308. Some nice and easy properties in a triangle.

by Virgil Nicula, Aug 8, 2011, 2:10 AM

PP1. Let an acute $\triangle ABC$ with orthocenter $H$ . Let midpoints $E$ , $F$ of $[AC]$ , $[AB]$ . Prove that $AH$ is radical axis of circles with diameters $[BE]$ , $[CF]$ .

Let the circles $\left\{\begin{array}{c} y=C\left(Y,\frac {m_b}{2}\right)\\\\ z=C\left(Z,\frac {m_c}{2}\right)\end{array}\right\|$ with diameters $[BE]$ , $[CF]$ , where $Y$ , $Z$ are the midpoints of $[BE]$ , $[CF]$ . Let $\left\{\begin{array}{c} N\in BH\cap AC\\\\ P\in CH\cap AB\end{array}\right\|$ and the power $p_w(X)$ of $X$ w.r.t. $w$ .

Proof 1. $\left\{\begin{array}{c} N\in y\implies p_y(H)=\overline{HB}\cdot \overline{HN}\\\\ P\in z\implies p_z(H)=\overline{HC}\cdot \overline{HP}\end{array}\right\|$ $\implies$ $p_y(H)=p_z(H)=\overline{HA}\cdot \overline{HM}$ , $M\in AH\cap BC$ and $\left\{\begin{array}{ccc} N\in y\implies p_y(A)=\overline{AN}\cdot \overline{AF}=c\cdot\cos A\cdot\frac b2=\frac {bc\cdot\cos A}{2}\\\\ P\in z\implies p_z(A)=\overline{AP}\cdot \overline{AF}=b\cdot\cos A\cdot\frac c2=\frac {bc\cdot\cos A}{2}\end{array}\right\|$

$\implies$ $p_y(A)=p_z(A)=\frac {bc\cdot\cos A}{2}$ . Thus, $A$ and $H$ belong to the radical axis of $y$ , $z$ , i.e. $AH$ is the radical axis of $y$ , $z\ .$

Proof 2. $4p_y(H)=4\cdot HY^2-m_b^2=$ $2\cdot \left(HB^2+HE^2-m_b^2\right)=$ $2\left[HB^2+HN^2-\left(m_b^2-NE^2\right)\right]=$ $2\left(HB^2+HN^2-BN^2\right)=$

$2\left[\left(HB+HN\right)^2-2\cdot HB\cdot HN-BN^2\right]=-4\cdot HB\cdot HN\implies$ $p_y(H)=-HB\cdot HN$ . Prove analogously that $p_z(H)=-HC\cdot HP$ .

Since $HB\cdot HN=HC\cdot HP$ obtain that $H$ belongs to the radical axis of $y\ ,\ z$ . On the other side $4\cdot p_y(A)=4\cdot AY^2-m_b^2=$ $2\left(c^2+\frac {b^2}{4}-m_b^2\right)=$

$2c^2+\frac {b^2}{2}-\frac {2\left(a^2+c^2\right)-b^2}{2}=$ $2c^2+b^2-a^2-c^2\implies$ $p_y(A)=\frac 14\cdot\left(b^2+c^2-a^2\right)$ . Prove analogously that $p_z(A)=\frac 14\cdot\left(b^2+c^2-a^2\right)$ . Since

$p_y(A)=p_z(A)=\frac {bc\cdot\cos A}{2}$ obtain that $A$ belongs to the radical axis of $y$ , $z$ . Thus, $AH$ is the radical axis of $y$ , $z$ .

An easy extension. In $\triangle ABC$ let $\{E,N\}\subset (AC)$ and $\{F,P\}\subset (AB)$ so that $EF\parallel BC$ and $BCNP$

is cyclically. Let $L\in BN\cap CP$ . Prove that $AL$ is the radical axis of the circumcircles for $\triangle BEN$ and $\triangle CFP$ .

Proof. Denote the circumcircles $y$ , $z$ of $\triangle BEN$ , $\triangle CFP$ and the circumcircle $w$ of $BCNP$ . Observe that $\left\{\begin{array}{ccc} N\in y & \implies & p_y(L)=\overline{LB}\cdot \overline{LN}\\\\ P\in z & \implies & p_z(L)=\overline{LC}\cdot \overline{LP}\end{array}\right\|$ $\implies$

$p_y(L)=p_z(L)=p_w(L)$ . Thus, $L$ is the radical center of $y$ , $z$ and $w$ . Particularly, $L$ belongs to the radical axis of $y$ and $z$ . Let $p_w(L)=\overline{AN}\cdot \overline{AC}=\overline{AP}\cdot \overline{AB}=k$

and the common ratio $\frac {AE}{b}=\frac{AF}{c}=l$ . Thus, $\left\{\begin{array}{ccc} N\in y & \implies & p_y(A)=\overline{AN}\cdot \overline{AE}=\frac kb\cdot AE\\\\ P\in z & \implies & p_z(A)=\overline{AP}\cdot \overline{AF}=\frac kc\cdot AF\end{array}\right\|$ $\implies$ $p_y(A)=p_z(A)=kl$ . In conclusion, $A$ and $L$ belong

to the radical axis of the circles $y$ , $z$ , i.e. the line $AL$ is the radical axis of the circles $y$ , $z$ .

PP2 (Cono Sur M.O. - 2007). Let $ABC$ be an acute triangle with the orthocenter $H$ and the orthic triangle $DEF$ , where $D\in BC$ , $E\in CA$ , $F\in AB$ .

Denote midpoint $M$ of $[BC]$ , the circumcircle $w$ of $\triangle AEF$ , $\{A,N\}=AM\cap w$ , $Q\in BN\cap AD$ and $P\in AM\cap CF$ . Prove that $PQ\parallel BC$ .

Proof 1 (synthetic - Sunken Rock). Denote the diameter $[AS]$ in the circumcircle $w=C(O,R)$ of $\triangle ABC$ and the reflection $A^{\prime}$ of $A$ about the midpoint $M$ of $[BC]$ . Observe that

$ABA^{\prime}C$ is evidently a parallelogram, the quadrilateral $BA^{\prime}CH$ is cyclically, where $HA^{\prime}$ ia a diameter of its circumcircle. Since $AH$ and $AS$ are isogonal w.r.t. $\angle BAC$ obtain that

$AHA^{\prime}D$ is a parallelogram as well. So $\angle NBH=\angle NA^{\prime}H=$ $\angle SAM=$ $\angle CAM-\angle CAS=$ $\angle CAM-\angle HAB\ \ (1)$ . From $\triangle ABQ\ :\ \angle AQN=$ $\angle HQN=$

$\angle BAH+\angle ABQ=$ $\angle BAH+$ $\angle ABH+\angle HBN$ . With relation $(1)$ obtain that $\angle HQN=$ $\angle BAH+90^\circ-$ $A+$ $\angle CAM-$ $\angle BAH=$ $\angle CAM+$ $90^\circ-A \ \ (2)$ . Similarly, from

$\triangle ACP\ :\ \angle APH=$ $\angle NPH=$ $\angle CAM+$ $90^\circ-A=$ $\angle NQH$ . So the quadrilateral $HQPN$ is cyclically, but $HDMN$ is cyclic as well. Hence $PQ\parallel BC$ .

Proof 2 (metric). Prove easily that $ME$ and $MF$ are the tangents from the point $M$ to the circle $w$ and $ME=\frac a2$ . From the power $p_w(M)$ of $M$ w.r.t. $w$ obtain that $MN\cdot MA=ME^2$

i.e. $MN=\frac {a^2}{4m_a}$ and $NA=m_a-MN=\frac {4m_a^2-a^2}{4m_a}=\frac {2\left(b^2+c^2-a^2\right)}{4m_a}$ $\implies$ $\frac {NA}{NM}=\frac {2\left(b^2+c^2-a^2\right)}{a^2}\ (1)$ .Apply the Menelaus' theorem to the transversal $\overline {BQN}$ and

$\triangle ADM\ :\ \frac {BD}{BM}\cdot \frac {NM}{NA}\cdot\frac {QA}{QD}=1$ $\iff$ $\frac {QA}{QD}=$ $\frac {BM}{BD}\cdot \frac {NA}{NM}\stackrel{(1)}{=}$ $\frac {\frac a2}{c\cdot\cos B}\cdot\frac {2\left(b^2+c^2-a^2\right)}{a^2}\implies$ $\boxed{\frac {QA}{QD}=\frac {2\left(b^2+c^2-a^2\right)}{a^2+c^2-b^2}}\ (2)$ . Denote the point $S$ for which $SA\parallel BC$ .

Observe that $\frac {SA}{BC}=\frac {FA}{FB}\implies$ $\frac {SA}{a}=\frac {b\cdot\cos A}{a\cdot\cos B}\implies$ $SA=\frac {b\cdot\cos A}{\cos B}$ . Therefore, $\frac {PA}{PM}=\frac {SA}{MC}=$ $\frac {\frac {b\cdot\cos A}{\cos B}}{\frac a2}=\frac {2b\cdot\cos A}{a\cdot\cos B}\cdot \frac {2c}{2c}\implies$ $\boxed{\frac {PA}{PM}=\frac {2\left(b^2+c^2-a^2\right)}{a^2+c^2-b^2}}\ (3)$ .

In conclusion, from the relations $(2)$ and $(3)$ obtain that $PQ\parallel BC$ .

PP3. Let $\triangle ABC$ with circumcircle $\mathbb C(O,R)$ . Its bisectors meet again $C(O,R)$ in $A'$ , $B'$ , $C'$ , i.e. its

incenter $I\in AA'\cap BB'\cap CC'$ . Prove that $\frac{1}{[BA'C]}+\frac{1}{[CB'A]}+\frac{1}{[AC'B]} \ge\ \frac{9}{[ABC]}$ .

Proof. I"ll use the well-known relations $\left\{\begin{array}{cc} [BA'C]=\frac {ra^2}{4(s-a)} & (1)\\\\ \sum a^2(s-a)=4sr(R+r) & (2)\\\\ \frac {s^2}{r}\ge 16R-5r\ge 9(R+r) & (3)\end{array}\right|$ . Thus, $\sum\frac {1}{[BA'C]}\ \stackrel{(1)}{=}\ \frac 4r$ $\cdot\sum\frac {s-a}{a^2}\ =\ \frac 4r$ $\cdot\sum\frac {(s-a)^2}{a^2(s-a)}\ \stackrel{\mathrm{(C.B.S)}}{\ge }$

$\frac 4r$ $\cdot\frac {s^2}{\sum a^2(s-a)}\ \stackrel{(2)}{=}\ \frac 4r$ $\cdot \frac {s^2}{4sr(R+r)}\ =$ $\frac {s}{r^2(R+r)}\ =\ \frac {s^2}{9r(R+r)}$ $\cdot\frac 9S\ \stackrel{(3)}{\ge}\ \frac {16R-5r}{9(R+r)}$ $\cdot\frac 9S\ =\ \left(1+\frac 79\cdot\frac {R-2r}{R+r}\right)$ $\cdot\frac 9S\ \ge\ \frac 9S$ , where

$S=[ABC]$ is the area of $\triangle ABC$ . So $\boxed{\ \sum\frac {1}{[BA'C]}\ \ge\ \ \left(1+\frac 79\cdot\frac {R-2r}{R+r}\right)\cdot\frac {9}{[ABC]}\ \ge\ \frac {9}{[ABC]}\ }$ .

Remark. $\sum a^2(s-a)=4sr(R+r)\ \iff\ \sum\frac {a^2}{r_a}=4(R+r)$ . Using C.B.S. - inequality obtain that $4(R+r)\ge \frac {4s^2}{4R+r}$ , i.e. $s^2\le (R+r)(4R+r)$ .

PP4. Let $G$ be the centroid of $\triangle ABC$ and $S$ the symmedian point. Show that $\frac{AS}{AG}+\frac{BS}{BG}+\frac{CS}{CG}\le 3$ .

Proof. Denote the midpoint $M$ of the side $[BC]$ and $N\in AS\cap BC$ . Is well-known that $\frac {NB}{c^2}=\frac {NC}{b^2}=\frac {a}{b^2+c^2}$ . Apply van Aubel's relation to $S\ :\ \frac {AS}{b^2+c^2}=\frac {SN}{a^2}=\frac {AN}{a^2+b^2+c^2}$ .

Denote $AM=m_a\ ,\ AN=s_a\ ,\ m\left(\widehat{BAN}\right)=$ $m(\left(\widehat{CAM}\right)=\phi$ . Apply the Sinus' theorem to the triangles $:\ \left\{\begin{array}{cc} \triangle MAC\ : & \frac {MC}{\sin \phi}=\frac {m_a}{\sin C}\\\\ \triangle NAB\ : & \frac {s_a}{\sin B}=\frac {NB}{\sin\phi}\end{array}\right|\ \bigodot\ \implies\ \frac {s_a}{m_a}=\frac {2bc}{b^2+c^2}\implies$

$\frac {AS}{AG}=$ $\frac {\frac {s_a\left(b^2+c^2\right)}{a^2+b^2+c^2}}{\frac {2m_a}{3}}=$ $\frac {3\left(b^2+c^2\right)}{2\left(a^2+b^2+c^2\right)}\cdot\frac {s_a}{m_a}=$ $\frac {3\left(b^2+c^2\right)}{2\left(a^2+b^2+c^2\right)}\cdot\frac {2bc}{b^2+c^2}=$ $\frac {3bc}{a^2+b^2+c^2}$ a.s.o. $\implies$ $\sum\frac {AS}{AG}=\frac {3(ab+bc+ca)}{a^2+b^2+c^2}\le 3$ .

PP5. Let $\triangle ABC$ with circumcircle $w$ . Let $P\in AA\cap BC$ , the midpoint $M$ of $[AP]$ and $\{B,R\}=MB\cap w$ , $\{R,S\}=PR\cap w$ . Prove that $AP\parallel CS$ (V.N.).

Proof. $MP=MA\iff$ $MP^2=MA^2\iff MP^2=MB\cdot MR\iff$ $\triangle MPB\sim\triangle MRP\implies$

$\widehat{MPB}\equiv\widehat{MRP}\iff$ $\widehat{MPC}\equiv\widehat{BAS}\equiv$ $\widehat{BCS}\equiv\widehat{PCS}\implies$ $\widehat{MPC}\equiv$ $\widehat{PCS}\implies AP\parallel CS$

PP6. Let an isosceles $\triangle ABC$ with $b=c$ and a mobile $P\in BC$ so that $B\in (CP)$ . Let the incircle $w_1=C\left(I_1,r_{1}\right)$

of $\triangle APB$ and the $P$ -excircle $w_2=C\left(I_2,r_{2}\right)$ of $\triangle APC$ . Prove that the sum $r_{1}+r_{2}$ is constant.

Proof. Denote the incircle $w=C(I,r)$ of $\triangle ABC\ ,\ D\in w\cap BC$ and the tangent points $\left\{\begin{array}{cc} U\in w_1\cap BC\ ; & BU=u\\\\ V\in w_2\cap BC\ ; & CV=v\end{array}\right\|$ . Observe that

$\left\{\begin{array}{c} 2u=PB+c-PA\\\\ 2v=PA+b-PC\\\\ a=PC-PA\end{array}\right\|$ $\bigoplus\implies \boxed{u+v=s-a}\ (*)$ . Thus, $\left\{\begin{array}{ccccccc} \triangle BI_1U\sim\triangle IBD & \implies & \frac {BU}{ID}=\frac {I_1U}{BD} & \implies & \frac ur=\frac {r_1}{\frac a2} & \implies & 2rr_1=au\\\\ \triangle CI_2V\sim\triangle ICD & \implies & \frac {CV}{ID}=\frac {I_2V}{CD} & \implies & \frac vr=\frac {r_2}{\frac a2} & \implies & 2rr_2=av\end{array}\right\|$ $\bigoplus\stackrel{(*)}{\implies}$

$2r\left(r_1+r_2\right)=a(s-a)\implies$ $r_1+r_2=\frac {\frac a2}{\tan\frac A2}=\frac {BD}{\tan\frac A2}=AD\implies$ $r_1+r_2=h_a$ (constant)..

An easy extension. Let $\triangle ABC$ and a line $d$ so that $A\in d$ and $P\in BC\cap d\ ,\ B\in (PC)$ . Denote

the inradius $r_1$ of $\triangle APB$ and the $P$ -exradius $r_2$ of $\triangle APC$ . Prove that $\frac {r_1}{s-b}+\frac {r_2}{s-c}=\cot\frac A2$ .

Proof 1. Let the incircle $w=C(I,r)$ of $\triangle ABC\ ,\ D\in w\cap BC$ , the incircle $w_1=C\left(I_1,r_1\right)$ of $\triangle APB$ , the $P$ -excircle $w_2=C\left(I_2,r_2\right)$ of $\triangle APC$ and the

tangent points $\left\{\begin{array}{cc} U\in w_1\cap BC\ ; & BU=u\\\\ V\in w_2\cap BC\ ; & CV=v\end{array}\right\|$ . Thus, $\left\{\begin{array}{ccccc} \triangle I_1BU\sim\triangle BID & \implies & \frac {I_1U}{BD}=\frac {BU}{ID} & \implies & \frac {r_1}{s-b}=\frac ur\\\\ \triangle I_2CV\sim\triangle CID & \implies & \frac {I_2V}{CD}=\frac {CV}{ID} & \implies & \frac {r_2}{s-c}=\frac vr\end{array}\right\|$ $\implies$ $\frac {r_1}{s-b}+\frac {r_2}{s-c}=\frac {u+v}{r}\ (*)$ .

In conclusion, $\left\{\begin{array}{c} 2u=BX+BA-AX\\\\ 2v=AX+AC-CX\\\\ BC=CX-BX\end{array}\right\|\bigoplus$ $\implies$ $2(u+v)+BC=BA+AC\implies$ $u+v=s-a\stackrel{(*)}{\implies}$ $\frac {r_1}{s-b}+\frac {r_2}{s-c}=\frac {s-a}{r}\implies$

$r_1\tan\frac B2+r_2\tan\frac C2=r\cot\frac A2\implies$ $\boxed{\frac {r_1}{s-b}+\frac {r_2}{s-c}=\cot\frac A2}\implies$ $r_1(s-c)+r_2(s-b)=rs$ . Remark that $b=c\implies r_1+r_2=h_a$ .

Proof 2. The incircle $C(I_{1},r_{1})$ of $\triangle APB$ touches its sides in $X\in (AP)\ ,\ U\in (PB)\ ,\ R\in (AB)$ . The $P$ - exincircle $C(I_{2},r_{2})$ of $\triangle APC$ touches its sidelines in

$Y\in AP\ ,\ V\in PC\ ,\ S\in (AC)$ . Denote $AX=AR=x$ , $AY=AS=y$ . Thus, $BR=BU=c-x$ , $CS=CV=b-y$ and $UV=UB+BC+CV$ $\Longrightarrow$

$UV=(c-x)+a+(b-y)=$ $2s-(x+y)$ . From the relation $XY=UV$ obtain $x+y=2s-(x+y)$ , i.e. $\boxed{\ x+y=s\ }\ \ (1)$ . From the right $\triangle BI_{1}U$ obtain

$\tan\frac{B}{2}=\frac{RB}{RI_{1}}=\frac{c-x}{r_{1}}$ , i.e. $\boxed{\ x=c-r_{1}\tan\frac{B}{2}\ }\ \ (2)$ . From the right triangle $CI_{2}V$ obtain $\tan\frac{C}{2}=\frac{VC}{VI_{2}}=\frac{b-y}{r_{2}}$ , i.e. $\boxed{\ y=b-r_{2}\tan\frac{C}{2}}\ \ (3)$ . Adding $(2)\ ,\ (3)$

and using $(1)$ obtain $\boxed{\ r_{1}\tan\frac{B}{2}+r_{2}\tan\frac{C}{2}=s-a\ }\ \ (*)$ $\frac{p-c}{\tan \frac{B}{2}}=\frac{p-b}{\tan\frac{C}{2}}=\frac {S}{p-a}\iff$ $r_{1}(p-c)+r_{2}(p-b)=S$ $\Longrightarrow$ $\frac{r_{1}}{p-b}+\frac{r_{2}}{p-c}=\frac{S}{(p-b)(p-c)}\iff$

$\boxed{\ \frac{r_{1}}{p-b}+\frac{r_{2}}{p-c}=\cot\frac{A}{2}\ }$ .

PP7. Let $I$ be the incenter of $\triangle ABC$ and let $w_a$ be the $A$ -exincircle which touches $BC$ at $D$ . If $ID$ meets $w_a$ again at $S$ , then prove that $DS$ bisects $\angle BSC$ .

Proof. Suppose w.l.o.g. $b\ne c$ . Denote $L\in BC$ for which $IL\perp BC$ and $N\in BC$ so that $D$ , $N$ are harmonical conjugate w.r.t the pair $\{B,C\}$ , i.e. $\frac {NB}{NC}=\frac {DB}{DC}=\frac {s-c}{s-b}$ . Prove

easily that $DN=\frac {2(s-b)(s-c)}{|b-c|}$ and $\boxed{DN\cdot DL=2(s-b)(s-c)}\ (1)$ . Observe that $DI\cdot DS=ID\cdot (IS-ID)=$ $p_{w_a}(I)-ID^2=$ $II_a^2-r_a^2-r^2-(b-c)^2=$

$\frac {a^2}{s^2}\cdot AI_a^2-r_a^2-r^2-(b-c)^2=$ $\frac {a^2}{s^2}\cdot\left(s^2+r_a^2\right)-r_a^2-r^2-(b-c)^2=$ $4(s-b)(s-c)+$ $\frac {a^2r^2}{(s-a)^2}-$ $\frac {s^2r^2}{(s-a)^2}-r^2=$ $4(s-b)(s-c)-$

$\frac {r^2}{(s-a)^2}\cdot\left\{(s-a)^2+s^2-\left[s-(s-a)\right]^2\right\}=$ $4(s-b)(s-c)-\frac {r^2}{s-a}\cdot 2s=$ $4(s-b)(s-c)-2(s-b)(s-c)$ $\implies$ $\boxed{DI\cdot DS=2(s-b)(s-c)}\ (2)$ . Thus, from

$(1)$ , $(2)$ results that $DI\cdot DS=DN\cdot DL$ , i.e. $ILSN$ is cyclic. Thus, $SD\perp SN$ , i.e. the ray $[SD$ bisects the angle $\widehat{BSC}$ because the division $(B,D,C,N)$ is harmonically.

Remark. I used the power $p_{w_a}(I)$ of $I$ w.r.t. $w_a$ . If the circle $w_a$ touches $AC$ , $AB$ on $U$ , $V$ respectively, then $N\in UV$ . Indeed, apply the Menelaus' theorem to

the transversal $\overline{NUV}$ and $\triangle ABC\ :\ N\in UV\iff$ $\frac {VB}{VA}\cdot \frac {UA}{UC}\cdot\frac {NC}{NB}=1\iff$ $\frac {s-c}{s}\cdot\frac {s}{s-b}\cdot\frac {NC}{NB}=1\iff$ $\frac {NB}{NC}=\frac {s-c}{s-b}$ , what is truly.

PP8. Let $A_i\ ,\ i=\overline{1,n}$ be the points that divide the side $[BC]$ into $n+1$ equal segments such that $BA_1=A_1A_2=\ldots=A_{n-1}A_n=A_nC$ and also let $\Gamma_a=\sum_{i=1}^nAA_i^2$ .

Similarly we define $B_i$ and $C_i$ , where $i=\overline{1,n}$ as well as $\Gamma_b$ and $\Gamma_c$ . With the mentioned notations, we will prove that : $\boxed{\ \frac {\Gamma_a+\Gamma_b+\Gamma_c}{a^2+b^2+c^2}=\frac {n\left(5n+4\right)}{6\left(n+1\right)}\le\frac {5n}{6}\ }$ .

Proof. Apply the Stewart's theorem to the cevian $AA_k$ in the triangle $ABC$ for any $k\in\overline {1,n}$ . Since $\left\{\begin{array}{c} A_kB=\frac {ka}{n+1}\\\\ A_kC=\frac {(n+1-k)a}{n+1}\end{array}\right|$ , so $AA_k^2=-\frac {k(n+1-k)}{(n+1)^2}\cdot a^2+\frac {k}{n+1}\cdot b^2+$

$\frac {n+1-k}{n+1}\cdot c^2\implies$ $\Gamma_a\equiv\sum_{k=1}^nAA_k^2=$ $-\frac {a^2}{(n+1)^2}\cdot\left[\frac {n(n+1)^2}{2}-\frac {n(n+1)(2n+1)}{6}\right]+$ $\frac {b^2}{n+1}\cdot \frac {n(n+1)}{2}+\frac {c^2}{n+1}\cdot \left[n(n+1)-\frac {n(n+1)}{2}\right]\implies$

$\Gamma_a=-\frac {na^2}{n+1}\cdot\left[\frac {n+1}{2}-\frac {2n+1}{6}\right]+$ $b^2\cdot \frac {n}{2}+c^2\cdot \left[n-\frac {n}{2}\right]\implies$ $\boxed{\Gamma_a=-\frac {n(n+2)}{6(n+1)}\cdot a^2+\left(b^2+c^2\right)\cdot \frac n2}\ \ (*)$ . In conclusion, from the relation $(*)$ obtain that

$\frac {\Gamma_a+\Gamma_b+\Gamma_c}{a^2+b^2+c^2}=-\frac {n(n+2)}{6(n+1)}+n=\frac {n\left(5n+4\right)}{6\left(n+1\right)}$ .

PP9. Let $ABC$ be a triangle with the circumcircle $w$ . The $A$ -symmedian of $\triangle ABC$ meet again the circle $w$ at $D$ . Denote the midpoint $E$ of $[AD]$ . Prove that $m\left(\widehat{BEC}\right)=2A$ .

Proof. Denote the midpoints $M$ , $N$ , $P$ of $[BC]$ , $[CA]$ , $[AB]$ respectively and the intersection $S\in AD\cap BC$ .

Thus, $\left\{\begin{array}{ccc} \triangle ABD\sim\triangle AMC & \implies & m(\angle BED)=m(\angle MNC)=A\\\\ \triangle ACD\sim\triangle AMB & \implies & m(\angle CED)=m(\angle MPB)=A\end{array}\right|\implies$ $m(\angle BEC)=2A$ .

An easy extension. Let $ABC$ be a triangle with the circumcircle $w$ . Consider two points $\{M,S\}\subset (BC)$ so that $S\in (BM)$ and $\widehat{SAB}\equiv\widehat {MAC}$ . Denote

$AS\cap w=\{A,D\}$ and the points $\{E,F\}\subset (AS)$ so that $\frac {EA}{ED}=\frac {MB}{MC}=\frac {FD}{FA}$ . Denote $K\in BE\cap CF$ . Prove that $KE=KF$ and $m(\angle BKC)=2A$ .

PP10. Let $ABC$ be an acute triangle with the circumcircle $w=C(O,R)$ and the orthocenter $H$ . Denote the diameter

$[AS]$ of circle $w$ , the point $D\in AH\cap BC$ and $\left\{\begin{array}{c} K\in (AB)\ ,\ BD\\\ L\in (AC)\ ,\ CL=CD\end{array}\right|$ . Prove that $SK=SL$ .

Proof 1 (luisgeometria, nice). Denote the reflections $U$ , $V$ of $K$ , $L$ w.r.t. $B$ , $C$ respectively, i.e. $SB$ and $SC$ are the perpendicular bisectors of $[KU]$

and $[LV]$ respectively. The linee $AD$ is the common tangent of the circles $w_b=C(B,BH)$ and $w_c=C(C,CH)$ $\implies$ $AD$ is their radical axis $\Longrightarrow$

$p_{w_b}(A)=p_{w_c}(A)$ $\Longrightarrow$ $AK \cdot AU=AL \cdot AV$ $\Longrightarrow$ $KLVU$ is a cyclical quadrilateral with circumcenter $S$ $\Longrightarrow$ $SK=SL$ as desired.

Proof 2 (metric). $\triangle SBK$ is right $\implies$ $SK^2+AD^2=\left(BK^2+BS^2\right)+AD^2=$ $BS^2+\left(DB^2+DA^2\right)=$

$BS^2+BA^2=AS^2\implies$ $SK^2=4R^2-h_a^2$ (symmetrically w.r.t. $b$ , $c$ ). In conclusion, $\boxed{SK=SL=\sqrt {4R^2-h_a^2}}$ .

Proof 3 (metric). $HBSC$ is a parallelogram $\iff$ $\left\{\begin{array}{c} HB=SC\\\ HC=SB\end{array}\right|$ . Thus, $HD\perp BC\iff$ $HB^2-HC^2=DB^2-DC^2\iff$

$SC^2-SB^2=BK^2-CL^2\iff$ $BS^2+BK^2=CS^2+CL^2\iff$ $SK^2=SL^2\iff$ $SK=SL$ .

Proof 4 (trigonometric). $BK=BD=c\cdot \cos B=2R\cdot\sin C\cos B$ and $BS=2R\cdot \cos C\implies$ $SK=BK^2+BS^2=$

$4R^2\cdot \left(\sin^2C\cos^2B+\cos^2C\right)=$ $4R^2\cdot \left(\sin^2C\cos^2B+1-\sin^2C\right)=$ $4R^2\cdot \left[1-\sin^2C\left(1-\cos^2B\right)\right]=$ $4R^2\cdot \left(1-\sin^2B\sin^2C\right)=$

$4R^2- b\sin C\cdot c\sin B=$ $4R^2-h_a^2\implies$ $SK^2=4R^2-h_a^2$ (symmetrically w.r.t. $b$ , $c$ ). In conclusion, $\boxed{SK=SL=\sqrt {4R^2-h_a^2}}$ .

PP11. $ABC$ is an equilateral triangle and $d$ is a line tangent to its circumcircle $w=C(O,R)$ . Let the distances from

$A$ , $B$ , $C$ onto $d$ be $d_1$ , $d_2$ , $d_3$ . Prove that one of the values $\sqrt{d_1}$ , $\sqrt{d_2}$ , $\sqrt{d_3}$ equals the sum of the other two.

Proof. Suppose that $d$ is tangent to $w$ at $P$ and $P'$ is the antipode of $P$ . If $AX\perp d$ , where $X\in d$ , then $AX=d_1$ and $\Delta PAX\equiv \Delta P'PA$ , i.e. $\frac {PA}{AX}=\frac {P'P}{PA}$ . So $PA^2=d_1\cdot 2R$

So $\sqrt {d_1}=\frac {PA}{\sqrt {2R}}$ . So from the Pompeiu's theorem the result follows. Apply the Ptolemy's theorem to the cyclical $PABC$ and obtain that one of the values $PA$ , $PB$ , $PC$ is equally to

the sum of the other two, i.e. one of the values $\sqrt{d_1}$ , $\sqrt{d_2}$ , $\sqrt{d_3}$ is equally to the sum of the other two.

PP12. Let $\triangle ABC$ with the incircle $w=C(I,r)$ and the $A$ -exincircle $w_a$ . The circle $w$ touches $\triangle ABC$ in $D\in (BC)$ , $E\in (CA)$ and $F\in (AB)$ .

Denote $\{D,Q\}=DI\cap w$ , $M\in BC\cap w_a$ , $N\in AB\cap w_a$ and $P\in QE\cap MN$ . Prove that $CP\perp BC$ .

Proof. Denote $\left\{\begin{array}{cc} X\in MN\ ; & XC\perp BC\\\\ Y\in EQ\ : & YC\perp BC\end{array}\right\|$ , i.e. $C\in XY$ and $\overline {XYC}\perp BC$ . Prove easily that $\left\{\begin{array}{ccccccc} DB=MC & \mathrm{and} & MN\parallel BI & \implies & \triangle DBI\equiv\triangle CMX & \implies & CX=r\\\\ IQ\parallel CY & \mathrm{and} & QE\parallel CI & \implies & CIQY\ \mathrm{is\ parallelogram} & \implies & CY=r\end{array}\right\|$ $\implies$

$X\equiv Y\equiv P\implies$ $CP\perp BC$ . Remark. Denote $R\in AC\cap w_a$ and $S\in QF\cap RM$ . Prove analogously that $SB\perp BC$ .

PP13. Let $A_1$ be the center of the rectangle $MNPR$ inscribed in acute triangle $ABC$ with $\{M,N\}\subset BC$ and $P\in (AC)$ , $R\in (AB)$ so that $\frac {PN}{PR}=k$ (constant).

The points $B_1$ and $C_1$ are defined in a similar way for inscribed rectangles with two vertices on $AC$ and $AB$ respectively. Prove that $AA_1\cap BB_1\cap CC_1\ne\emptyset$ .

Proof 1. Let $X\in AA_1\cap PR\ ,\ A_2\in AA_1\cap BC$ and $m\left(\widehat{PRN}\right)=\phi$ . Thus, $A_1P=A_1R$ , $\tan\phi =k$ and $\frac {A_2B}{A_2C}=$ $\frac {XR}{XP}=$ $\frac {[ARA_1]}{[APA_1]}=$ $\frac {RA\cdot RA_1\cdot\sin\widehat{ARA_1}}{PA\cdot PA_1\cdot\sin\widehat{APA_1}}=$

$\frac {AR}{AP}\cdot \frac {\sin (B+\phi )}{\sin (C+\phi )}$ $\implies$ $\boxed{\frac {A_2B}{A_2C}=\frac cb\cdot\frac {\sin (B+\phi )}{\sin (C+\phi )}}$ . Get analogously that $\frac {B_2C}{B_2A}=\frac ac\cdot\frac {\sin (C+\phi )}{\sin (A+\phi )}$ and $\frac {C_2A}{C_2B}=\frac ba\cdot\frac {\sin (A+\phi )}{\sin (B+\phi )}$ . Apply the Ceva's theorem

to the cevians $AA_2$ , $BB_2$ , $CC_2$ and obtain $\frac {A_2B}{A_2C}\cdot\frac {B_2C}{B_2A}\cdot\frac {C_2A}{C_2B}=1\iff$ $AA_2\cap BB_2\cap CC_2\ne\emptyset$ , i.e. $AA_1\cap BB_1\cap CC_1\ne\emptyset$ .

Proof 2. Let $X\in AA_1\cap PR\ ,\ A_2\in AA_1\cap BC$ , $MN=PR=u$ , $MR=NP=v$ and $m\left(\widehat{PRN}\right)=\phi$ . Thus, $\tan\phi =\frac vu=k$ (constant), $XR=A_2N$ , $XP=A_2M$ ,

$BM=v\cdot \cot B$ and $CN=v\cdot \cot C$ . Thus, $\frac {A_2B}{A_2C}=$ $\frac {XR}{XP}=$ $\frac {A_2N}{A_2M}=$ $\frac {A_2B+A_2N}{A_2C+A_2M}=$ $\frac {BM+u}{CN+u}=$ $\frac {v\cdot\cot B+u}{v\cdot \cot C+u}=$ $\frac {\frac vu+\tan B}{\frac vu+\tan C}\implies$ $\boxed{\frac {A_2B}{A_2C}=\frac {k+\tan B}{k+\tan C}}$ a.s.o.

PP14. Let an acute $\triangle ABC$ with orthocentre $H$ and circumcircle $w=C(O)$ so that $c<a$ . For $P\in (AC)$ define $\left\{\begin{array}{ccc} R\in (BP)\ : & OR\parallel AC\\\ Q\in (AC)\ : & HQ\parallel BP\end{array}\right\|$ . Prove that $RP=RQ$ .

Remark. For $P\in BO\cap AC$ obtain the Yasinsky's problem P414 from "Words of mathematics" (see here).

Proof. Let $\{B,H\}\cap w=\{B,K\}$ and $Q'\in KR\cap AC$ , $E\in BH\cap AC$ . Thus, $EH=EK$ and $BE\perp AC$ $\implies Q'H=Q'K\implies$ $\widehat{Q'HK}\equiv$ $\widehat{Q'KH}\equiv$ $\widehat{RKB}$ and

$RB=RK\implies$ $\widehat{RKB}$ $\equiv\widehat{RBK}$ . Thus, $\widehat{Q'HK}\equiv\widehat{RBK}$ , i.e. $HQ'\parallel BP\implies$ $Q'\equiv Q$ . Hence $\widehat{RQP}\equiv\widehat{EQK}\equiv\widehat{EQH}\equiv\widehat{QPR}\implies$ $\widehat{RQP}\equiv\widehat{QPR}\implies$ $RP=RQ$ .

Lemma. Let $\triangle ABC$ with $AB<AC$ , the median $AM$ and the bisector $AD$ , where $\{M,D\}\subset BC$ . Prove that $\tan\widehat{MAD}=\frac {b-c}{b+c}\cdot\tan\frac A2=\tan^2\frac A2\tan\frac {B-C}{2}$ .

Proof 1. Let $\phi =m\left(\widehat{MAD}\right)$ and prove easily that $MD=\frac {a(b-c)}{2(b+c)}$ . Apply an well-known property to the ray $[AM$ in the triangle $ACD\ :\ \frac {MD}{MC}=$ $\frac {AD}{AC}\cdot\frac {\sin\widehat{MAD}}{\sin\widehat{MAC}}$ $\implies$ $\frac {b-c}{b+c}=$

$\frac {2bc\cdot \cos\frac A2}{b(b+c)}\cdot \frac {\sin\phi}{\sin \left(\frac A2-\phi \right)}$ $\implies$ $\frac {b-c}{2c}=$ $\frac {\cos\frac A2\sin\phi}{\sin\frac A2\cos\phi -\cos\frac A2\sin\phi}$ $\implies$ $\frac {b-c}{2c}=$ $\frac {\tan\phi}{\tan\frac A2-\tan\phi}$ $\implies$ $\boxed{\tan\phi =\frac {b-c}{b+c}\cdot\tan\frac A2}$ . Prove easily that $\frac {b-c}{b+c}=$ $\tan\frac A2\tan\frac {B-C}{2}$ .

Indeed, $\frac {b-c}{b+c}=\frac {\sin B-\sin C}{\sin B+\sin C}=$ $\frac {\sin\frac {B-C}{2}\cos\frac {B+C}{2}}{\sin\frac {B+C}{2}\cos\frac {B-C}{2}}=$ $\cot\frac {B+C}{2}\tan\frac {B-C}{2}=$ $\tan\frac A2\tan\frac {B-C}{2}$ . Thus, $\boxed{\tan\phi =\tan^2\frac A2\tan\frac {B-C}{2}}$ .

Proof 2 (similar). Let $\phi =m\left(\widehat{MAD}\right)$ and prove easily that $MD=\frac {a(b-c)}{2(b+c)}$ . Apply an well-known property to the ray $[AM$ in $\triangle ABD\ :$ $\frac {MD}{MB}=$ $\frac {AD}{AB}\cdot\frac {\sin\widehat{MAD}}{\sin\widehat{MAB}}$ $\iff$

$\frac {\frac {a(b-c)}{2(b+c)}}{\frac a2}=$ $\frac {\frac {2bc\cdot \cos\frac A2}{(b+c)}}{c}\cdot \frac {\sin\phi}{\sin \left(\frac A2+\phi \right)}$ $\iff$ $\frac {b-c}{2b}=\frac {\cos\frac A2\sin\phi}{\sin\frac A2\cos\phi +\sin\phi \cos\frac A2}$ $\iff$ $\frac {2b}{b-c}=1+\tan\frac A2\cot\phi\iff$ $\frac {b+c}{b-c}=\tan\frac A2\cot\phi\iff$ $\boxed{\tan\phi =\frac {b-c}{b+c}\cdot \tan\frac A2}$ .

PP15. Give $\triangle ABC$ , medians $AM$ , $AN$ and bisector of $ABC$ . The line passing $N$ which perpendicular with $AN$ meet

$AB$ , $AM$ at $P$ , $Q$ . The line pasing through $P$ which perpendicular with $AB$ meet $AN$ at $O$ . Prove that $OQ \bot BC$ .

Proof 1 (metric). Suppose w.l.o.g. $c<a$ and denote $AQ=l$ , $\phi =m\left(\widehat{MAN}\right)$ and the projections $U$ , $V$ of $Q$ , $O$ on $BC$ . Observe that $m\left(\widehat{QNC}\right)=$ $\frac {B-C}{2}$

and $m\left(\widehat{ONC}\right)=$ $90^{\circ}-\frac {B-C}{2}$ . Thus, $NQ=AQ\cdot\sin\widehat{MAN}=$ $l\sin\phi$ $\implies$ $NU=NQ\cdot\cos\widehat{QNC}=$ $l\sin\phi\cos\frac {B-C}{2}$ $\implies$ $\boxed{NU=l\sin\phi\cos\frac {B-C}{2}}\ (1)$ .

Analogously $AN=AQ\cos\widehat{MAN}=$ $l\cos\phi$ $\implies$ $PN=AN\cdot\tan\frac A2=$ $l\cos\phi\tan\frac A2$ $\implies$ $PN^2=NA\cdot NO$ $\implies$ $ON=l\cos\phi\tan^2\frac A2$ $\implies$

$NV=ON\cos\widehat{ONC}=$ $l\cos\phi\tan^2\frac A2\sin\frac {B-C}{2}$ $\implies$ $\boxed{NV=l\cos\phi\tan^2\frac A2\sin\frac {B-C}{2}}\ (2)$ . Thus, $OQ \bot BC\iff$ $NU=NV\stackrel{1\wedge 2}{\iff}$

$l\sin\phi\cos\frac {B-C}{2}=l\cos\phi\tan^2\frac A2\sin\frac {B-C}{2}\iff$ $\boxed{\tan\phi=\tan^2\frac A2\tan\frac {B-C}{2}}$ (see here).

Remark. Let $S\in AC\cap PQ$ and Menelaus' theorem to $:\ \overline{NQS}/\triangle ACM\ : \frac {NM}{NC}$ $\cdot\frac {SC}{SA}\cdot\frac {QA}{QM}=1\iff$ $\frac {\frac {a(b-c)}{2(b+c)}}{\frac {ab}{b+c}}\cdot \frac {\frac {b(b-c)}{b+c}}{\frac {2bc}{b+c}}\cdot\frac {QA}{QM}=1\iff$ $\boxed{\frac {QA}{4bc}=\frac{QM}{(b-c)^2}=\frac {m_a}{(b+c)^2}}$ .

Proof 2. Menelaus' theorem to $:\ \overline{AQM}/\triangle PBN\ : \frac {QN}{QP}$ $\cdot\frac {AP}{AB}\cdot\frac {MB}{MN}=1\iff$ $\frac {QN}{QP}\cdot\frac {\frac {2bc}{b+c}}{c}\cdot\frac {\frac a2}{\frac {a(b-c)}{2(b+c)}}=1\iff$ $\boxed{\frac {QN}{b-c}=\frac {QP}{2b}=\frac {PN}{b+c}}\ (*)$ . Let $\left\{\begin{array}{c} D\in BC\\\ AD\perp BC\end{array}\right\|$ . Thus,

$OQ\perp BC\iff$ $OQ\parallel AD\iff$ $\triangle ADN\sim\triangle ONQ\iff$ $\frac {AD}{DN}=\frac {ON}{NQ}\iff$ $\frac {AD}{DN}=\frac {ON}{PN}\cdot\frac {PN}{NQ}\stackrel{(*)}{\iff}$ $\cot\frac {B-C}{2}=\tan\frac A2\cdot \frac {b+c}{b-c}$ , what is true.

PP16. Let an $A$ -isosceles $\triangle ABC$ and a mobile $P\in BC$ so that $B\in (PC)$ . Let $r_{1}$ be inradius of $\triangle APB$ and let $r_{2}$ be $P$ -exradius of $\triangle APC$ . Prove that the sum $r_{1}+r_{2}$ is constant.

Proof. Apply the Stewart's relation to the cevian $AB$ in $\triangle APC\ :\ AP^2\cdot BC+AC^2\cdot PB=AB^2\cdot PC+PB\cdot PC\cdot BC\iff$ $\boxed{PA^2=PB\cdot PC+AB^2}\ (*)$ .

Let the midpoint $M$ of $[BC]$ , $AM=h$ , $x$ - the length of the inradius for $\triangle APB$ and $y$ - the length of the $P$ -exradius for $\triangle APC$ . Therefore, $x+y=h\iff$ $\frac xh+\frac yh=1\iff$

$\frac {PB}{PA+PB+AB}+$ $\frac {PC}{PA+PC-AC}=1$ $\stackrel{(\mathrm{AB=AC})}{\iff}$ $PB(PA+PC-AB)+\underline{PC}(PA+PB+AB)=$ $(PA+PB+AB)(PA+\underline{PC}-AB)$ $\iff$

$\underline{PB}(PA+PC-AB)=$ $(PA+\underline{PB}+AB)(PA-AB)\iff$ $PB\cdot PC=(PA+AB)(PA-AB)\iff$ $PA^2=PB\cdot PC+AB^2\iff (*)$ .

Remark. Other proof of the relation $(*)\ :\ AM\perp PB\iff$ $AP^2-AB^2=MP^2-MB^2\stackrel{(\mathrm{MB=MC})}{\iff}$ $AP^2-AB^2=(MP-MB)(MP+MC)\iff$

$PA^2=PB\cdot PC+AB^2$ .Observe that an interesting property: if $x'$ is the length of the inradius for $\triangle PAC$ and $y'$ is the length of the $P$ -exradius for $\triangle PAB$ ,

then $x'+y'=h$ . Its proof is symmetrically w.r.t. $B$ and $C$ .

PP17. Let $\triangle ABC$ with circumcircle $w=C(O,R)$ and the exincircle $w_a=C(I_a,r_a)$ . Let the feet $E\in AC$ , $F\in AB$ of bisectors from $B$ and $C$ . Prove that $OI_{a} \perp EF$ .

Proof 1. Suppose w.l.o.g. $c<b$ and denote $L\in EF\cap BC$ , $T\in BC\cap w_a$ .

$\blacktriangleright\ m\left(\widehat{CLE}\right)=\phi\implies$ $\frac {LB}{\sin (B-\phi )}=\frac {BF}{\sin \phi}\iff$ $\frac {ac}{(b-c)\sin (B-\phi )}=\frac {ac}{(a+b)\sin\phi}\iff$ $(b-c)\sin (B-\phi )=(a+b)\sin\phi\iff$

$(b-c)(\sin B-\cos B\tan\phi )=(a+b)\tan\phi\iff$ $\tan\phi =\frac {(b-c)\sin B}{(b-c)\cos B+(a+b)}\iff$ $\boxed{\tan\phi =\frac {(b-c)\sin B}{b(1+\cos B+\cos C)}}\ (1)$ . I used the identity $a=b\cos C+c\cos B$ .

$\blacktriangleright\ m\left(\widehat{AI_aT}\right)=\psi$ . Denote $S\in I_aT$ so that $OS\perp I_aT\implies$ $\tan\psi =\frac {OS}{I_aS}\implies$ $\boxed{\tan\psi =\frac {b-c}{2(r_a+R\cos A)}}\ (2)$ . Therefore, $OI_a\perp EF\iff \widehat{CLE}\equiv \widehat{AI_aT}\iff$

$\tan\phi =\tan\psi\stackrel{(1\wedge 2)}{\iff}$ $\frac {\sin B}{b(1+\cos B+\cos C)}=\frac {1}{2(r_a+R\cos A)}\iff$ $R(1+\cos B+\cos C)=r_a+R\cos A\iff$ $\boxed{r_a=R(1+\cos B+\cos C-\cos A)}$ , what is well

known or can show easily. Indeed, $(1-\cos A)+(\cos B+\cos C)=$ $2\sin^2\frac A2+2\sin\frac A2\cos \frac {B-C}{2}=$ $2\sin \frac A2\left(\cos\frac {B+C}{2}+\cos\frac {B-C}2\right)=$ $4\sin \frac A2\cos\frac B2\cos\frac C2=$

$4\sqrt{\frac {(s-b)(s-c)}{bc}\cdot\frac {s(s-b)}{ac}\cdot\frac {s(s-c)}{ab}}=$ $\frac {4s(s-b)(s-c)}{abc}=$ $\frac {4S^2}{4RS(s-a)}=\frac {S}{R(s-a)}=\frac {r_a}{R}$ .

Proof 2. Suppose w.l.o.g. $c<b$ . Is well-known the equivalence $\boxed{OI_{a} \perp EF \iff I_aF^2-I_aE^2=OF^2-OE^2}\ (*)$ .

$\blacktriangleright\ \left\{\begin{array}{c} I_aF^2=r_a^2+\left(s-\frac {bc}{a+b}\right)^2\\\\ I_aE^2=r_a^2+\left(s-\frac {bc}{a+c}\right)^2\end{array}\right|\implies$ $I_aF^2-I_aE^2=\left (2s-\frac {bc}{a+b}-\frac {bc}{a+c}\right)\left(\frac {bc}{a+c}-\frac {bc}{a+b}\right)\implies$ $I_aF^2-I_aE^2=\frac {abc(b-c)\left[a^2+2a(b+c)+\left(b^2+bc+c^2\right)\right]}{(a+b)^2(a+c)^2}$

$\blacktriangleright\ \left\{\begin{array}{c} R^2-OF^2=FA\cdot FB\\\\ R^2-OE^2=EA\cdot EC\end{array}\right|\implies$ $OF^2-OE^2=EA\cdot EC-FA\cdot FB=$ $\frac {bc}{a+c}\cdot\frac {ba}{a+c}-\frac {bc}{a+b}\cdot\frac {ca}{a+b}=$ $\frac {abc\left[b(a+b)^2-c(a+c)^2\right]}{(a+b)^2(a+c)^2}=\implies$

$OF^2-OE^2=\frac {abc(b-c)\left[a^2+2a(b+c)+\left(b^2+bc+c^2\right)\right]}{(a+b)^2(a+c)^2}$ . So $I_aF^2-I_aE^2=$ $OF^2-OE^2$ $\stackrel{(1\wedge 2)}{=}\frac {abc(b-c)\left[a^2+2a(b+c)+\left(b^2+bc+c^2\right)\right]}{(a+b)^2(a+c)^2}$ $\stackrel{(*)}{\implies}$ $OI_a\perp EF$ .

Lemma. Let $ABC$ be a triangle with the orthocenter $H$ , the circumcircle $w=C(O,R)$ and the orthic triangle $DEF$ , where $\left\{\begin{array}{c} D\in BC\ ;\ AD\perp BC\\\ E\in CA\ ;\ BE\perp CA\\\ F\in AB\ ;\ CF\perp AB\end{array}\right|$ .

Denote $\left\{\begin{array}{c} X\in EF\cap BC\\\ Y\in FD\cap CA\\\ Z\in DE\cap AB\end{array}\right|$ . Then $Z\in XY$ , where $d=\overline {XYZ}$ is the orthic axis of $\triangle ABC$ and $OH\perp d$ , i.e. the Euler's line $OH$ is perpendicular on the orthic axis.

Proof. Suppose w.l.o.g. $c<b$ . Thuse, $\frac {XB}{XC}=\frac {DB}{DC}\implies$ $\frac {XB}{c\cdot \cos B}=\frac {XC}{b\cdot \cos C}=$ $\frac {a}{b\cdot\cos C-c\cdot\cos B}\ (*)$ . Thus, $XO^2-XH^2=$ $XB\cdot XC+R^2-\left(XD^2+DH^2\right)=$

$R^2+XB\cdot XC-(XB+BD)^2-\left(HB^2-BD^2\right)=$ $R^2+XB\cdot XC-XB^2-2\cdot XB\cdot BD-HB^2=$ $R^2+XB(DC-DB)-HB^2=$

$R^2+XB(b\cdot \cos C-c\cdot\cos B)-4R^2\cos^2B\stackrel{(*)}{=}$ $R^2-4R^2\left(1-\sin^2B\right)+ac\cdot\cos B=$ $-3R^2+b^2+\frac 12\cdot\left(a^2+c^2-b^2\right)\implies$

$\boxed{XO^2-XH^2=\frac {a^2+b^2+c^2}{2}-3R^2}$ , what is a symmetrical form w.r.t. $\{a,b,c\}$ Thus, $XO^2-XH^2=YO^2-YH^2=ZO^2-ZH^2$ , i.e. $Z\in XY$ and $OH\perp \overline {XYZ}$ .

Proof 3 (synthetic). Observe that for the triangle $II_bI_c$ we have: $ABC$ is its orthic triangle; $EF$ is its orthic axis; $OI_a$ is its Euler's line. In conclusion, $OI_a\perp EF$ .

PP18. Let $\triangle ABC$ with $B=60^{\circ}$ , its incenter $I$ and $\left\{\begin{array}{cc} F\in (AC)\ ; & FC=2\cdot FA\\\\ E \in (AB)\ ; & IE\parallel BC\end{array}\right|$ . Prove that $m\left(\widehat{AEF}\right)=\frac A2$ .

Proof 1 (metric - own). I"ll use the well-known property: $\triangle ABC\implies$ $\boxed{A=2B\ \iff\ a^2=b(b+c)}\ (*)$ . Thus, $\frac {EA}{EB}=$ $\frac {IA}{ID}=$ $\frac {b+c}{a}$ , where $D\in AI\cap BC$ . Hence

$\frac {EA}{b+c}=\frac {EB}{a}=\frac {c}{a+b+c}\implies$ $\boxed{EA=\frac {c(b+c)}{a+b+c}}$ . Observe that $AF=\frac b3$ and $B=60^{\circ}\iff \boxed{b^2=a^2+c^2-ac}\ (1)$ . Apply the property $(*)$ to $\triangle AEF\ :$

$m\left(\widehat{EAF}\right)=$ $2\cdot m\left(\widehat{AEF}\right)\iff$ $EF^2=AF\cdot(AF+AE)\iff$ $EA^2+AF^2-2\cdot EA\cdot AF\cdot\cos A=AF^2+AF\cdot AE\iff$ $EA=(1+2\cos A)\cdot AF\iff$

$\frac {3c(b+c)}{b(a+b+c)}=1+2\cos A\iff$ $\frac {3c(b+c)}{b(a+b+c)}=1+\frac {b^2+c^2-a^2}{bc}\iff$ $3c^2(b+c)=(a+b+c)\left(b^2+c^2+bc-a^2\right)\stackrel{(1)}{\iff}$

$3c(b+c)=[b+(a+c)][b+(2c-a)]\stackrel{(1)}{\iff}$ $3c(b+c)=\left(a^2+c^2-ac\right)+3bc+(a+c)(2c-a)\iff$ $3c^2=a^2+c^2-ac+ac-a^2+2c^2$ , what is truly.

Proof 2 (Sunken Rock). $AI$ will intersect the arc $BC$ of the circle $(ABC)$ at $N$ , $CI$ the arc $AB$ at $M$ , $MN$ will intersect $AB$ at $E$ and $BC$ at $D$ . Next take $P$ as symmetrical of $C$

about $D$ and $Q$ the symmetrical of $A$ about $E$ . $M$ and $N$ being the midpoints of the arcs $AB$ and $BC$ respectively, it's easy to see that a $60^\circ$ rotation around $I$ will map $E$ to $D$ and $A$ to $M$

so $DM = AE$ , similarly $NE = CD$ (actually, the triangles $MAI$ and $CNI$ are equilateral). It's also easy to see that $MEIA$ and $NCID$ are cyclic and, as previously mentioned

since the two triangles are equilateral, we shall easily get $AE = ME + EI$ and $CD = ND + DI$ . We have also $BD=DI=IE=BE=DE$ . From all these we get

$ND=BP\ ( 1 )$ and $ME=BQ\ ( 2 )$ . From the two cyclic quads we shall get that $\triangle DNI \sim \triangle EIM$ , and from their similarity: $\frac {DI}{ND} = \frac {ME}{EI}\ ( 3 )$ . From $(1)$ , $(2)$ and $(3)$ we get

$\frac {BP}{BD} = \frac {BE}{BQ}$ , i.e. $PE \parallel DQ$ . If $F' = PE \cap AC$ and $G = QD \cap AC$ , from $E$ - midpoint of $AQ$ and $D$ - midpoint of $CP$ we get $AF' = GF' = CG$ , hence $F' \equiv F$ . Last, seeing

for instance, that $\triangle DNI \equiv \triangle BPE$ (s.a.s.), we get the required angle equality as well. Even easier using the above proof, one can show the following: $\frac {1}{CD} + \frac {1}{AE} = \frac {1}{DE}$ .

This post has been edited 253 times. Last edited by Virgil Nicula, Nov 20, 2015, 2:40 PM

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You seem to have solved every existing problem.
by Goutham, May 2, 2011, 10:59 AM
I love your posts, man (those which I understand)! It's interesting how you punish integrals with recurrence relations and sometimes make seemingly unrelated integrals evaluate each other!
by Caspar, Apr 11, 2011, 2:36 PM
Thank you a lot dear Virgil for an interesting and instructive blog.
by vittasko, Mar 12, 2011, 9:42 AM
Interesting Blog
by ShahinBJK, Mar 12, 2011, 5:07 AM
Interesant, d-le Nicula.
by silent_control, Mar 3, 2011, 6:00 PM
Yours is the largest blog full of math equations as far as I have known! Congrats!
by Goutham, Jan 30, 2011, 2:06 PM
15001th view.
by fries4guys, Jan 20, 2011, 4:57 PM
Thanks to all !
by Virgil Nicula, Dec 11, 2010, 8:32 PM
El_Ectric:
You can either copy some parts from the blog post, paste it at "Search Blogs" (which is lower on the page, after "Blog Stats") and find the whole post.

You're welcome.
by Thomas_, Dec 8, 2010, 6:42 PM
Thanks Thomas_.
by El_Ectric, Dec 8, 2010, 4:21 PM
El_Ectric: Try to open this blog with Internet Explorer (browser). If it still doesn't work, just copy the post you're interested in, and than paste it into Word Document, or almost any kind of reading documents and you'll have all the details.

Nice blog!
by Thomas_, Nov 24, 2010, 4:59 PM
Dear Virgil !

Congratulations for this blog !
Babis stergiou - Greece
by stergiu, Nov 12, 2010, 6:15 AM

72 shouts

My (math)blog

308. Some nice and easy properties in a triangle.

by Virgil Nicula, Aug 8, 2011, 2:10 AM

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