumcircle $ \rho = C(O,R)$. The circles $ c_{1}= C\left(P,r_{1}\right)$ and $ c_{2}= \mathbb C\left(Q,r_{2}\right)$ are tangent internally to $ \rho$ in the same point $ A$.

The circle $ w$ is tangent externally to the circle $ c_{1}$ and is tangent internally to the circle $ c_{2}$. Prove that $PQ =\frac{a^{2}(p-a)}{4S}$, where $p$ is the semiperimeter and $ S=[ABC]$ is the area for$ \triangle ABC$.


Remark. Prove easily that $\left \{\begin{array}{ccccc}
IP = r+r_{1} & ; & IQ = r_{2}-r\\\\
PO = R-r_{1} & ; & PA = r_{1}\\\\
QO = R-r_{2} & ; & QA = r_{2}\\\\
OA = R & ; & PQ = r_{2}-r_{1}\end{array}\right\|$. The relations $\left \{\begin{array}{ccc}
IO^{2}= R(R-2r) & ; & IA^{2}=\frac{bc(p-a)}{p}=\frac{4Rr(p-a)}{a}\\\\
IA^{2}-r^{2}= (p-a)^{2} & ; & IA^{2}+4Rr = bc\\\\
p(p-a)+(p-b)(p-c) = bc.& ; & p(p-a)(p-b)(p-c) = S^2\end{array}\right\|$ are well-known.

Proof 1. Denote $\left\{\begin{array}{ccc}
IO & = & m\\\
 IA & = & n\end{array}\right\|$. Apply the Stewart's theorem in $ \triangle OIA$ for the cevian-rays $ [IP$ and $ [IQ\ :\ \left\{\begin{array}{ccc}
m^{2}r_{1}^{2}+n^{2}(R-r_{1}) & = & R(r+r_{1})^{2}+Rr_{1}(R-r_{1})\\\\
m^{2}r_{2}+n^{2}(R-r_{2}) & = &  R(r_{2}-r)^{2}+Rr_{2}(R-r_{2})\end{array}\right\|$ $ \implies$

$ r_{1}(R^{2}+2Rr+n^{2}-m^{2}) = R(n^{2}-r^{2}) = r_{2}(R^{2}-2Rr+n^{2}-m^{2})$ $ \implies$ $ r_{1}(R^{2}+2Rr+bc-4Rr-R^{2}+2Rr) =$ $ R(p-a)^{2}=$ $ r_{2}(R^{2}-2Rr+bc-4Rr-R^{2}+2Rr)$ $ \implies$

$ r_{1}bc = R(p-a)^{2}= r_{2}(bc-4Rr)$ $ \implies$ $ PQ = r_{2}-r_{1}=$ $ \frac{4R^{2}r(p-a)^{2}}{bc(bc-4Rr)}=$ $ \frac{4R^{2}pr(p-a)^{2}}{b^{2}c^{2}(p-a)}=$ $ \frac{4R^{2}pra^{2}(p-a)^{2}}{16R^{2}p^{2}r^{2}(p-a)}$ $ \implies$ $ \boxed{\ PQ =\frac{a^{2}(p-a)}{4S}\ }$.

Proof 2. Apply the Pythagoras' theorem in the triangles $:\ \left\{\begin{array}{ccccc}
\triangle\ IOP & \implies & (r+r_{1})^{2}= (R-r_{1})^{2}+(R^{2}-2Rr)-2(R-r_{1})\cdot \cancel {OI} \cdot\frac{(R^{2}-2Rr)+R^{2}-\frac{4Rr(p-a)}{a}}{2R\cdot \cancel {IO}} & \implies & \boxed{r_{1}=\frac{R(p-a)^{2}}{bc}}\\\\
\triangle\ IOQ & \implies & (r_{2}-r)^{2}= (R-r_{2})^{2}+(R^{2}-2Rr)-2(R-r_{2})\cdot \cancel{IO}\cdot \frac{(R^{2}-2Rr)+R^{2} -\frac{4Rr(p-a)}{a}}{2R\cdot \cancel{IO}} & \implies & \boxed{r_{2}=\frac{Rp(p-a)}{bc}}\end{array}\right\|\ .$

Therefore, $ \frac{r_{1}}{p-a}=\frac{r_{2}}{p}=\frac{R(p-a)}{bc}=\frac{PQ}{a}$, i.e. $ \boxed{PQ =\frac{aR(p-a)}{bc}}=\frac{a^{2}(p-a)}{4S}$ $\implies\frac{r_{1}}{r}=\frac{r_{2}}{r_{a}}$, where $ r_{a}$ is the $ A$- exinradius of $ \triangle ABC$.

Remark. Prove easily that two more interesting relations : $ \boxed{\sum PQ = R+r\ }$ and $ r_{1}= R-\frac{a(4R+r)}{4p}$ , $ \boxed{r_{2}=\frac{ r_{b}+r_{c}}{4}}$.

.
This post has been edited 118 times. Last edited by Virgil Nicula, Aug 12, 2017, 8:57 AM

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