439. Properties of medians in triangle ABC and applications.

by Virgil Nicula, Jan 23, 2016, 2:10 PM

An fantastic inequality. $\boxed{4s^2-28Rr+29r^2\le \left(m_a+m_b+m_c\right)^2\le 4s^2-16Rr+5r^2}\ (*)\ .$

Proof.

P1 (Borislav Mirchev). Prove that in any acute or right triangle $ABC$ there is the inequality $\boxed{m_a+m_b+m_c\ge 2R+5r}\ (2)\ .$

Proof. Apply the Walker's inequality: $\boxed{s^2\ge 2R^2+8Rr+3r^2}\ (1)\ :\ \left(m_a+m_b+m_c\right)^2\ \stackrel{(*)}{\ge}\ 4s^2-28Rr+$ $29r^2\ \stackrel{(1)}{\ge}\ 8R^2+$ $32Rr+12r^2-28Rr+29r^2=$

$8R^2+4Rr+41r^2$ . Observe that $8R^2+4Rr+41r^2\ge (2R+5r)^2\iff$ $4R^2-16Rr+16r^2\ge 0\iff 4(R-2r)^2\ge 0$ , what is true. Hence required inequality is true.

P2 (own). Prove that in any acute or right triangle $ABC$ there is the inequality $\boxed{m_a+m_b+m_c\ge 5r+2\sqrt{2Rr}}\ (3)\ .$
Proof. Denote the midpoints $M$ and $S$ (south !) of $[BC]$ and the smaller arc $\overarc{BC}$ . I"ll use the well known (prove easily!) relations $\left\{\begin{array}{cc} \cos\frac {B-C}2\ge\sqrt{\frac {2r}R} & (4)\\\\ MS=\frac {r_a-r}2=\frac {ar}{2(s-a)} & (5)\\\\ am_a+bm_b+cm_c\ge 6sr & (6)\end{array}\right\|$ .

Apply in $\triangle AMS$ the inequality $AS\le MA+MS\iff$ $2R\cos\frac {B-C}2\le m_a+2R(1-\cos A)\ \stackrel{4\wedge 5}{\implies}$ $2R\sqrt{\frac {2r}R}\le m_a+\frac {ar}{2(s-a)}\iff$

$ar+2(s-a)m_a\ge 4(s-a)\sqrt{2Rr}\iff$ $ar+2sm_a\ge 2am_a+4(s-a)\sqrt{2Rr}\ \stackrel{\sum}{\implies}\ 2sr+2s\sum m_a\ge 2\sum am_a+4s\sqrt{2Rr}\ \stackrel{(6)}{\iff}$

$2sr+2s\sum m_a\ge 12sr+4s\sqrt{2Rr}\iff$ $r+\sum m_a\ge 6r+2\sqrt{2Rr}\iff$ $m_a+m_b+m_c\ge 5r+2\sqrt{2Rr}\ .$

P3 (own). Prove that in any triangle $ABC$ there is the inequality $\boxed{4m_bm_c\le 2a^2+bc}\ (7)\ .$

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

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

$(b-c)^2\left[(b+c)^2-a^2\right]\ge 0\iff$ $4s(s-a)(b-c)^2\ge 0\iff (b-c)^2\ge 0$ what is truly. I have equality iff $b=c$ .

P4. Prove that in any acute triangle $ABC$ there is the inequality $\boxed{b^2+c^2\le 4Rm_a}\ (8)\ .$ .

Proof 1. Denote the diameter $[AR]$ of circumcircle $w=\mathbb C(O,R)$ and $\{A,K\}=AM\cap w$ . Is well known that $M\in HR$ and $MH=MR$ , where $H$ is the

orthocenter. In the isosceles triangle $AKO$ apply the inequality $AK\le 2\cdot AO\iff$ $MA+MK\le 2R$ . Since $MA\cdot MK=MB\cdot MC\iff$ $MK=\frac {a^2}{4m_a}$

obtain that $m_a+\frac {a^2}{4m_a}\le 2R\iff$ $4m_a^2+a^2\le 2Rm_a\iff$ $2\left(b^2+c^2\right)-a^2+a^2\le 8Rm_a\iff$ $b^2+c^2\le 4Rm_a$ .

Remark. Apply the theorem of median in $\triangle BCR\ :\ 4\cdot MR^2=2\left(RB^2+RC^2\right)-BC^2=$

$2\left[\left(4R^2-c^2\right)+\left(4R^2-b^2\right)\right]-a^2=$ $16R^2-2\left(b^2+c^2\right)-a^2=$ $16R^2-4m_a^2-2a^2\implies$ $\boxed{MR^2=4R^2-m_a^2-\frac {a^2}2}\ (9)\ .$

Proof 2. Apply in $\triangle AMO$ the inequality $MA\le OM+OA\iff$ $m_a\le R(1+\cos A)\iff$ $\left(m_a-R\right)^2\le R^2\left(1-\sin^2A\right)\iff$

$m_a^2-2Rm_a+R^2\sin^2A\le 0\iff$ $2\left(b^2+c^2\right)-a^2-8Rm_a+a^2\le 0\iff$ $b^2+c^2\le 4Rm_a$ .

Proof 3. Apply in $\triangle MAR$ the inequality $AR\le MA+MR\iff$ $2R\le m_a+MR\iff$ $\left(2R-m_a\right)^2\le MR^2\iff$ $4R^2-4Rm_a+m_a^2\le$

$4R^2-m_a^2-\frac {a^2}2\iff$ $2m_a^2-4Rm_a+\frac {a^2}2\le 0\iff$ $2\left(b^2+c^2\right)-a^2-8Rm_a+a^2\le 0\iff$ $b^2+c^2\le 4Rm_a$ .

P5. The incircle of $\triangle ABC$ touches $BC$ , $CA$ , $AB$ at $(D,E,F)$ respectively. The median $AM$ meet again the circumcircle $\rho =\mathbb C(O,R)$ at $K$ , where $M\in BC$ .

Let $L\in EF\cap DI$ and $\{A,K\}=\{A,M\}\cap\rho$ . Prove that $L\in AM$ and $\left\{\begin{array}{ccccc} \frac {LE}c & = & \frac {LF}b & = & \frac {EF}{b+c}\\\\ \frac {LD}{a+b+c} & = & \frac {LI}a & = & \frac r{b+c}\end{array}\right\|$ and $\left\{\begin{array}{ccccc} \frac {LM}a & = & \frac {LA}{b+c-a} & = & \frac {m_a}{b+c}\\\\ \frac 1{4m_a} & = & \frac {MK}{a^2} & = & \frac {AK}{2\left(b^2+c^2\right)}\end{array}\right\|$ .

P6. Prove that in any $\triangle ABC$ exists the inequality $\sum\frac {m_a}{a}\ge \frac {s}{R}\ .$ (standard notations).

Proof.

P7. Let $\triangle ABC$ with the lengths $\{m_a,m_b,m_c\}$ of its medians. Find $\{a,b,c\}$ .

Proof 1. Denote the centroid $G$ of $\triangle ABC$ , the midpoint $M$ of $[BC]$ and the symmetrical point $S$ of $G$ w.r.t. $M$ . Observe that $[BM]$ is the median in $\triangle BGS\ ,$

$BM=\frac a2$ and $\left\{\begin{array}{c} BG=\frac 23\cdot m_b\\\\ GS=\frac 23\cdot m_a\\\\ SB=\frac 23\cdot m_c\end{array}\right|$ . Thus, $4\cdot BM^2=2\left(BG^2+BS^2\right)-GS^2$ $\iff$ $a^2=\frac 49\cdot \left[2\left(m_b^2+m_c^2\right)-m_a^2\right]\iff$ $a=\frac 23\cdot\sqrt {2\left(m_b^2+m_c^2\right)-m_a^2}$ a.s.o.

Proof 2. Solve the system $\left\{\begin{array}{ccc} -a^2+2b^2+2c^2 & = & 4m_a^2\\\\ 2a^2-b^2+2c^2 & = & 4m_b^2\\\\ 2a^2+2b^2-c^2 & = & 4m_c^2\end{array}\right|$ . Obtain that $\Delta =\left|\begin{array}{ccc} -1 & 2 & 2\\\\ 2 & -1 & 2\\\\ 2 & 2 & -1\end{array}\right|=27$ and $\Delta_a =4\cdot \left|\begin{array}{ccc} m_a^2 & 2 & 2\\\\ m_b^2 & -1 & 2\\\\ m_c^2 & 2 & -1\end{array}\right|=$

$12\cdot\left[2\left(m_b^2+m_c^2\right)-m_a^2\right]\implies$ $a^2=\frac {\Delta_a}{\Delta}=\frac 49\cdot \left[2\left(m_b^2+m_c^2\right)-m_a^2\right]\implies$ $a=\frac 23\cdot\sqrt {2\left(m_b^2+m_c^2\right)-m_a^2}$ a.s.o.

P8. Let $ABC$ be a triangle so that $b\ne c$ . The medians $AD$ , $BE$ , $CF$ ( where $D\in BC$ , $E\in CA$ , $F\in AB$ )

intersect again the circumcircle $w$ of $ABC$ at $L$ , $M$ , $N$ respectively. Prove that $LM = LN\ \Longleftrightarrow\ b^2 + c^2 = 2a^2$ .

Proof. Apply the power of the points $D$ , $E$ , $F$ w.r.t. circle $w$ :

$\left\{\begin{array}{ccccc} DL\cdot DA = DB\cdot DC & \implies & LD = \frac {a^2}{4m_a} & \implies & LA = LD + m_a = \frac {b^2 + c^2}{2m_a} \\ \\ EM\cdot EB = EC\cdot EA & \implies & ME = \frac {b^2}{4m_b} & \implies & MB = ME + m_b = \frac {c^2 + a^2}{2m_b} \\ \\ FN\cdot FC = FA\cdot FB & \implies & NF = \frac {c^2}{4m_c} & \implies & NC = NF + m_c = \frac {a^2 + b^2}{2m_c}\end{array}\right\}$

Apply the equivalence of areas and Ptolemeu's therem to the quadrilaterals $ABLC$ , $BCMA$ , $CANB$ :

$\left\{\begin{array}{cccccc} \left\|\begin{array}{c} c\cdot LB = b\cdot LC \\ \\ b\cdot LB + c\cdot LC = a\cdot LA\end{array}\right\| & \implies & \left\|\begin{array}{c} LB = \frac {ab}{b^2 + c^2}\cdot LA \\ \\ LC = \frac {ac}{b^2 + c^2}\cdot LA\end{array}\right\| & \implies & \left\|\begin{array}{c} LB = \frac {ab}{2m_a} \\ \\ LC = \frac {ac}{2m_a}\end{array}\right\| \\ \\ \left\|\begin{array}{c} a\cdot MC = c\cdot MA \\ \\ c\cdot MC + a\cdot MA = b\cdot MB\end{array}\right\| & \implies & \left\|\begin{array}{c} MC = \frac {bc}{c^2 + a^2}\cdot MB \\ \\ MA = \frac {ba}{c^2 + a^2}\cdot MB\end{array}\right\| & \implies & \left\|\begin{array}{c} MC = \frac {bc}{2m_b} \\ \\ MA = \frac {ba}{2m_b}\end{array}\right\| \\ \\ \left\|\begin{array}{c} b\cdot NA = a\cdot NB \\ \\ a\cdot NA + b\cdot NB = c\cdot NC\end{array}\right\| & \implies & \left\|\begin{array}{c} NA = \frac {ca}{a^2 + b^2}\cdot NC \\ \\ NB = \frac {cb}{a^2 + b^2}\cdot NC\end{array}\right\| & \implies & \left\|\begin{array}{c} NA = \frac {ca}{2m_c} \\ \\ NB = \frac {cb}{2m_c}\end{array}\right\|\end{array}\right\}$

Apply Ptolemeu's theorem to the quadrilaterals $LBNC$ , $LCMB$ :

$\left\{\begin {array}{c} a\cdot LN = LB\cdot NC + LC\cdot NB = \frac {ab(a^2 + b^2 + c^2)}{4m_am_c} \\ \\ a\cdot LM = LB\cdot MC + LC\cdot MB = \frac {ac(a^2 + b^2 + c^2)}{4m_am_b}\end{array}\right\}$ . Therefore, $LN = LM$ $\Longleftrightarrow$ $bm_b = cm_c$ $\Longleftrightarrow$ $b^2 + c^2 = 2a^2$ .

Generalization. Let $ABC$ be a triangle, $b\ne c$ with the circumcircle $w$ . For an interior point $P(x,y,z)$ denote the second intersections $L$ , $M$ , $N$ of the

circle $w$ with $AP$ , $BP$ , $CP$ respectively. Prove that $LM = LN\ \Longleftrightarrow\ x\left[\left(zb^4 - yc^4\right) + (y - z)b^2c^2\right] = a^2\left[z(x + z)b^2 - y(x + y)c^2\right]$ .

Remark. I denoted by $P(x,y,z)$ the point $P$ with the the barycentrical coordinates $(x,y,z)$ w.r.t. the triangle $ABC$ . Here are two particular cases.

$1.\ \blacktriangleright\ \ P = G(1,1,1)$ - centroid : $LM = LN\ \Longleftrightarrow\ b^2 + c^2 = 2a^2$ .

$2.\ \blacktriangleright\ \ P = I(a,b,c)$ - incenter : $LM = LN\ \Longleftrightarrow\ \emptyset$ .

a.s.o.

$bm_b=cm_c\iff$ $b^2\cdot 4m_b^2=c^2\cdot 4m_c^2\iff$ $b^2\cdot\left[2\left(a^2+c^2\right)-b^2\right]=c^2\cdot\left[2\left(a^2+b^2\right)-c^2\right]\iff$

$b^2\cdot\left(2a^2-b^2\right)=c^2\cdot\left(2a^2-c^2\right)\iff$ $2a^2\left(b^2-c^2\right)=b^4-c^4\iff$ $2a^2=b^2+c^2$ because $b\ne c$ .

Lemma (well-known). Let $\triangle ABC$ and $M\in AB$ , $N\in AC$ so that $BM=CN$ . Denote the middlepoints $X$ , $Y$ of $[BC]$ , $[MN]$ respectively. Prove that :

If $BC$ doesn't separate $M$ , $N$ then $XY$ is parallel with the internal bisector of $\widehat{BAC}\ ;$ if $BC$ separates $M$ , $N$ then $XY$ is parallel withe the external bisector of $\widehat{BAC}$ .

Proof 1. Denote $\{D,D'\}\subset BC$ so that the rays $[AD$ , $[AD'$ are respectively the internal and external bisectors of the angle $\widehat{B AC}$ . Define

the reflection $P$ of the point $N$ w.r.t. the point $X$ . Observe that $MP\parallel XY$ , $PB\parallel AC$ , $MB=NC=PB$ , i.e. $\triangle BMP$ is $B$ - isosceles.

$\left\{\begin{array}{cccccccc}1\blacktriangleright & m(\widehat{MBP})=B+C & \implies & m(\widehat{BMP})=\frac{A}{2}& \implies & MP\parallel AD & \implies & XY\parallel AD\ .\\\\ 2\blacktriangleright & m(\widehat{MBP})=A & \implies & m(\widehat{BMP})=\frac{B+C}{2}& \implies & MP\parallel AD' & \implies & XY\parallel AD'\ .\end{array}\right\|$

Remark. Let mobile $M\in AB$ , $N\in AC$ so that $BM=CN$ . Then the geometrical locus of the midpoint of $[MN]$ is $AD\cup AD'$ .

Proof 2. Let $\left\{\begin{array}{c}\{M,M'\}\subset AB\\\\ \{N,N'\}\subset AC\end{array}\right\|$ so that $\left\{\begin{array}{c} M\in (AB)\ ,\ N\in (AC)\\\\ BM=BM'\ ,\ CN=CN'\end{array}\right\|$ . Denote the midpoints $X$ , $Y$ , $U$ , $V$ $Z$ of $[BC]$ , $[MN]$ , $[M'N]$ , $[MN']$ , $[M'N']$ respectively.

Prove easily that $UYVZ$ is a rhombus ( $UY\parallel VZ\parallel AB$ , $YV\parallel UZ\parallel AC$ ) and $X\in UV\cap YZ$ . Thus, $\overline{UXV}\perp \overline{YXZ}$ , i.e. $\overline{YXZ}\parallel AD$ and $\overline{UXV}\parallel AD'$ .

Remark. Let $ABCD$ be a convex quadrilateral so that $E\in AB\cap CD$ and $AB=CD$ . Denote the midpoints $X,Y,U,V$ of $[BC]$ , $[AD]$ , $[AC]$ , $[BD]$ respectively.

Then $XY$ , $UV$ are parallel respectively to the internal and external bisectors of $\widehat{AED}$ .

P9. Let $ABC$ be an acute triangle with the centroid $G$ and the circumcircle $w=C(O,R)$ . Denote $\{A,S\}=\{A,G\}\cap w$

and the diameter $[SN]$ of $w$ . Prove that $\boxed{\sqrt{a^2+b^2+c^2}\le 3R\cos\widehat{ASN}}\le 3R$ . Have the equality iff $b^2+c^2=2a^2$ .

Proof. Let the midpoint $M$ of $[BC]$ and $m\left(\widehat{ASN}\right)=\phi$ .Thus, $AS=2R\cos\phi$ and from the power $p_w(G)$ of $G$ w.r.t. $w$ obtain that

$GA\cdot GS=$ $\frac {a^2+b^2+c^2}{9}\implies$ $2R\cos\phi =AS=GA+GS\ge$ $2\sqrt{GA\cdot GS}=$ $2\cdot \sqrt{\frac {a^2+b^2+c^2}{9}}$ $\implies$ $\sqrt {a^2+b^2+c^2}\le$ $3R\cos\phi$ . Have

the equality iff $GA=GS\iff$ $-p_w(G)=GA^2\iff$ $\frac {a^2+b^2+c^2}{9}=\frac {4m_a^2}{9}\iff$ $a^2+b^2+c^2=2\left(b^2+c^2\right)-a^2\iff$ $b^2+c^2=2a^2$ .

P10. $\boxed{\mathrm{Prove\ that}\ (\forall )\ \triangle\ ABC\ \mathrm{exists\ the\ inequality}\ 4m_bm_c\le 2a^2+bc}$ (notatii standard).

Proof 1. $4m_bm_c\le 2a^2+bc\iff$ $4m_b^2\cdot 4m_c^2\le \left(2a^2+bc\right)^2\iff$ $\left[2\left(a^2+c^2\right)-b^2\right]\cdot \left[2\left(a^2+b^2\right)-c^2\right]\le 4a^4+4a^2bc+b^2c^2\iff$ $4\left(a^2+b^2\right)\left(a^2+c^2\right)-2b^2\left(a^2+b^2\right)-$

$2c^2\left(a^2+c^2\right)\le 4a^2\left(a^2+bc\right)\iff$ $2\left(a^2+b^2\right)\left(a^2+c^2\right)-b^2\left(a^2+b^2\right)-c^2\left(a^2+c^2\right)\le 2a^2\left(a^2+bc\right)\iff$ $\cancel 2a^2\left(b^2+c^2\right)+2b^2c^2-\cancel{a^2\left(b^2+c^2\right)}-\left(b^4+c^4\right)\le 2a^2bc\iff$

$a^2\left(b^2+c^2-2bc\right)\le b^4+c^4-2b^2c^2\iff$ $a^2(b-c)^2\le \left(b^2-c^2\right)^2=(b+c)^2(b-c)^2\iff$ $(b-c)^2\left[(b+c)^2-a^2\right]\ge 0\iff$ $4s(s-a)(b-c)^2\ge 0\iff$ $(b-c)^2\ge 0\ ,$ what is true.

Proof 2. Let $\left\{\begin{array}{ccc} N\in AC & ; & NA=NC\\\ P\in AB & ; & PA=PB\end{array}\right\|$ and apply the Ptolemy's inequality in $BPNC\ :\ BN\cdot CP\le BC\cdot PN+BP\cdot CN\iff$ $m_bm_c\le a\cdot \frac a2+\frac c2\cdot\frac b2\iff \boxed{4m_bm_c\le 2a^2+bc}\ .$

P11 (Adil Abdullayev). Prove that $(\forall )\ \triangle\ ABC$ exists the inequality $\left(m_a+m_b+m_c\right)^2+4r(R-2r)\le (4R+r)^2\ .$ (notatii standard).

Proof. I"ll use the remarkable inequality $\boxed{s^2\le 4R^2+4Rr+3r^2}\ (*)$ and the conclusion of the previous problem $:\ \boxed{4m_bm_c\le 2a^2+bc}\ (1)\ .$ Thus $:\ 4\cdot\sum m_bm_c\ \stackrel{(1)}{\le}\ 2\cdot \sum a^2+\sum bc=$

$4\left(s^2-r^2-4Rr\right)+$ $\left(s^2+r^2+4Rr\right)=$ $5s^2-3r^2-12Rr\ \stackrel{(*)}{\le}\ 5\left(4R^2+4Rr+3r^2\right)-3r^2-12Rr=$ $20R^2+8Rr+12r^2$ $\implies$ $\boxed{\sum m_bm_c\le 5R^2+2Rr+3r^2}\ (2)\ .$ Hence $:$

$\underline{\underline{4\left(\sum m_a\right)}}^2=$ $4\sum m_a^2+2\cdot 4\sum m_bm_c\ \stackrel{(2)}{\le}\ 3\sum a^2+8\left(5R^2+2Rr+3r^2\right)=$ $6\left(s^2-r^2-4Rr\right)+8\left(5R^2+2Rr+3r^2\right)\ \stackrel{(*)}{\le}\ 6\left[\left(4R^2+\cancel{4Rr}+3r^2\right)-\left(r^2+\cancel{4Rr}\right)\right]+$

$8\left(5R^2+2Rr+3r^2\right)=$ $\underline{\underline{64R^2+16Rr+36r}}^2\implies$ $\left(\sum m_a\right)^2\le 16R^2+4Rr+9r^2=$ $(4R+r)^2-4r(R-2r)\implies$ $\boxed{\left(\sum m_a\right)^2+4r(R-2r)\le (4R+r)^2}\ .$

P12 (Adil Abdullayev). Prove $(\forall )\ \triangle ABC$ there is the following inequality $:\ \boxed{\frac {\cot A}{s-a}+\frac {\cot B}{s-b}+\frac {\cot C}{s-c}\le\frac 1r-\frac {R-2r}{2Rr}}\ (*)\ .$ (standard notations).

An interesting substitution in the geometry of a triangle. Denote $:\ \left\|\begin{array}{ccccccc} s-a=x & ; & a=y+z & \implies & x+y+z & = & s\\\\ s-b=y & ; & b=z+x & \implies & xy+yz+zx & = & r(4R+r)\\\\ s-c=z & ; & c=x+y & \implies & xyz & = & sr^2\end{array}\right\|\ .$ For example $:$

$\blacktriangleright\ \left\|\begin{array}{ccccccc} 2bc\cdot\cos A & = & b^2+c^2-a^2=(x+z)^2+(x+y)^2-(y+z)^2=2[s(s-a)-(s-b)(s-c)] & \implies & s(s-a)-(s-b)(s-c) & = & bc\cdot \cos A\\\\ bc & = & (x+z)(x+y)=x^2+x(y+z)+yz=x(x+y+z)+yz=s(s-a)+(s-b)(s-c) & \implies & s(s-a)+(s-b)(s-c) & = & bc\end{array}\right\|\implies$

$\left\{\begin{array}{ccccccc} 2s(s-a)=bc(1+\cos A) & \implies & \cancel 2s(s-a)=\cancel 2bc\cdot\cos^2\frac A2 & \implies & \cos\frac A2 & = & \sqrt{\frac {(s(s-a)}{bc}}\\\\ 2(s-b)(s-c)=bc(1-\cos A) & \implies & \cancel 2(s-b)(s-c)=\cancel 2bc\cdot \sin^2\frac A2 & \implies & \sin\frac A2 & = & \sqrt{\frac {(s-b)(s-c)}{bc}}\end{array}\right\|\ .$

$\blacktriangleright\ X\equiv \sum\frac {yz}x=\frac 1{xyz}\cdot \sum(yz)^2=\frac 1{sr^2}\cdot\left[\left(\sum yz\right)^2-2xyz(x+y+z)\right]=$ $\frac {\cancel {r^2}(4R+r)^2-2s^2\cancel{r^2}}{s\cancel{r^2}}\implies$ $X=\boxed{\sum\frac {yz}x=\frac {(4R+r)^2-2s^2}s}\ (*)\ .$

Proof. $\sum\frac {\cot A}{s-a}=\sum\frac {b^2+c^2-a^2}{4S(s-a)}=\frac 1{4sr}\cdot \sum\frac {2[x(x+y+z)-yz]}x=$ $\frac 1{2sr}\cdot\sum \left(x+y+z-\frac {yz}x\right)=$ $\frac 1{2sr}\cdot (3s-X)\ \stackrel{(*)}{=}\ \frac 1{2sr}\cdot\left[5s-\frac {(4R+r)^2}s\right].$ So $\sum\frac {\cot A}{s-a}\le \frac {R+2r}{2Rr}\iff$

$\frac 1{\cancel 2s\cancel r}\cdot\left[5s-\frac {(4R+r)^2}s\right]\le \frac {R+2r}{\cancel 2R\cancel r}$ $\iff$ $R\left[5s^2-(4R+r)^2\right]\le s^2(R+2r)\iff$ $s^2(4R-2r)\le R(4R+r)^2\iff$ $\boxed{s^2\le \frac {R(4R+r)^2}{2(2R-r)}}\ ,$ what is Blundon & Gerretsen's inequality.

P13 (Adil Abdullayev). Prove $(\forall )\ \triangle ABC$ there is the following inequality $:\ \boxed{S\le \frac {s^2}{\sqrt{27+k}}\ ,\ \mathrm{where}\ k=\frac {R-2r}{R-r}}\ .$ (standard notations).

Proof. $S\le \frac {s^2}{\sqrt{27+k}}$ $\iff$ $S^2(27+k)\le s^4\iff$ $r^2(27+k)\le s^2\iff$ $r^2\left(27+\frac{R-2r}{R-r}\right)\le s^2\iff$ $r^2\left(28R-29r\right)\le s^2(R-r)\iff$ $\boxed{s^2\ge \frac{r^2(28R-29r)}{R-r}}\ (1)\ .$ Suppose that is

well-known the inequality $\boxed{s^2\ge 16Rr-5r^2}\ (*)\ .$ I"ll prove $\boxed{16Rr-5r^2\ge \frac {r^2(28R-29r)}{R-r}}\ (2)\ .$ Indeed, $16R\cancel r-5r\cancel {^2}\ge \frac {r\cancel{^2}(28R-29r)}{R-r}\iff$ $\left(16R-5r\right)(R-r)\ge r(28R-29r)\ \stackrel{R=tr}{\iff}$

$(16t-5)(t-1)\ge 28t-29\iff$ $16t^2-49t+34\ge 0\iff$ $(t-2)(16t-17)\ge 0\iff t\ge 2\iff R\ge 2r\ ,$ what is true. From the relations $(*)$ and $(2)$ obtain the required inequality $(1)\ .$

P14. https://scontent.fotp3-3.fna.fbcdn.net/v/t1.0-9/21150439_10155064914552833_6318925499382620265_n.jpg?oh=71590900f4f7dd485f267982266a6b8e&oe=5A59174B.

Proof. I''l apply the following simple inequalities $:\ \left\{\begin{array}{cccc} (a+b)^2 & \le & 2\left(a^2+b^2\right) & (1)\\\\ 4ab & \le & (a+b)^2 & (2)\\\\ 2s & \le & 3R\sqrt 3 & (3)\end{array}\right\|\ .$ Hence $\frac {ab}{\sqrt {a^2+b^2}}\ \stackrel{(1)}{\le}\ \frac {ab\sqrt 2}{a+b}\ \stackrel {(2)}{\le}\ \frac {(a+b)\sqrt 2}4\implies$ $\sum\frac {ab}{\sqrt {a^2+b^2}}\le s\sqrt 2\ \stackrel{(3)}{\le}\ \frac {3R\sqrt 6}2\ .$

This post has been edited 145 times. Last edited by Virgil Nicula, Sep 6, 2017, 12:30 PM

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Here is the file with the proof of the "fantastic" inequality.

Attachments:

10020203.pdf (107kb)

This post has been edited 1 time. Last edited by Virgil Nicula, Jan 23, 2016, 3:33 PM

by Virgil Nicula, Jan 23, 2016, 2:16 PM

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In document (10020203.pdf) is proof only for right inequality. Proof for left inequality is in
" Research in Inequalities " (in Chinese language) by Xi.-G.chu and Xu.-Z.Yang.
In document left inequality is given without proof. It is remark1.

Attachments:

by bitrak, Jan 24, 2016, 5:43 PM

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Here is my proof from 21 Jan 2016 of PP1 (Borislav Mirchev)
https://www.facebook.com/photo.php?fbid=10153866447611552&set=p.10153866447611552&type=3&theater
or
http://artofproblemsolving.com/community/q1h1186527p5786758

by bitrak, Jan 24, 2016, 5:51 PM

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459.Working page V.

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457. Relatii metrice si inegalitati geometrice.

456. Probleme de Algebra.

July 2017

456* Integrals.

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454. Mathematical note: "trig. ineg."- GMB 6/1992, page 201.

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453. Working page III.

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452. Working page II.

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451. ABC with C(O,R), C(I,r) and N-Nagel point ==> ON=R-2r.

450. Nota matematica: "Asupra unei inegalitati remarcabile".

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449. Working page I.

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448. Extensions or generalizations.

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417. The incircle of ABC and other two incircles.

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357. Morley theorem.

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330. A triangle and an its transversal through a point.

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280. Common roots of two polynomial equations.

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276. Chinese TST 2009 and Second Round 2006.

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271. Relations concerning Fermat points.

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270. Morrocan M.O. 2010 and Croatia TST 2009.

269. Some nice and easy properties in a trapezoid.

268. Saraghin contest, 2011.

267. Germany 2002, Russia 2003, IMO 2003, Aus.-Polish 1992.

266. Rectangles erected on sides of a triangle. Concurrence.

265. Really nice problem !

264. Identities with complex numbers.

263. Newton's sums. Applications.

262. IMO Shortlist 2005, Geometry 6th.

261. IMO 2009, problem 2 and JBMO 2007, problems 1 & 2.

260. Some nice, simple identities and their applications.

259. ABC equilateral triangle ==> DEF equilateral triangle.

258. Some difficult and nice problems with complex numbers.

257. Seeking roots of 2th equation with complex coef.

256. An extremum problems with complex numbers.

March 2011

255. A line which is parallelly with OH.

254. An interesting C. Mateescu's geometrical inequality.

253. An old algebraic inequality.

252. OI=f(A,R) in certain conditions.

251. Problems of Analytical Geometry.

250. An application of the Euler's relation.

249. An usual and nice metrical relations in quadrilateral.

248. Pagina instructiva.

247. An remarkable inequality with median.

246. R1-R2=OI (nice).

245. An inequality in an acute triangle.

244. Some properties of cyclic orthodiagonal quadrilaterals.

243. Nice ! (a+b) is constant.

242. A general geometrical problems.

241. Areas in a triangle.

240. Some interesting geometrical inequalities.

239. Algebraic marathon.

238. Some metrical problems.

237. Some problems with ... problems.

236. A locus with the metrical relation.

235. Proofs of some geometry problems with complex numbers.

234. Some simple and nice problems from contests.

233. An interesting identity and an easy & nice inequality.

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232. Some geometrical inequalities (own).

Pagina libera

231. A conditioned minimum value.

230. Some usual synthetical properties in a triangle.

229. Concurs online MATEFBC, ed. 5 -subiect 2.

228. OJM - Romania, 2010 problem 3.

227. IMO ShortList 2003 (problem 5).

226. Some inequalities from Calculus (Real Analysis).

225. An interesting problems from Kunny.

224. Some nice and easy inequalities in a triangle.

223. An inequality with complex numbers-RMO shortlist 2010.

222. An extension of probl. 2 from IMO 2009.

221. Some nice and simple "slicing" problems.

220. Some nice geometry problems from contest.

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219. A very nice geometrical inequality.

218. Some problems from the Suiss IMO Selection Team 2006.

217. A nice and interesting Murray S. Klamkin's problem.

216. Nice ! Find the length of hypotenuse (Belarus - 2010)

215. Limit of the unique root for polynominal equations.

214. A range of some functions.

213. Another classical "slicing" problems.

212. A simple applications of the Casey's theorem.

211. Pretty hard problem !

210. Tangential circle of the triangle ABC.

209. Nice problem, nice proof !

208. An interesting geometrical inequality.

207. Some interesting problems of Toshio Seimiya, Japan.

206. Four problems with extremum in a quadrilateral.

205. A "slicing" problem with 11 methods.

204. Saraghin 2010 (three geometry problems).

203. An application of the Pascal's permutation theorem.

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202. Concurrent lines in a right triangle.

201. Applications of Desarques' and Pappus' theorems.

200. Some nice applications of the inversion.

199. Some properties of symmedian. Two extensions.

198. Generalized Carnot's lemma.

197. Triangles outside of a triangle and concurrence.

195. Semicircle. Nice application of harmonical division.

194. The Lalescu's sequence. Properties.

193. An inequality with n! .

192. An easy and nice problem of Valentin Vornicu.

191. Six nice problems with the incircle of a triangle.

190. Very nice ! Coculescu's Contest seniors 2010.

189. Pascal's theorem and some nice and easy applications.

188. Some problems from vectorial geometry.

187. Distancies.

186. Without the l'Hospital's rules.

185. The Durrande's relation.

184. An interesting geometry problem from RMO 2010.

183. A remarkable characterization of equilateral triangle.

182. Another two nice problems with complex numbers (own).

181. Two special inequalities from CRUX (own).

180. Own proposed problem from CRUX (with complex numbers).

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179. The range of |z| , |z^2+a|=|bz+c|. Particular cases.

178. Some nice and simple geometry problems.

177. A nice proof of one identity in a triangle.

176. Another nice problems for middle school.

175. A Geometric Proof (without trig.) of Heron's Formula.

174. Some interesting properties of modulus (middle school).

173. An interesting trigonometric equations.

172. The Zelinsky's problem.

171. Another "slicing" proposed problems (middle school).

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170. A nice "slicing" proposed problem for middle school.

169. Some metrical problems in a circle (III).

168. Some properties of the Lemoine's point.

167. Nice problems - OIM, 1995 and ... Toshio Seimyia.

166. Some constant geometrical expressions.

165. Some metrical problems (5) in a circle (II).

164. Some metrical problems (5) in a circle.

163. Applications of Ptolemy's inequality. Ex. - China,1998.

162. Generalization of proposed problem from Mongolia, 2000.

161. A difficult inequality (barycentrical coordinates)

160. A very nice "slicing" problem.

159. Multiple derivatives in a point.

158. Very nice and difficult limits.

157. Steinbart's theorem. Applications.

156. Some problems with perpendicularities.

155. A constant ratio in a triangle ABC.

154. Difficult inequality (without derivatives eventually).

153. Geometrical minimum/maximum.

152. Concurrence, cyclically, metrics, symmedian a.s.o.

151. Miscellaneous problems for the admission at math.

150. Some (5) problems with fixed points.

149. Some (8) problems with collinearity or concurrence.

148. A special class of systems.

147. A problem with two polynominals.

146. Synthetic, metric, trigonometric and analitycal proofs.

145. Spain MO P5 - Point on median.

144. A conditioned concurrency in a triangle.

143. Assess of a ratio (nice and difficult problem).

142. Inscribed/circumscribed quadrilateral w.r.t. ABC.

141. N is interior <==> 1/3<a/b<3 (CRUX Mathematicorum).

140. A synthetical property <==> linear/angled relation.

139. A caracterization of a right angle in a triangle.

138. Generalized "slicing" problems (own).

137. Stewart's relation.

136. A geometrical interpretation of a system (own).

135. Three strong geometrical inequalities.

September 2010

134. Some problems with projections/perpendiculars (2).

133. Some problems with projections/perpendiculars (1).

132. Asupra unei inegalitati intr-un triunghi.

131. A stronger Form of the Steiner-Lehmus Theorem (own).

130. A triangle and a fixed point.

129. BX=CY and three properties.

128. Value of IG and two inequalities in neighbourhood of 0.

127. Pagina educativa.

126. Routh's theorem.

125. I like very much this Darij Grinberg's message.

124. A nice and easy problem with a right-angled triangle.

123. My extension and its proof.

122. An identity for an equilateral triangle. Extension.

121. Viete's relations with cos A , cos B , cos C.

120. Two equal neighbouring angles/sides in a quadrilateral.

119. Inspired by a Miculita's problem.

118. An usual remarkable geometrical locus.

117. Some applications of harmonic division/quadrilateral.

116. The length of AE (nine methods !).

115. An difficult equation with one radical.

114. A property of tangential triangle for ABC.

113. Range for a function.

112. The equivalence relation ".s.s."

111. Simple, but interesting ...

110. German TST 2004 - Exam V , Problem 1

109. Two equal areas ==> find A.

108. Old, short and nice proof of Euler's relation.

107. XY parallel BC and X,Y are isogonal/izotomic points.

106. OLM - Bucuresti (geometrie).

105. A problem with three circles.

104. Concursul Revistei de Matematica din Galati (RMG).

103. Metrical usual relations in triangle. Applications.

102. Characterization of a metrical relation in a triangle.

101. O problema cu numere complexe.

100. Bobiller's theorem.

99. Some simple and nice geometry problems.

98. Symmedian & median (metrical relations).

97. Problem of butterfly.

August 2010

96. Integrals.

95. Three circles in an angle.

94. Equations of angle bisectors (interior and exterior).

93. The Appolonius' construct problems.

92. CA , CB tangent to parabola - OIMU 2008 Problem 4.

91. JBMO 2010, Problem 3.

90. Nice and difficult inequality in ABC (own).

89. Similarity (parallel/antiparallel).

88. A nice geometric locus conditioned metrically.

87. An interesting vectorial identity.

86. A very nice maxmin.

85. Problem "slicing" : 60-24-12.

84. IRAN national math olympiad - 2010

83. A metrical characterization of concurrence.

82. AA' , BN , CM concur in Gergonne point.

81. Sawayama and Thebault’s theorem.

80. Three concurrent perpendiculars (the Carnot's lemma).

79. A quadrilateral with many perpendicularities.

78. Jakobi's theorem.

77. Two equivalent perpendicularities.

76. A parallelogram and a circle (nice !).

75. Extended Steinbart Theorem.

74. Inegalitatea Erdos-Mordell.

73. A nice problems with a square and a perpendicularity.

July 2010

72. Colinearity in a triangle - H , P , M .

70. Colinearity with pole/polar - H\in PQ .

71. An angle questions (synthetic/trigonometric proofs).

69*1. Slicing problem 3.

69. Position of x1, x2 w.r.t. a given real number k.

68. Some nice applications of "pole/polar".

67. Remarkable algebraic/geometrical inequalities (1).

66. Remarkable perpendicularities in a triangle.

65. Parallel to BC through I .

64. Problems of the type "instant" (at most one minute).

63. A nice problem with radical center (livetolove212).

62. Nice and easy problem with a right triangle.

61. IMO Shortlist 2009 (very nice problem and its proof).

60. Mediteranean M.C. 2010 Problem 3.

59. IMO 2010, problem 4.

58. A cevian and two incircles.

57. Opinii si comentarii relativ la inv. rom. II

56. A extremum problem with areas.

55. Two important circles - mixtilinear incircle/excircle.

54. Cinci grade de libertate.

53. An remarkable concurrency.

52. Harmonical division.

51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.

51. A nice problem with perpendicularity from Jaime.

50. IMAC 2010 seniors, the first day.

49. The midpoint of a cevian in an acute triangle.

June 2010

48. An inequality in a triangle : h_a+max{b,c} ...

47. Interesting and difficult identities in a triangle.

46. Outside of a quadrilateral.

45. \cos B+\cos C=1 <==> nice property.

44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.

43. A "slicing" problem with a parameter.

42. IMAC 2010 Problema 2.

41. ONM 2010 Calarasi Problema 3.

May 2010

40. Lema lui Haruki.

39. A nice and remarkable problem with Brocard's point.

38. Minimum area of triangle touching with semicircle.

37. Own extension of the Steiner-Lehmus' problem.

36. ONM (juniori) - 2010 Iasi, problema 4.

35. Nice problem from Arqady (Eduard_Tur).

34. IMO Shortlist 2002 geometry problem 7.

April 2010

32. Linkuri diverse.

31.Some nice metrical problems.

29. Inegalitate integrala de la Kunny.

28. A fost odată ...

27.Problem of Darij Grinberg (I - centroid of triangle DEF).

26. Outside of a triangle.

25. Geometric equations.

24. A fine and cool inequalities.

23. A very nice exponential inequality (own - from CRUX).

22. Some interesting limits (sequence/function).

21 bis. Some problems with complex numbers.

21. Probleme de geometrie (Yetty & I).

20. O ineg.cu suma de AG/AI .

19. Own proposed problems in CRUX Mathematicorum - Canada.

18. O problema a lui Toshio Seimiya.

17. O inegalitate "tare" intr-un triunghi.

16. Length of the (interior) angle bisector (10 proofs).

15. Inequalities stronger than Panaitopol's.

14. Opinii si comentarii relativ la inv. rom.

13. Inegalitatea de la Sorin Borodi.

12. Inegalitatea I. Maftei & S. Radulescu (2).

11. Inegalitatea I. Maftei & S. Radulescu (1).

10. Walker's inequality and its extension.

9. Inegalitatea HO+HA+HB+HC >= 3R .

8. The first problem Shortlist ONM 2010 Romania.

7. Some interesting "slicing" problems.

6. Interesting problems from JBMO.

5. Theoretical preliminary for inequalities.

4. Remarkable geometrical inequalities (2).

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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!
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Thank you a lot dear Virgil for an interesting and instructive blog.
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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

439. Properties of medians in triangle ABC and applications.

by Virgil Nicula, Jan 23, 2016, 2:10 PM

3 Comments