283. Some nice and easy problems.

by Virgil Nicula, Jun 4, 2011, 5:51 PM

P1. Let $ABC$ be an $A$ -right triangle with incenter $I$ . Denote $\left\{\begin{array}{ccc} E & \in & BI\cap AC\\\\ F & \in & CI\cap AB\end{array}\right\|\ .$ and the midpoint $M$ of $[EF]$ . Prove that $MI\perp BC$ . The drawing is here.

Proof 1 (synthetic). Denote the projections $X$ , $Y$ on $BC$ of $F$ , $E$ respectively. Observe that $EA=EY$ and $FA=FX$ . Therefore, $IA=IY$

and $IA=IX$ , i.e. $I$ is the circumcenter of $\triangle XAY$ . In conclusion, $MI$ is middleline in the right trapezoid $EFXY$ $\implies$ $MI\perp BC$ .

Remark. Prove easily that the point $I$ is the $B$ -exincenter of $\triangle BFX$ and is the $C$ -exincenter of $\triangle CEY$ . Thus $IX\perp IY$ and $m\left(\widehat{XAY}\right)=45^{\circ}$ .

Proof 2 (metric). I"ll use the equivalence $MI\perp BC\ \iff\ MB^2-MC^2=IB^2-IC^2$ and the well-known relations $\left\{\begin{array}{c} BE^2=ac-EA\cdot EC\\\\ CF^2=ab-FA\cdot FB\end{array}\right\|\ (1)$ .

Observe that $\left\{\begin{array}{ccc} BF=\frac {ac}{a+b}=\frac {a(a-b)}{c}\\\\ CE=\frac {ab}{a+c}=\frac {a(a-c)}{b}\end{array}\right\|\implies$ $\boxed{c\cdot BF-b\cdot CE=a(c-b)}\ (2)$ . Apply the theorem of the median to :

$\left\{\begin{array}{ccc} BM/\triangle EBF & \implies & 4\cdot BM^2=\left(BF^2+BE^2\right)-EF^2\\\\ CM/\triangle ECF & \implies & 4\cdot CM^2=\left(CE^2+CF^2\right)-EF^2\end{array}\right\|$ $\implies$ $2\cdot\left(MB^2-MC^2\right)=$ $\left(BF^2-CE^2\right)+BE^2-CF^2\stackrel{(1)}{=}$

$\left(BF^2-CE^2\right)+\left(ac-EA\cdot EC\right)-\left(ab-FA\cdot FB\right)=$ $a(c-b)+BF(BF+FA)-CE(CE+EA)=$ $a(c-b)+c\cdot BF-b\cdot CE\stackrel{(2)}{=}$

$2a(c-b)\implies$ $\boxed{MB^2-MC^2=a(c-b)}\ (3)$ . On other hand, $IB^2-IC^2=\frac {ac(s-b)}{s}-\frac {ab(s-c)}{s}\implies$ $\boxed{IB^2-IC^2=a(c-b)}\ (4)$ .

From the last relations $(3)$ and $(4)$ obtain that $MB^2-MC^2=IB^2-IC^2$ , i.e. $MI\perp BC$ .

Proof 3 (metric). Denote the point $N\in EF$ for which $NI\perp BC$ . I"ll show that $N\equiv M$ . Using an well-known property $\frac {NF}{NE}=\frac {IF}{IE}\cdot\frac {\sin\widehat{NIF}}{\sin\widehat{NIE}}=$

$\frac cb\cdot\frac {CF}{BE}\cdot\frac {\cos\frac C2}{\cos\frac B2}=$ $\frac {a+c}{a+b}\cdot \frac {\cos^2\frac C2}{\cos^2\frac B2}=$ $\frac {c(a+c)(a+b-c)}{b(a+b)(a+c-b)}=$ $\frac {bc(a+c)+c\left(a^2-c^2\right)}{bc(a+b)+b\left(a^2-b^2\right)}=$ $\frac {bc(a+c)+b^2c}{bc(a+b)+bc^2}=1\implies$ $NE=NF\implies N\equiv M$ .

Remark. Prove easily that in any triangle $ABC$ exists the relation $MB^2-MC^2=a(c-b)\cdot\left[1+\frac {bc\cdot \cos A}{(a+b)(a+c)}\right]=a(c-b)\cdot\frac {(a+b+c)^2}{2(a+b)(a+c)}$ .

Observe that $A=90^{\circ}\iff b^2+c^2=a^2$ $\iff (a+b+c)^2=$ $2(a+b)(a+c)\iff$ $c(a+c)(a+b-c)=b(a+b)(a+c-b)\iff MI\perp BC\iff NE=NF$ .

Proof 4. Let $D\in (BC)$ so that $ID\perp BC.$ Thus, $DB^2-DC^2=(s-b)^2-(s-c)^2=a(c-b)\implies$ $\boxed{DB^2-DC^2=a(c-b)}\ (1)\ .$ Apply the theorem of median to $:$

$\left\{\begin{array}{cc} BM/\triangle EBF\ : & 4\cdot MB^2=2\cdot \left(BF^2+BE^2\right)-EF^2\\\\ CM/\triangle FCE\ : & 4\cdot MC^2=2\cdot \left(CE^2+CF^2\right)-EF^2\end{array}\right\|\ \stackrel{(-)}{\implies}\ 2\cdot$ $\left(MB^2-MC^2\right)=\left(BF^2-CE^2\right)+\left(BE^2-CF^2\right)=$ $\left[\left(\frac {ac}{a+b}\right)^2-\left(\frac {ab}{a+c}\right)^2\right]+$

$\left[ac-\frac {ab^2c}{(a+c)^2}-ab+\frac {abc^2}{(a+b)^2}\right]=$ $\left[\underline{\frac {a^2(a-b)}{a+b}}-\underline{\underline{\frac {a^2(a-c)}{a+c}}}\right]+\left[a(c-b)-\underline{\underline{\frac {ac(a-c)}{a+c}}}+\underline{\frac {ab(a-b)}{a+b}}\right]=$ $a(c-b)+\left(a^2+ab\right)\cdot \frac {a-b}{a+b}-\left(a^2+ac\right)\cdot\frac {a-c}{a+c}=$

$a (c-b)+a(a-b)-a(a-c)=$ $2a(c-b)\implies$ $\boxed{MB^2-MC^2=a(c-b)}\ (2)\ .$ From $(1)$ and $(2)$ get $MB^2-MC^2=DB^2-DC^2\iff$ $MD\perp BC\iff$ $M\in ID\ .$

An easy extension. Let $\triangle ABC$ with the incircle $w=\mathbb C(I,r)$ and $\left\{\begin{array}{ccccccc} E & \in & BI\cap AC & ; & F & \in & CI\cap AB\\\\ D & \in & BC\cap w & ; & M & \in & EF\cap ID\end{array}\right\|\ .$ Prove that $\boxed{\frac {ME}{MF}=\frac {b(s-b)(a+b)}{c(s-c)(a+c)}}\ (*)\ .$

Proof 1.

$\blacktriangleright\ \mathrm{Pr}_{BC}(F)=X \implies BX=BF\cdot\cos B=\frac {2ac\cdot \cos B}{2(a+b)}=$ $\frac {a^2+c^2-b^2}{2(a+b)}\implies$ $DX=DB-BX=(s-b)-\frac {a^2+c^2-b^2}{2(a+b)}=$

$\frac {(a+b)(a+c-b)-\left(a^2+c^2-b^2\right)}{2(a+b)}=$ $\frac {\cancel a^2+a(c-\cancel b)+\cancel{ab}+b(c-\cancel b)-\cancel a^2-c^2+\cancel b^2}{2(a+b)}=\frac {c(a+b-c)}{2(a+b)}\implies$ $\boxed{DX=\frac {c(s-c)}{a+b}}\ (1)\ .$

$\blacktriangleright\ \mathrm{Pr}_{BC}(E)=Y \implies$ $CY=CE\cdot\cos C=\frac {2ab\cdot \cos C}{2(a+c)}=$ $\frac {a^2+b^2-c^2}{2(a+b)}\implies$ $DY=DC-CY=(s-c)-\frac {a^2+b^2-c^2}{2(a+c)}=$

$\frac {(a+c)(a+b-c)-\left(a^2+b^2-c^2\right)}{2(a+c)}=$ $\frac {\cancel a^2+a(b-\cancel c)+\cancel{ac}+c(b-\cancel c)-\cancel a^2-b^2+\cancel c^2}{2(a+c)}=\frac {b(a+c-b)}{2(a+c)}\implies$ $\boxed{DY=\frac {b(s-b)}{a+c}}\ (2)\ .$

In conclusion, $FX\parallel MD\parallel EY\iff$ $\frac {ME}{MF}=\frac {DY}{DX}\ \stackrel{(1\wedge 2)}{=}\ \frac {\frac {b(s-b)}{a+c}}{\frac {c(s-c)}{a+b}}\implies$ $\frac {ME}{MF}=\frac {b(s-b)(a+b)}{c(s-c)(a+c)}\ .$

Proof 2. Denote $U\in AC\cap DI$ and apply the Menelaus' theorem to the transversals $:\ \left\{\begin{array}{cccc} \overline{DIU}/\triangle BCE\ : & \frac {UE}{UC}\cdot\frac {DC}{DB}\cdot \frac {IB}{IE}=1\\\\ \overline{UMI}/\triangle EFC\ : & \frac {UC}{UE}\cdot \frac {ME}{MF}\cdot \frac {IF}{IC}=1\end{array}\right\| \bigodot\implies$

$\frac {ME}{MF}\cdot\frac {DC}{DB}\cdot\frac {IB}{IE}\cdot\frac {IF}{IC}=1\implies$ $\frac {ME}{MF}\cdot\frac {s-c}{s-b}\cdot\frac {a+c}{b}\cdot\frac {c}{a+b}=1\implies$ $\frac {ME}{MF}=\frac {b(s-b)(a+b)}{c(s-c)(a+c)}\ .$

Particular case $:\ ME=MF\iff$ $b(s-b)(a+b)=c(s-c)(a+c)\iff$ $b(a+c-b)(a+b)=c(a+b-c)(a+c)\iff$ $bc(a+b)+b\left(a^2-b^2\right)=$ $bc(a+c)+$

$c\left(a^2-c^2\right)\iff$ $bc(b-c)+a^2(b-c)=$ $b^3-c^3\iff$ $(b-c)\left(a^2+bc\right)=$ $(b-c)\left(b^2+bc+c^2\right)\iff$ $b=c\ \vee\ b^2+c^2=a^2\iff$ $ABC$ is $A$ -isosceles or $A$ -right.

Generalization. Let $\triangle ABC\ ,$ $D\in (BC)\ ,$ an interior point $P$ and $\left\{\begin{array}{ccc} E & \in & BP\cap AC\\\\ F & \in & CP\cap AB\\\\ M & \in & EF\cap PD\end{array}\right\|\ .$ Prove that $\boxed{\frac {ME}{MF}=\frac {DB}{DC}\cdot\frac {PE}{PB}\cdot\frac {PC}{PF}}\ (*)\ .$

Proof. Denote $U\in AC\cap DP$ and apply the Menelaus' theorem to the transversals $:\ \left\{\begin{array}{cccc} \overline{DPU}/\triangle BCE\ : & \frac {UE}{UC}\cdot\frac {DC}{DB}\cdot \frac {PB}{PE}=1\\\\ \overline{UMP}/\triangle EFC\ : & \frac {UC}{UE}\cdot \frac {ME}{MF}\cdot \frac {PF}{PC}=1\end{array}\right\| \bigodot\implies$

$\frac {ME}{MF}\cdot\frac {DC}{DB}\cdot\frac {PB}{PE}\cdot\frac {PF}{PC}=1\implies$ $\frac {ME}{MF}=\frac {DB}{DC}\cdot\frac {PE}{PB}\cdot\frac {PC}{PF}\ .$ Study some particular cases $:\ P\in \{O,H,L,N,\Gamma\}$ a.s.o. (standard notations).

P2. Let $\triangle ABC,$ an interior $P$ for which $\left\{\begin{array}{ccc} D & \in & AP\cap BC\\\\ E & \in & BP\cap CA\\\\ F\ & \in & CP\cap AB\end{array}\right|$ and $M\in (BC)$ for which $\left\{\begin{array}{ccc} K & \in & MP\cap EF\\\\ N\ & \in & AK\cap BC\end{array}\right|\ .$ Prove that $\frac {MB}{MC}\cdot\frac {NB}{NC}=\left(\frac {DB}{DC}\right)^2\ (*)\ .$ See here.

P3. Let $A_1A_2A_3A_4A_5A_6A_7$ be a regular heptagon and let $X\in A_1A_3\cap A_2A_5$ . Prove that $\frac {1}{A_1A_2}=\frac {1}{A_1A_3}+\frac {1}{A_1A_4}$ and find $m(\angle A_1XA_7)$ .

Proof. Apply the Ptolemy's theorem to the quadrilateral $A_1A_2A_3A_5$ and obtain the required relation. Remark that $\frac {1}{\sin\frac {\pi}{7}}=\frac {1}{\sin\frac {2\pi}{7}}+\frac {1}{\sin\frac {3\pi}{7}}$ .

Prove easily that $m\left(\widehat{A_1XA_2}\right)=m\left(\widehat{A_1A_2X}\right)=\frac {3\pi}{7}$ . Thus, $A_1A_7=A_1A_2=A_1X\implies$

$\triangle XA_1A_7$ is $A_1$ -isosceles. Since $m\left(\widehat{XA_1A_7}\right)=\frac {4\pi}{7}$ obtain that $m\left(\widehat{A_1XA_7}\right)=\frac 12\cdot\left(\pi -\frac {4\pi}{7}\right)=\frac {3\pi}{14}$ .

P4. Show that $\log_{a}bc+\log_bca+\log_cab \ge 4(\log_{ab}c+\log_{bc}a+\log_{ca}b)$ for all $a$ , $b$ , $c$ greater than $1$ .

Proof. Denote $x = \lg a$ , $y = \lg b$ and $z = \lg c$ , where $x$ , $y$ , $z$ are positive. Our statement comes to proving

$\frac{x + y}{z} + \frac{y + z}{x} + \frac{z + x}{y} \ge 4\left( \frac{x}{y + z} + \frac{y}{z + x} + \frac{z}{x + y} \right)$ . However, this follows from $\frac{x}{y} + \frac{x}{z} \ge \frac{4x}{y + z}$ .

P5. If the incircle of $\triangle ABC$ touches its sides in the points $X\in (BC)$ , $Y\in (CA)$ and $Z\in (AB)$ , then $XY^2+YZ^2+ZX^2=\frac {2r^2}{R}\cdot (4R+r)\le \frac {s^2}{3}$ .

Proof. $\boxed{XY^2+YZ^2+ZX^2}\equiv\sum YZ^2=$ $\sum \left(2r\cdot\cos\frac A2\right)^2=$ $\sum 4r^2\cdot\frac {s(s-a)}{bc}=$ $\frac {4r^2s}{abc}\cdot \sum a(s-a)=$ $\boxed{\frac {2r^2}{R}\cdot (4R+r)}\le$

$\frac 2R\cdot \frac {s^2}{27}\cdot (4R+r)=$ $\frac {s^2}{3}\cdot\frac {2(4R+r)}{9R}\le \boxed{\frac {s^2}{3}}$ . I used the well-known relations $\sum a(s-a)=2r(4R+r)$ , $3r\sqrt 3\le s$ , $2r\le R$ .

P6. IF $a$ , $b$ , $c$ , $d$ are the roots of the equations $P(x)\equiv x^4-7x^3+8x^2-5x+10=0$ , THEN find the value of $\prod\left(a^2+2\right)$ .

Proof. $P(X)=X^4-7X^3+8X^2-5X+10=\prod (X-a)\implies$ $P(X)\cdot P(-X)=$ $\left(8X^2+10\right)^2-\left(X^7-7X^3-5X\right)^2=$

$X^8-33X^6+14X^4+135X^2+100=$ $\prod\left(X^2-a^2\right)\implies$ $\prod\left(a^2+2\right)=q(-2)=166$ ,

where $q(t)=\prod\left(t-a^2\right)=t^4-33t^3+14t^2+135t+100$ . Thus, $\prod\left(a^2+2\right)=166$ . Remark that $\prod\left(a^2+2\right)=P\left(i\sqrt 2\right)\cdot P\left(-i\sqrt 2\right)$ .

P7. Find the maximum area of $\triangle ABC$ with $a=4$ and $c=3b$ .

Proof 1. Suppose w.l.o.g. $b=x$ , $c=3x$ , where $x>1$ . Observe that $s=2(x+1)$ , $s-a=2(x-1)$ , $s-b=2+x$ , $s-c=2-x$

and $S^2=4\left(x^2-1\right)\left(4-x^2\right)$ . Thus, $S$ is maximum $\iff$ $\left(x^2-1\right)\left(4-x^2\right)$ is maximum $\iff$ $x^2-1=4-x^2=\frac 32\iff$

$x^2=\frac 52\iff$ $x=\frac {\sqrt {10}}{2}$ because $\left(x^2-1\right)+\left(4-x^2\right)=3$ (constant). In this case $S^2_{\mathrm{max}}=4\cdot\left(\frac 32\right)^2$ , i.e. $\boxed{\ S_{\mathrm{max}}=3\ }$ .

Proof 2. Denote $D\in (BC)$ and $C\in (DE)$ so that $3=\frac {DB}{DC}=\frac {EB}{EC}$ , i.e. the rays $[AD$ , $[AE$ are the interior and exterior $A$ -angled

bisectors of $\triangle ABC$ and the circle $w$ with diameter $[DE]$ is the Appolonius' circle which has the radius $\frac 32$ because prove easily that

$CE=2$ , i.e. $DE=3$ . Thus, for any $M\in w$ we have $\frac {MB}{MC}=3$ . In conclusion, $S_{\mathrm{max}}=\frac 12\cdot 4\cdot \frac 32$ $\implies$ $S_{\mathrm{max}}=3$ .

P8. Let $ABC$ be a triangle for which denote $m=\sin2A$ , $n= \sin2B$ and $p= \sin2C$ .

Show that $s(s - m)(s - n)(s -p) \geq 0$ , where $2s=m+n+p$ . When is the equality ?

Proof. Prove easily that $\boxed{\alpha +\beta+\gamma=\pi\implies\sin 2\alpha +\sin 2\beta+\sin 2\gamma=4\sin\alpha\sin\beta\sin\gamma}$ . Therefore,

$\left\|\begin{array}{ccc} A+B+C=\pi & \implies & 2s=\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C\\\\ \left(\frac {\pi}{2}-B\right)+\left(\frac {\pi}{2}-C\right)+(\pi -A)=\pi & \implies & 2(s-m)=\sin 2B+\sin 2C-\sin 2A=4\cos B\cos C\sin A\\\\ \left(\frac {\pi}{2}-C\right)+\left(\frac {\pi}{2}-A\right)+(\pi -B)=\pi & \implies & 2(s-n)=\sin 2C+\sin 2A-\sin 2B=4\cos C\cos A\sin B\\\\ \left(\frac {\pi}{2}-A\right)+\left(\frac {\pi}{2}-B\right)+(\pi -C)=\pi & \implies & 2(s-p)=\sin 2A+\sin 2B-\sin 2C=4\cos A\cos B\sin C\end{array}\right\|\ \bigodot\implies$

$16s(s-m)(s-n)(s-p)=$ $4\cdot\left[\prod (2\sin A\cos A)\right]^2=$ $4\cdot \prod\sin ^22A\ge 0$ . The equality is hold iff $\frac {\pi}{2}\in \left\{A,B,C\right\}$ .

P9. Suppose that for every $k \in\left\{ 1,2,...,n\right\}$ , where $n\in\mathbb N$ , we have given a complex number $z_k$

for which $\left|z_{k}\right| = 1$ . Assume that $\sum_{k = 1}^{n}z_{k} = 0$ . Prove that for $(\forall )\ z\in\mathbb{C}$ we have $\sum_{k=1}^{n}\left|z - z_{k}\right|\ge n$ .

Proof. Suppose w.l.o.g. that $z\ne 0$ . Then $\sum_{k=1}^n |z-z_k| =\sum_{k=1}^n \left|\overline{z}_k\right|\cdot \left|z-z_k\right| =$

$\sum_{k=1}^n \left|\overline{z}_k\cdot z-1\right| \ge$ $\left|\sum_{k=1}^n \left(\overline{z}_k\cdot z-1\right)\right|=$ $\left|z\cdot\sum_{k=1}^n\overline{z}_k-n\right|=$ $\left|z\cdot\overline{\sum_{k=1}^nz_k}-n\right|=n$ .

P10. Prove that in any triangle $ABC$ exists the inequality $m_a+m_b+m_c\ge \frac {a^2+b^2+c^2}{2R}$ , where $R$ is the length of the circumradius .

Proof 1. Denote the midpoint $M$ of $[BC]$ . The line $AM$ meet again the circumcircle of $\triangle ABC$ in $X$ . Since $MA\cdot MX=MB\cdot MC$

obtain that $MX=\frac {a^2}{4m_a}$ . Thus, $AX\le 2R\iff m_a+MX\le 2R\iff$ $m_a+\frac {a^2}{4m_a}\le 2R\iff$ $a^2+4m_a^2\le 8Rm_a$ .

Thus, $2\left(b^2+c^2\right)\le 8Rm_a\iff$ $b^2+c^2\le 4Rm_a$ a.s.o. In conclusion, $\sum\left(b^2+c^2\right)\le 4R\sum m_a\iff$ $\sum m_a\ge \frac {\sum a^2}{2R}$ .

Proof 2. The line $BO$ meet again the circumcircle $C(O,R)$ of $\triangle ABC$ in the point $E$ . Observe that $CE=2\cdot OM$ , $CE^2=4R^2-a^2$

and $\left|m_a-R\right|\le OM\iff$ $2\cdot\left|m_a-R\right|\le CE\iff$ $4\left(m_a-R\right)^2\le 4R^2-a^2\iff$ $8Rm_a\ge 4m_a^2+a^2\iff$

$4Rm_a\ge b^2+c^2\ (*)$ a.s.o. We have the equality in the relation $(*)$ iff $b=c$ or $AB\perp AC$ .

Proof 3. A simpler way uses the well-known inequality $\boxed{\ h_a\, \le\, s_a\ }$ , where $s_a$ represent the length of the $A$ -symmedian. Indeed,

since $h_a=\frac {bc}{2R}$ and $s_a=\frac {2bc}{b^2+c^2}\cdot m_a$ we obtain $h_a\le s_a\iff\frac {bc}{2R}\le\frac {2bc}{b^2+c^2}\cdot m_a\implies\boxed{\ m_a\, \ge\, \frac {b^2+c^2}{4R}\ }$ a.s.o.

P11. Prove that in any $\triangle ABC$ exists the inequality $m_bm_c\le \frac {a^2}{2}+\frac {bc}{4}$ .

Proof 1. Denote the midpoints $M$ , $N$ of the sides $[AB]$ , $[AC]$ respectively. Apply the Ptolemy's inequality in the trapezoid

$MBCN\ :\ BN\cdot CM\le$ $BM\cdot CN+BC\cdot MN\iff$ $m_bm_c\le \frac c2\cdot \frac b2+a\cdot\frac a2\iff$ $m_bm_c\le \frac {a^2}{2}+\frac {bc}{4}$ .

Proof 2. $4m_am_b\le 2a^2+bc\iff$ $\left[2\left(a^2+c^2\right)-b^2\right]\cdot\left[2\left(a^2+b^2\right)-c^2\right]\le\left(2a^2+bc\right)^2\iff$ $4\left(a^2+b^2\right)\left(a^2+c^2\right)+$ $b^2c^2-$ $2c^2\left(a^2+c^2\right)-$

$2b^2\left(a^2+b^2\right)\le$ $4a^4+b^2c^2+$ $4a^2bc\iff$ $2a^2\left(b^2+c^2\right)+$ $4b^2c^2-2\left(b^4+c^4\right)$ $\le 4a^2bc\iff$ $2a^2(b-c)^2\le$ $2\left(b^2-c^2\right)^2\iff$ $(b-c)^2\ge 0$ .

Remark.

$\blacktriangleright\ \sum\frac {m_bm_c}{a}\le \frac 12\cdot\sum a+\frac 14\cdot abc\sum\frac {1}{a^2}\le$ $s+Rsr\cdot\frac {1}{4r^2}\le s+\frac {Rs}{4r}\implies$ $\boxed{\frac {m_bm_c}{a}+\frac {m_cm_a}{b}+\frac {m_am_b}{c}\le s\left(1+\frac {R}{4r}\right)}$ . I used the chain

$\boxed{\frac {1}{R^2}\le\sum\frac {1}{a^2}\le\frac {1}{4r^2}}$ . Indeed, $\frac {1}{R^2}\le \frac {1}{2Rr}=$ $\frac {2s}{4RS}=\frac {a+b+c}{abc}=$ $\sum\frac {1}{bc}=$ $\frac {\sum (ab)(ac)}{(abc)^2}\le$ $\sum\frac {(bc)^2}{a^2b^2c^2}=$ $\sum\frac {1}{a^2}\le$ $\frac {1}{4(s-b)(s-c)}=$ $\frac {1}{4r^2}$ .

Otherwise, $\sum\frac {1}{a^2}\ge\frac {9}{\sum a^2}\ge \frac {1}{R^2}$ . I used the well-known inequalities $\left\{\begin{array}{c} R\ge 2r\\\\ a^2+b^2+c^2\le 9R^2\\\\ \sum (bc)^2\ge abc(a+b+c)\\\\ a^2\ge 4(s-b)(s-c)\end{array}\right\|$ and the identities $\left\{\begin{array}{c} abc=4RS=4Rsr\\\\ \sum\frac {1}{(s-b)(s-c)}=\frac {1}{r^2}\end{array}\right\|$ .

$\blacktriangleright\ \sum\frac {m_bm_c}{bc}\le\sum\frac {2a^2+bc}{4bc}=\frac 12\cdot \frac {\sum a^3}{abc}+\frac 34$ . Observe that $\frac {\sum a^3}{abc}=$ $\frac {s^2-3r^2-6Rr}{2Rr}\le$ $\frac {4R^2-2Rr}{2Rr}=$ $\frac {2R}{r}-1\implies$ $\boxed{\ \sum\frac {m_bm_c}{bc}\le\frac Rr+\frac 14\ }$ .

P12. Let $ABC$ be a triangle. Denote the feet $D$ , $E$ of the altitudes from $A$ and $C$ respectively. Let $F$ be the second

point of intersection of the circumcircles of $ABC$ and $AEH$ . Show that $F\in DE\iff DC=3\cdot DB$ .

Proof. Denote the midpoint $M$ of $[BC]$ , the circumcircle $w$ of $\triangle ABC$ and the second intersections $L$ , $A'$ , $F'$ with $w$ of the lines $AH$ , $AO$ , $A'H$ respectively.

Since $HL=2\cdot HD$ , $HA'=2\cdot HM$ and $HA\cdot HL=HA'\cdot HF'$ $\implies$ $HA\cdot HD=HM\cdot HF'$ $\implies$ the quadrilateral $AF'DM$ is cyclically $\implies$

$AF'\perp HF'$ $\implies$ $F'\equiv F$ . Denote $G\in BH\cap AC$ . From $HB\cdot HG=HD\cdot HA=HM\cdot HF$ obtain that $HB\cdot HG=HM\cdot HF$ ,

i.e. $BMGF$ is cyclical quadrilateral. Thus, $DC=3\cdot DB\iff$ $D$ is the midpoint of $[BM]\iff$ $\widehat{GBM}\equiv\widehat{FMB}\iff$ $FG\parallel BC\iff$

$\widehat{BED}\equiv\widehat{BCG}\equiv$ $\widehat{AGF}\equiv\widehat{AEF}\iff$ $\widehat{BED}\equiv\widehat{AEF}\iff$ $F\in DE$ .

P13.The equation $f_a(x)\equiv x^{2}+(a+2)x+a^{2}-a+2=0\ (*)$ has real roots, where $a\in\mathbb R^*$ . Find the range of these roots.

Proof. $r$ is a root of $(*)\iff$ $f_a(r)=\underline r^{2}+(a+2)\underline r+a^{2}-a+2=0$ , i.e. $\underline a^2+(r-1)\cdot \underline a+\left(r^2+2r+2\right)=0$ .

$\blacktriangleright\ r\in\mathbb R\iff$ $\Delta_r' (a)\equiv (a+2)^2-4\left(a^2-a+2\right)\ge 0\iff$ $3a^2-8a+4\le 0\iff$ $a\in\left[\frac 23,2\right]$ .

$\blacktriangleright\ a\in\mathbb R^*\iff$ $\Delta_a^{\prime}(r)\equiv (r-1)^2-4\left(r^2+2r+2\right)\ge 0\iff$ $3r^2+10r+7\le 0\iff$ $\boxed{\ r\in\left[-\frac 73,-1\right]\ }$ .

P14. Let $ABC$ be a triangle with the incenter $I$ and the circumcircle $w=C(O,R)$ . Denote $B_1\in BI\cap AC$ ,

$C_1\in CI\cap AB$ and $\{M,N\}=B_1C_1\cap w$ . Prove that the circumradius of $\triangle MIN$ is equally to $2R$ .

Proof. Denote the exincenters $I_a$ , $I_b$ , $I_c$ of $\triangle ABC$ . Then $I$ is the orthocenter of $\triangle I_aI_bI_c$ and $w$ is nine-point circle of $\triangle I_aI_bI_c$ , i.e. the lengths of the circumradius

for the triangles $I_aI_bI_c$ and $II_bI_c$ are equally to $2R$ . We will show that $M$ , $N$ belong to the circumcircle of $\triangle II_bI_c$ . Suppose w.l.o.g. $C_1\in (MB_1)$ . Thus, $I_cAIB$

and $I_bAIC$ are cyclically $\Longrightarrow$ $\begin{array}{c} C_1I\cdot C_1I_c=C_1A\cdot C_1B=C_1M\cdot C_1N \Longrightarrow I_cMIN\ \mathrm{is\ cyclically}\\\ B_1I\cdot B_1I_b=B_1A\cdot B _1C=B_1M\cdot B_1N \Longrightarrow I_bNIM\ \mathrm{is\ cyclically}\end{array}\implies$ $M$ , $N$ belong to the circumcircle of $\triangle MIN$ .

P15. Find $x \in R$ such $\left[x+\frac{1}{2}\right] + \left[x-\frac{1}{2}\right] = [2x]$ .

Proof. Using the Hermite's identity $[x]+\left[x+\frac 12\right]=[2x]$ obtain that $\left[x+\frac{1}{2}\right] + \left[x-\frac{1}{2}\right] = [2x]\iff$ $[x]=\left[x-\frac 12\right]=z\in\mathrm Z\iff$

$\left\{\begin{array}{c} z\le x<z+1\\\ z\le x-\frac 12<z+1\end{array}\right\|\iff$ $\left\{\begin{array}{c} z\le x<z+1\\\ z+\frac 12\le x<z+\frac 32\end{array}\right\|\iff$ $z+\frac 12\le x<z+1\ ,\ z\in\mathbb Z$ . In conclusion, $x\in\bigcup_{z\in\mathbb Z}\left[z+\frac 12,z+1\right)$ .

Remark. $\left[x+\frac{1}{2}\right] + \left[x-\frac{1}{2}\right] = [2x]\iff$ $x\in\bigcup_{z\in\mathbb Z}\left[z+\frac 12,z+1\right)\ \iff\ \left[2\cdot\left\{x\right\}\right]=1$ .

P16. Prove that if $\triangle ABC$ is acute, then $\frac {1}{\cos A\cos B} + \frac {1}{\cos B\cos C} + \frac {1}{\cos C\cos A}\geq12$ .

Proof I. $\boxed {a = b\cdot\cos C + c\cdot\cos B}$ $\implies$ $a^2\ge 4bc\cdot \cos B\cos C$ $\implies$ $\sum\frac {1}{\cos B\cos C}\ge 4\cdot \sum \frac {bc}{a^2}\ge 12\cdot \sqrt [3]{\frac {bc}{a^2}\cdot \frac {ca}{b^2}\cdot\frac {ab}{c^2}} = 12$ .

Proof II. $\sum \frac {1}{\cos B\cos C} =$ $\sum \frac {2}{\cos (B - C) + \cos (B + C)}\ge$ $\sum \frac {2}{1 - \cos A} =$ $\sum\frac {1}{\sin^2\frac A2} =$

$\sum\frac {bc}{(s - b)(s - c)} =$ $\frac {abc}{pr^2}\cdot \sum \frac {s - a}{a} =$ $\frac {4R}{r}\cdot\left(s\cdot\sum \frac 1a - 3\right)\ge$ $\frac {4R}{r}\cdot\left(\frac 92 - 3\right) =$ $\frac {6R}{r}\ge 12$ .

Observe that $\boxed {\ \sum \frac {1}{\cos B\cos C}\ \ge\ \frac {36R^2}{s^2 + r^2 - 4R^2}\ \ge\ \frac {6R}{r}\ \ge\ 12\ }$ .

P17. Let $ABCD$ be a rectangle with $AD=1$ and let $P\in (AB)$ be the point

for which the rays $[DP$ , $[DB$ trisect $\angle ADC$ . Find the perimeter of $\triangle BDP$ .

Proof. Denote $DB=x>1$ . Observe that $AB=\sqrt {x^2-1}$ and $\frac {PA}{DA}=\frac {PB}{DB}=\frac {AB}{DA+DB}\implies$

$\frac {PA}{1}=\frac {PB}{x}=\frac {\sqrt {x^2-1}}{1+x}$ .Since $AB\parallel CD$ obtain that $\widehat{PDB}\equiv\widehat{CDB}\equiv$ $\widehat{PBD}\implies$

$\widehat{PDB}\equiv\widehat{PBD}\implies$ $PD=PB=\frac {x\sqrt{x^2-1}}{x+1}$ and $PA=\frac {\sqrt {x^2-1}}{1+x}$ . Since $\triangle ADP\sim\triangle CDB$

obtain that $\frac {AD}{CD}=\frac {AP}{CB}\implies$ $\frac {1}{\sqrt{x^2-1}}=\frac {\sqrt {x^2-1}}{x+1}\iff$ $x^2-1=x+1\iff$ $x=2$ .

In conclusion, $BD=2$ and $PB=PD=\frac {2\sqrt 3}{3}$ $\implies$ the perimeter of the triangle $BDP$ is $2+\frac {4\sqrt 3}{3}$ .

P18 (from my dear Kunny). Given is a circle $w$ with the diameter $[AB]$ . For a mobile point $M\in w$ define $N\in (AB)$

for which $AN=MB$ . Find the position of the point $M$ so that the area $[AMN]$ of the triangle $AMN$ is maximum.

Proof. Denote $AB=a$ , $AM=u$ , $MB=AN=v$ and $\phi =m\left(\widehat{MAB}\right)$ . Suppose w.l.o.g. that $a=1$ . Thus, $u^2+v^2=1$ , $\sin \phi =\frac {MB}{AB}=v$

and $[ABE]=\frac 12\cdot AM\cdot AN\cdot\sin\phi =\frac {uv^2}{2}$ . In conclusion our problem becomes $\max_{u^2+v^2=1}uv^2\iff$ $u^2v^4$ is max. Observe that $u^2+\frac {v^2}{2}+\frac {v^2}{2}=1$

(constant). Thus $uv^2$ is max. $\iff$ $u^2=\frac {v^2}{2}=\frac 13\iff$ $u=\frac {\sqrt 3}{3}$ , $v=\frac {\sqrt 6}{3}$ and the maximum of the area of the triangle $AMN$ is $\frac {\sqrt 3}{9}$ .

Here are three similar problems.

SP1. Find the rotation cylinder with the constant volume and with the minimum total surface ("problem of the instant coffee can").

Proof. Denote the length $r$ of the circle from base and the length $g$ of the generatrix. Then its volume is $v(r,g)=\pi r^2g$ and its total surface is $s(r,g)=2\pi rg+2\pi r^2=$

$2\pi r(r+g)$ . Therefore, our problem is equivalently with $\min_{r^2g=k}f(r,g)$ , where $f(r,g)=r^2+rg$ . Observe that $r^2g$ is constant $\iff$ $\frac {r^4g^2}{4}$ is constant $\iff$

$r^2\cdot\frac {rg}{2}\cdot\frac {rg}{2}$ is constant. Thus, $r^2+rg=$ $r^2+\frac {rg}{2}+\frac {rg}{2}$ $\implies$ $r^2+rg$ is min. $\iff$ $r^2=\frac {rg}{2}\iff$ $g=2r=\sqrt[3]{4k}$ , i.e. the axial section is a square.

SP2. Given is a circle $w$ with the diameter $[AB]$ . For a mobile point $M\in w$ define $N$ of $(AB)$ so

that $MN\perp AB$ . Find the position of the point $M$ so that the sum $MB+MN$ is maximum.

Proof. Observe that $b+h_a$ is max. $\iff$ $b+\frac {bc}{a}$ is max. $\iff$ $b(a+c)$ is max. $\iff$ $b^2(a+c)^2$ is max. $\iff$ $\left(a^2-c^2\right)(a+c)^2$ is max. $\iff$ $(a-c)(a+c)^3$

is max. Since $(a-c)+\frac {a+c}{3}+\frac {a+c}{3}+\frac {a+c}{3}=2a$ (constant) obtain that $b+h_a$ is max. $\iff$ $a-c=\frac {a+c}{3}=\frac a2$ , i.e. $c=\frac a2$ , $b=\frac {a\sqrt 3}{2}$ , $h_a=\frac {a\sqrt 3}{4}$

and the maximum of the sum $b+h_a$ is equally to $\frac {3a\sqrt 3}{4}$ . With other words, in any $A$ -right triangle $ABC$ there is the inequality $\boxed{\ h_a+\max\{b,c\}\le \frac {3a\sqrt 3}{4}\ }$ .

SP3. A plane figure is the reunion between a rectangle $ABCD$ and a semicircle $\mathbb S$ with the diameter $[DC]$ such

that $[DC]=[ABCD]\cap \mathbb S$ . Suppose that the perimeter of this figure is constant. Find the maximum of its area.

Proof. Denote $AB=CD=2x$ and $AD=BC=y$ . Thus, the perimeter of the figure is $(2+\pi )x+2y=k$ (constant) and the area of the plane figure is

$a(x,y)=2xy+\frac {\pi x^2}{2}=x\left(2y+\frac {\pi x}{2}\right)$ is max. $\iff$ the product $\left(2+\frac {\pi}{2}\right)x\cdot \left(2y+\frac {\pi x}{2}\right)$ is max. Observe that the sum $\left(2+\frac {\pi}{2}\right)x+\left(2y+\frac {\pi x}{2}\right)=k$

(constant). In conclusion, the area of the figure is maximum $\iff$ $\left(2+\frac {\pi}{2}\right)x= 2y+\frac {\pi x}{2}=\frac k2$ , i.e. $x=y=\frac {k}{4+\pi}$ (the rectangle is the reunion of two squares !).

P19. Let $ABC$ be a triangle. Consider the points $K\in (BC)$ , $L\in (CA)$ , $M\in (AB)$ so that $KLMB$ is

a parallelogram. Suppose that $[BMK]=S_1$ and $[KLC]=S_2$ . Find the area of the parallelogram $MKLA$ .

Proof. Denote $[AMK]=[ALK]=x$ . Since $\triangle BMK\sim\triangle KLC\ (*)$ obtain that $\frac {x+S_1}{x+S_2}=\frac {KB}{KC}\stackrel{(*)}{=}\sqrt{\frac {S_1}{S_2}}\iff$ $\frac {x+S_1}{x+S_2}=\sqrt{\frac {S_1}{S_2}}\iff$

$(x+S_1)\sqrt {S_2}=$ $(x+S_2)\sqrt {S_1}\iff$ $\left(x-\sqrt{S_1S_2}\right)\left(\sqrt{S_1}-\sqrt{S_2}\right)=0\iff$ $S_1=S_2\ \vee\ x=\sqrt{S_1S_2}\iff$ $[MKLA]=2\sqrt{S_1S_2}$ .

P20. Prove that the points the points $A\left(a,a^2\right)$ , $B\left(b,b^2\right)$ , $C\left(c,c^2\right)$ , $D\left(d,d^2\right)$ are concyclically $\iff$ $a+b+c+d=0$ .

Proof. Let the equation of the circle be $x^2+y^2+mx+ny+k=0$ . Now put $y=x^2$ to get another equation which

$a,b,c,d$ satisfy. The coefficient of $x^3$ in that equation is $0$ , therefore the sum of its roots is $0$ , i.e. $a+b+c+d=0$ .

P21. Let $\triangle ABC$ be a triangle such that $m(\angle BAC)=120 ^\circ$ . Denote $O$ as circumcenter, and $H$ as orthocenter. Prove that $OH=AB+AC$ .

Proof (metric). $m(\angle BAC)=120 ^\circ\implies$ $3R^2=a^2=b^2+c^2+bc$ and $OH=b+c\iff$

$9R^2-\sum a^2=9\cdot OG^2=OH^2=(b+c)^2\iff$ $3R^2=a^2+b^2+bc$ , what is truly.

P22. Let $\{P,Q\}\subset (BC)$ be two points so that $PC=QB$ . Prove that $AQ\ge AP\iff (c-b)(PB-PC)\ge 0$ .

Proof. Apply the Stewart's relation to the cevians $AP$ and $AQ$ using that $\left\{\begin{array}{c} PC=QB\\\ PB=QC\end{array}\right\|$ :

$\left\{\begin{array}{c} a\cdot AP^2+a\cdot PB\cdot PC=c^2\cdot PC+b^2\cdot PB\\\\ a\cdot AQ^2+a\cdot PB\cdot PC=c^2\cdot PB+b^2\cdot PC\end{array}\right\|\implies$ $a\cdot \left(AQ^2-AP^2\right)=\left(c^2-b^2\right)(PB-PC)$ .

In conclusion, $AQ\ge AP\iff$ $(c-b)(PB-PC)\ge 0$ .

Particular case. If the point $P$ is the foot of the (interior) $A$ -bisector of $\triangle ABC$ , then $\frac {PB}{c}=\frac {PC}{b}=\frac {a}{b+c}=\frac {PB-PC}{c-b}\implies$

$PB-PC=\frac {a(c-b)}{b+c}$ . In this case $AQ\ge AP\iff$ $(c-b)(PB-PC)\ge 0\iff$ $\frac {a(c-b)^2}{b+c}\ge 0$ , what is truly.

P23. Calculate the following sums : $\sum_{k=1}^nk2^{k-1}\ \ ;\ \sum _{k=1}^nkC_n^k$ .

Method 1. $\boxed{S_1\equiv \sum_{k=1}^n k2^{k-1}}\implies$ $S_1=2S_1-S_1=\sum_{k=1}^n k2^k-\sum_{k=1}^n k2^{k-1}=$ $\sum_{k=2}^{n+1} (k-1)2^{k-1}-\sum_{k=1}^n k2^{k-1}=$

$n2^n+\sum_{k=1}^n(k-1)2^{k-1}-\sum_{k=1}^n k2^{k-1}=$ $n2^n-\sum_{k=1}^n 2^{k-1}=$ $n2^n-\left(2^n-1\right)\implies$ $\boxed{S_1=(n-1)2^n+1}$ .

I used the formal property of the $\sum\ :\ \boxed{\sum_{k=m}^nf(k)=\sum_{k=m+p}^{n+p}f(k-p)}\ ,\ \{m,n,p\}\subset \mathbb Z$ and $m\le n$ .

Method 2. For $x\ne 1$ obtain that $\sum_{k=1}^nkx^k=\sum_{m=1}^n\sum_{k=m}^nx^k=$ $\sum_{m=1}^n\frac {x^{n+1}-x^m}{x-1}=$ $\frac {1}{x-1}\cdot \left(nx^{n+1}-\sum_{m=1}^nx^m\right)=$

$\frac {1}{x-1}\cdot \left(nx^{n+1}-\frac {x^{n+1}-x}{x-1}\right)\implies$ $\boxed{\sum_{k=1}^nkx^{k-1}=\frac {nx^{n+1}-(n+1)x^{n}+1}{(x-1)^2}}$ .

In the our particular case $x=2$ obtain that $S_1=\sum_{k=1}^nk2^{k-1}=$ $n2^{n+1}-(n+1)2^n+1$ , i.e. $S_1=(n-1)2^n+1$ .

Method 3 (with derivatives). $\sum_{k=1}^nkx^{k-1}=$ $\left(\sum_{k=1}^nx^k\right)'=$ $\left(\frac {x^{n+1}-x}{x-1}\right)'\implies$ $\sum_{k=1}^nkx^{k-1}=\frac {nx^{n+1}-(n+1)x^{n}+1}{(x-1)^2}$ .

Remark. For $x>0$ obtain that $nx^{n+1}+1\ge (n+1)x^n$ . For $x:=\sqrt [n+1]a\ ,\ a>0$ obtain that $\frac {1+na}{1+n}\ge \sqrt [n+1]{a^n}$

with equality for $a=1$ . This inequality means the Cauchy's inequality for $a_1=1$ and $a_k=a$ for any $k\in\overline{2,n}$ .

Here are some formal properties of the $\sum$ .

$\blacktriangleright\ \sum_{k=1}^n\left(a_k+b_k\right)=\sum_{k=1}^n a_k+\sum_{k=1}^nb_k\ \ ;\ \ \sum_{k=1}^n\lambda b_k=\lambda \cdot\sum_{k=1}^nb_k$ .

$\blacktriangleright\ \sum_{k=1}^nf(k)=\sum_{k=1}^nf(n+1-k)\ \ ;\ \ \sum_{k=m}^nf(k)=\sum_{k=m+p}^{n+p}f(k-p)$ .

$\blacktriangleright\ \sum_{p=1}^m\sum_{k=1}^nf(k,p)=\sum_{k=1}^n\sum_{p=1}^mf(k,p)\ \ ;\ \ \sum_{p=1}^m\sum_{k=1}^na_kb_p=\sum_{k=1}^na_k\cdot \sum_{p=1}^mb_p$ .

$\blacktriangleright\ \sum_{m=1}^n\sum_{k=1}^mf(k)=\sum_{m=1}^n\sum_{k=m}^nf(n+1-k)=\sum_{k=1}^n(n+1-k)f(k)=\sum_{k=1}^nkf(n+1-k)$ .

Method 1. $k\cdot C_n^k=\frac {n!k}{k!(n-k)!}=\frac {(n-1)!n}{(k-1)!(n-k)!}=n\cdot C_{n-1}^{k-1}\ ,\ k\in\overline{1,n}$ $\implies$ $S_2=n\cdot\sum_{k=1}^n C_{n-1}^{k-1}\implies$ $\boxed{S_2=n\cdot 2^{n-1}}$ .

Method 2. $\sum_{k=1}^nkC_n^k=\sum_{k=1}^n(n+1-k)C_n^{n+1-k}=$ $\sum_{k=1}^n(n+1-k)C_n^{k-1}=$ $\sum_{k=1}^nnC_n^{k-1}-\sum_{k=1}^n(k-1)C_n^{k-1}=$

$n\left(2^n-1\right)-\sum_{k=1}^{n-1}kC_n^k=$ $n\left(2^n-1\right)-\left(S_2-n\right)\implies$ $S_2=n2^n-S_2$ , i.e. $\boxed{S_2=n2^{n-1}}$ .

Method 3. $\sum_{k=1}^nkC_n^k=\sum_{m=1}^n\sum_{k=m}^nC_n^k=$ $\sum_{m=1}^n\left(2^n-\sum_{k=0}^{m-1}C_n^k\right)=$ $n2^n-\sum_{m=1}^n\sum_{k=0}^{m-1}C_n^k=$

$n2^n-\sum_{k=1}^nkC_n^k$ , i.e. $S_2=n2^n-S_2$ from where obtain $\boxed{S_2=n2^{n-1}}$ .

P24. $\left\{\begin{array}{c} \sin x+\sin y+\sin z=0\\\\ \cos x+\cos y+\cos z=0\end{array}\right|\implies$ $\left\{\begin{array}{c} \cos (y-z)=-\frac{1}{2}\\\\ \phi\in\mathbb R\ \implies\ \sum \sin(\phi \pm x)=\sum\cos (\phi \pm x)=0\\\\ \sum\cos 2x=\sum\sin 2x=0\\\\ \sin^2x+\sin^2y+\sin^2z=\frac 32\end{array}\right|$ .

Proof. $1=\sin^2x+\cos^2x=\left[-(\sin y+\sin z)\right]^2+\left[-(\cos y+\cos z)\right]^2=$ $2+2\cos (y-z)\implies\cos (y-z)=-\frac{1}{2}$ .

$\sum \sin (\phi \pm x)=$ $\sum\left(\sin\phi\cos x\pm\cos\phi \sin x\right)=$ $\sin\phi\sum \cos x\pm\cos\phi\sum\sin x=0$ .

$\sum \sin (\phi \pm x)=$ $\sum\left(\cos\phi\cos x\mp\sin\phi \sin x\right)=$ $\cos\phi\sum \cos x\mp\sin\phi\sum\sin x=0$ .

Particularly, for $\phi :=x+y+z$ obtain that $\sum\cos (y+z)=\sum\sin (y+z)=0$ . Thus, $\sum\sin^2x-\sum\cos^2x=$

$\left(\sum\sin x\right)^2-2\cdot\sum\sin y\sin z-$ $\left(\sum\cos x\right)^2+2\cdot\sum\cos y\cos z=$ $2\cdot\sum(\cos y\cos z-\sin y\sin z)\implies$

$-\sum\cos 2x=$ $\sum\sin^2x-\sum\cos^2x=2\cdot\sum\cos (y+z)=0$ . In conclusion, $\sum\cos 2x=0$ and $\sum\sin^2x=\sum\cos^2x=\frac 32$ .

$\sum\sin 2x=\sum\left[(\sin x+\cos x)^2-1\right]=$ $-3+\left(\sum\sin x+\sum\cos x\right)^2-2\cdot\sum (\sin y+\cos y)(\sin z+\cos z)=$

$-3-2\cdot \sum\left[(\cos (y-z)+\sin (y+z)\right]=$ $-3-2\cdot \left(-\frac 32\right)=0$ . In conclusion, $\sum\sin 2x=0$ .

Otherwise. Denote $\left\{\begin{array}{c} \cos x+i\cdot\sin x=u\\\ \cos y+i\cdot\sin y=v\\\ \cos z+i\cdot\sin z=w\end{array}\right|\implies$ $u+v+w=0\iff$ $\overline {\sum u}=\sum\overline u=0\iff$ $\sum\frac {1}{u}=0\iff$

$uv+vw+wu=0\ \implies\ u^2+v^2+w^2=0\ \implies$ $\left\{\begin{array}{c} \sum\cos 2x=0\\\\ \sum\sin 2x=0\end{array}\right|$ $\implies \sum\left(1-2\sin^2x\right)=0\implies$ $\sum\sin^2x=\frac 32$ .

P25. Let $\{u, v\}\subset C^*$ and $L _{u,v}=\left \{\ z \in C \mid \; z +u \overline z + v = 0\ \right\}$ . What are the necessary and sufficient conditions for $L_{u,v}$ is a line ?

Proof. Proof. $z +u \overline z + v = 0\ \iff\ \overline z+\overline uz+\overline v=0$ . Eliminate $\overline z$ between the equivalent equations $\left\{\begin{array}{c} z +u\cdot \overline z + v = 0\\\ \overline u\cdot z+\overline z+\overline v=0\end{array}\right|\implies$

$\left(1-|u|^2\right)\cdot z+\left(v-u\overline v\right)=0$ , a relation what is verified by an infinitude of points $z\in L _{u,v}$ . In conclusion, $|u|=1\ \ \wedge\ \ u\overline v=v$ .

Otherwise. $\frac {1}{\overline u}=\frac u1=\frac {v}{\overline v}\iff$ $|u|=1\ \ \wedge\ \ u\overline v=v$ $\iff$ $u\overline v=v$ because $|u\overline v|=|v|$ and $v\ne 0\implies |u|=1$ .

P26. Prove that $\sin x = k\cdot\sin (a-x)\ \iff\ \tan \left(x-\frac a2\right) = \frac{k-1}{k+1}\cdot\tan\frac a2$ .

Proof. $\sin x = k\cdot\sin (a-x)\iff$ $\frac k1=\frac{\sin x}{\sin (a-x)}\iff$ $\frac {k-1}{k+1}=\frac {\sin x-\sin (a-x)}{\sin x+\sin (a-x)}\iff$

$\frac {k-1}{k+1}=\frac {2\cdot\sin\left(x-\frac a2\right)\cos\frac a2}{2\cdot\sin\frac a2\cos\left(x-\frac a2\right)}\iff$ $\frac {k-1}{k+1}=\frac {\tan\left(x-\frac a2\right)}{\tan\frac a2}\iff$ $\tan \left(x-\frac a2\right) = \frac{k-1}{k+1}\cdot\tan\frac a2$ .

Remark. For $x:=B\ ,\ a:=-A\ ,\ k:=-k$ obtain the Mollweide's identity in $\triangle ABC\ :\ \sin B=k\cdot\sin C\ \iff$

$\tan\frac {B-C}{2}=$ $\frac {k-1}{k+1}\cdot \cot\frac A2\ \iff\ \tan\frac {B-C}{2}=$ $\frac {b-c}{b+c}\cdot \cot\frac A2$ because $k=\frac {\sin B}{\sin C}=\frac bc$ .

P27. Let $ABCD$ be a convex quadrilateral such that $AC\perp BD$ and let $P\in AC\cap BD$ . Prove that the reflections of $P$ w.r.t. $AB$ , $BC$ , $CD$ , $DA$ are concyclic. Let $ABCD$

be a cyclic onvex quadrilateral such that $AC\perp BD$ . Prove that the line through the intersection of diagonals and the midpoint of any sides is perpendicular to the opposite side.

P28. Solve the system $\left\{\begin{array}{ccc} xy+x & = & 2\\\\ yz+y & = & 3\\\\ zx+z & = & 4\end{array}\right\|$ , where $\{x,y,z\}\subset\mathbb R$ .

Proof 1. Denote $t\equiv xyz\ne 0$ . Thus, $\left\{ \begin{array}{ccc} xy+x & = & 2\\\\ yz+y & = & 3\\\\ zx+z & = & 4\end{array}\right|$ $\left|\begin{array}{ccccccc} \odot\ z & \implies & t+zx=2z & \implies & t+(4-z)=2z & \implies & z=\frac {t+4}3\\\\ \odot\ x & \implies & t+xy=3x & \implies & t+(2-x)=3x & \implies & x=\frac {t+2}4\\\\ \odot\ y & \implies & t+yz=4y & \implies & t+(3-y)=4y & \implies & y=\frac {t+3}5\end{array}\right\|$ $\bigodot$ $\implies$

$(t+2)(t+3)(t+4)=60t\iff$ $t^3+9t^2-34t+24=0\odot\begin{array}{ccccc} \nearrow & t=1 & \implies & \left(\ \frac 34\ ,\ \frac 45\ ,\ \frac 53\ \right)\in\emptyset & \searrow\\\\ \rightarrow & t=2 & \implies & \left(\ 1\ ,\ 1\ ,\ 2\ \right) & \rightarrow \\\\ \searrow & t=-12 & \implies & \left(\ -\frac 52\ ,\ -\frac 95\ ,\ -\frac 83\ \right) & \nearrow\end{array}\odot$ .

Remark. Denote $s_1=x+y+z\ ,\ s_2=xy+yz+zx$ . Therefore, $\boxed{s_1+s_2=9}\ (*)$ and $\left\{\begin{array}{c} x(y+1)=2\\\\ y(z+1)=3\\\\ z(x+1)=4\end{array}\right\|\ \bigodot\ \implies$

$xyz(x+1)(y+1)(z+1)=24\implies$ $t\left(t+s_2+s_1+1\right)=24$ $\stackrel{(*)}{\implies}$ $t(t+10)=24\implies$ $t^2+10t-24=0\begin{array}{ccc} \nearrow & 2 & \searrow\\\\ \searrow & -12 & \nearrow\end{array}\odot$ a.s.o.

Proof 2. $\left\{\begin{array}{cccc} x(y+1)=2 & \implies & y=\frac {2-x}x & (1)\\\\ z(x+1)=4 & \implies & z=\frac 4{x+1} & (2)\\\\ y(z+1)=3 & \implies & \frac {2-x}x\cdot\left(\frac 4{x+1}+1\right)=3 & (3)\end{array}\right\|\stackrel{(3)}{\implies}$ $(2-x)(x+5)=3x(x+1)\implies$ $2x^2+3x-5=0\begin{array}{cc} \nearrow & x=1\stackrel{(1)}{\implies}y=1\stackrel{(2)}{\implies} z=2\\\\ \searrow & x=-\frac 52\stackrel{(1)}{\implies}y=-\frac 95\stackrel{(2)}{\implies}z=-\frac 83\end{array}$ .

This post has been edited 305 times. Last edited by Virgil Nicula, May 5, 2017, 11:10 AM

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by mathMagicOPS, Jan 9, 2025, 3:40 AM
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391345 views moment
by ryanbear, May 9, 2023, 6:10 AM
We need virgil nicula to return to aops, this blog is top 10 all time.
by OlympusHero, Sep 14, 2022, 4:44 AM
blog
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Amazing blog.
by OlympusHero, May 13, 2021, 10:23 PM
the visits tho
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Bro this blog is ripped
by samrocksnature, Apr 14, 2021, 5:16 AM
Holy- Darn this is good. shame it's inactive now
by the_mathmagician, Jan 17, 2021, 7:43 PM
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by OlympusHero, Dec 30, 2020, 6:08 PM
long blog
by MrMustache, Nov 11, 2020, 4:52 PM
372554 views!
by mrmath0720, Sep 28, 2020, 1:11 AM
wow... i am lost.

369302 views!

-piphi
by piphi, Jun 10, 2020, 11:44 PM
That was a lot! But, really good solutions and format! Nice blog!!!!
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impressive
awesome. 358,000 visits?????
by OlympusHero, May 14, 2020, 8:43 PM
Nice blog.
The element of Mathematics is dense here.
I love the style of your posts.
Thanks for writing the blog! I can learn Geometry from here too.
by epsilonist, Jun 27, 2011, 6:00 AM
Thank you very much for this blog.
by Affine, May 7, 2011, 6:20 PM
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

283. Some nice and easy problems.

by Virgil Nicula, Jun 4, 2011, 5:51 PM

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