461. Problems for the relaxation in ... weekend (II).

by Virgil Nicula, Jan 26, 2018, 7:36 PM

P1. Sa se rationalizeze fractia $\boxed{\ F=\frac 1{\sqrt[3]4+\sqrt [3]{10}+\sqrt[3]{25}}\ }$ (clasa a VII - a).

Proof. Daca notam $\sqrt[3]2=a$ si $\sqrt [3]5=b\ ,$ atunci fractia $F$ devine $\frac 1{a^2+ab+b^2}\ ,$ adica $F=\frac {b-a}{(b-a)\left(b^2+ab+a^2\right)}=$ $\frac {b-a}{b^3-a^3}=\frac{\sqrt [3]5-\sqrt [3]2}{5-2}\implies$ $\boxed{\ F=\frac{\sqrt [3]5-\sqrt [3]2}3\ }\ .$

P2 (Adil ABDULLAYEV, Baku). Prove that in any $\triangle ABC$ the following relationship holds $\ :\ \frac {\left(r_a+r_b\right)\left(r_b+r_c\right)\left(r_c+r_a\right)}{8r_ar_br_c}=\frac R{2r}\ge \frac{h_a^2+h_b^2+h_c^2}{h_ah_b+h_bh_c+h_ch_a}\ge 1$ (standard notations).

Proof. Prove easily that $rr_ar_br_c=S^2=s^2r^2\ ,$ i.e. $\boxed{ r_ar_br_c=s^2r\ }\ (1)$ and $\frac {\left(r_a+r_b\right)\left(r_b+r_c\right)\left(r_c+r_a\right)}{8r_ar_br_c}=$ $\prod\frac{r_b+r_c}{2r_a}=$ $\frac 18\cdot\prod\left(\frac{r_b}{r_a}+\frac{r_c}{r_a}\right)=$ $\frac 18\cdot\prod\left(\frac{s-a}{s-b}+\frac{s-a}{s-c}\right)=$

$\frac 18\cdot\prod\frac{a(s-a)}{(s-b)(s-c)}=$ $\frac {abc}{8(s-a)(s-b)(s-c)}=$ $\frac{4R\cancel s\cancel r}{8\cancel sr^{\cancel 2}}=\frac R{2r}\ ,$ i.e. $\boxed{\ \frac {\left(r_a+r_b\right)\left(r_b+r_c\right)\left(r_c+r_a\right)}{8r_ar_br_c}=\frac R{2r}\ }\ (2)\ .$ Observe that $\boxed{\ ah_a=2sr\ }\ \mathrm{a.s.o.}\iff$ $\boxed{\ h_a=\frac {2sr}a\ }\ \mathrm{a.s.o.}\implies$

$h_bh_c=\frac {2sr}{b}\cdot\frac {2sr}{c}=\frac {4s^2r^2}{bc}\ \mathrm{a.s.o.}\implies$ $\sum \left(h_bh_c\right)=4s^2r^2\cdot\sum\frac 1{bc}=$ $4s^2r^2\cdot\frac{a+b+c}{abc}=$ $4s^2r^2\cdot\frac{2s}{4Rrs}=$ $4s^2r^2\cdot\frac{2s}{4Rrs}=$ $\frac {2s^2r}{R}\implies$ $\boxed{\ h_ah_b+h_bh_c+h_ch_a=\frac {2s^2r}R\ }\ (3)\ .$

Otherwise. $\boxed{\ bc=2Rh_a\ }\ \mathrm{a.s.o.}\implies$ $\boxed{\ h_a=\frac {bc}{2R}\ }\ \mathrm{a.s.o.}\implies$ $h_bh_c=\frac {ac}{2R}\cdot\frac {ab}{2R}=a\cdot \frac {abc}{4R^2}=\frac {a\cdot 4RS}{4R^2}=\frac {aS}R\implies$ $\boxed{\ h_bh_c=\frac {aS}R\ }\implies$ $\sum h_bh_c=\frac SR\cdot \sum a=\frac {2s\cdot sr}R\implies$

$\boxed{\ \sum h_bh_c=\frac {2s^2r}R\ }\ .$ From $a^2\ge a^2-(b-c)^2=(a+b-c)(a-b+c)=4(s-b)(s-c)\ \mathrm{a.s.o.}$ obtain that $\boxed{\ a^2\ge 4(s-b)(s-c)\ }\ \mathrm{a.s.o.}\implies$ $4s(s-a)(s-b)(s-c)=$

$(2S)^2=\left(ah_a\right)^2=a^2h_a^2\ge 4(s-b)(s-c)\cdot h_a^2\implies$ $\cancel 4s(s-a)\cancel{(s-b)(s-c)}\ge \cancel 4\cancel{(s-b)(s-c)}\cdot h_a^2\implies$ $\boxed{\ h_a^2\le s(s-a)\ }\implies$ $\sum h_a^2\le s\sum (s-a)\implies$

$\boxed{\ h_a^2+h_b^2+h_c^2\le s^2\ }\ (4)\ .$ Remain to prove $\frac {\sum h_a^2}{\sum \left(h_bh_c\right)}\le \frac R{2r}\ .$ Indeed, from the relation $(3)$ obtain that $\left(h_a^2+h_b^2+h_c^2\right)\cdot\frac {\cancel R}{\cancel 2s^2\cancel r}\le \frac {\cancel R}{\cancel 2\cancel r}\ ,$ i.e. the true relation $(4)\ .$

Remark. Prove easily that $h_ah_b+h_bh_c+h_ch_a\le 3S\sqrt 3\le 3r(4R+r)\le s^2\ .$

P3 (Θεόδωρος Σαμπάς, Greece). Prove that the following exercise (<= click).

Proof. $\prod\cos A=\frac 18\iff$ $4\cos A\cdot 2\cos B\cos C=1\iff$ $4\cos A[\cos (B-C)-\cos A]=1\iff$ $4\cos^2A-4\cos A\cos (B-C)+1=0\iff$

$[2\cos A-\cos (B-C)]^2+\sin^2(B-C)=0\iff$ $B=C$ and $2\cos A=\cos (B-C)=\cos 0=1\ ,$ i.e. $A=60^{\circ}$ and $A=B=C\ .$

P4 (Θεόδωρος Σαμπάς, Greece). Prove that the following identity $:\ \left(\frac bc+\frac cb\right)\cdot\cos A+\left(\frac ca+\frac ac\right)\cdot\cos B+\left(\frac ab+\frac ba\right)\cdot \cos C=3\ .$

Proof. Prove easily that $\boxed{\ \sum_{\mathrm{cyc}}\frac{b\cdot\cos A}c=\sum_{\mathrm{cyc}}\frac{a\cdot\cos C}b\ }\ (*)$ and I"ll use the well known identities $\boxed{\ b\cdot \cos C+c\cdot\cos B=a\ }\ (1)\ .$ Thus, $\sum_{\mathrm{cyc}}\left(\frac bc+\frac cb\right)\cdot\cos A=$

$\sum_{\mathrm{cyc}}\frac {b\cdot \cos A}c+\sum_{\mathrm{cyc}}\frac {c\cdot \cos A}b\ \stackrel{(*)}{=}\ \sum_{\mathrm{cyc}}\frac{a\cdot\cos C}b+\sum_{\mathrm{cyc}}\frac {c\cos A}b=$ $\sum_{\mathrm{cyc}}\frac{a\cdot\cos C+c\cdot\cos A}b\ \stackrel{(1)}{=}\ \sum_{\mathrm{cyc}}\frac bb=3\implies$ $\sum_{\mathrm{cyc}}\left(\frac bc+\frac cb\right)\cdot\cos A=3\ .$

P5 (Nguyen Viet Hung). Prove the following inequality (<= click) $:\ \frac{h_bh_c}{h_ar_a^2}+\frac{h_ch_a}{h_br_b^2}+\frac{h_ah_b}{h_cr_c^2}\le\frac 1r\ .$

Proof. $\frac {h_bh_c}{h_ar_a^2}=\frac{2\cancel S}b\cdot\frac {\cancel 2\cancel S}c\cdot\frac a{\cancel 2S}\cdot \frac {(s-a)^2}{\cancel{S^2}}=$ $\frac {2a(s-a)^2}{bcS}=$ $\frac {2a^2(s-a)^2}{abcS}=$ $\frac {a^2(s-a)^2}{2RS^2}\implies$ $\frac {h_bh_c}{h_ar_a^2}=\frac {a^2(s-a)^2}{2RS^2}\ .$ Hence $\boxed{\ \sum \frac {h_bh_c}{h_ar_a^2}=\frac 1{2RS^2}\cdot \sum a^2(s-a)^2\ }\ (1)\ .$

In conclusion, the required inequality $\boxed{\ \sum\frac{h_bh_c}{h_ar_a^2}\le\frac 1r\ }\ (*)$ is equivalent with the inequality $\sum a^2(s-a)^2\le 2Rs^2r\iff$ $\boxed{\ \sum a^2(b+c-a)^2\le abc(a+b+c)\ }\ (2)\ .$

Prove easily that $\sum a^2(s - a)^2 = 2r^2\left[\left(4R + r\right)^2 - s^2\right]$ starting from the development of $\left[\sum a(s-a)\right]^2 = \ldots$ The inequality becomes $2r^2\left[(4R+r)^2-s^2\right]\le 2Rs^2r\iff$

$r\left[(4R+r)^2-s^2\right]\le Rs^2\iff$ $r(4R+r)^2\le s^2(R+r)\iff$ $\boxed{\ s^2\ge \frac {r(4R+r)^2}{R+r}\ }\ .$ Is well known $s^2\ge 16Rr-5r^2$ and observe that $16Rr-5r^2\ge \frac {r(4R+r)^2}{R+r}\iff$

$(16R-5r)(R+r)\ge (4R+r)^2\iff$ $16R^2+16Rr-5Rr-5r^2\ge 16R^2+8Rr+r^2\iff$ $3Rr\ge 6r^2\iff$ $R\ge 2r\ .$ Hence $s^2\ge 16Rr-5r^2\ge \frac {r(4R+r)^2}{R+r}\implies$

$s^2\ge \frac {r(4R+r)^2}{R+r}\ .$ In conclusion, our inequality $(*)$ is true. Very nice!

P6 (G. Baron). Solve the following system from here.

Proof. Denote $x+y=S\ ,$ $xy=P\ .$ Thus, $\left\{\begin{array}{ccc} x^2+x & = & y^3-y\\\\ y^2+y & = & x^3-x\end{array}\right\|\implies$ $\left(y^3-\cancel y\right)+\left(y^2+\cancel y\right)=\left(x^2+\cancel x\right)+\left(x^3-\cancel x\right)\iff$

$y^3+y^2=x^3+x^2\iff$ $\left(y^3-x^3\right)+\left(y^2-x^2\right)=0\iff$ $(y-x)\left(y^2+xy+x^2+y+x\right)=0\ .$ Appear two cases $:$

$1.\blacktriangleright\ x=y\iff x^2+x=x^3-x\iff$ $x^3-x^2-2x=0\iff$ $x(x+1)(x-2)=0\iff$ $x=y\in\{-1,0,2\}\ .$

$2.\blacktriangleright\ x\ne y\iff\left(x^2+y^2\right)+(x+y)+xy=0\iff$ $S^2-2P+S+P=0\iff$ $\boxed{\ S^2+S=P\ }\ (1)\ .$ Observe that

$\left(x^2+x\right)+\left(y^2+y\right)=\left(y^3+x^3\right)-(x+y)\iff$ $\left(x^2+y^2\right)+2(x+y)=\left(x^3+y^3\right)\iff$ $\left(S^2-2P\right)+2S=$ $S^3-3PS\iff$

$P(3S-2)=S^3-S^2-2S\ \stackrel{(1)}{\iff}\ S(S+1)(3S-2)=S(S+1)(S-2)\iff$ $S\in\{0,-1\}$ or $3S-2=S-2\ ,$ i.e. $S=0\ .$

Therefore: for $S=0$ obtain $P=0\ ,$ i.e. $x=y=0\ ;$ for $S=-1$ obtain $P=0\ ,$ i.e. $(\ x=0\ \wedge\ y=-1\ )\ \vee\ (\ x=-1\ \wedge\ y=0\ )\ .$

In conclusion, the solutions of this system are $:\ (x,y)\ \in\ \left\{\ (-1,-1)\ ;\ (0,0)\ ;\ (2,2)\ ;\ (0,-1)\ ;\ (-1,0)\ \right\}\ .$

P7 (Marian URSARESCU). Let $\triangle ABC$ with the incircle $w=\mathbb C(I,r)\ .$ The lines $AI\ ,$ $BI$ and $CI$ meet again $w$ in $D\ ,$ $E$ and $F$ respectively. Prove that the inequality $\boxed{\ \frac{ID}{IA}+\frac {IE}{IB}+\frac{IF}{IC}\ge 3\ }\ (*)\ .$

Proof. $\sum\frac {ID}{IA}=$ $\sum\frac {ID^2}{IA\cdot ID}=$ $\sum\frac {ID^2}{p_w(I)}=$ $\frac 1{2Rr}\cdot\sum BD^2=$ $\frac 1{2Rr}\cdot\sum \left(2R\sin\frac A2\right)^2=$ $\frac {\cancel 2R^{\cancel 2}}{\cancel 2\cancel Rr}\cdot\sum \left(2\sin^2\frac A2\right)=$ $\frac Rr\cdot\sum \left(1-\cos A\right)=$ $\frac Rr\cdot\left[3-\left(1+\frac rR\right)\right]=$ $\frac {\cancel R}r\cdot \frac {2R-r}{\cancel R}=$

$\frac {2R-r}r\ge 3$ because $2R-r\ge 3r\iff R\ge 2r\ ,$ what is true. In conclusion, $\sum\frac {ID}{IA}\ge 3\ .$ I used the well known relations $DI=DB$ and the power of $I$ w.r.t. $w$ is $p_w(I)=-2Rr\ .$

P8 (M. URSARESCU). Let the Gergonne's point $\Gamma$ of $\triangle ABC$ and $A\Gamma\ ,$ $B\Gamma\ ,$ $C\Gamma$ meet again $BC\ ,$ $CA\ ,$ $AB$ at $D\ ,$ $E\ ,$ $F$ respectively. Prove that $\boxed{\ a\cdot AD^2+b\cdot BE^2+c\cdot CF^2\ge 54r^3\sqrt 3\ }\ (*)\ .$

Proof. Prove easily the following identities $:\ \left\{\begin{array}{cccc} \sum a(s-b)(s-c) & = & 2sr(2R-r) & (1)\\\\ \sum a^2(s-a) & = & 4sr(R+r) & (2)\end{array}\right\|\ .$ Apply the Stewart's relation to the cevian $AD\ ,$ where $DB=s-b$ and $DC=s-c\ :$

$a\cdot AD^2+a(s-b)(s-c)=b^2(s-b)+c^2(s-c)\implies$ $\sum a\cdot AD^2+\sum a(s-b)(s-c)=$ $\sum\left[b^2(s-b)+c^2(s-c)\right]\ \stackrel{(1\wedge 2)}{\iff}\sum a\cdot AD^2+2sr(2R-r)=$ $2\cdot 4sr(R+r)\iff$

$\sum a\cdot AD^2=2sr\left[4(R+r)-(2R-r)\right]=2sr(2R+5r)\implies$ $\boxed{\ \sum a\cdot AD^2=2sr(2R+5r)\ }\ (3)\ .$ Apply the well known inequalities $3r\sqrt 3\le s$ and $R\ge 2r\ :$

$\ 2sr(2R+5r)\ge 2\cdot 3r^2\sqrt 3(2\cdot 2r+5r)=6r^2\sqrt 3\cdot 9r=54r^3\sqrt 3\ .$ In conclusion, $2sr(2R+5r)\ge 54r^3\sqrt 3\ \stackrel{(3)}{\implies}\ \sum a\cdot AD^2\ge 54r^3\sqrt 3\ .$

Remark.

$1.\blacktriangleright\ \sum a(s-b)(s-c)=\sum[s(s-b)(s-c)-(s-a)(s-b)(s-c)]=sr(4R+r)-3sr^2=$ $sr[(4R+r)-3r]=sr(4R-2r)=\boxed{\ 2sr(2R-r)\ }\ .$

$2.\blacktriangleright\ \sum a^2(s-a)=\sum x(y+z)^2=$ $\sum \left[x\left(y^2+z^2\right)+2xyz\right]=$ $\sum [yz(x+y+z)-xyz+2xyz]=$ $(x+y+z)(xy+yz+zx)+3xyz=$ $sr(4R+r)+3sr^2=\boxed{\ 4sr(R+r)\ }\ .$

Denoted $s-a=x\ ,$ $s-b=y$ and $s-c=z\ .$ Hence $x+y+z=(s-a)+(s-b)+(s-c)=s\ ,$ $xy+yz+zx=\sum (s-b)(s-c)=r(4R+r)$ and $xyz=\prod (s-a)=sr^2\ .$

Otherwise. $\sum a^2(s-a)=s\cdot \sum a^2-\sum a^3=$ $2s\left(s^2-r^2-4Rr\right)\left[(a+b+c)^3-3(a+b+c)(ab+bc+ca)+3abc\right]=$ $2s^3-2sr^2-8Rsr-8s^3+6s\left(s^2+r^2+4Rr\right)-$

$12Rsr=$ $4sr^2+4Rsr=\boxed{\ 4sr(R+r)\ }\ .$ I used the notations and the identities $\left\{\begin{array}{ccccc} s_1\ & = & a+b+c & = & 2s\\\ s_2 & = & ab+bc+ca & = & s^2+r^2+4Rr\\\ s_3\ & = & abc & = & 4Rsr\end{array}\right\|$ and $\left\{\begin{array}{ccccc} S_2 & = & \sum a^2 & = & 2\left(s^2-r^2-4Rr\right)\\\\ S_3 & = & \sum a^3 & = & s_1^3-3s_1s_2+3s_3\end{array}\right\|\ .$

P9 (Hung Nguyen Viet).Prove that $\ (\forall )\ \triangle ABC$ there is the followung inequality $:\ \boxed{\ (a+b+c)\left(ab+bc+ca+R^2\right)\ge a^3+b^3+c^3+7abc\ }\ (*)\ .$

Proof. I"ll use the notations and the identities $\left\{\begin{array}{ccccc} s_1\ & = & a+b+c & = & 2s\\\ s_2 & = & ab+bc+ca & = & s^2+r^2+4Rr\\\ s_3\ & = & abc & = & 4Rsr\end{array}\right\|$ and $\left\{\begin{array}{ccccc} S_2 & = & \sum a^2 & = & 2\left(s^2-r^2-4Rr\right)\\\\ S_3 & = & \sum a^3 & = & s_1^3-3s_1s_2+3s_3\end{array}\right\|\ ,$

where prove easily $\boxed{\ s_1^3-3s_1s_2+3s_3=2s\left(s^2-3r^2-6Rrs\right)\ }\ (1)\ .$ Therefore, the required inequality $(*)$ becomes $2s\cdot \left(s^2+r^2+4Rr\right)+2sR^2\ \stackrel{((1)}{\ge}$

${2s}\left(s^2-3r^2-6Rr\right)+28Rsr\iff$ $s^2+r^2+4Rr+R^2\ge s^2-3r^2-6Rr+14Rr\iff$ $R^2-4Rr+4r^2\ge 0\iff$ $(R-2r)^2\ge 0\ ,$ what is true.

P10 (Mehmet Sahin). Prove the identity $\boxed{\ \frac {h_a}{AI_a^2}+\frac {h_b}{AI_b^2}+\frac {h_c}{AI_c^2}=\frac 1{2R}\ }\ (*)$ (standard notations).

Proof 1. Prove easily that $\triangle ABI_a\sim \triangle AIC\ ,$ i.e. $\frac {AB}{AI}=\frac {AI_a}{AC}\iff \boxed{\ AI\cdot AI_a=bc\ }\ (1)\ .$ Therefore, $\sum \frac {h_a}{AI_a^2}=\frac 1{2R}\iff$

$\sum \frac {2Rh_a}{AI_a^2}=1\iff$ $\sum \frac {bc}{AI_a^2}= 1 \stackrel{(1)}{\iff}\ \sum \frac {AI\cdot AI_a}{AI_a^2}= 1\iff$ $\sum \frac {AI}{AI_a}= 1\iff$ $\sum \frac r{r_a}=1\iff$ $\sum \frac 1{r_a}=\frac 1r\ ,$ what is true.

Proof 2. I"ll apply the identities $\boxed{\ \frac{AI_a^2}{bc}=\frac s{s-a}=\frac{bc}{AI^2}\ }\ ,$ $\boxed{\ bc=2Rh_a\ }\ ,$ $\sum (s-b)(s-c)=r(4R+r)$ and $(s-a)(s-b)(s-c)=sr^2\ .$ Hence $\ :$

$\blacktriangleright\ \sum\frac {h_a}{AI^2}=$ $\sum \frac{2Rh_a}{2R\cdot AI^2}=$ $\frac 1{2R}\cdot\sum \frac{\cancel {bc}s}{\cancel {bc}(s-a)}=$ $\frac s{2R}\cdot\sum \frac 1{s-a}=$ $\frac s{2R}\cdot \frac{\sum (s-b)(s-c)}{(s-a)(s-b)(s-c)}=$ $\frac {\cancel s}{2R}\cdot \frac{\cancel r(4R+r)}{\cancel sr^{\cancel 2}}=$ $\frac{4R+r}{2Rr}\implies$ $\boxed{\sum\frac {h_a}{AI^2}=\frac{4R+r}{2Rr}\ }\ .$

$\blacktriangleright\ \sum\frac {h_a}{AI_a^2}=$ $\sum \frac{2Rh_a}{2R\cdot AI^2}=$ $\frac 1{2R}\cdot\sum \frac{\cancel{bc}(s-a)}{\cancel{bc}s}=$ $\frac1{2Rs}\cdot\sum(s-a)=\frac 1{2R\cancel s}\cdot \cancel s=$ $\frac 1{2R}\implies \boxed{\sum\frac {h_a}{AI_a^2}=\frac 1{2R}\ }\ .$ Remark. $\boxed{\ \left(r_a+r_b+r_c\right)\cdot\sum\frac{h_a}{AI_a^2}=r\cdot\sum\frac{h_a}{AI^2}\ }\ .$

P11 (Miguel Ochoa Sanchez). Let $\triangle ABC$ and the midpoints $(D,E,F)$ of its sides $([BC],[CA],[AB])$ respectively. Prove that $\boxed{\ BE\perp CF\implies \cos A\ge\frac 45\ }\ .$

Proof. $BE\perp CF\iff BFEC$ is orthodiagonal $\iff$ $BC^2+EF^2=BF^2+CE^2\iff$ $a^2+\frac{a^2}4=\frac{c^2}4+\frac {b^2}4\iff$

$\boxed{\ b^2+c^2=5a^2\ }\ (*)\ .$ Hence $\cos A=\frac{b^2+c^2-a^2}{2bc}\ \stackrel{(*)}{=}\ \frac{4a^2}{2bc}\ge \frac {4a^2}{b^2+c^2}=\frac {4a^2}{5a^2}=\frac 45\implies \cos A\ge \frac 45\ .$

Generalization (own). Let $\triangle ABC$ and the points $E\in (AC)\ ,$ $F\in (AB)$ so that $\frac {EA}{EC}=m$ and $\frac {FA}{FB}=n\ .$ Prove that

the chain of the relations $BE\perp CF\ \implies \cos A\ge \frac {2\sqrt{mn(m+1)(n+1)}}{mn+(m+1)(n+1)}\ \iff\ \sin A\le \frac {m+n+1}{2mn+m+n+1}\ .$

Proof. $\left\{\begin{array}{ccccc} \frac{EA}m=\frac {EC}1=\frac b{m+1}\\\\ \frac {FA}n=\frac {FB}1=\frac c{n+1}\end{array}\right\|\implies$ $EF^2=AE^2+AF^2-2\cdot AE\cdot AF\cdot\cos A\implies$ $\boxed{\ EF^2=\frac{m^2b^2}{(m+1)^2}+\frac{n^2c^2}{(n+1)^2}-\frac{2mnbc\cdot\cos A}{(m+1)(n+1)}\ }\ (*)\ .$ Therefore, $BE\perp CF\iff$

$BFEC$ is orthodiagonal $\iff BF^2+CE^2=BC^2+EF^2\iff$ $\frac{c^2}{(n+1)^2}+\frac{b^2}{(m+1)^2}=a^2+\frac{m^2b^2}{(m+1)^2}+\frac{n^2c^2}{(n+1)^2}-\frac{2mnbc\cdot\cos A}{(m+1)(n+1)}\iff$ $\frac{c^2\left(n^2-1\right)}{(n+1)^2}+\frac{b^2\left(m^2-1\right)}{(m+1)^2}+$

$\left(b^2+c^2-2bc\cdot\cos A\right)=\frac{2mnbc\cdot\cos A}{(m+1)(n+1)}\iff$ $\frac{c^2\left(n-1\right)}{(n+1)}+\frac{b^2\left(m-1\right)}{(m+1)}+b^2+c^2=2bc\cdot\cos A\cdot\left[1+\frac{mn}{(m+1)(n+1)}\right]\implies$ $\cos A=\frac {n(m+1)\cdot \frac cb+m(m+1)\cdot\frac bc}{mn+(m+1)(n+1)}\ge$

$\frac {2\sqrt{mn(m+1)(n+1)}}{mn+(m+1)(n+1)}\ \iff\ \sin A\le \frac {m+n+1}{2mn+m+n+1}\ .$ In the particular case $m=n=1$ obtain that $\cos A\ge \frac 45\iff \sin A\le \frac 35\ ,$ i.e. the proposed problem P11.

P12 (Mehmet Sahin). Prove that $(\forall )\ \triangle ABC$ there is the following inequality $:\ \boxed{\ \left[I_aBC\right]+\left[I_bCA\right]+\left[I_cAB\right]\ge 3[ABC]\ }\ .$

Proof. I"ll use the identity $\boxed{\ ar_a=s\left(r_a-r\right)\ }\ (*)\ .$ Therefore, $\sum\left[I_aBC\right]=\frac 12\cdot\sum ar_a\ \stackrel{(*)}{=}\ \frac 12\cdot \sum s\left(r_a-r\right)=$ $\frac s2\cdot\sum \left(r_a-r\right)=$ $\frac s2\cdot (4R+r-3r)=$ $\frac s2\cdot (4R-2r)=$

$s(2R-r)\implies$ $\boxed{\ \sum\left[I_aBC\right]=s(2R-r)\ }\ (1)\ .$ Observe that $\sum\left[I_aBC\right]\ge 3S\ \stackrel{(1)}{\iff}\ \cancel s(2R-r)\ge 3\cancel sr\iff$ $2R-r\ge 3r\iff$ $2R\ge 4r\iff$ $R\ge 2r\ .$ what is true.

P13 (Marian Ursarescu). Prove that $(\forall )$ an acute $\ \triangle ABC$ the following relationship holds $:\ \boxed{\ (b+c)m_a+(c+a)m_b+(a+b)m_c\le 6sR\ }$ (standard notations).

Proof. Let the midpoint $M$ of $[BC]\ ,$ the circumcenter $O$ of $\triangle ABC$ and the midpoint $S$ of $I_aI\ .$ Observe that $m_a=AM\le OA+OM=$ $OA+(OS-MS)=$

$(OA+OS)-MS=$ $2R-\frac {r_a-r}2=\frac {4R+r-r_a}2=$ $\frac{r_b+r_c}2\implies$ $\boxed{\ m_a\le\frac{4R+r-r_a}2\ }\ (1)\ .$ Thus, $\sum (b+c)m_a\ \stackrel{(1)}{\le}\ (b+c)\cdot\frac {(4R+r)-r_a}2=$

$\sum\frac {4R+r}2\cdot (b+c)-\sum\frac {r_a(b+c)}2=$ $\frac{4R+r}2\cdot 4s-\sum \frac {r_a(2s-a)}2=$ $2s(4R+r)-s\sum r_a+\sum\frac{ar_a}2=$ $2s(4R+r)-s(4R+r)+$

$\sum\frac{s\left(r_a-r\right)}2=$ $s(4R+r)+\frac s2\cdot \sum (r_a-r)=$ $s(4R+r)+\frac s2\cdot (4R+r-3r)=$ $s(4R+r)+s(2R-r)=6sR\implies$ $\sum (b+c)m_a\le 6sR\ .$

P14 (Mehmet Sahin). Prove that $\boxed{\ \{a,b,c\}\subset\mathbb R^*_+\ \implies \ \sum\frac 1{a^3+b^3+abc}\le\frac 1{abc}\ }$

Proof. Prove easily that $a^2-ab+b^2\ge ab\iff$ $(a+b)\left(a^2-ab+b^2\right)\ge ab(a+b)\iff$ $\boxed{a^3+b^3\ge ab(a+b)\ }\ (*)\ .$ Hence $:$

$\sum\frac 1{a^3+b^3+abc}\le\sum\frac 1{ab(a+b)+abc}=\sum\frac 1{ab(a+b+c)}=\frac 1{a+b+c}\cdot\sum\frac 1{ab}=\frac 1{a+b+c}\cdot \frac {a+b+c}{abc}=\frac 1{abc}\ .$

P15 (Daniel Florescu). Sa se rezolve ecuatia irationala $\boxed{\ 2\sqrt[3]{2x-1}=x^3+1\ }\ (*)\ .$

Proof. Notam $:\ \left\{\begin{array}{ccc} x^3+1=2y\\\\ y^3+1=2x\end{array}\right\|\ .$ Se observa ca prin diferenta celor doua relatii se obtine $x^3-y^3=2(y-x)\iff$ $(x-y)\left(x^2+xy+y^2\right)+2(x-y)=0\iff$
$(x-y)\left(x^2+y^2+xy+2\right)=0\iff$ $x=y\iff$ $x^3+1=2x\iff$ $x^3-2x+1=0\iff$ $(x-1)\left(x^2+x-1\right)=0\iff$ $x\in\left\{1,\frac {-1\pm\sqrt 5}2\right\}\ .$

P16 (Seyran Ibrahimov). Prove that $(\forall )\ \triangle ABC$ there is the inequality $\boxed{\ \sum\frac {\left(am_a\right)^2}{b+c}\ge 9sS\ }\ (*)$ (standard notations).

Proof. Prove easily that $\boxed{\ 6sr\le \sum am_a\le 2s(R+r) \ }\ (1)\ .$ Therefore, $\sum\frac {\left(am_a\right)^2}{b+c}\ \stackrel{C.B.S}{\ge}\ \frac {\left(\sum am_a\right)^2}{\sum (b+c)}\ge$ $\frac {36s^2r^2}{4s}=9sr^2\implies$ $\sum\frac {\left(am_a\right)^2}{b+c}\ge 9rS\ .$

Remark. I"ll prove the chain $(1)\ :\ h_a\le m_a\implies ah_a\le am_a\implies 2S\le am_a\implies \boxed{\ 6sr\le \sum am_a\ }\ ;\$ Let $M$ be the midpoint of $[BC]$ and the circumcircle $w=\mathbb C(O,R)\ .$ Thus,

if $\triangle ABC$ is acute, then $AM\le OA+OM\iff \boxed{m_a\le R(1+\cos A)}\iff$ $am_a\le R(a+a\cdot \cos A)\implies$ $\sum am_a\le R\cdot\left(2s+\sum a\cos A\right)=$ $2sR+2R^2\sum\sin A\cos A=$

$2sR+R^2\sum \sin 2A=$ $2sR+4R^2\prod\sin A=$ $2sR+2S=2sR+2sr=2s(R+r)\implies$ $\boxed{\ am_a+bm_b+cm_c\le 2s(R+r)\ }\ (1)\ .$ Otherwise. $\boxed{m_a\le R(1+\cos A)}\iff$

$am_a\le 2aR\cdot\frac {s(s-a)}{bc}=$ $\cancel 2a\cancel R\cdot\frac {s(s-a)}{\cancel {2R}h_a}=$ $\frac {a^2\cancel s(s-a)}{2\cancel sr}\implies$ $\boxed{am_a\le \frac {a^2(s-a)}{2r}}\implies$ $\sum am_a\le \frac 1{2r}\cdot\sum a^2(s-a)=\frac 1{2r}\cdot 4sr(R+r)=2s(R+r)\implies$ $\boxed{\ \sum am_a\le 2s(R+r)\ }\ .$

P17 (Mehmet Sahin). Prove that $(\forall )\ \triangle ABC$ there is the inequality $\boxed{\ 1+\frac{r_a}{m_a}+\frac {r_b}{m_b}+\frac{r_c}{m_c}\le \frac {2R}r\ }\ (*)$ (standard notations).

Proof. I"ll apply the well known inequality $\boxed{\ 2sr\le am_a\ }$ and the identity $\boxed{\ ar_a=s\left(r_a-r\right)\ }\ .$ Hence $:\ \frac{r_a}{m_a}=\frac {ar_a}{am_a}\le\frac {s\left(r_a-r\right)}{2sr}=\frac {r_a-r}{2r}\implies$ $\frac{r_a}{m_a}\le \frac {r_a-r}{2r}\implies$

$\sum\frac{r_a}{m_a}\le \sum\frac {r_a-r}{2r}=\frac{4R+r-3r}{2r}=\frac {2R-r}r\implies 1+\sum \frac{r_a}{m_a}\le 1+\frac {2R-r}r=\frac {2R}r\ .$ In conclusion, $1+\frac{r_a}{m_a}+\frac {r_b}{m_b}+\frac{r_c}{m_c}\le \frac {2R}r\ .$

Proposed problem (test) P18.

$1.\blacktriangleright$ Find the solutions (zeroes) of the following equations $\ :\ \left[\frac {x+2}3\right]=\frac {x-1}4\ ;\ \left[\frac {x+1}3\right]=\frac {x-1}2\ ;\ \left[\frac {4x+1}{5x+3}\right]=\frac {x}{x+8}\ .$

$2.\blacktriangleright$ Let $\{a,b\}\subset\mathbb R$ so that $0<a\ne 1$ and $0<b\ne 2\ .$ Prove that $\log_a\frac a2=\log_{\frac b2}b\iff ab=2\ .$

$3.\blacktriangleright$ Find $S_{2015}=z_1^{2015}+z_2^{2015}\ ,$ where $\{z_1,z_2\}$ are the roots of the equation $z^2-z+1=0\ .$

(for my friend ema_manole2010).

Proofs.

$1.1\blacktriangleright=\frac {x-1}4=\left[\frac {x+2}3\right]=z\iff$ $\boxed{\ x=4z+1\ }$ and $z\le \frac {x+2}3<z+1\iff$ $3z\le (4z+1)+2<3(z+1)\iff$

$3z\le 4z+3<3z+3\iff$ $-3\le z<0\iff$ $z\in [-3,0)\cap\mathbb Z\iff$ $z\in\{-3,-2,-1\}\iff$ $\boxed{\ x\in\{-11,-7,-3\} }\ .$

$1.2\blacktriangleright\ \frac {x-1}2=\left[\frac {x+1}3\right]=z\in\mathbb Z\implies$ $\boxed{\ x=2z+1\ }$ and $z\le \frac {x+1}3<z+1\iff$ $3z\le (2z+1)+1<3(z+1)\iff$

$3z\le 2z+2<3z+3\iff$ $-1<z\le 2\iff$ $z\in (-1,2]\cap \mathbb Z\iff$ $z\in\{0,1,2\}\iff$ $\boxed{\ x\in \{1,3,5\}\ }\ .$

$1.3\blacktriangleright\ \frac x{x+8}=\left[\frac {4x+1}{5x+3}\right]=z\in \mathbb Z\iff$ $\boxed{\ x=\frac {8z}{1-z}\ }$ and $z\le\frac {4x+1}{5x+3}<z+1\iff$ $z\le\frac {4\cdot\frac {8z}{1-z}+1}{5\cdot \frac {8z}{1-z}+3}<z+1\iff$ $z\le\frac {31z+1}{37z+3}<z+1\iff$

$\left(z-\frac {31z+1}{37z+3}\right)\left[(z+1)-\frac {31z+1}{37z+3}\right]\le 0$ and $\frac {31z+1}{37z+3}\ne (z+1)\iff$ $\left(37z^2-28z-1\right)\left(37z^2+9z+2\right)\le 0\iff$ $f(z)=37z^2-28z-1\le 0$ and $z\in\mathbb Z\iff \boxed{\ z=0\ }\ .$

$2\blacktriangleright\ \log_a\frac a2=\log_{\frac b2}b\iff$ $\frac {\ln\frac a2}{\ln a}=\frac {\ln b}{\ln\frac b2}\iff$ $(\ln a-\ln 2)(\ln b-\ln 2)=\ln a\ln b\iff$ $\cancel{\ln 2}(\ln a+\ln b)=\ln^{\cancel 2}2\iff$ $\ln (ab)=\ln 2\iff$ $\boxed{\ ab=2\ }\ .$

$3\blacktriangleright\ \boxed{\ z_1+z_2=1\ }\ (*)$ and $(\forall )\ k\in\{1,2\}\ ,$ $z_k^2-z_k+1=0\implies$ $\boxed{\ z_k^2=z_k-1\ }\implies$ $z_k^3=z\cdot z_k^2=z_k(z_k-1)=z_k^2-z_k=(z_k-1)-z_k=-1\implies$ $\boxed{\ z_k^3=-1\ }\implies$ $z_k^{2015}=$

$z_k^{3\cdot 671+2}=$ $\left(z_k^3\right)^{671}\cdot z_k^2=$ $(-1)^{671}\cdot z_k^2=-z_k^2=-(z_k-1)=1-z_k\implies$ $\boxed{\ z_k^{2015}=1-z_k\ }\ , (\forall )\ k\in \{1,2\}\implies$ $S_{2015}=$ $z_1^{2015}+z_2^{2015}=$ $(1-z_1)+(1-z_2)\ \stackrel{(*)}{=}\ 1$ $\implies$ $\boxed{\ S_{2015}=1\ }\ .$

Remark. $\boxed{\ (\forall )\ x>0\ ,\ \ln_a{\frac ax}=\ln_{\frac bx}b\iff\ \frac {\ln\frac ax}{\ln a}=\frac {\ln b}{\ln\frac bx}\iff x\in \{1,ab\}\ }$

P19. Prove that $(\forall )\ \{a,b,c\}\subset\mathbb R$ there is the following inequality $:\ \boxed{\ (a+b+c)\left(a^3+b^3+c^3+3abc\right)\ge 2(ab+bc+ca)\left(a^2+b^2+c^2\right)\ }$

Proof. Let $\left\{\begin{array}{ccccc} a+b+c & = & s_1 & = & 2s\\\\ ab+bc+ca & = & s_2 & = & s^2+r^2+4Rr\\\\ abc & = & s_3 & = & 4Rrs\end{array}\right\|\ .$ Thus, $:\ S_2=\sum a^2=s_1^2-2s_2=4s^2-2\left(s^2+r^2+4Rr\right)=2s^2-2r^2-8Rr\implies$ $\boxed{\ \sum a^2=2\left[s^2-r^2(4R+r)^2\right]\ }\ ;$

$S_3=\sum a^3=s_1^3-3s_1s_2+3s_3=8s^3-6s\left(s^2+r^2+4Rr\right)+12Rrs=$ $2s^3-6sr^2-12Rrs=$ $2s\left(s^2-3r^2-6Rr\right)\implies$ $3abc+\sum a^3=\cancel {12Rrs}+2s\left(s^2-3r^2-\cancel {6Rr}\right)=$

$2s\left(s^2-3r^2\right)\implies$ $\boxed{\ 3abc+\sum a^3=2s\left(s^2-3r^2\right)\ }\ .$ Hence, $(a+b+c)\left(a^3+b^3+c^3+3abc\right)\ge 2(ab+bc+ca)\left(a^2+b^2+c^2\right)\iff$ $\cancel 4s^2\left(\cancel {s^2}-3r^2\right)\ge \cancel 4\left[\cancel {s^4}-r^2(4R+r)^2\right]\iff$

$r^2(4R+r)^2\ge 3s^4\iff$ $\boxed{\ r(4R+r)\ge s\sqrt 3\ }\ ,$ what is true. Indeed, $3\left(r_ar_b+r_br_c+r_cr_a\right)\le\left(r_a+r_b+r_c\right)^2\iff$ $3s^2\le (4R+r)^2\iff$ $s\sqrt 3\le 4R+r\ .$

Remark. $LHS-RHS=\frac{1}{2}((a-b)^2(a+b-c)^2+(b-c)^2(b+c-a)^2+(c-a)^2(c+a-b)^2)\ge 0\iff$ $\frac16\sum \left[(a-c)^2+b(a+c-2b)\right]\ge 0\iff$

$\sum a^4+abc(a+b+c)-\sum a^3(b+c)\ge 0\iff$ which is fourth degree Schur inequality and holds for all reals.

P20 (Thanos KALOGERAKIS, Greece). Let an acute $\triangle ABC$ with the circumcircle $w=\mathbb C(O,R)$ and the midpoints $(M,N,P)$ of the

arcs $\left(\overarc{BC},\overarc {CA},\overarc {AB}\right)$ respectively. Prove that the area of the convex hexagon $APBMCN$ is equally to $sR\ ,$ where $2s=a+b+c\ .$

Proof. Let $(M,N,P)$ be the midpoints of $\left(\ [BC]\ ,\ [CA]\ ,\ [AB]\ \right)$ respectively. Prove easily that $\boxed{\ ar_a=s\left(r_a-r\right)\ }\ (*)$ a.s.o. and $\boxed{\ XM=\frac {r_a-r}2\ }\ (1)$

a.s.o. where $\left\{r_a,r_b,r_c\right\}$ are the exradii of $\triangle ABC\ .$ Thus, $[APBMCN]=S+\sum [BMC]=S+\frac 12\cdot \sum (BC\cdot XM)\ \stackrel{(1)}{=}\ sr+\sum \frac {a\left(r_a-r\right)}4\ \stackrel{(*)}{=}$

$sr+\frac 14\cdot\left[\sum s\left(r_a-r\right)-2sr\right]=$ $sr+\frac s4\cdot\left[\sum \left(r_a-r\right)-2r\right]=$ $sr+\frac s4\cdot \left(4R-4r\right)=$ $sr+sR-sr=sR\implies$ $\boxed{\ [APBMCN]=sR\ }\ .$

P21 (<= click). Proof. Denote $S\in DE\cap AC$ and apply the Menelaus' theorem for the transversal $BM$ to the triangles $\triangle ADS$ and $\triangle CES$ respectively: $\frac {BA}{BD}\cdot\frac {GD}{GS}\cdot\frac {MS}{MA}=$

$\frac {BC}{BE}\cdot\frac {GE}{GS}\cdot\frac {MS}{MC}=1\ (*)\ .$ From the relations $BD=BE\ ,$ $MA=MC$ and $(*)$ obtain $BA\cdot GD=BC\cdot GE\ .$ There is generally the identity $\frac{GD}{GE}\cdot\frac {BA}{BC}=\frac {MA}{MC}\cdot\frac {BD}{BE}\ .$

P22 (Valentin VORNICU). Let $ABC$ be an acute triangle with $AB \neq AC$ . Let $D$ be the foot of the altitude from $A$ and $\omega$ the circumcircle

of the triangle. Let $\omega_1$ be the circle tangent to $AD$ , $BD$ and $\omega$ . Let $\omega_2$ be the circle tangent to $AD$ , $CD$ and $\omega$ . Let $\ell$ be the interior common

tangent to both $\omega_1$ and $\omega_2$ , different from $AD$ . Prove that $\ell$ passes through the midpoint of $BC$ if and only if $2BC = AB + AC$ .

Lemma 1. let $w_1$ be a circle which is interior tangent in the point $A$ to the circle $w$ . For a point $M\in w_1$ , $M\ne A$ denote the intersections

$X,Y$ between the circle $w$ and the tangent-line in the point $M$ to the circle $w_1$ . Then the ray $[AM$ is the bisector of the angle $\widehat {XAY}$ .

Lemma 2. For the incenter $I$ and the centroid $G$ of the triangle $ABC$ we have

$\blacktriangleright IG\parallel BC\Longleftrightarrow b+c=2a\ .$

$\blacktriangleright IG\perp BC\Longleftrightarrow b=c\ \ \vee\ \ b+c=3a\ .$

Indication for proof the proposed problem. The circles $w_1,w_2$ are interior tangent to the circle $w$ . Can

apply the above remarkable lemma for each of the circles $w_1,w_2$ and for each of the lines $BC,AD,l$ .

This post has been edited 306 times. Last edited by Virgil Nicula, May 10, 2018, 7:53 AM

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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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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
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Holy- Darn this is good. shame it's inactive now
by the_mathmagician, Jan 17, 2021, 7:43 PM
godly blog. opopop
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!!!!
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

461. Problems for the relaxation in ... weekend (II).

by Virgil Nicula, Jan 26, 2018, 7:36 PM

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