301. Some trigonometrico-geometrical inequalities.

by Virgil Nicula, Jul 22, 2011, 9:32 PM

PP1. $\triangle\, ABC\ \ \implies\ \ \boxed{\ \cos A\cos B\cos C\le \frac {r^2}{2R^2}\ }$ .

Proof. Prove easily an interesting and nice identity $2\cdot \sum bc-\sum a^2=\sum (a+c-b)(a+b-c)$ .

Apply this identity in $\triangle ABC\ :\ 16S^2=\left\{\begin{array}{c} 2\sum b^2c^2-\sum a^4\\\\ \sum \left(a^2+c^2-b^2\right)\left(a^2+b^2-c^2\right)\\\\ 4abc\cdot \sum a\cdot\cos B\cos C\end{array}\right\|\implies$ $\sum a\cdot\cos B\cos C=\frac SR$ . Therefore,

$\frac {S}{R\cos A\cos B\cos C}=$ $\sum\frac {a}{\cos A}=$ $\sum\frac {a^2}{a\cos A}\ge\frac {4s^2}{\sum a\cos A}=$ $\frac {4s^2}{4R\sin A\sin B\sin C}=$ $\frac {2Rs^2}{S}$ $\implies$ $\cos A\cos B\cos C\le\frac {r^2}{2R^2}$ .

Otherwise. Prove easily or is well-known that the powers of $H$ (orthocenter) and $G$ (centroid) w.r.t. the circumcircle $w$ of $\triangle ABC$ are

$p_w(H)=OH^2-R^2=-8R^2\cdot \prod\cos A$ and $p_w(G)=OG^2-R^2=-\frac {a^2+b^2+c^2}{9}$ . Thus, $OH^2=9OG^2\implies$

$8R^2\cdot\prod\cos A=\sum a^2-8R^2=2s^2-2(2R+r)^2\implies$ $\boxed{\prod\cos A=\frac {s^2-(2R+r)^2}{4R^2}}$ . I used $\sum bc=s^2+r^2+4Rr$ .

Using this well-known identity $\prod\, \cos A=\frac {s^2-\left(2R+r\right)^2}{4R^2}$ , our inequality becomes $\frac {s^2-\left(2R+r\right)^2}{4R^2}\, \le\, \frac {r^2}{2R^2}$

i.e. $s^2$ $-\left(2R+r\right)^2\, \le\, 2r^2\iff$ $\boxed{s^2\, \le\, 4R^2+4Rr+3r^2}$ which is just Gerretsen's inequality. Nice and easy inequality !

Otherwise. If $A\ge 90^{\circ}$ , then $\prod\cos A\le 0<\frac {r^2}{2R^2}$ . Can suppose that $\triangle ABC$ is acute. In this case

$\prod\tan A=\sum\tan A=$ $2\cdot\sum\frac {\sin^2A}{\sin 2A}\stackrel{\mathrm{(C.B.S)}}{\ \ge\ }$ $\frac {2\cdot\left(\sum \sin A\right)^2}{\sum\sin 2A}=$ $\frac {2\cdot\left(\sum \sin A\right)^2}{4\prod \sin A}=$ $\frac {s^2}{R^2}\cdot \frac {R^2}{S}$ $\implies$

$\boxed{\prod\tan A\ge \frac sr}\ (*)$ . Therefore, $\prod\cos A=\frac {\prod\sin A}{\prod\tan A}\ \stackrel{(*)}{\le}\ \frac rs\cdot\prod\sin A=$ $\frac rs\cdot \frac {S}{2R^2}\implies$ $\prod\cos A\le \frac {r^2}{2R^2}$ . See and inequalities

PP2. Prove that in any triangle $ABC$ there are the following inequalities :

(click ==>) here $\boxed{\frac {R^2-r^2}{R^2}\le \cos^2A+\cos^2B+\cos^2C\le 3\left(1-\frac rR\right)^2}\ (1)$

(click ==>) here $\boxed{\frac {2R^2-2Rr-r^2}{4R^2}\le\sin^{4}\frac{A}{2}+\sin^{4}\frac{B}{2}+\sin^{4}\frac{C}{2}\le\frac{(2R-3r)(2R-r)}{4R^{2}}}\ (2)$ .

Proof. Prove easily that $(2)\iff (1)\iff$ $3\left[1-\left(1-\frac rR\right)^2\right]\le$ $\sum\sin^2A\le 3-\frac {R^2-r^2}{R^2}\iff$ $12\left(2Rr-r^2\right)\le\sum a^2\le 4\left(2R^2+r^2\right)\iff$

$6\left(2Rr-r^2\right)\le s^2-r^2-4Rr\le 2\left(2R^2+r^2\right)\iff$ $\boxed{16Rr-5r^2\le s^2\le 4R^2+4Rr+3r^2}$ - the Gerrretsen's inequality.

Remark. Observe that $4\cdot\sum\sin^4\frac A2=\sum\left(2\sin^2\frac A2\right)^2=$ $\sum\left(1-\cos A\right)^2=$ $3-2\cdot\sum\cos A+\sum\cos^2A=$

$6-\frac {2(R+r)}{R}-\sum\sin^2A=$ $\frac {2(2R-r)}{R}-\frac {\sum a^2}{4R^2}=$ $\frac {2(2R-r)}{R}-\frac {s^2-r^2-4Rr}{2R^2}\implies$ $\boxed{\sum\sin^4\frac A2= \frac {8R^2+r^2-s^2}{8R^2}}$ .

Thus, $\frac {2R^2-2Rr-r^2}{4R^2}\le\sin^{4}\frac{A}{2}+\sin^{4}\frac{B}{2}+\sin^{4}\frac{C}{2}\le\frac{(2R-3r)(2R-r)}{4R^{2}}\iff$ $2\left(2R^2-2Rr-r^2\right)\le$

$\boxed{8R^2+r^2-s^2\le 2\left(4R^2-8Rr+3r^2\right)}\iff$ $16Rr-5r^2\le s^2\le 4R^2+4Rr+3r^2$ - the Gerrretsen's inequality.

PP3. $\triangle\, ABC\ \ \implies\ \ \boxed{\ \sum\frac{\sin B\cdot \sin C}{\sin^2 \frac{A}{2}}\ge 9\ \ \ (1)}$ and $\boxed{\ \sum \frac {\cos^2\frac A2}{\sin^3A}\ge\, \frac 9{4\sin A\sin B\sin C}\ \ \ (2)}$ .

Proof. Show easily that the inequality $(2)$ is equivalently with the inequality $(1)$ .

$\blacktriangleleft (1)\blacktriangleright\ \sum\frac{\sin B\cdot \sin C}{\sin^2 \frac{A}{2}}=$ $\sum\frac {bc}{4R^2}\cdot\frac {bc}{(s-b)(s-c)}\cdot\frac {s(s-a)}{s(s-a)}=$ $\sum\frac {b^2c^2s(s-a)}{4R^2S^2}\cdot\frac {4a^2}{4a^2}=$

$\sum\frac {(abc)^2\cdot 4s(s-a)}{(abc)^2\cdot a^2}=$ $\sum\frac {(b+c)^2-a^2}{a^2}\ge \sum\left(\frac {4bc}{a^2}-1\right)=$ $4\sum\frac {bc}{a^2}-3\ge 4\cdot 3-3=9$ .

$\blacktriangleleft (2)\blacktriangleright$ Using the well known identities $\left\{\begin{array}{c} \cos\frac A2=\sqrt{\frac {s(s-a)}{bc}} \\ \\ a=2R\sin A\end{array}\right\|$ the given inequality $(2)$ becomes $\sum\, \left[\frac {s(s-a)}{bc}\cdot\frac 1{\frac {a^3}{8R^3}}\right]\, \ge\, \frac 9{4\cdot\frac {abc}{8R^3}}$

$\iff \sum\, \frac {s-a}{a^2}\, \ge\, \frac 9{4s}$ . By Cauchy-Schwarz inequality one has : $\sum\, \frac {s-a}{a^2}=\sum\, \frac {\left(s-a\right)^2}{(s-a)a^2}\, \stackrel{C.S.}{\ge}\, \frac {s^2}{s\sum\, a^2-\sum\, a^3}$

and taking into account that : $\left\{\begin{array}{c} a^2+b^2+c^2=2(s^2-4Rr-r^2) \\ \\ a^3+b^3+c^3=2s(s^2-6Rr-3r^2)\end{array}\right|$ the last inequality is now equivalent to the following :

$\sum\, \frac {s-a}{a^2}\, \ge\, \frac s{4r(R+r)}$ . Consequently, it suffices to show that : $\frac s{4r(R+r)}\, \ge\, \frac 9{4s}\iff s^2\, \ge\, 9Rr+9r^2$ .

But this inequality is true because it is weaker than Gerretsen's inequality i.e. $\ s^2\, \ge\, 16Rr-5r^2$ .

PP4. Prove that the remarkable inequality $\boxed{\ \sum a\cdot\sin\frac A2\ge s\ }$ .

Proof. I"ll use the inequality $\boxed{\sin\frac{A}{2}\leq\frac{a}{b+c}}\ (*)$ and the identity $a=b\cdot\cos C+c\cdot\cos B$ . So, we get $\sum a\cdot \sin\frac{A}{2}\geq\sum (b+c)\sin^{2}\frac{A}{2}=$

$\sum\frac {b+c}{2}\cdot (1-\cos A)=$ $2s-\frac 12\cdot\sum (b+c)\cdot\cos A=$ $2s-\frac 12\cdot \sum (b\cdot\cos C+c\cdot\cos B)=$ $2s-\frac 12\cdot\sum a=2s-s=s$ .

Application. Let $ABC$ be a triangle. Its excircles touch sides $BC$ , $CA$ , $AB$ at $D$ , $E$ , $F$

respectively. Prove that the perimeter of triangle $ABC$ is at most twice that of triangle $DEF$ .

Proof 1 (Luisgeometria). Denote the projections $X$ , $Y$ of $E$ , $F$ on exterior $A$ -bisector of $\triangle ABC$ . Observe that

$AX=(s-c)\sin\frac A2$ , $AY=(s-b)\sin\frac A2$ and $EF\ge XY=AX+AY=[(s-b)+(s-c)]\cdot \sin \frac A2=$

$a\cdot\sin\frac A2$ . Using the inegality from the proposed problem PP4 obtain that $\sum EF\ge \sum a\cdot\sin\frac A2\ge s$ .

Proof 2 (Zephyredx). Denote $\{U,V\}\subset (BC)$ for which $EU$ , $FV$ are parallely with the (interior) $A$ -bisector of $\triangle ABC$ . Prove easily

that $EF\ge UV$ , $CU=BV=\frac {a(s-a)}{b+c}$ and $UV=\frac {a^2}{b+c}$ . In conclusion, $\sum EF\ge \sum\frac {a^2}{b+c}\stackrel{(C.B.S.)}{\ge}\frac {\left(\sum a\right)^2}{\sum (b+c)}=\frac {4s^2}{4s}=s$ .

Remark. Observe that $\boxed{EF\ge UV=\frac {a^2}{b+c}\stackrel{(*)}{\ge} a\cdot\sin\frac A2=XY}$ and $\sum EF\ge\sum UV=\sum\frac {a^2}{b+c}\ge \sum XY=\sum a\cdot\sin\frac A2\ge s$ .

$EF^2=(s-b)^2+(s-c)^2-2(s-b)(s-c)\cos A=$ $[(s-b)-(s-c)]^2+2(s-b)(s-c)(1-\cos A)=$ $(b-c)^2+\frac {4(s-b)^2(s-c)^2}{bc}$

$\implies$ $\boxed{EF\ge\frac {2(s-b)(s-c)}{\sqrt {bc}}}$ . If $M$ , $L$ are the midpoints of $[BC]$ , $[EF]$ , then $ML$ is parallely with $A$ -bisector of $\triangle ABC$ and $2\cdot LM=EU+FV$ .

PP5. Prove that in any triangle there is the following chain of the equivalencies

$\boxed {\ \cos\ (B - C)\ \le\ \frac {2bc}{b^2 + c^2} \ \Longleftrightarrow\ A\ \le\ 90^{\circ}\ \ \vee\ \ B = C\ \Longleftrightarrow\ \frac rR\ \le\ \frac {2a(p - a)}{b^2 + c^2}\ \Longleftrightarrow\ \frac {2a(p - b)(p - c)}{p\left(b^2 + c^2\right)}\ \le\ \frac rR\ }$ .

Proof 1.

Show easily that $\cos (B-C)=\frac {b\cdot\cos B+c\cdot\cos C}{a}$ . Therefore, $\boxed {\ \cos (B-C)\le\frac {2bc}{b^2+c^2}\ }\Longleftrightarrow$

$\frac {b^2\cdot 2ac\cdot\cos B+c^2\cdot 2ab\cdot\cos C}{2a^2bc}\le \frac {2bc}{b^2+c^2}$ $\Longleftrightarrow (b^2+c^2)\left[b^2(a^2+c^2-b^2)+c^2(a^2+b^2-c^2)\right]\le 4a^2b^2c^2$ . Denote

$\boxed {\ a^2=x\ ,\ b^2=y\ ,\ c^2=z\ }$ . Thus, $(y+z)\left[x(y+z)+2yz-(y^2+z^2)\right]\le 4xyz\Longleftrightarrow$ $x(y+z)^2-(y+z)(y-z)^2\le 4xyz$ $\Longleftrightarrow$

$x(y-z)^2\le (y+z)(y-z)^2\Longleftrightarrow y=z\ \vee\ x\le y+z$ $\Longleftrightarrow b=c\ \vee\ a^2\le y^2+z^2\Longleftrightarrow \boxed {\ B=C\ \vee\ A\le 90^{\circ}\ }$ .

Proof 2.

$\boxed{\ \cos (B - C)\le \frac {2bc}{b^2 + c^2}\ }$ . Add $1$ left/right and use the relation $1 + \cos x = 2\cos^2\frac x2$ .

$\Longleftrightarrow\ \left\|\ 2\cos^2\frac {B - C}{2}\le \frac {(b + c)^2}{b^2 + c^2}\ \right\|$ . Multiply by $2\sin ^2\frac {B + C}{2} = 2\cos^2\frac A2 > 0$ .

Use the relations $2\sin\frac {B + C}{2}\cos\frac {B - C}{2} = \sin B + \sin C$ si $\cos^2\frac A2 = \frac {p(p - a)}{bc}$ .

$\Longleftrightarrow\ \left\|\ (\sin B + \sin C)^2\le \frac {(b + c)^2}{b^2 + c^2}\cdot\frac {2p(p - a)}{bc}\ \right\|$ . But $\sin B = \frac {b}{2R}$ , $\sin C = \frac {c}{2R}$ .

$\Longleftrightarrow\ \left\|\ bc\left(b^2 + c^2\right)\le 4R^2\cdot 2p(p - a)\ \right\|$ . Multiply by $r$ and use the relation $4Rpr = abc$ .

$\Longleftrightarrow\ \left\|\ rbc\left(b^2 + c^2\right)\le R\cdot abc\cdot 2(p - a)\ \right\|$ . Simplify by $bc > 0$ and $2(p - a) = b + c - a$ .

$\Longleftrightarrow\ \boxed {\frac rR\le\frac {a(b + c - a)}{b^2 + c^2}\ }$ $\Longleftrightarrow\ \frac {4(p - a)(p - b)(p - c)}{abc}\le \frac {2a(p - a)}{b^2 + c^2}$

$\Longleftrightarrow\ 2\left(b^2 + c^2\right)(p - b)(p - c)\le a^2bc$ $\Longleftrightarrow\ \left(b^2 + c^2\right)\left[a^2 - (b - c)^2\right]\le 2a^2bc$

$\Longleftrightarrow\ a^2\left[\left(b^2 + c^2\right) - 2bc\right]\le\left(b^2 + c^2\right)(b - c)^2$ $\Longleftrightarrow\ a^2(b - c)^2\le \left(b^2 + c^2\right)(b - c)^2$

$\Longleftrightarrow\ \boxed {\ B = C\ \ \vee\ \ A\ \le\ 90^{\circ}\ }$ because $a^2\le b^2 + c^2\ \Longleftrightarrow\ A\ \le\ 90^{\circ}$ .

Remark. For the last one on right side : $\boxed {\frac {2a(p-b)(p-c)}{p\left(b^2+c^2\right)}\le\frac rR}\Longleftrightarrow$

$\frac {2a(p - b)(p - c)}{p\left(b^2 + c^2\right)}\le\frac {4(p - a)(p - b)(p - c)}{abc}\Longleftrightarrow$ $2a^2bc\le 4p(p - a)(b^2 + c^2)\Longleftrightarrow$

$2a^2bc\le \left[(b + c)^2 - a^2\right](b^2 + c^2)$ $\Longleftrightarrow a^2(b + c)^2\le (b + c)^2(b^2 + c^2)\Longleftrightarrow a^2\le b^2 + c^2$ $\Longleftrightarrow \boxed {\ A\le 90^{\circ}\ }$ .

PP6. Let $D$ , $E$ , $F$ be points on the sides $BC$ , $CA$ , $AB$ respectively of a triangle $ABC$ (distinct from the vertices).
If the quadrilateral $AFDE$ is cyclic, prove that $\frac{ 4\cdot [DEF] }{[ABC] } \leq \left( \frac{EF}{AD} \right)^2$ (Balkan MO - 1992, Greece).

Proof. Denote the second intersection $L$ of $AD$ with the circumcircle $w$ of $\triangle ABC$ . Observe that $DA\cdot DL=DB\cdot DC\implies$

$4\cdot DA\cdot DL\le (DB+DC)^2=BC^2\ (*)$ and $\left\{\begin{array}{ccc} \widehat{DEF}\equiv\widehat{DAF}\equiv\widehat{LAB}\equiv\widehat{LCB} & \implies & \widehat{DEF}\equiv\widehat{LCB}\\\\ \widehat{DFE}\equiv\widehat{DAE}\equiv\widehat{LAC}\equiv\widehat{LBC} & \implies & \widehat{DFE}\equiv\widehat{LBC}\end{array}\right\|\implies$

$\triangle DEF\sim\triangle LCB\implies$ $\frac {4\cdot [DEF]}{[ABC]}=\frac {4\cdot [DEF]}{[LCB]}\cdot$ $\frac {[LCB]}{[ABC]}=$ $\frac {4\cdot EF^2}{BC^2}\cdot\frac {LD}{AD}=$ $\left(\frac {EF}{AD}\right)^2\cdot\frac {4\cdot DA\cdot DL}{BC^2}\stackrel{(*)}{\le}$ $\left(\frac {EF}{AD}\right)^2$ .

PP7. Prove that in any triangle $ABC$ there is thetrigonometric inequality $\frac {\sin B}{\sin ^2\frac C2}+\frac {\sin C}{\sin^2\frac B2}\ge\frac {4\cos\frac A2}{1-\sin\frac A2}$ .

Proof 1. $\frac {\sin B}{\sin ^2\frac C2}+\frac {\sin C}{\sin^2\frac B2}\ge\frac {4\cos\frac A2}{1-\sin\frac A2}\iff$ $\frac{\frac{2S}{ac}}{\frac{(p-b)(p-a)}{ab}}+\frac{\frac{2S}{ab}}{\frac{(p-c)(p-a)}{ac}}\geq\frac{4\cdot\sqrt{\frac{p(p-a)}{bc}}}{1-\sqrt{\frac{(p-b)(p-c)}{bc}}}\iff$

$\frac{2S}{p-a}\cdot\left[\frac{b}{c(p-b)}+\frac{c}{b(p-c)}\right]\geq\frac{4\cdot\sqrt{p(p-a)}}{\sqrt{bc}-\sqrt{(p-b)(p-c)}}\iff$

$\frac{\sqrt{(p-b)(p-c)}}{p-a}\cdot\left[\frac{b}{c(p-b)}+\frac{c}{b(p-c)}\right]\geq\frac{2\left[\sqrt{bc}+\sqrt{(p-b)(p-c)}\right]}{p(p-a)}\iff$

$p\sqrt{(p-b)(p-c)}\cdot\left[\frac{b}{c(p-b)}+\frac{c}{b(p-c)}\right]\geq2\left[\sqrt{bc}+\sqrt{(p-b)(p-c)}\right]$ , what is the product of these

simple inequalities $p\ge\sqrt{bc}+\sqrt{(p-b)(p-c)}$ and $\sqrt{(p-b)(p-c)}\cdot\left[\frac{b}{c(p-b)}+\frac{c}{b(p-c)}\right]\ge 2$ .

PP8. Let $ABCD$ be a square with circumcircle $\Gamma$ . Let $M$ be a point on the minor arc of $AB$ . Prove that $\boxed{\ MC\cdot MD\, \ge\, \left(3+2\sqrt 2\right)\cdot MA\cdot MB\ }\ \ (*)$ .

Proof. Suppose w.l.o.g. $AB=1$ and note $\left\{\begin{array}{c} m\left(\widehat{MCB}\right)=x\\\\ m\left(\widehat{MCA}\right)=y\\\\ x+y=45^{\circ}\end{array}\right|$ . Thus, $\left\{\begin{array}{ccc} MA=\sqrt 2\cdot \sin y & ; & MB=\sqrt 2\cdot \sin x\\\\ MC=\sin x+\cos x & ; & MD=\sin y+\cos y\end{array}\right|$ .

In conclusion, $\left\{\begin{array}{c} MA\cdot MB=\cos (x-y)-\frac {\sqrt 2}{2}\\\\ MC\cdot MD=\cos (x-y)+\frac {\sqrt 2}{2}\end{array}\right|\ \implies$ $\boxed{MA\cdot MB+\sqrt 2=MC\cdot MD}$ - an interesting identity.

Observe that $MC\cdot MD\ge \left(3+2\sqrt 2\right)\cdot MA\cdot MB\iff$ $\left[\cos (x-y)+\frac {\sqrt 2}{2}\right]\ge\left(3+2\sqrt 2\right)\left[\cos (x-y)-\frac{\sqrt 2}{2}\right]$

$\iff$ $\cos (x-y)\le 1$ , what is evidently. In my opinion and $\boxed{\frac {MC}{MB}+\frac{MD}{MA}\ge 2\left(1+\sqrt 2\right)}$ is an interesting inequality.

PP9. Let $ABC$ be a triangle and let $\lambda\in\mathbb{R}$ . Prove that the following inequalities hold :

$(1)\blacktriangleright\ \boxed{\cos A+\lambda\left(\cos B+\cos C\right)\, \le\, 1+\frac {\lambda^2}2}$ with equality iff $B=C$ and $\sin\frac A2=\frac {\lambda}{2}$ only for $|\lambda|\le 2$ .

$(2)\blacktriangleright\ \boxed{\cos 2A+\lambda\left(\cos 2B+\cos 2C\right)\, \ge\, -1-\frac {\lambda^2}2}$ with equality iff $B=C$ and $\cos A=\frac {\lambda}{2}$ only for $|\lambda|\le 2$ .

Proof.

$\blacktriangleright (1)\iff$ $1-\cos A-\lambda\left(\cos B+\cos C\right)+\frac {\lambda^2}2\ge 0\iff$ $2\sin^2\frac A2-2\lambda\cos\frac {B-C}{2}\sin\frac A2+\frac {\lambda^2}2\ge 0\iff$ os

$2\left(\sin\frac A2-\frac{\lambda}{2}\cos\frac {B-C}{2}\right)^2+\frac {\lambda^2}{2}\sin^2\frac {B-C}{2}\ge 0$ , what is truly. Observe that have equality if and only if $B=C$ and $\sin \frac A2=\frac {\lambda}{2}$ .

$\blacktriangleright (2)\iff$ $1+\cos 2A+\lambda\left(\cos 2B+\cos 2C\right)+\frac {\lambda^2}2\ge 0\iff$ $2\cos^2A-2\lambda\cos (B-C)\cos A+\frac {\lambda^2}2\ge 0\iff$

$2\left[\cos A-\frac{\lambda}{2}\cos (B-C)\right]^2+\frac {\lambda^2}{2}\sin^2(B-C)\ge 0$ , what is truly. Observe that have equality if and only if $B=C$ and $\cos A=\frac {\lambda}{2}$ .

Remark 1. For $\lambda =1$ obtain the well-known inequalities

$\left\{\begin{array}{c} \cos A+\cos B+\cos C\le\frac 32=1+4\sin\frac A2\sin\frac B2\sin\frac C2=1+\frac {4(s-a)(s-b)(s-c)}{abc}=1+\frac {4sr^2}{4srR}=1+\frac rR\le\frac 32\\\\ \cos 2A+\cos 2B+\cos 2C=2\cos^2A-1-2\cos (B-C)\cos A=-1-4\cos A\cos B\cos C\ge -\frac 32\end{array}\right\|$ .

Remark 2. If the triangle $ABC$ is acute, then the second inequality rezults from the first by the substitution $\left\{\begin{array}{c} A:=\pi -2A\\\ B:=\pi -2B\\\ C:=\pi -2C\end{array}\right\|$ .

PP10. Prove that $(\forall )\ \{a,b\}\subset\mathbb R\ , \left|\frac{(a+b)(1-ab)}{(1+a^2)(1+b^2)}\right|\le\frac 12\ (*)$ .

Proof 1. For any $\{a,b\}\subset\mathbb R$ exist unique $\{x,y\}\subset \left(-\frac {\pi}{2},\frac {\pi}{2}\right)$ so that $\left\{\begin{array}{c} a=\tan x\\\ b=\tan y\end{array}\right\|$ . Our inequality $(*)$ becomes $|\sin 2(x+y)|\le 1$ .

Proof 2. We have $\frac{1}{2}-\frac{(a+b)(1-ab)}{(1+a^2)(1+b^2)}=\frac{1}{2}\frac{(ab+b+a-1)^2}{(1+a^2)(1+b^2)}\geq0$ and $\frac{(a+b)(1-ab)}{(1+a^2)(1+b^2)}+\frac{1}{2}=\frac{1}{2}\frac{(ab-b-a-1)^2}{(1+a^2)(1+b^2)}\geq0$ .

Another similar exercises. $\left\{\begin{array}{ccccc} 1\blacktriangleright & 4\left|x\left(x^2-1\right)\right|\le \left(x^2+1\right)^2 & , & (\forall ) & x\in\mathbb R\ .\\\\ 2\blacktriangleright & 8\left|x\left(x^4-6x^2+1\right)\right|\le \left(x^2+1\right)^4 & , & (\forall ) & x\in\mathbb R\ .\\\\ 3\blacktriangleright & \left|x\left(x^2-1\right)\left(16x^4-20x^2+5\right)\right|\le 1& , & (\forall ) & |x|\le 1\ .\end{array}\right\|$

PP11. Prove that in a triangle $ABC$ thee is the inequality $a^2 + b^2 + c^2 \geq \frac{36}{35}\cdot \left(\frac{abc}{s} + s^2\right)$ , where $2s=a+b+c$ .

Proof 1. Since $abc=4Rrs$ and $a^2+b^2+c^2=2\cdot\left(s^2-r^2-4Rr\right)$ obtain that $a^2 + b^2 + c^2 \geq \frac{36}{35}\cdot \left(\frac{abc}{s} + s^2\right)\iff$

$35\cdot \left(s^2-r^2-4Rr\right)\ge 18\cdot \left(4Rr + s^2\right)\iff$ $\boxed{17s^2\ge 212Rr+35r^2}$ . Using the well-known inequality $s^2\ge 16Rr-5r^2$

remain to show $17\left(16Rr-5r^2\right)\stackrel{?}{\ge} 212Rr+35r^2$ . Indeed, $17\left(16Rr-5r^2\right)\ge 212Rr+35r^2\iff R\ge 2r$ , what is true.

Proof 2 (for any positive numbers a,b,c). $a^2 + b^2 + c^2 \geq \frac{36}{35}\cdot \left(\frac{abc}{s} + s^2\right)\iff$ $35(a+b+c)\left(a^2+b^2+c^2\right)\ge 72abc+9(a+b+c)^3$ .

$\left\{\begin{array}{ccc} (1) & 3(a+b+c)\left(a^2+b^2+c^2\right)\ge (a+b+c)^3 & \odot \ \ 9\\\\ (2) & (a+b+c)\left(a^2+b^2+c^2\right)\ge 9abc & \odot\ \ 8\end{array}\right\|\ \bigoplus \implies$ $35(a+b+c)\left(a^2+b^2+c^2\right)\ge 72abc+9(a+b+c)^3$ .

The inequality $(2)$ is evidently and the ineguality $(1)$ is equivalently with $3\left[\sum a^3+\sum a^2(b+c)\right]\ge \sum a^3+3\prod (b+c)\iff$

$2\sum a^3+3\sum a^2(b+c)\ge 3\left[2abc+\sum a^2(b+c)\right]\iff$ $a^3+b^3+c^3\ge 3abc$ , what is evidently by A.M-G.M.

PP12. Let $\{x,y,z\}$ be three positive numbers so that $x^2 + y^2 + z^2 = 1$ . Prove that $\frac{x}{x^2+1} + \frac{y}{y^2+1} + \frac{z}{z^2+1}\le\frac {3\sqrt 3}{4}$ .

$\blacktriangleright$ Proof 1. Consider the function $f(t)=\frac {\sqrt t}{t+1}\ ,\ t\in (0,1)$ . Since $f''(t)=\frac {3t^2-6t-1}{4t(t+1)^3\sqrt t}<0$ for any $t\in (0,1)$ obtain

that the function $f$ is concave $\stackrel{(\mathrm{Jensen})}{\implies} \sum\frac {x}{x^2+1}=\sum f\left(x^2\right)\le 3\cdot f\left(\frac {x^2+y^2+z^2}{3}\right)=$ $3\cdot f\left(\frac 13\right)=\frac {3\sqrt 3}{4}$ .

PP13. Prove that $\{x,y,z\}\subset \mathbb R^*_+\implies \left(x + y + z\right)^2 \left(yz + zx + xy\right)^2 \leq$ $3\left(y^2 + yz + z^2\right)\left(z^2 + zx + x^2\right)\left(x^2 + xy + y^2\right)$ .

Proof 1. Exist a triangle $ABC$ for $BC=a,CA=b,AB=c$ and point $M$ in it such that $MA=x$ , $MB=y$ , $MC=z$ and $\widehat{AMB}=$

$\widehat{BMC}=\widehat{CMA}=120^{0}$ . We have $c^{2}=x^{2}+y^{2}+xy$ , $b^{2}=$ $z^{2}+x^{2}+zx$ , $a^{2}=$ $y^{2}+z^{2}+yz$ and $xy+yz+zx=\frac{4S}{\sqrt{3}}$

and $(x+y+z)^{2}=\frac{1}{2}\cdot \left(a^{2}+b^{2}+c^{2}+4S\sqrt{3}\right)$ . Where $S$ is area of triangle $ABC$ . By $4RS=abc$ we have the relation

$(*)\Leftrightarrow a^{2}+b^{2}+c^{2}+4S\sqrt{3}\leq 18R^{2}$ . But this is truly because $a^{2}+b^{2}+c^{2}\leq 9R^{2}$ and $4S\sqrt{3}\leq a^{2}+b^{2}+c^{2}$ .

Remark. Problem is trivial using inherent properties of the point $M$ (Fermat's point) using $c^{2}=x^{2}+y^{2}+xy$ , $b^{2}=z^{2}+x^{2}+zx$ ,

$a^{2}=y^{2}+z^{2}+yz$ and $xy+yz+zx=\frac{4S}{\sqrt{3}}$ and $4RS=abc$ . The inequality gets reduced to $x+y+z \leq 3R$ , i.e. if $O$ is

the circumcentre, then we have to prove that $MA+MB+MC \leq OA+OB+OC$ , which is true by original definition of

the Fermat's point (a point $P$ inside triangle $ABC$ (with each angle less than $120^{\circ}$ ) such that $PA+PB+PC$ is minimum).

Proof 2. We have $4\left(x^{2} + xy + y^{2}\right) - 3\left(x + y\right)^{2} = \left(x - y\right)^{2}\ge 0$ . Multiplying the inequalities $\left\{\begin{array}{c} 4\left(x^{2} + xy + y^{2}\right)\ge 3\left(x + y\right)^2\\\\ 4\left(y^{2} + yz + z^{2}\right)\ge 3\left(y + z\right)^2\\\\ 4\left(z^{2} + zx + x^{2}\right)\ge 3\left(z + x\right)^2\end{array}\right\|$

we get $64\left(y^{2} + yz + z^{2}\right)\left(z^{2} + zx + x^{2}\right)\left(x^{2} + xy + y^{2}\right)\ge 27\left(x + y\right)^{2}\left(y + z\right)^{2}\left(z + x\right)^{2}$ .

Thus, it suffices to show that $81\left(x + y\right)^{2}\left(y + z\right)^{2}\left(z + x\right)^{2}\ge 64\left(x + y + z\right)^{2}\left(yz + zx + xy\right)^{2}$ $\Longleftrightarrow$

$9\left(x + y\right)\left(y + z\right)\left(z + x\right)\ge 8\left(x + y + z\right)\left(ab + bc + ca\right)$ $\Longleftrightarrow$ $x\left(y - z\right)^{2} + y\left(z - x\right)^{2} + z\left(x - y\right)^{2}\ge 0$ .

Proof 3. Be cause of homogeneity normalize it as $z=1$ . Need to prove that $3\left(x^2+y^2+xy\right)\left(1+x^2+x\right)\left(1+y^2+y\right)>(1+x+y)^2(x+y+xy)^2$ .

Put $x+y=u$ and $xy=v$ . On expanding it becomes $3\left(u^2+u^4-2u^2v+u^3+u^2v^2+u^3v-v+v^2-uv-v^3-uv^2\right)\ge$

$\left(u^2+u^4+2u^3+v^2+u^2v^2+2uv^2+2uv+2u^3v+4u^2v\right)\iff$ $\left(2u^4-8u^2v\right)+$ $\left(u^3-4uv\right)+\left(2u^2-8v\right)+$

$\left(u^3v-4uv^2\right) +\left(2u^2v^2-8v^3\right)-$ $\left(2u^2v-8v^2\right)+$ $\left(uv^2+uv+5v+5v^3-6v^2\right)\ge 0\iff$

$\left(u^2-4v\right)\left[2\left(u^2-4v\right)+6v+u+2+uv+2v^2\right]+\left(uv^2+uv\right)+5\left( v3/2-v1/2\right)^2+4v^2\ge 0$ , what is true.

PP14. Let $ABC$ be a triangle. Prove that $\sin A+\sin B+\sin C\le \frac {3\sqrt 3}{2}$ .

Proof 1 (trigonometric). Prove easily that if $\sin\frac {x+y}{2}\ge 0$ , then $\sin x+\sin y=2\cdot\sin\frac {x+y}{2}\cos\frac {x-y}{2}\le 2\cdot\sin\frac {x+y}{2}$ , i.e. $\frac {\sin x+\sin y}{2}\le\sin\frac {x+y}{2}$ .

Therefore, $\left\{\begin{array}{c} \sin B+\sin C\le 2\cdot\sin \frac{B+C}{2}\\\\ \sin A+\sin 60^{\circ} \le 2\cdot \sin \frac{A+60^{\circ}}{2}\\\\ \sin \frac{B+C}{2}+\sin \frac{A+60^{\circ}}{2}\le 2\cdot\sin \frac{A+B+C+60^{\circ}}{4}\end{array}\right\|\implies$ $(\sin B+\sin C)+\left(\sin A+\sin 60^{\circ}\right)\le$

$2\cdot\sin \frac{B+C}{2}+2\cdot \sin \frac{A+60^{\circ}}{2}\le$ $4\cdot \sin\frac {A+B+C+\frac {\pi}{3}}{4}=4\cdot\sin 60^{\circ}\implies$ $\sin A+\sin B+\sin C\le 3\cdot \sin 60^{\circ}=\frac {3\sqrt 3}{2}$ .

Proof 2 (algebraic/geometric). From the relations $a=2R\sin A$ a.s.o. obtain that $\sum\sin A\le\frac {3\sqrt 3}{2}\iff 2s\le 3R\sqrt 3$ .

Therefore, $\left\{\begin{array}{cccc} \sum (s-a)\ge 3\cdot\sqrt [3]{\prod (s-a)}\iff s^3\ge 27sr^2 & \iff & \boxed{3r\sqrt 3\le s} & (1)\\\\ 3s^2=3\sum r_br_c\le \left(\sum r_a\right)^2=(4R+r)^2 & \iff & \boxed{s\sqrt 3\le 4R+r} & (2)\end{array}\right\|$ . In conclusion, from the

upper remarkable inequalities obtain the chain $s\sqrt 3\stackrel{(2)}{\le}$ $4R+r\stackrel{(1)}{\le} 4R+\frac {s}{3\sqrt 3}\implies$ $9s\le 12R\sqrt 3+s\implies$ $2s\le 3R\sqrt 3$ .See here.

PP15. Let $ABC$ be a triangle. Prove that $\sum\left(\frac{c}{b}+\frac{b}{c}\right)\cdot\cos A=3$ and deduce that $\left\{\begin{array}{ccc} \cos C+\cos B+\cos A & \le & \frac{3}{2}\\\\ \sin \frac A2\sin\frac B2\sin\frac C2 & \le & \frac{1}{8}\end{array}\right\|$ .

Proof. From the well-known identity $b\cdot\cos C+c\cdot\cos B=a$ obtain that $\frac ba\cdot \cos C+\frac ca\cdot\cos B=1$ .Therefore, obtain similarly $\left\{\begin{array}{c} \frac ba\cdot \cos C+\frac ca\cdot\cos B=1\\\\ \frac cb\cdot \cos A+\frac ab\cdot\cos C=1\\\\ \frac ac\cdot \cos B+\frac bc\cdot\cos A=1\end{array}\right\|\bigoplus\implies$ $\sum\left(\frac{c}{b}+\frac{b}{c}\right)\cdot\cos A=3$ .

If $ABC$ is nonobtuse, then from $\frac bc+\frac cb\ge 2$ obtain that $3= \sum\left(\frac{c}{b}+\frac{b}{c}\right)\cdot\cos A\ge 2\sum\cos A\implies$ $\sum\cos A\le \frac 32$ . If $ABC$ is obtuse, then you must find another proof. For example, see lower remark,

where we used only the remarkable inequality $2r\le R$ . Thus, $(-1+\cos A)+(\cos B+\cos C)=$ $-2\sin^2\frac A2+2\cos\frac {B+C}{2}\cos\frac {B-C}{2}=$ $2\sin\frac A2\left(\cos\frac {B-C}{2}-\cos\frac {B+C}{2}\right)=$

$4\sin\frac A2\sin\frac B2\sin\frac C2\implies$ $4\sin\frac A2\sin\frac B2\sin\frac C2=-1+\sum\cos A\le -1+\frac 32=\frac 12\implies$ $\sin\frac A2\sin\frac B2\sin\frac C2\le\frac 18$ .

Remark. $\prod\sin\frac A2=\prod\sqrt{\frac {(s-b)(s-c)}{bc}}=$ $\frac {(s-a)(s-b)(s-c)}{abc}=\frac {sr^2}{4Rsr}=\frac {r}{4R}\le\frac 18$ $\implies$ $\sum\cos A=$ $1+4\prod\sin\frac A2=1+\frac rR\le 1+\frac 12=\frac 32$ .

PP16 Let $ABC$ be a triangle. Prove that $\sum \cos{A}+\sum \cos{2A}=0\iff A=B=C$ .

Proof. Is evidently the implication $A=B=C\implies\sum\cos A+\sum\cos 2A=0$ . Suppose that the relation $\sum \cos A+\sum\cos 2A=0$ is truly. I"ll use

the relations $\boxed{\sum\cos A=1+\frac rR}$ and $\boxed{\sum \cos 2A=-1-4\prod\cos A}$ . I"ll show that $\boxed{\prod\cos A=\frac {s^2-(2R+r)^2}{4R^2}}$ . So $\cos 2A=1-2\sin^2A=$

$1-2\left(\frac {a}{2R}\right)^2\implies$ $\cos 2A=1-\frac {a^2}{2R^2}$ . Thus, $\sum\cos 2A=3-\frac {s^2-r^2-4Rr}{R^2}$ and $4\prod\cos A=$ $-1-\sum\cos 2A=$ $-4+\frac {s^2-r^2-4Rr}{R^2}=$

$\frac {s^2-(2R+r)^2}{R^2}\implies$ $\sum\cos 2A=-1- \frac {s^2-(2R+r)^2}{R^2}$ . In conclusion, $\sum\cos A+\sum\cos 2A=0\implies$ $1+\frac rR-1-\frac {s^2-(2R+r)^2}{R^2}=0$

$\implies$ $s^2=(4R+r)(R+r)$ . Since $s^2\le 4R^2+4Rr+3r^2$ get that $4R^2+5Rr+r^2\le 4R^2+4Rr+3r^2 \iff$ $R\le 2r\implies A=B=C$ .

PP17 Prove that $x+y+z=\frac {\pi}{2}\implies \sin x\sin y\sin z\le\frac 18$ .

Proof 1. $\sin x\sin y\sin z\le\frac 18\iff$ $1-8\sin x\sin y\sin z\ge 0\iff$ $1-4\sin x\left[\cos (y-z)-\sin x\right]\ge 0\iff$ $4\sin^2x-4\sin x\cos (y-z)+1\ge 0\iff$

$\left[2\sin x-\cos (y-z)\right]^2+\sin^2(y-z)\ge 0$ , what is truly. We have equality if and only if $\sin (y-z)=0$ and $2\sin x=\cos (y-z)$ $\iff$ $y=z$ and $x=\frac {\pi}{6}$ , i.e. $x=y=z=\frac {\pi}{6}$ .

Proof 2. $\sin x\sin y\sin z\le \left(\frac {\sin x+\sin y+\sin z}{3}\right)^3\le$ $\sin^3\frac {x+y+z}{3}=\sin^3\frac {\pi}{6}=\frac 18$ . I used AM-GM and the concavity of the function SIN on $\left[0,\frac {\pi}{2}\right]$ .

PP18. Prove that $\boxed{\begin{array}{ccccc} \cos A+\cos B+\cos C & = & 1+\frac rR & \le & \frac 32\\\\ (*)\ \ \ \ \ \cos A\cos B\cos C & = & \frac {a^2+b^2+c^2}{8R^2}-1 & \le & \frac 18\\\\ \cos 2A+\cos 2B+\cos 2C & = & -1-4\prod\cos A & \ge & -\frac 32\end{array}}$ .

Proof 1. $1-8\prod\cos A=1-4\cos A[\cos (B+C)+\cos (B-C)]=$ $1-4\cos A[-\cos A+\cos (B-C)]=$ $4\cos^2A-4\cos A\cos (B-C)+1=$

$[2\cos A-\cos (B-C)]^2+\sin^2(B-C)\ge 0$ with equality iff $\left\{\begin{array}{c} \sin (B-C)=0\\\\ 2\cos A=\cos (B-C)\end{array}\right|\iff$ $B=C$ and $\cos A=\frac 12\iff$ $ABC$ is an equilateral triangle.

Proof 2. $\sum a^2\left(b^2+c^2-a^2\right)=16S^2\iff$ $\sum a^2\cdot 2bc\cos A=16S^2\iff$ $2abc\sum a\cos A=16S^2\iff$ $8RS=\sum a\cos A=16S^2\iff$ $\boxed{\sum a\cos A=\frac {2S}{R}}$ .

On the other hand, $\left\{\begin{array}{c} a=b\cos C+c\cos B\\\ b=c\cos A+a\cos C\\\ c=a\cos B+b\cos A\end{array}\right|\bigoplus\implies$ $\left\{\begin{array}{c} \sum (b+c)\cos A=2s\\\\ \sum a\cos A=\frac {2S}{R}\end{array}\right|\bigoplus\implies$ $2s\sum\cos A=2s\left(1+\frac rR\right)\iff$ $\boxed{\sum\cos A=1+\frac rR}\le \frac 32$

because $R\ge 2r$ . Now suppose w.l.o.g. that $ABC$ isn't obtuse. Then $\prod\cos A\le \left(\frac {\sum\cos A}{3}\right)^3\le \left(\frac {\frac 32}{3}\right)^3\implies$ $\cos A\cos B\cos C\le\frac 18$ .

Proof 3. . Is well-known the inequality $xyz\ge (y+z-x)(z+x-y)(x+y-z)$ , where $\{x,y,z\}\subset\mathbb R^*_+$ . Suppose w.l.o.g. that

$ABC$ isn't obtuse. For $\left\{\begin{array}{c} x:=a^2\\\ y:=b^2\\\ z:=z^2\end{array}\right|$ obtain that $a^2b^2c^2\ge\prod \left(b^2+c^2-a^2\right)\iff$ $a^2b^2c^2\ge \prod 2bc\cos A\iff$ $(*)$ .

Proof 4. Observe that $\sum \cos 2A=\cos 2A+2\cos (B+C)\cos (B-C)=$ $2\cos^2A-1-2\cos A\cos (B-C)=$ $-1-2\cos A[(\cos (B+C)+\cos (B-C)]\implies$

$\boxed{\sum\cos 2A=-1-4\prod\cos A}$ . Therefore, $\frac{a^2+b^2+c^2}{2R^2}=$ $2\sum\sin^2A=\sum(1-\cos 2A)=$ $4+4\prod\cos A$ . Since $\boxed{OG^2=R^2-\frac {a^2+b^2+c^2}{9}\ge 0}\implies$

$a^2+b^2+c^2\le 9R^2$ obtain that $\boxed{1+\prod\cos A=\frac {a^2+b^2+c^2}{8R^2}}\le\frac 98$ . In conclusion, $\cos A\cos B\cos C\le\frac 18$ .

Remarks. $\boxed{\sum\cos 2A=-1-4\prod\cos A\ge -\frac 32}$ and $\cos A\cos B\cos C\le \frac 2{27}\left(\frac S{R^2}\right)^2$ $=\prod\frac {1-\cos 2A}{3}\le\frac 12$ .

This post has been edited 130 times. Last edited by Virgil Nicula, Nov 21, 2015, 7:43 AM

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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.

February 2011

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.

January 2011

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.

December 2010

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).

November 2010

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).

October 2010

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).

Submit

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391345 views moment
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blog
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the visits tho
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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!!!!
by CSPAL, May 27, 2020, 4:17 PM
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

301. Some trigonometrico-geometrical inequalities.

by Virgil Nicula, Jul 22, 2011, 9:32 PM

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