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

by Virgil Nicula, Jul 20, 2010, 4:58 PM

Cezar . 296.060 . GIL

$\bf\color{red}Problems\ of\ the\ type\ "instant"$ $\bf\color{black}===================$
P0.0. $(\forall )\ \{x,y\}\subset\mathbb R\ ,\ x^2+y^2=2\implies -2\le xy(x+y)\le 2\ .$

Proof. $\left\{\begin{array}{ccc} 2|xy|=2|x||y|\le |x|^2+|y|^2=x^2+y^2=2 & \implies & 0\le |xy|\le 1\\\\ |x+y|^2=(x+y)^2\le 2(x^2+y^2)=4 & \implies & 0\le |x+y|\le 2\end{array}\right\|$ $\bigodot\implies$ $|xy(x+y)|\le 2$ , i.e. $-2\le xy(x+y)\le 2\ .$ Remark. $|x|=\max \{x,-x\}\ .$

P0.1. Let $\{a,b\}\subset\mathbb R^*_+$ so that $a^4\le b^3\ .$ Prove that the equation $x^4-ax+b=0$ hasn't real roots, i.e. for any $x\in\mathbb R\ ,\ x^4+b>ax\ .$

Proof 1. $\left\{\begin{array}{cccc} (\forall ) & x\le \frac ba & \Longrightarrow & x^4-ax+b=x^4+(-ax+b)>0\\\\ (\forall ) & \sqrt[3] a \le x & \Longrightarrow & x^4-ax+b=x\left(x^3-a\right)+b>0\end{array}\right\|\ .$ Thus, $a^4\le b^3\implies \sqrt [3] a\le \frac ba\ .$ In conclusion, $(\forall )\ x\in\mathbb R\ ,\ x^4-ax+b>0\$

Proof 2. $x^4+b=x^4+\frac b3+\frac b3+\frac b3\ge$ $4\cdot \sqrt[4]{\frac {3b^3x^4}{81}}=$ $\frac {4|x|}3\cdot\sqrt[4]{3b^3}\ge$ $\frac {4|x|}3\cdot\sqrt[4]{3a^4}=$ $\frac {4\sqrt[4]3}3\cdot a|x| >a|x|\ge$ $ax\implies$ $x^4-ax+b>0$ for any $x\in\mathbb R\ .$

P0.2.Solve over $\mathbb R$ the equation $x^2+\left(\frac x{x-1}\right)^2=8\ .$
Proof 1. $\left(x^2-8\right)(x-1)^2+x^2=0\iff$ $\left(x^2-8\right)\left(x^2 2x+1\right)+x^2=0\iff$ $x^4-2x^3-6x^2+16x-8=0\ \stackrel{\mathrm{Horner}}{\iff}\ (x-2)^2$ $\left(x^2+2x-2\right)=0$ $\begin{array}{cc} \nearrow & \ \stackrel{double}{2}\\\\ \rightarrow & -1-\sqrt 3\\\\ \searrow & -1+\sqrt 3\end{array}\ .$
Proof 2. $(x-1)^2+\left(\frac x{x-1}\right)^2=$ $9-2x\iff$ $\left[(x-1)+\frac x{x-1}\right]^2=9$ $\implies$ $\left|\begin{array}{c} (x-1)+\frac x{x-1}=3\implies x^2-4x+4=0\\\\ (x-1)+\frac x{x-1}=-3\implies x^2+2x-2=0\end{array}\right|\ \stackrel{x\ne 1}{\implies}\ x$ $\in\left\{2\ ;\ -1\pm\sqrt 3\right\}\ .$

P0.3 (Dan Stefan Marinescu & Viorel Cornea, shortlist 2008). Prove that the inequality $x^{n+1}+\frac{1}{x^{n+1}} \geq$ $x^n+\frac{1}{x^n},\ \forall\ x > 0,\ \forall\ n \in \mathbb{N}\ .$

Proof. Define the relation $X\ \underline{.s.s.}\ Y \ \iff$ $\ X=Y=0\ \vee\ X\cdot Y>0$ , i.e. $X$ and $Y$ have "same sign". Therefore, $x^{n+1}+\frac{1}{x^{n+1}} \geq$ $x^n+\frac{1}{x^n}$ $\iff$

$\left(x^{n+1}+\frac 1{x^{n+1}}\right)-\left(x^n+\frac 1{x^n}\right)\ \underline{.s.s.}\ \left(x^{2n+2}+1\right)-x\left(x^{2n}+1\right) \underline{.s.s.}\ x^{2n+1}(x-1)-(x-1)\ \underline{.s.s.}\ (x-1)\left(x^{2n+1}-1\right)\ \underline{.s.s.}\ (x-1)^2\ge 0\ .$

P0.4. Let $\triangle ABC$ with the circumcircle $w=C(O,R)$ . Denote the midpoint $M$ of $[BC]$ , the tangent $AA$ to $w$ at $A\in w$ , $D\in BC$ so that $AD\perp BC$ and $\{A,N\}=\{A,M\}\cap w$ .

Prove that $m\left(\widehat { MAC}\right)= 90^{\circ}-B\iff$ $\boxed{O\in AM}\iff\ AB=AC\ \ \vee\ \ AB\perp AC\ \iff\widehat {DAB}\equiv\widehat {MAC}\iff$ $\left\{\begin{array}{ccc} AA\perp AM & \iff & BN\perp BA\\\\ \sin 2B=\sin 2C & \iff & [OAB]=[OAC]\end{array}\right\|$

P0.5. Let $\triangle ABC$ with the circumcircle $\mathbb C(O,R)\ .$ Solve the geometrical equation $x^2-4Rx+a^2=0 .$ See here.

Proof 1. I"ll show that $x^2-4Rx+a^2=0$ $\begin{array}{ccccc} \nearrow & r_b+r_c & \searrow\\\\ \searrow & r_a-r & \nearrow\end{array}\odot$ Indeed, $S=\left(r_b+r_c\right)+\left(r_a-r\right)=\left(r_a+r_b+r_c\right)-r=(4R+r)-r=4R\implies$ $\boxed{S=4R}\ .$

$P\equiv x_1x_2=$ $\left(r_b+r_c\right)\cdot\left(r_a-r\right)=$ $\left(\frac S{s-b}+\frac S{s-c}\right)\cdot\left(\frac S{s-a}-\frac Ss\right)=$ $\frac {aS}{(s-b)(s-c)}\cdot\frac {aS}{s(s-a)}=a^2\implies$ $\boxed{P=a^2}$ because $S^2=s(s-a)(s-b)(s-c)\ .$

Proof 2. Let midpoint $M$ of $[BC]$ and diameter $[NS]$ so that $M\in NS$ (south, north), i.e. $MN+MS=2R\ .$ Prove easily that $\left\{\begin{array}{cccc} \mathrm{If} & A\le 90^{\circ} & \implies & \odot\begin{array}{cc} \nearrow & MN=\frac {r_b+r_c}2\\\\ \searrow & MS=\frac {r_a-r}2\end{array}\\\\ \mathrm{If} & A>90^{\circ} & \implies & \odot\begin{array}{cc} \nearrow & MN=\frac {r_a-r}2\\\\ \searrow & MS=\frac {r_b+r_c}2\end{array}\end{array}\right\|$

P0.6. Prove that in any triangle $ABC$ there is the inequality $\frac a{b+c-a}+\frac b{c+a-b}+\frac c{a+b-c}\ge 3\ .$

Proof 1. $\sum\frac a{b+c-a}=\sum\frac a{2(s-a)}=$ $\frac {\sum a(s-b)(s-c)}{2(s-a)(s-b)(s-c)}=$ $\frac {2sr(2R-r)}{2sr^2}=\frac {2R-r}{r}=$ $2\cdot\frac Rr-1\ \stackrel{R\ge 2r}{\ge}\ 2\cdot 2-1=3\implies$ $\sum\frac a{b+c-a}\ge 3\ .$

Proof 2. Exist $\{x,y,z\}\subset\mathbb R^*_+$ so that $\left\{\begin{array}{ccc} a & = & y+z\\\ b & = & z+x\\\ c & = & x+y\end{array}\right\|\ .$ The our inequality becomes $\sum \frac {y+z}{2x}=\frac 12\cdot\sum \left(\frac yz+\frac zy\right)\ge \frac 12\cdot\sum 2=3\implies$ $\sum\frac a{b+c-a}\ge 3\ .$

Proof 3. $\sum\frac a{b+c-a}=$ $\sum\frac {a^2}{a(b+c-a)}\ \stackrel{\mathrm{C.B.S.}}{\ge}\frac {(a+b+c)^2}{2\sum a(s-a)}=\frac {4s^2}{4r(4R+r)}=\frac {s^2}{r(4R+r)}\ge \frac {16Rr-5r^2}{r(4R+r)}=\frac {16R-5r}{4R+r}=4-\frac {9r}{4R+r}\ge 3\ .$

Proof 4. I"ll show that $\boxed{\sum\frac a{b+c-a}\ge 2+\frac {2R}r-\left(\frac {4R+r}s\right)^2}\ (*)\ .$ I found from first proof that $\sum\frac a{b+c-a}=\frac {2R-r}{r}\ .$ Thus, $\frac {2R-r}{r}\ge 2+\frac {2R}r-\left(\frac {4R+r}s\right)^2\iff$

$\left(\frac {4R+r}s\right)^2\ge 3\iff$ $4R+r\ge s\sqrt 3$ what is true. I used $\sum\frac {y+z}x= \sum\left(\frac yz+\frac zy\right)$ and $\left|\begin{array}{ccc} \sum a(s-a) & = & 2r(4R+r)\\\\ \sum a(s-b)(s-c) & = & 2sr(2R-r)\\\\ (s-a)(s-b)(s-c) & = & sr^2\end{array}\right|\ \ \wedge\ \ \left|\begin{array}{ccc} R & \ge & 2r\\\\ 4R+r & \ge & s\sqrt 3\\\\ 16Rr-5r^2 & \le & s^2\end{array}\right|$

P0.7 (Ruben Dario). Let a rectangle $BCFE$ , $A\in (EF)$ so that $AB\perp AC$ . The incircles of $ABE$ , $ACF$ touch $AB$ , $AC$ at $M$ , $N$ respectively. Prove that $[MAN]=[BMNC].$

Proof. Let the incircle $w_1=\mathbb C\left(I_1,r_1\right)$ of $\triangle ABE$ and the incircle $w_2=\mathbb C\left(I_2,r_2\right)$ of $\triangle ACF\ .$ Denote $\left\{\begin{array}{ccccc} P\in EF\cap w_1 & ; & m\left(\widehat {BAE}\right) & = & \theta\\\\ R\in EF\cap w_2 & ; & m\left(\widehat {CAF}\right) & = & \phi\end{array}\right\|$ . Observe that

$\theta +\phi =\frac {\pi}2$ and $\cot\frac {\theta}2=t\iff \cot\frac {\phi}2=\frac {t+1}{t-1}\ .$ Therefore $\odot$ $\begin{array}{ccccccc} \nearrow & PM\parallel BE & \implies & \frac {AM}{AB}=\frac {AP}{AE}=\frac {r_1\cot\frac {\theta}2}{r_1\left(1+\cot\frac {\theta}2\right)}=\frac t{1+t} & \implies & \frac {AM}{AB}=\frac t{1+t} & \searrow\\\\ \searrow & RN\parallel CF & \implies & \frac {AN}{AC}=\frac {AR}{AF}=\frac {r_2\cot\frac {\phi}2}{r_1\left(1+\cot\frac {\phi}2\right)}=\frac {\frac {t+1}{t-1}}{1+\frac {t+1}{t-1}} & \implies & \frac {AN}{AC}=\frac {t+1}{2t} & \nearrow\end{array}\odot\implies$

$\frac {[AMN]}{[ABC]}=\frac {AM}{AB}\cdot\frac {AN}{AC}=\frac t{1+t}\cdot\frac {t+1}{2t}=\frac 12\implies$ $[MAN]=[BMNC]$ (Daniel Dan).

P0.8 (Cristinel Mortici, G.M.B. 9/2005). Prove that $(\forall )\ \{x,y,z\}\subset\mathbb R^*_+$ there is the inequality $:\ \frac{9}{x+y+z}-\frac{1}{xyz}\le 2\ .$

Proof 1. Denote $\sqrt[3]{xyz}=t$ and apply $\mathrm{A.M.}\ge \mathrm{G.M.} :\ \ \mathrm{LHS}\le \frac{9}{3\sqrt[3]{xyz}}-\frac 1{xyz}=$ $\frac{3}{t}-\frac{1}{t^3}\le 2$ because $(t-1)^2(2t+1)\ge 0\ .$

Proof 2. $\left\|\ \begin{array}{ccc} 3\cdot\sqrt[3]{1\cdot xyz\cdot xyz} & \le & 1+2xyz\\\\ 3\cdot\sqrt[3]{(xyz)} & \le & x+y+z\end{array}\ \right\|\ \bigodot\ \Longrightarrow\ 9xyz\le (1+2xyz)(x+y+z)\implies$ $\frac 9{x+y+z}\le \frac 1{xyz}+2\implies$ $\frac{9}{x+y+z}-\frac{1}{xyz}\le 2\ .$ Vezi si aici.

P0.9 (Luis Saavedra). Let an $A$ -right $\triangle ABC\ .$ Denote the bisector $BE$ where $E\in (AC)\ .$ Prove that $2\cdot AB=BC+CE\iff$ $AB=AC\ .$

Proof 1. $\frac {EA}{c}=\frac {EC}a=\frac b{a+c}\implies \boxed{EC=\frac {ab}{a+c}}\ (*)\ .$ Thus, $2\cdot AB=BC+CE\ \stackrel{(*)}{\iff}\ 2c=$ $a+\frac {ab}{a+c}\iff$ $(2c-a)(a+c)=ab\iff$ $2c^2+ac=a^2+ab\iff$

$2c^2+ac=\left(b^2+c^2\right)+ab\iff$ $c^2+ac=b^2+ab\iff$ $\left(b^2-c^2\right)+a(b-c)=0\iff$ $(b-c)(a+b+c)=0\iff b=c\iff AC=AB\ .$

Proof 2 ("without words"). Construct the points $:\ \left\{\begin{array}{ccccc} F\in AB & \mathrm{so\ that} & A\in (BF) & \mathrm{and} & AF=AB\\\ G\in BC & \mathrm{so\ that} & C\in (BG) & \mathrm{and} & CG=CE\end{array}\right\|\ .$ Observe that $BF=BG$ and $EB=EF=EG$ , i.e. $\triangle ECG$ is

$C$ -isosceles, $\triangle FBG$ is $B$ -isosceles and its circumcenter is the point $E\ .$ In conclusion, $C=m\left(\widehat{CEG}\right)+m\left(\widehat{CGE}\right)=$ $2\cdot m\left(\widehat{CGE}\right)=$ $2\cdot m\left(\widehat{CBE}\right)=B\iff B=C\ .$

P0.10. In $\triangle{ABC}$ bisectoarea unghiului $\widehat{BAC}$ intersecteaza latura $(BC)$ in punctul $D\ .$ Aratati ca $AD^2<bc$ si $AM^2\ge s(s-a)\ ,$ unde $M$ este mijlocul lui $[BC]\ .$

Dem. Notam cel de-al doilea punct $S$ unde bisectoarea $[AD$ taie cercul circumscris $w$ al triunghiului dat. Se observa ca $\triangle ABS\sim\triangle ADC\ ,$ adica $\frac {c}{AD}=\frac {AS}{b}$ de unde obtinem

$AS\cdot AD=bc\ ,$ adica $AD\cdot (AD+DS)=bc\ .$ Folosind puterea punctului $D$ in raport cu cercul $w\ ,$ adica $DA\cdot DS=DB\cdot DC\ ,$ obtinem ca $\boxed{AD^2=bc-DB\cdot DC}<bc\ .$

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P1. Find the value of the sum $S=\frac 1{1+x+xy }+\frac 1{1+y+yz}+\frac 1{1+z+zx}$ where $xyz=1\ .$

Proof. $\left\{\begin{array}{ccccccc} \frac 1{1+y+yz} & = & \frac x{x(1+y+yz)} & = &\frac x{x+xy+\underbrace{\text{xyz}}_{\text{1}}} & = & \frac x{1+x+xy}\\\\ \frac 1{1+z+zx} & = & \frac {xy}{xy(1+z+zx)} & = & \frac {xy}{xy+\underbrace{\text{xyz}}_{\text{1}}+\underbrace{\text{xyz}}_{\text{1}}\cdot x} & = & \frac {xy}{1+x+xy}\end{array}\right\|\bigoplus\implies$ $S=\frac 1{1+x+xy}+\frac x{1+x+xy}+\frac {xy}{1+x+xy}=\frac {1+x+xy}{1+x+xy}\implies S=1\ .$

P2. Prove two implications over $\mathbb R\ :\ \left\{\begin{array}{ccc} \begin{array}{cccc} b(a+b)\le c(a+b) & (1) & \searrow\\\\ c(b+c)\le a(b+c) & (2) & \nearrow\end{array} & \implies & a(c+a)\ge b(c+a)\ (3)\\\\ \begin{array}{cccc} a^2+b^2\le c(a+b) & (5) & \searrow\\\\ a^2+c^2\le b(a+c) & (6) & \nearrow\end{array} & \implies & b^2+c^2\ge a(b+c)\ (7)\end{array}\right\|$

Proof. Suppose a.a.r. (against all reason) $a(c+a)< b(c+a)\ (4)\ .$ Then $\left\{\begin{array}{cccc} b(a+b) & \le & c(a+b) & (1)\\\\ c(b+c) & \le & a(b+c) & (2)\\\\ a(c+a) & < & b(c+a) & (4)\end{array}\right\|\ \bigoplus\ \implies a^2+b^2+c^2<ab+bc+ca$ is falsely $\implies (3)$ is truly.

Suppose a.a.r. (against all reason) $b^2+c^2< a(b+c)\ (8)\ .$ Then $\left\{\begin{array}{cccc} a^2+b^2 & \le & c(a+b) & (5)\\\\ a^2+c^2 & \le & b(a+c) & (6)\\\\ b^2+c^2 & < & a(b+c) & (8)\end{array}\right\|\ \bigoplus\ \implies a^2+b^2+c^2<ab+bc+ca$ is falsely $\implies (7)$ is truly.

P3. Let numbers $\{a,b,c,d\}\subset\mathbb R^*_+$ so that $ab=1$ and $ac+bd=2\ .$ Prove that $cd\le 1$ (contest in Hungary, 2015).

Proof. $2=ac+bd\ge 2\sqrt{abcd}=2\sqrt {cd}\implies cd\le 1\ .$

P4. Prove that $\{x,y,z\}\subset\mathbb R$ and $\left\{\begin{array}{ccc} 2x+y+z & = & 0\\\\ x(y+z)+y(z-y) & = & 0\end{array}\right\|\ \implies\ x=y=z=0\ .$

Proof. $\left\{\begin{array}{ccc} x+y+z & = & -x\\\\ xy+yz+zx & = & y^2\end{array}\right\|\ \implies$ $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\implies$ $x^2=x^2+y^2+z^2+2y^2\implies$ $3y^2+z^2=0\implies x=y=z=0\ .$

P5. Solve over $\mathbb R$ the inequation $1<\frac {2x+1}{3x-2}<3\ .$

Proof. $x\ne\frac 23$ and $1<\frac {2x+1}{3x-2}<3\iff$ $\left(\frac {2x+1}{3x-2}-1\right)\left(\frac {2x+1}{3x-2}-3\right)<0\iff$ $(-x+3)(-7x+7)<0\iff$ $(x-3)(x-1)<0\iff$ $x\in \left(1,\frac 23\right)\cup\left(\frac 23,3\right)\ .$

P6. Solve over $\mathbb R$ the equations $:\ \left\{\begin{array}{cccccc} (1) & \left|x^2+x-2\right| & + & \left|x^2+x-6\right| & = & 4\\\\\ (2) & \left|x^2-x-2\right| & - & \left|x^2-9\right| & = & |x-7|\end{array}\right\|\ .$

Proof. $x\in [a,b]\cup [b,a]\iff |x-a|+|x-b|=|a-b|$ and $\left\{\begin{array}{ccccccccccc} |a|+|b| & = & |a+b| & \iff & ab\ge 0 & ; & |a|+|b| & = & |a-b| & \iff & ab\le 0\\\\ |a|-|b| & = & |a-b| & \iff & b(a-b)\ge 0 & ; & |a|-|b| & = & |a+b| & \iff & b(a+b)\le 0\end{array}\right\|\ (*)$

$\begin{array}{cccccccccc} \blacktriangleright\ (1)\ : & \odot\begin{array}{ccccc} \nearrow & x^2+x-2 & = & a & \searrow\\\\ \searrow & x^2+x-6 & = & b & \nearrow\end{array}\odot & \implies & |a|+|b|=|a-b| & \stackrel{(*)}{\iff} & ab\le 0 & \iff & \left(x^2+x-2\right)\left(x^2+x-6\right)\le 0 & \iff & x\in [-3,-2]\cup [1,2]\ .\\\\ \blacktriangleright\ (2)\ : & \odot\begin{array}{ccccc} \nearrow & x^2-x-2 & = & a & \searrow\\\\ \searrow & x^2-9 & = & b & \nearrow\end{array}\odot & \implies & |a|-|b|=|a-b| & \stackrel{(*)}{\iff} & b(a-b)\ge 0 & \iff & \left(x^2-9\right)\left(7-x\right)\ge 0 & \iff & x\in (-\infty ,-3]\cup [3,7]\ .\end{array}$

P7. Denote $\triangle ABC$ with $A=120^{\circ}$ and the incircle $w=\mathbb C(I,r)$ . Denote $\left\{\begin{array}{ccc} D & \in & (BC)\cap AI\\\\ E & \in & (CA)\cap BI\\\\ F & \in & (AB)\cap CI\end{array}\right\|\ .$ Prove that $DE\perp DF$ .

Proof. $\left|\begin{array}{cc} X\in DF\cap AC\implies & \widehat {BAX}\equiv\widehat {BAD}\implies\mathrm{[AB\ is\ bisector\ of\ \widehat{DAX}}\implies\mathrm{F\ is\ C-excenter\ of\ \triangle ADC}\implies\mathrm{[DF\ is\ bisector\ of\ \widehat{ADB}}\\\\ Y\in DE\cap AB\implies & \widehat {CAY}\equiv\widehat {CAD}\implies\mathrm{[AC\ is\ bisector\ of\ \widehat{DAY}}\implies\mathrm{E\ is\ B-excenter\ of\ \triangle ADB}\implies\mathrm{[DE\ is\ bisector\ of\ \widehat{ADC}}\end{array}\right|$ $\implies DE\perp DF$

P8. Let an $A$ -right-angled $\triangle ABC$ with $B=54^{\circ}$ . Denote $\left\{\begin{array}{ccc} M\in (BC) & ; & MB=MC\\\\ E\in (AM) & ; & \widehat{ECA}\equiv\widehat{ECB}\end{array}\right\|$ . Prove that $CE=AB$ .

Proof 1 (own). Let $N\in (BM)$ so that $AN=AM$ . Prove easily that $\triangle CEM\equiv \triangle ABN$ from where get $CE=AB$ .

Proof 2 (own). Let $P\in (MC)$ so that $AP=AB$ . Prove easily that $\triangle ACE\equiv \triangle CAP$ from where get $CE=AP\implies CE=AB$ .

P9. Let a convex $ABCD$ and the incircles $w_a$ , $w_C$ of $\triangle ABD$ , $\triangle CDB$ . Prove that $ABCD$ is circumscribed $\iff BD$ is tangent to $w_a$ , $w_c$ at the same point.

Proof. $\left\{\begin{array}{c} E\in w_a\cap BD\implies BE=AB+BD-AD\\\\ F\in w_c\cap BD\implies BF= CB+BD-CD\end{array}\right\|$ . Thus, $BE=BF\iff AB-AD=CB-CD\iff$ $AB+CD=AD+BC\iff$ $ABCD$ is tangential
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P10. Let $ABC$ be a triangle and the tangent points $E$ , $F$ of the incircle $w=C(I,r)$ with $AC$ , $AB$ .

Let the intersections $P$ , $Q$ of the line $EF$ with $BI$ , $CI$ . Prove that $PB\perp PC$ si $QC\perp QB$ .

Proof. $m(\widehat{PEC})=m(\widehat{AEF})=$ $90-\frac A2=$ $m(\widehat{PIC}) \implies$ $m(\widehat{PEC})=$ $m(\widehat{PIC})$ $\implies$ $ICPE$ is cyclically $\implies$ $m(\widehat{CPI})=$

$m\widehat{<CEI})=90$ $\implies$ $PC\perp PI\implies PC\perp PB$ . Prove analogously $QC\perp QB$ . Observe that $BCPQ$ is inscribed in the circle with the diameter $[BC]$ .

P11. Let $\triangle ABC$ with the incircle $w=\mathbb C(I,r)\ .$ Prove that $IB\cdot IC=r\cdot IA \iff b+c=3a\iff IG\perp BC\ .$

Prove easily that the first equivalence. Indeed, $r\cdot BC=2\cdot [BIC]=IB\cdot IC\cdot\sin\left(\frac {\pi}2+\frac A2\right)\iff$ $ar=IB\cdot IC\cdot\cos\frac A2$ . Therefore, $IB\cdot IC=r\cdot IA\iff$ $a=IA\cos\frac A2$

$\iff$ $a=(s-a)\iff$ $2a=b+c-a\iff b+c=3a\ .$ Hence $\boxed{IB\cdot IC=r\cdot IA\iff b+c=3a}\ .$ For the second equivalence I"ll present three proofs.

Proof 1. Let the midpoint $M$ of $[BC]$ , the projections $D$ , $U$ , $V$ of $A$ , $I$ , $G$ respectively on $BC$ and $R\in AD\cap MI$ . Prove easily that $AR=r$ . So $IG\perp BC\iff$ $U\equiv V\iff$

$\frac {MV}{MD}=\frac {MU}{MD}\iff$ $\frac {MG}{MA}=\frac {IU}{RD}\iff$ $\frac 13=\frac r{h_a-r}\iff$ $h_a=4r\iff$ $ah_a=4ar\iff$ $2sr=4ar\iff b+c=3a$ . In conclusion, $\boxed{IG\perp BC\iff b+c=3a}\ .$

Proof 2. $IB^2-IC^2=\frac {ac(s-b)}s-\frac {ab(s-c)}s=a(c-b)$ , i.e. $\boxed{IB^2-IC^2=a(c-b)}$ and $GB^2-GC^2=\frac 49\cdot\left(m_b^2-m_c^2\right)=$ $\frac{c^2-b^2}3\implies$ $\boxed{GB^2-GC^2=\frac {c^2-b^2}3}\ .$

In conclusion, $IG\perp BC\iff$ $IB^2-IC^2=GB^2-GC^2\iff$ $3a(c-b)=c^2-b^2\iff$ $3a=b+c\ .$ Hence $\boxed{IG\perp BC\iff b+c=3a}\ .$

Proof 3. Denote the midpoint $M$ of $[BC]$ , the projection $D$ of $A$ on $BC$ , i.e. $\left\{\begin{array}{ccc} AD & = & h_a\\\\ MD & = & \frac {\left|b^2-c^2\right|}{2a}\end{array}\right\|$ and the distance $\delta (X)$ of the point $X$ to the $A$ -altitude $AD\ .$

Thus, $IG\perp BC\iff$ $IG\parallel AD\iff$ $\delta(I)=\delta (G)\ .$ But the distance $\delta(G)=\frac 23\cdot MD\implies$ $\boxed{\delta (G)=\frac {\left|b^2-c^2\right|}{3a}}$ and the distance $\delta (I)=IA\sin\frac {|B-C|}2=$

$\frac {(s-a)\sin\frac {|B-C|}2}{\cos\frac A2}=$ $\frac {2(s-a)\sin\frac {|B-C|}2\sin\frac A2}{2\cos\frac A2\sin\frac A2}=$ $\frac {(s-a)|\sin B-\sin C|}{\sin A}=$ $\frac {(s-a)|b-c|}a\implies$ $\boxed{\delta (I)=\frac {(s-a)|b-c|}a}\ .$ In conclusion, $IG\perp BC\iff$

$\delta (G)=\delta (I)\iff$ $\frac {\left|b^2-c^2\right|}{3a}=\frac {(s-a)|b-c|}a\iff$ $\frac {b+c}3=s-a\iff$ $b+c=3(s-a)\iff 2(b+c)=3(b+c-a)\iff$ $b+c=3a\ .$

Remark. Prove easily that $\boxed{IG\parallel BC\iff b+c=2a}\ .$ Indeed, $IG\parallel BC\iff \delta_{BC}(G)=\delta_{BC}(I)\iff$ $\frac{h_a}3=r\iff$ $ah_a=3ar\iff$ $2sr=3ar\iff$ $b+c=2a\ .$

P12. Solve the trigonometrical equation $\sin\ \left(\ 45^{\circ}\ +\ x\ \right)\ \cdot\ \sin\ 15^{\circ}=\sin\ x\ \cdot\ \sin\ 30^{\circ}$ , where $x\ \in\ \left(\ 0\ ,\ \frac {\pi}{2}\ \right)$ .

Proof. $\sin\ \left(\ 45^{\circ}\ +\ x\ \right)\ \cdot\ \sin\ 15^{\circ}=\sin\ x\ \cdot\ \sin\ 30^{\circ}\Longleftrightarrow$ $\sin(45+x)=2\sin x\cos 15\Longleftrightarrow$ $\sin(45+x)=\sin (x+15)+\sin (x-15)\Longleftrightarrow$

$\sin(45^{\circ}+x)-\sin (x-15^{\circ})=\sin (x+15^{\circ})\Longleftrightarrow$ $2\sin 30^{\circ}\cos (x+15^{\circ})=\sin (x+15^{\circ})\Longleftrightarrow$ $\tan (x+15^{\circ})=1\Longleftrightarrow x+15^{\circ}=45^{\circ}\Longleftrightarrow x=30^{\circ}$ .

P13. Let $\triangle ABC$ with the area $S=[ABC]$ , the incircle $w=\mathbb (I,r)$ and the excircles $w_a=\mathbb C\left(I_a,r_a\right)$ , $w_b=\mathbb C\left(I_b,r_b\right)$ , $w_c=\mathbb C\left(I_c,r_c\right)$ .

Prove that $S=rs=r_a(s-a)=r_b(s-b)=r_c(s-c)$ and $S^2=rr_ar_br_c=s(s-a)(s-b)(s-c)$ , where $2s=a+b+c$ .

Proof 1. $[ABC]=\left\{\begin{array}{ccccc} [AIB]+[BIC]+[CIA] & = & \frac 12\cdot (cr+ar+br) & \implies & S=rs\\\\ \left[AI_aB\right]+\left[AI_aC\right]-\left[BI_aC\right] & = & \frac 12\cdot \left(cr_a+br_a-ar_a\right) & \implies & S=r_a(s-a)\end{array}\right\|$ $\implies S^2=rr_as(s-a)\ (*)$ . Denote $:$

$\left\{\begin{array}{c} \begin{array}{cccc} U\in AB & \mathrm{and} & IU\perp AB & \searrow\\\\ V\in AB & \mathrm{and} & I_aV\perp AB & \nearrow\end{array}\odot\implies\triangle BUI\sim\triangle I_aVB\implies\frac {BU}{I_aV}=\frac {UI}{VB}\implies\frac {s-b}{r_a}=\frac {r}{s-c}\implies rr_a=(s-b)(s-c)\end{array}\right\|\stackrel{(*)}{\implies}$ $S^2=s(s-a)(s-b)(s-c)\ .$

Proof 2. $[ABC]=\left\{\begin{array}{ccccc} \left[BI_bC\right]+\left[BI_bA\right]-\left[CI_bA\right] & = & \frac 12\cdot \left(ar_b+cr_b-br_b\right) & \implies & S=r_b(s-b)\\\\ \left[CI_cA\right]+\left[CI_cB\right]-\left[AI_cB\right] & = & \frac 12\cdot \left(br_c+ar_c-cr_c\right) & \implies & S=r_c(s-c)\end{array}\right\|$ $\implies S^2=r_br_c(s-b)(s-c)\ (*)$ . Denote $:$

$\left\{\begin{array}{c} \begin{array}{cccc} X\in BC & \mathrm{and} & I_cX\perp BC & \searrow\\\\ Y\in BC & \mathrm{and} & I_bY\perp BC & \nearrow\end{array}\odot\implies\triangle BXI_c\sim\triangle I_bYB\implies\frac {BX}{I_bY}=\frac {XI_c}{YB}\implies\frac {s-a}{r_b}=\frac {r_c}{s}\implies r_br_c=s(s-a)\end{array}\right\|\stackrel{(*)}{\implies}$ $S^2=s(s-a)(s-b)(s-c)\ .$

Remark. $A=90^{\circ}\iff$ $\tan\frac A2=1\iff$ $\sqrt{\frac {(s-b)(s-c)}{s(s-a)}}=1\iff$ $S=s(s-a)=(s-b)(s-c)=rr_a=r_br_c\ .$

P14. Prove the trigonometrical identity $\sin (x+y)=\sin x\cos y+\sin y\cos x$ for any $\{x,y\}\subset \left(0,\frac{\pi}2\right)$ using only the Ptolemy's theorem.

Proof. Let a convex $ABCD$ what is inscribed in the circle $w=\mathbb C(O,R)$ with the diameter $[AC]$ for which exist $\{x,y\}\subset \left(0,\frac{\pi}2\right)$ so that $\left\{\begin{array}{ccc} m\left(\widehat{ABD}\right) & = & x\\\\ m\left(\widehat{CBD}\right) & = & y\end{array}\right\|\ .$

The Ptolemy's theorem is $\boxed{ac+bd=ef}\ (*)\ ,$ where $\left\{\begin{array}{ccccccc} a=AB=2R\cos x & ; & b=BC=2R\cos y & ; & c=CD=2R\sin y\\\\ d=DA=2R\sin x & ; & e=AC=2R\sin (x+y) & ; & f=BD=2R\end{array}\right\|\ .$ In conclusion,

the relation $(*)$ becomes $4R^2\cos x\sin y+4R^2\cos y\sin x=4R^2\sin (x+y)\ ,$ i.e. the required trigonometrical identity $\sin (x+y)=\sin x\cos y+\sin y\cos x\ .$

P15. Let $\triangle ABC$ with $:\ \left\{\begin{array}{cccc} \mathrm{incircle} & w & = &\mathbb C\left(I,r\right)\\\\ \mathrm{A-excircle} & w_a & = & \mathbb C\left(I_a,r_a\right)\end{array}\right\|\ ;$ $\left\{\begin{array}{ccc} M\in (BC) & , & MB=MC\\\\ D\in BC & , & AD\perp BC\end{array}\right\|\ ;$ $\left\{\begin{array}{c} E\in AD\cap MI\\\\ F\in AD\cap MI_a\end{array}\right\|$ . Prove that $\left\{\begin{array}{ccc} AE & = & r\\\\ AF & = & r_a\end{array}\right\|\ .$

Proof. Suppose w.l.o.g. $b>c$ . Let $AD=h_a$ and $\left\{\begin{array}{ccc} S\in BC\cap w & ; & AE=x\\\\ T\in BC\cap w_a & ; & AF=y\end{array}\right\|$ . Is well-known that $MS=MT=\frac {b-c}2$ and $MD=\frac {b^2-c^2}{2a}$ . Therefore $:$

$\blacktriangleright\ DE\parallel IS\iff$ $\frac {DE}{IS}=\frac {MD}{MS}\iff$ $\frac {h_a-x}{r}=\frac {\frac {b^2-c^2}{2a}}{\frac {b-c}2}=\frac {b+c}a\iff$ $ah_a-ax=r(b+c)\iff$ $(b+c+a)r-ax=r(b+c)\iff x=r$ .

$\blacktriangleright\ DF\parallel I_aT\iff$ $\frac {DF}{I_aT}=\frac {MD}{MT}\iff$ $\frac {h_a+y}{r_a}=\frac {\frac {b^2-c^2}{2a}}{\frac {b-c}2}=\frac {b+c}a\iff$ $ah_a+ay=r_a(b+c)\iff$ $(b+c-a)r_a+ay=r_a(b+c)\iff y=r_a$ .

P16. $(\forall )\ \{x,y,z\}\cap \pi \mathbb Z=\emptyset\ ,\ \cot\frac x2\cot\frac y2+\cot\frac y2\cot\frac z2+\cot\frac z2\cot\frac x2=\tan\frac x2\tan\frac y2+\tan\frac y2\tan\frac z2+\tan\frac z2\tan\frac x2$ $\iff \sin (x+y)+\sin (y+z)+\sin (z+x)=0\ .$

Proof. I"ll use an simple and well-known identities $:\ \tan\frac x2=\frac {1-\cos x}{\sin x}$ and $\cot\frac x2=\frac {1+\cos x}{\sin x}\ .$ Therefore, $\cot\frac y2\cot\frac z2-\tan\frac y2\tan\frac z2=$ $\frac {1+\cos y}{\sin y}\cdot \frac {1+\cos z}{\sin z} -$

$\frac {1-\cos y}{\sin y}\cdot \frac {1-\cos z}{\sin z}=$ $\frac{2(\cos y+\cos z)}{\sin y\sin z}\implies$ $\cot\frac y2\cot\frac z2-\tan\frac y2\tan\frac z2= \frac{2(\cos y+\cos z)}{\sin y\sin z}\implies$ $\sum\left(\cot\frac y2\cot\frac z2-\tan\frac y2\tan\frac z2\right)=2\cdot \sum \frac{\cos y+\cos z}{\sin y\sin z}=$

$\frac{\sin x(\cos y+\cos z)+\sin y(\cos z+\cos x)+\sin z(\cos x+\cos y)}{\sin x\sin y\sin z}=$ $\frac {(\sin x\cos y+\sin y\cos x)+(\sin y\cos z+\sin z\cos x)+(\sin z\cos x+\sin x\cos z)}{\sin x\sin y\sin z}=$

$\frac {\sin (x+y)+\sin (y+z)+\sin (z+x)}{\sin x\sin y\sin z}\ .$ Hence $\sum\left(\cot\frac y2\cot\frac z2-\tan\frac y2\tan\frac z2\right)=\frac {\sin (x+y)+\sin (y+z)+\sin (z+x)}{\sin x\sin y\sin z}\ .$

In conclusion, $\sum\cot\frac y2\cot\frac z2=\sum\tan\frac y2\tan\frac z2\iff$ $\sin (x+y)+\sin (y+z)+\sin (z+x)=0\ .$

P17. $(\forall )\ \{x,y,z\}\subset\left(0,\frac {\pi}2\right)\ ,\ \tan B\tan C+\tan C\tan A+\tan A\tan B\ \ge\ \frac 1{\cos A}+\frac 1{\cos B}+\frac 1{\cos C}+3\ .$

Proof. $\tan A+\tan B+\tan C=\tan A\tan B\tan C\iff$ $\boxed{\tan B\tan C=1+\frac{\tan B+\tan C}{\tan A}}\ (1)\ .$ Observe that $\frac {\tan B+\tan C}{\tan A}=$ $\frac {\frac {\sin (B+C)}{\cos B\cos C}}{\frac{\sin A}{\cos A}}=$ $\frac {\sin A\cos A}{\cos B\cos C\sin A}=$

$\frac {\cos A}{\cos B\cos C}\implies$ $\boxed{\frac {\tan B+\tan C}{\tan A}=\frac {\cos A}{\cos B\cos C}}\ (2)\ .$ Therefore, $\sum \tan B\tan C\ \stackrel{(1\wedge 2)}{=}\ 3+\sum \frac {\cos A}{\cos B\cos C}=$ $3+ \frac {\sum \cos^2A}{\prod \cos A}\ge$ $3+\frac {\sum\cos B\cos C}{\prod \cos A}=3+\sum \frac 1{\cos A}$

P18. $(\exists )\ \{m,n\}\subset\mathbb C$ so that $a^4+ma^3+na=b^4+mb^3+nb=c^4+mc^3+nc=1\ \implies\ ab+bc+ca=\frac 1{ab}+\frac 1{bc}+\frac 1{ac}\ .$

Proof $\left\{\begin{array}{ccc} a^4+ma^3+na-1 & = & 0\\\\ b^4+mb^3+nb-1 & = & 0\\\\ c^4+mc^3+nc-1 & = & 0\end{array}\right\|\implies$ $x^4+mx^3+nx-1=0\begin{array}{ccccc} \nearrow & a & \searrow\\\\ \rightarrow & b & \rightarrow\\\\ \searrow & c & \nearrow\end{array}\odot\implies$ $(\exists )\ d$ so that $\boxed{abcd=-1\ \wedge\ ab+cd+(a+b)(c+d)=0}\implies$

$d=-\frac 1{abc}$ and $ab-c\cdot\frac 1{abc}+(a+b)\left(c-\frac 1{abc}\right)=0\implies$ $ab-\frac 1{ab}+c(a+b)-\frac 1{bc}-\frac 1{ac}=0\implies$ $ab+bc+ca=\frac 1{ab}+\frac 1{bc}+\frac 1{ac}\ .$

P19. Prove that in any triangle $ABC$ there is the inequality $\sum\frac {(a+b)^2}{a+b-c}\le$ $\frac {2R}r\cdot (a+b+c)\ .$ (standard notation).

Proof. I"ll use the well known identities $:\ \left\{\begin{array}{ccc} a+b=4R\cos\frac C2\cos\frac {A-B}2 & ; & a+b-c=8R\cos\frac C2\sin\frac A2\sin\frac B2\\\\ \sin A+\sin B+\sin C=\frac sR & ; & \sin\frac A2\sin\frac B2\sin\frac C2=\frac r{4R}\end{array}\right\|$ . Thus, $\sum {\frac{(a+b)^2}{a+b-c}}=$

$\sum {\frac{16R^2\cos^2\frac{C}{2}\cos^2\frac{A-B}{2}}{8R\cos\frac{C}{2}\sin\frac{A}{2}\sin\frac{B}{2}}}\le$ $2R\cdot \sum \frac{\cos\frac{C}{2}}{\sin\frac{A}{2}\sin\frac{B}{2}}=$ $2R\cdot \frac{\sum {\sin\frac{A}{2}\cos\frac{A}{2}}}{\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}=$ $R\cdot \frac{\sin A+\sin B+\sin C}{\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}=$ $\frac{2R}{r}\cdot(a+b+c)\ .$

P20. Prove that the equivalence $\boxed{1+\cos B+\cos C=\frac {r_a}R\ \iff 2R+r=r_a\ \iff\ r_b+r_c=2R\ \iff\ A=90^{\circ}}\ .$

Proof 1. I"ll use the identities $:\ \left\{\begin{array}{cccc} \cos A+\cos B+\cos C & = & 1+\frac rR & (1)\\\\ s(s-a)+(s-b)(s-c) & = & bc & (2)\\\\ s(s-a)(s-b)(s-c) & = & S & (3)\end{array}\right\|$ . Thus, $1+\cos B+\cos C=\frac {r_a}R\ \stackrel{(1)}{\iff}\ 1+\left(1+\frac rR\right)-\cos A=\frac {r_a}R\iff$

$1+(1-\cos A)=\frac {r_a-r}R\iff$ $1+2\sin^2\frac A2=\frac {r_a-r}R\iff$ $1+\frac {2(s-b)(s-c)}{bc}=\frac {\frac S{s-a}-\frac Ss}R\iff$ $bc+2(s-b)(s-c)=\frac {abcS}{Rs(s-a)}\ \stackrel{(3)}{\iff}$

$bc+2(s-b)(s-c)=$ $4(s-b)(s-c)\iff$ $\boxed{bc=2(s-b)(s-c)}\ (4)$ $\iff$ $2bc=a^2-(b-c)^2\iff$ $a^2=b^2+c^2\ .$ Otherwise. If I"ll use the

relation $(2)$ , then the relation $(4)$ becomes $:\ s(s-a)+(s-b)(s-c)=2(s-b)(s-c)$ $\iff$ $\tan^2\frac A2=$ $\frac {(s-b)(s-c)}{s(s-a)}=1\iff$ $A=90^{\circ}\ .$

Proof 2. Denote the diameter $[NS]$ of the circumcircle $w=\mathbb C(O,R)$ for which the midpoint $M$ of $[BC]$ belongs to $NS$ and $BC$ separates $A$ , $S$ (N-north, S-south). Are well-known the

relations $\left\{\begin{array}{ccc} MS & = & \frac {r_a-r}2\\\\ MN & = & \frac {r_b+r_c}2\\\\ 2R & = & \frac {r_a-r}2+\frac {r_b+r_c}2\end{array}\right\|\ .$ Thus, $1+\cos B+\cos C=\frac {r_a}R$ $\iff$ $1+\left(1+\frac rR\right)=$ $\frac {r_a}R+\cos A$ $\iff$ $2R+r=$ $r_a+R\cos A\iff$ $2R+r=$ $r_a+R-\frac {r_a-r}2$

$\iff$ $2R+2r=$ $2r_a-r_a+r\iff$ $\boxed{2R+r=r_a}$ $\iff$ $2R-\frac {r_a-r}2=\frac {r_a-r}2$ $\iff$ $NS-MS=MS\iff$ $MN=MS\iff$ $M\equiv O$ $\iff$ $A=90^{\circ}\ .$

P21. Prove that in any triangle $ABC$ there are the inequalities $\boxed{\frac {a}{r_a}+\frac {b}{r_b}+\frac {c}{r_c}\ge \frac {a+b+c}{2R - r}}\ (*)$ and $\boxed{\frac {a}{R+r_a}+\frac {b}{R+r_b}+\frac {c}{R+r_c}\ge \frac {a+b+c}{3R - r}}\ (**)\ ,$ where $a+b+c=2s\ .$

Proof. I"ll use the well known identity $\sum ar_a=2s(2R-r)\ (*)\ .$ Therefore, $\left\{\begin{array}{ccc} \sum \frac a{r_a}= \sum \frac {a^2}{ar_a}\stackrel{\mathrm{C.B.S.}}{\ge}\frac{(a+b+c)^2}{ar_a+br_b+cr_c}=\frac{2s}{2R-r} & \implies & \boxed{\sum\frac a{r_a}\ge\frac {2s}{2R-r}}\\\\ \sum \frac a{R+r_a}= \sum \frac {a^2}{aR+ar_a}\stackrel{\mathrm{C.B.S.}}{\ge}\frac{4s^2}{2sR+2s(2R-r)}=\frac{2s}{3R-r} & \implies & \boxed{\sum\frac a{R+r_a}\ge\frac {2s}{2R-r}}\end{array}\right\|$

Remark. $\sum\frac a{r_a}=\sum\frac {(s-b)+(s-c)}{r_a}=$ $\sum\left(\tan\frac C2+\tan\frac B2\right)=$ $2\sum\tan\frac A2\ge$ $2\sqrt{3\sum\left(\tan\frac B2\tan\frac C2\right)}=2\sqrt 3$ $\implies$ $\boxed{\sum\frac a{r_a}\ge2\sqrt 3}\ .$

The inequality $s\sqrt 3\le 4R+r$ is well known and $4R+r\le 3(2R-r)\iff$ $R\ge 2r$ what is true. Thus, $s\le \sqrt 3(2R-r)$ $\implies\boxed{\frac {2s}{2R-r}\le 2\sqrt 3\le \sum\frac a{r_a}}\ .$

P22. Let $n\in\mathbb N\ ,\ n\ge 1$ , $a,b\in (0,\infty)$ and $x\in R$ so that $\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b}$ . Prove that $\boxed{\frac{\sin^{2n}x}{a^{n-1}}+\frac{\cos^{2n}x}{b^{n-1}}=\frac {1}{(a+b)^{n-1}}}\ .$

Proof. I"ll use the identities $\left\{\begin{array}{c} 4\sin^4x=(2\sin^2x)^2=(1-\cos 2x)^2\\\\ 4\cos^4x=(2\cos^2x)^2=(1+\cos 2x)^2\end{array}\right\|$ . Thus, $\boxed{\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b}}$ $\Longleftrightarrow$ $b(a+b)(1-\cos 2x)^2+a(a+b)(1+\cos 2x)^2=4ab$ $\Longleftrightarrow$ $(a+b)^2\cos^22x+2(a^2-b^2)\cos 2x+(a-b)^2=0$ $\Longleftrightarrow$ $|(a+b)\cos 2x+(a-b)|=0$ $\Longleftrightarrow$ $\cos 2x=\frac {b-a}{b+a}$ $\implies$ $\left|\begin{array}{c} \sin^2x=\frac {a}{a+b}\\\\ \cos^2x=\frac {b}{a+b}\end{array}\right|$ $\Longrightarrow$ $\frac{\sin^{2n}x}{a^{n-1}}+\frac{\cos^{2n}x}{b^{n-1}}=\frac {1}{(a+b)^{n-1}}$

P23. Let an acute $\triangle ABC$ with the medians $BE$ , $CF$ where $E$ , $F$ are the midpoints of $[AC]$ , $[AB]$ respectively and $BE\perp CF$ . Prove that $\boxed{\tan A\le \frac 34}$ with equality iff $b=c$ .

Proof. Prove easily that $BE\perp CF\ \Longleftrightarrow\ b^2+c^2=5a^2$ . Apply the generalized Pythagoras' theorem : $a^2=b^2+c^2-2bc\cdot\cos A$ $\iff$ $a^2=5a^2-2bc\cdot\cos A$ $\iff$

$\boxed{\cos A=\frac {2\left(b^2+c^2\right)}{5bc}}\ .$ Hence $\tan A$ is $\max .\iff$ $\cos A$ is $\min .\iff$ $\frac {\left(b^2+c^2\right)}{bc}$ is $\min .\iff$ $\frac bc+\frac cb$ is $\min .\iff$ $b=c$ . In conclusion, $\cos A\ge \frac 45\iff\tan A\le\frac 34$ .

Extension. Let an acute $\triangle ABC$ and $\left\{\begin{array}{cc} F\in(AB)\ : & \frac {FA}{FB}=n\\\\ E\in (AC)\ : & \frac {EA}{EC}=m\end{array}\right\|$ where $\{m,n\}\subset\mathbb R^*_+$ and $|m-n|<1\ .$ Prove that $BE\perp CF\implies \tan A\le \frac {m+n+1}{2\sqrt{mn(m+1)(n+1)}}\ .$

Proof. Denote $P\in BE\cap CF$ and $D\in AP\cap BC\ .$ Apply the Ceva's theorem $:\ \frac {DB}{DC}\cdot\frac {EC}{EA}\cdot\frac {FA}{FB}=1\iff$ $\frac {DB}{DC}\cdot \frac nm=1\iff$ $\frac {DB}m=\frac {DC}n=\frac a{m+n}\ .$

Apply the Van Aubel's relation $:\ \left\{\begin{array}{ccccccc} \frac {PB}{PE}=\frac {DB}{DC}+\frac {FB}{FA}=\frac mn+\frac 1n=\frac {m+1}n & \implies & \frac {PB}{PE}=\frac {m+1}n & \implies & \frac {PB}{m+1}=\frac {PE}n=\frac {BE}{m+n+1} & \implies & \boxed{PB=\frac {m+1}{m+n+1}\cdot BE}\\\\ \frac {PC}{PF}=\frac {DC}{DB}+\frac {EC}{EA}=\frac nm+\frac 1m=\frac {n+1}m & \implies & \frac {PC}{PF}=\frac {n+1}m & \implies & \frac {PC}{n+1}=\frac {PF}m=\frac {CF}{m+n+1} & \implies & \boxed{PC=\frac {n+1}{m+n+1}\cdot CF}\end{array}\right\|\ .$

Apply the Stewart's relation to the cevians $BE$ and $CF\ :\ \left\{\begin{array}{ccc} BE^2+\frac {mb^2}{(m+1)^2}=\frac {a^2m}{m+1}+\frac {c^2}{m+1} & \implies & (m+1)^2\cdot BE^2=\left(ma^2+c^2\right)(m+1)-mb^2\\\\ CF^2+\frac {nc^2}{(n+1)^2}=\frac {a^2n}{n+1}+\frac {b^2}{n+1} & \implies & (n+1)^2\cdot CF^2=\left(na^2+b^2\right)(n+1)-nc^2\end{array}\right\|\ .$ Therefore,

$PB\perp PC\iff$ $PB^2+PC^2=BC^2\iff$ $(m+1)^2\cdot BE^2+(n+1)^2\cdot CF^2=(m+n+1)^2\cdot a^2\iff$ $\left(ma^2+c^2\right)(m+1)-mb^2+\left(na^2+b^2\right)(n+1)-nc^2=$

$a^2(m+n+1)^2\iff$ $(n-m+1)b^2+(m-n+1)c^2=(2mn+m+n+1)a^2$ , where $-1<m-n<1$ , i.e. $|m-n|<1\ .$ Apply the generalized Pythagoras' theorem $:$

$2bc\cdot\cos A=b^2+c^2-a^2\iff$ $2bc(2mn+m+n+1)\cdot \cos A=$ $(2mn+m+n+1)\left(b^2+c^2\right)-(2mn+m+n+1)a^2=$ $(2mn+m+n+1)\left(b^2+c^2\right)-$

$(n-m+1)b^2-(m-n+1)c^2=$ $2m(n+1)b^2+2n(m+1)c^2\implies$ $\boxed{\cos A=\frac {m(n+1)b^2+n(m+1)c^2}{bc(2mn+m+n+1)}}\ (*)\ .$ Hence $\tan A$ is $\max .\iff$ $\cos A$ is $\min .$ Observe that

$\cos A\ge \frac {2\sqrt{mn(m+1)(n+1)}}{2mn+m+n+1}$ with equality iff $\frac bc=\sqrt{\frac {n(m+1)}{m(n+1)}}\ .$ In conclusion, $\tan A\le \frac {m+n+1}{2\sqrt{mn(m+1)(n+1)}}\ .$

P24. Let $\triangle ABC$ with the circumcircle $\mathbb C(O,R)$ sand the incircle $C(I,r)\ .$ Prove that $a^2=4r(2R-r)\ \iff\ IG\parallel BC\ \iff\ b+c=2a\ \iff\ \cos A=1-\frac rR\ .$

Proof. $\boxed{IG\parallel BC}\iff h_a=3r\iff$ $ah_a=3ar\iff$ $2sr=3ar\iff$ $2s=3a\iff$ $a+b+c=3a\iff$ $\boxed{b+c=2a}\iff$ $\sin B+\sin C=2\sin A\iff$

$2\sin\frac {B+C}2\cos\frac {B-C}2=2\cdot2\sin\frac A2\cos\frac A2\iff$ $\cos\frac {B-C}2=2\sin\frac A2\iff$ $2\cos\frac {B-C}2\cos\frac {B+C}2=2\cdot 2\sin^2\frac A2$ $\iff$ $\cos B+\cos C=2(1-\cos A)\iff$

$\sum\cos A=2-\cos A\iff$ $1+\frac rR=2-\cos A\iff$ $\boxed{\cos B+\cos C=\frac {2r}R\ \ \wedge\ \ \cos A=1-\frac rR}\iff$ $a^2=(2R\sin A)^2=4R^2\left(1-\cos^2A\right)\iff$

$a^2=4R^2\left[1-\left(1-\frac rR\right)^2\right]\iff$ $a^2=4R^2-4(R-r)^2\iff$ $\boxed{a^2=4r(2R-r)}\ .$

Lema. Sa se arate ca in $\triangle ABC$ avem $B=2\cdot C\ \Longleftrightarrow \ a=c\ \ \vee\ \ b^2=c(c+a)\ .$

P25. Fie $\triangle ABC$ si $M\in (BC)$ astfel incat $m\left(\widehat {AMB}\right)=\frac B2\ .$ Sa se arate ca $MB=c$ sau $a\cdot MC+c(a+c)=b^2\ .$ In particular, $MC=\frac a4\implies$ $a=2(b-c)\ .$

Dem. Aplicam relatia Stewart cevienei $[AM$ in $\triangle ABC\ :$ $AM^2\cdot a+MB\cdot MC\cdot a=c^2\cdot MC+b^2\cdot MB\ (*)\ .$ Folosind lema precedenta in $\triangle ABM$ si se obtine

$MB=c\ \ \vee\ \ AM^2=c(c+MB)\ .$ Daca $AM^2=c(c+MB)\ ,$ atunci tinand seama de relatia $(*)$ obtinem $ac^2+ac(a-MC)+a(a-MC)MC=$

$c^2MC+b^2(a-MC)\Longleftrightarrow$ $a\cdot MC^2+\left[\left(c^2+ac-b^2\right)-a^2\right]$ $\cdot MC-a\left(c^2+ac-b^2\right)=0$ $\Longleftrightarrow$ $\left(MC-a\right)\left[aMC+c(a+c)-b^2\right]=0$ $\Longleftrightarrow$

$MC=a\ \ \vee\ \ a\cdot MC+c(a+c)=b^2\Longleftrightarrow$ $a\cdot MC+c(a+c)=b^2\ .$

P26. Consideram cinci puncte diferite intre ele $\{M,A,I,B,N\}$ astfel incat $\left\|\begin{array}{ccc} IA & = & IB\\\\ AM & \perp & AI\\\\ BN & \perp & BI\end{array}\right\|\ .$ Sa se arate ca $\boxed{\ IM\perp AN\ \Longleftrightarrow\ AM^2+BN^2=MN^2\ \Longleftrightarrow\ IN\perp BM\ }\ .$

Demonstratie. $\boxed{IM\perp AN}\iff$ $IN^2+AM^2=IA^2+MN^2\iff$ $IN^2-IA^2+AM^2=MN^2\iff$ $IN^2-IB^2+AM^2=MN^2$

$\iff$ $\boxed{AM^2+BN^2=MN^2}\iff$ $IM^2-IA^2+BN^2=MN^2\iff$ $MI^2+BN^2=IB^2+MN^2\iff$ $\boxed{IN\perp BM}\ .$

P27. Sa se rezolve ecuatiile peste $\mathbb R\ :\ 1\ \odot\ \ x^4+4x=1\ \ ;\ \ 2\ \odot\ \ x^4+4x^3+6x^2+8x+4=0\ .$

Proof.

$1\ \odot\ x^4+4x=1\iff x^4+2x^2+1-2x^2+4x-2=0\iff (x^2+1)^2-2(x^2-2x+1)=0\iff (x^2+1)^2-\left( (x-1)\sqrt 2\right)^2=0$

$\iff (x^2+x\sqrt 2+1-\sqrt 2)(x^2-x\sqrt 2+1+\sqrt 2)=0$ cu solutiile reale $\boxed{x_{1,2}=-\frac{1}{\sqrt 2}\ \pm\ \sqrt{\frac 12(2\sqrt 2-1)}}\ .$

$2\ \odot\ x^4+4x^3+6x^2+8x+4=0\iff x^4+4x^2+4+4x^3+4x^2+8x-2x^2=0\iff (x^2+2x+2)^2-2x^2=0$

$\iff (x^2+x(2-\sqrt 2)+2)(x^2+x(2+\sqrt 2)+2)=0$ cu solutiile reale $\boxed{x_{1,2}=-1-\frac{1}{\sqrt 2}\ \pm\ \sqrt{\frac12(2\sqrt 2-1)}}\ .$

$\mathrm{END}$

This post has been edited 532 times. Last edited by Virgil Nicula, Jun 18, 2016, 6:46 PM

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462. Diverse probleme de geometrie.

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

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449. Working page I.

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392. Art of Problem Solving (geometry).

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390. Probleme de analiza matematica.

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387. Algebra (II).

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372. The distancies : OI , OI_a , ON a.s.o.

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309. Beautiful problem ! IRAN Selection Test for IMO, 2004.

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307. Very nice proofs !

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234. Some simple and nice problems from contests.

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

February 2011

232. Some geometrical inequalities (own).

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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.

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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 !

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209. Nice problem, nice proof !

208. An interesting geometrical inequality.

207. Some interesting problems of Toshio Seimiya, Japan.

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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.

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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).

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.

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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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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

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

by Virgil Nicula, Jul 20, 2010, 4:58 PM

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