234. Some simple and nice problems from contests.

by Virgil Nicula, Mar 2, 2011, 8:03 AM

PP1. Let $a$ , $b$ be two real positive numbers for which $a^4\le b^3$ . Prove that doesn't exist $x\in \mathbb R$ so that $x^4+b=ax$ .

Proof. Denote $f(x)=x^4-ax+b$ , $x\in\mathbb R$ . Observe that $f(0)=b>0$ and $f(x)\ne 0$ for any $x\in \mathbb R\ \implies\ f(x)$ has

a constant sign, i.e. $f(x)>0$ for any $x\in\mathbb R$ . With other words, in the mentioned hypothesis is sufficient to prove $f>0$ ,

i.e $x\in\mathbb{R}\ \Longrightarrow\ x^4+b>ax$ . Observe easily that $\left\{\begin{array}{ccc} (\forall )\ x\ge \sqrt[3] a & \Longrightarrow & x^4-ax+b=x\left(x^3-a\right)+b>0\\\\ (\forall )\ x\le \frac ba & \Longrightarrow & x^4-ax+b=x^4+(-ax+b)>0\end{array}\right\|$ $\implies$

$f(x)>0$ for any $x\in \left(-\infty , \frac ba\right)\cup\left(\sqrt [3]a,\infty\right)$ . Since $a^4\le b^3\Longleftrightarrow \sqrt[3] a\le \frac ba$ obtain in conclusion that $x^4-ax+b>0$ , $(\forall )\ x\in\mathbb R$ .


PP2. Ascertain the geometrical locus $\Lambda_k$ of the mobile interior point $L$ w.r.t. acute given triangle $ABC$ for which $\frac {\delta_{AB}(L)+\delta_{AC}(L)}{\delta_{BC}(L)}=k>0$ (constant) .

Show that the geometrical locus $\Lambda_k$ pass through a fixed point $F$ for any value of the constant $k$ . I denoted the distance $\delta_d(X)$ of the point $X$ to the line $d$ .


PP3. Prove that for the functions $f:[0,1]\rightarrow\mathbb R$ and $g:[0,1]\rightarrow\mathbb R$ exists $\{a,b\}\subset [0,1]$ so that $\left|f(a)+g(b)-ab\right|\ge\frac 14$ .

Proof. Suppose that $\left|f(a)+g(b)-ab\right|\ge\frac 14$ for any $\{a,b\}\subset [0,1]$ . Therefore,

$\left\{\begin{array}{ccc} a=0\ \wedge\ b=1 & \implies & -\frac 14<f(0)+g(1)<\frac 14\\\\ a=1\ \wedge\ b=0 & \implies & -\frac 14<f(1)+g(0)<\frac 14\end{array}\right\|\bigoplus\implies$ $-\frac 12<f(0)+f(1)+g(0)+g(1)<\frac 12\ \ \ (1)$ .

$\left\{\begin{array}{ccc} a=0\ \wedge\ b=0 & \implies & -\frac 14<f(0)+g(0)<\frac 14\\\\ a=1\ \wedge\ b=1 & \implies & \frac 34\ <f(1)+g(1)<\frac 54\end{array}\right\|\bigoplus\ \implies\ \frac 12<f(0)$ $+f(1)+g(0)+g(1)<\frac 32\ \ \ (2)$ .

From the relations $(1)$ , $(2)$ obtain that $-\frac 12<f(0)+f(1)+g(0)+g(1)<\frac 12<f(0)+f(1)+g(0)+g(1)<\frac 32$ , what is falsely.


PP4. Construct outside of the triangle $ABC$ the rhombus $ABDE$ , $ACFG$ . Denote $\left\{\begin{array}{c} I\in CD\cap BF\\\ H\in AB\cap CD\\\ J\in AC\cap BF\end{array}\right\|$ . Prove that $[AHIJ]=[BIC]$ .

Proof. $\left\{\begin{array}{c} AC=b\ ;\ AB=c\\\ AJ=x\ ;\ AH=y\end{array}\right\|$ $\implies$ $\left\{\begin{array}{ccccccc} AJ\parallel GF & \implies & \frac {AJ}{GF}=\frac {BA}{BG} & \implies & \frac xb=\frac {c}{c+b} & \implies & x=\frac {bc}{b+c}\\\\ AH\parallel ED & \implies & \frac {AH}{ED}=\frac {CA}{CE} & \implies & \frac yc=\frac {b}{c+b} & \implies & y=\frac {bc}{b+c}\end{array}\right\|$ $\implies$ $x=y=\frac {bc}{b+c}$ .

Observe that $[AHIJ]=[BIC]\iff$ $[ABJ]=[BCH]\iff$ $\left\{\begin{array}{c} cx=b(c-y)\\\ (x=y)\end{array}\right\|$ $\iff$ $x=y=\frac {bc}{b+c}$ , what is truly. Remark that $\boxed{\frac 1x=\frac 1b+\frac 1c}$ .


PP5. Let $\triangle ABC$ with incenter $I$ . Let $D$ , $E$ , $F$ be incenters of triangles $IBC$ , $ICA$ , $IAB$ respectively. Prove that $AD\cap BE\cap CF\ne\emptyset$ .

Proof (trigonometric). $\left\{\begin{array}{ccc} \triangle ADB & \implies & \frac{DB}{DA}=\frac{sin\widehat{DAB}}{sin\frac {3B}{4}}\\\\ \triangle ADC & \implies & \frac{DA}{DC}=\frac{sin\frac {3C}{4}}{sin\widehat{DAC}}\\\\ \triangle BDC & \implies & \frac {DC}{DB}=\frac {\sin\frac B4}{\sin\frac C4}\end{array}\right\|\ \bigodot\ \implies\ \frac{sin\widehat{DAB}}{sin\widehat{DAC}}=\frac{sin\frac{C}{4}}{sin\frac{3C}{4}}.\frac{sin\frac{3B}{4}}{sin\frac{B}{4}}$ a.s.o.

In conclusion, using the trigonometric form of the Ceva's theorem obtain that $AD\cap BE\cap CF\ne\emptyset$ .


PP6. Let $ABC$ be a triangle. For two points $\{L,S\}\subset (BC)$ define $M\in (AC)$ , $N\in (AB)$ for which

$LM\parallel AB$ , $LN\parallel AC$ . Denote $X\in AS\cap LN$ and $Y\in AS\cap LM$ . Prove that $BX\parallel CY$ .

Proof. From the implications $\left\{\begin{array}{ccc} XL\parallel AC & \implies & \frac {SX}{SA}=\frac {SL}{SC}\\\\ YL\parallel AB & \implies & \frac {SA}{SY}=\frac {SB}{SL}\end{array}\right\|$ obtain that $\frac {SX}{SY}=\frac {SB}{SC}$ , i.e. $BX\parallel CY$ .


PP7.Given are a rectangle $ MATE$ and two points $ S \in (ME)$ , $ R \in (MA)$ so that $ RS\perp MT$ , $ Q \in (RE)$ , $ MQ\perp RE$. Prove that $ QS\perp QT$.

Prove . Denote $ P\in MQ\cap AT$ . Observe that $ \left\|\begin{array}{ccc} AMP\sim MER & \implies & AP=MR\cdot\frac {AM}{ME}\\\\ MRS\sim MEA & \implies & MS=MR\cdot \frac {MA}{ME}\end{array}\right\|\implies AP=MS$ $ \implies PS\perp ME$ .

$ \left\|\begin{array}{ccccc} AP\perp RE & \implies & ARQP\ -\ \mathrm{ciclic} & \implies & MQ\cdot MP=MR\cdot MA\\\\ RS\perp AT & \implies & ARSE\ -\ \mathrm{ciclic} & \implies & MR\cdot MA=MS\cdot ME\end{array}\right\|$ $ \implies$ $ MQ\cdot MP=$ $ MS\cdot ME\implies PSQE\ -\ \mathrm{ciclic}$ .

Since $ PS\perp SE$ si $ TP\perp TE$ obtain that $ T$ , $ Q$ , $ S$ belong to the circle with the diameter $ PE$ . Since $ PE=ST$ obtain in conclusion that $ QS\perp QT$ .


PP8. Let $ ABC$ be a triangle with the incircle $ w=C(I)$ . Denote $ M\in BC\cap w$ . Consider the points $ D\in BC$ , $ E\in CA$ ,

$ F\in AB$ for which $ \frac {DB}{DC}=\frac {BF}{CE}$ . For a point $ P\in AI$ denote $ K\in DF\cap BP$ , $ L\in DE\cap CP$ . Show that $ KL\parallel EF$ .

Proof. Denote $ M\in AI\cap BC$ . From $ \left\|\begin{array}{c} \frac {KF}{KD}=\frac {PA}{PM}\cdot\frac {BF}{BD}\cdot\frac {BM}{BA}\\\\ \frac {LE}{LD}=\frac {PA}{PM}\cdot \frac {CE}{CD}\cdot\frac {CM}{CA}\end{array}\right\|$ obtain $ \frac {KF}{KD}=\frac {LE}{LD}$ , i.e. $ KL\parallel EF$ because $ \frac {BF}{BD}=\frac {CE}{CD}$ and $ \frac {BM}{AB}=\frac {CM}{CA}$ .


PP9. Consideram un paralelogram $ ABCD$ . Notam cercul $ w$ circumscris triunghiului $ ABC$ si punctele $ \{A,E\} = AD\cap w$ ,

$ \{C,F\} = CD\cap w$ . Sa se arate ca centrul $ O$ al cercului circumscris triunghiului $ DEF$ apartine cercului $ w$ si $ OD\perp AC$ .

Proof.


PP10. Given an $A$- right triangle $ABC$ with $b\le c$ , where $h_a$ , $w_a$ , $m_a$ are its altitude, bisector and median from vertex $A$ respectively. Calculate $\lim_{b\rightarrow c}\frac{m_a-h_a}{w_a-h_a}$ .
Proof. From the relations $2m_a=a$ , $ah_a=bc$ and $w_a=\frac {2bc\cdot\cos\frac A2}{b+c}=\frac {bc\sqrt 2}{b+c}$ obtain that $\frac{m_a-h_a}{w_a-h_a}=\frac {2am_a-2ah_a}{2aw_a-2ah_a}=$ $\frac {a^2-2bc}{\frac {2abc\sqrt 2}{b+c}-2bc}=$

$\frac {(b^2+c^2-2bc)(b+c)}{2bc\cdot\left[\sqrt {2\left(b^2+c^2\right)}-(b+c)\right]}=$ $\frac {(b-c)^2(b+c)\left[\sqrt {2\left(b^2+c^2\right)}+(b+c)\right]}{2bc(b-c)^2}\stackrel{(b\ne c)}{=}$ $\frac {(b+c)\left[\sqrt {2\left(b^2+c^2\right)}+(b+c)\right]}{2bc}$ $\implies$

$\lim_{b\nearrow c}\frac{m_a-h_a}{w_a-h_a}\stackrel{(t=\frac bc)}{=}\lim_{t\nearrow 1}\frac {(t+1)\left[\sqrt {2\left(t^2+1\right)}+(t+1)\right]}{2t}=4$ .


PP11 (INMO 1995). Let $ABC$ be an acute triangle with $A = 30^{\circ}$ , the orthocenter $H$ and the midpoint $M$ of $[BC]$ .

Consider the reflection $T$ of $H$ w.r.t. $M$ , i.e. $M\in HT$ , $MH=MT$ . Prove that $AT = 2\cdot BC$ .

Proof. Observe that $BHCT$ is a parallelogram $\implies$ $\left\{\begin{array}{ccc} CT\parallel BH\perp AC & \implies & CT\perp CA\\\ BT\parallel CH\perp AB & \implies & BT\perp BA\end{array}\right\|$ $\implies$ $[AT]$ is a diameter in the circumcircle $C(O,R)$

of $\triangle ABC$ . Since $m\left(\widehat{BOC}\right)=2\cdot A=60^{\circ}$ and $OB=OC$ obtain that $\triangle BOC$ is equilateral, i.e. $2\cdot BC=2\cdot OB=2R$ $\implies$ $2\cdot BC=AT$ .


PP12. Let $ABCD$ be a quadrilateral with $\left\{\begin{array}{c} m\left(\widehat{BAC}\right) = 30^\circ\ ;\ m\left(\widehat{CAD}\right) = 20^\circ\\\\ m\left(\widehat{ABD}\right) = 50^\circ\ ;\ m\left(\widehat{DBC}\right) = 30^\circ\end{array}\right\|$ . Prove that $P\in AC\cap BD\implies PC = PD$ .

Here is a short trigonometrical proof of this easily and nice proposed problem.


PP13. Let $OAB$ be a triangle for which $OA=OB$ . For a mobile point $M\in OB$ define $N$ for which $NB\parallel OA$, $NM\parallel AB$ .

Denote the second intersection $Q$ of $AM$ with the circumcircle of $\triangle BMN$ . Prove that the line $NQ$ pass always through a fixed point.

Proof. Denote $R\in AO\cap NQ$ . Prove easily that $\widehat {MBA}\equiv\widehat {MQB}$ and $OMQR$ is cyclically. Therefore, $AB^2=AM\cdot AQ=$

$AO\cdot AR\implies $$\boxed{AR=\frac {AB^2}{AO}}$ (constant). If $m\left(\widehat {AOB}\right)=\phi$ , then $AR=4\cdot AO\cdot\sin^2\frac {\phi}{2}$ , i.e. $OR=OA\cdot |1-2\cdot \cos\phi |$ .


PP14. Solve in $\mathbb R$ the equation $\sqrt[5]{x^3+2x}=\sqrt[3]{x^5-2x}$ .

Proof. $\sqrt[5]{x^3+2x}=\sqrt[3]{x^5-2x}=t\iff$ $ x^3+2x=t^5$ and $x^5-2x=t^3\implies$ $x^5+x^3=t^5+t^3$ . Since the function $u\rightarrow u^5+u^3$

on pe $\mathbb {R}$ is increasing obtain that $t=x$ . Therefore, $x^5-x^3-2x=0$ $\iff$ $x(x^2-2)(x^2+1)=0$ with the solution $\left\{0,\pm\sqrt{2}\right\}$ .


PP15. Find all real solutions of the equation $x^2 - 19 \lfloor x \rfloor + 88 = 0$ .

Proof. Denote $\lfloor x \rfloor =z\iff z\in\mathbb Z\ \wedge\ z\le x< z+1$ . Thus, the given equation becomes $x^2=19z-88\iff$

$\boxed{x=\sqrt {19z-88} }\ (*)$ , where $z\in\mathbb Z$ and $5\le z\le \sqrt {19z-88}<z+1\iff$ $z^2\le 19z-88<(z+1)^2\iff$

$\left\{\begin{array}{c} z\in \mathbb Z\ ,\ z\ge 5\\\ z^2-19z+88\le 0\\\ z^2-17z+89>0\end{array}\right\|$ $\iff$ $z\in\overline{8,11}\stackrel{(*)}{\iff}$ $x\in \left\{\sqrt {19z-88}\ |\ z\in\overline{8,11}\right\}\iff$ $x\in\left\{8,\sqrt {83},\sqrt {102},11\right\}$ . .


PP16. Find all values of $m\in\mathbb R$ such that the equation $ 4^{x^2 - 2x + 2} - m 2^{x^2 - 2x +3} + 3m -2 = 0$

have only four real zeroes such that there is the chain $ x_1 < -1 < x_2 < 1 < x_3 < 2 < x_4$ .

Proof.


PP17. Let $\triangle{ABC}$ be an acute triangle with the orthocenter $H$ . The circle $w$ with the diameter $[BC]$ intersects $AC$ and $AB$ at points $E$

and $F$ respectively, i.e. $H\in BE\cap CF$ . The tangents drawn to the circle $w$ through $E$ and $F$ intersect at $P$ . Show that $P\in AH$ .

Proof 1. Denote the orthocenter $H$ of $\triangle ABC$ , i.e. $H\in BE\cap CF$ and the midpoints $M$ , $N$ of the segments $[BC]$ , $[AH]$ respectively.

Thus, $\left\{\begin{array}{ccc} NF=NH & \implies & m\left(\widehat{NFC}\right)=B\\\ NE=NH & \implies & m\left(\widehat{NEB}\right)=C\end{array}\right\|$ . Since $\left\{\begin{array}{c} m\left(\widehat{PFC}\right)=B\\\ m\left(\widehat{PEB}\right)=C\end{array}\right\|$ obtain that $P\in FN\cap EN\implies$ $P\equiv N\in AH$ .

Proof 2. Apply the Pascal's theorem to the cyclical degenerated hexagon $BFFCEE$ and obtain that $\left\{\begin{array}{ccc} A\in BF\cap CE\\\ P\in FF\cap EE\\\ H\in FC\cap EB\end{array}\right\|\implies P\in AH$ .

Proof 3. Let $L\in EF\cap BC$ and polar $l$ of $L$ $\iff$ $A\in l\ \wedge\ l\perp BC$ $\iff$ $l\equiv AH$ . Know that $EF$ is polar $p$ of $P$ and $L\in p$ $\iff$ $P\in l$ $\iff$ $P\in AH$ .

Proof 4. Denote the orthocenter $H$ of $\triangle ABC$ , i.e. $H\in BE\cap CF$ and the midpoints $M$ , $N$ of the segments $[BC]$ , $[AH]$ respectively. Observe that

the point $N$ belongs to the circumcircle of $\triangle EMF$ - the Euler's circle of $\triangle ABC$ with the diameter $[MN]$ and $\left\{\begin{array}{c} m\left(\widehat{NFH}\right)=m\left(\widehat{NHF}\right)=B\\\ m\left(\widehat{NEH}\right)=m\left(\widehat{NHE}\right)=C\end{array}\right\|$ $\implies$

The line $NF$ is tangent in the point $F\in w$ to the circle $w$ with diameter $[BC]$ and the line $NE$ is tangent in the point $E\in w$ to the circle $w$ , i.e. $P\equiv N\in AH$ .


PP18. Let $p\in \mathbb C[X]$ be a polynomial of degree $n\in \mathbb N^*$ such that $p(k) = \frac{1}{k}$ for any $k\in\overline{1,n + 1}$ . Find $p (n + 2)$ .

Proof. Observe that the polynomial $Q(X)=X\cdot P(X)-1$ of degree $(n+1)$ has the roots $k\in\overline{1,n + 1}$ . Therefore we can write

$Q(X)=a\cdot (X-1)(X-2)\cdots(X-n)[X-(n+1)]$ . Thus, $-1=Q(0)=a(-1)^{n+1}(n+1)!$ , i.e. $a\cdot (n+1)!=(-1)^n$ .

In conclusion, $Q(n+2)=a(n+1)!=(-1)^n=(n+2)P(n+2)-1$ $\implies$ $P(n+2)=\frac {1+(-1)^n}{n+2}$ .


PP19. Let $w=C(O,r)$ be a circle and let $l$ be a line which is tangent in a fixed point $A\in w$ to the circle $w$ .

For a mobile point $R\in w$ denote its projection $Q$ to given line $l$ . Determine the maximum area of $\triangle PQR$ .

Proof 1. Denote $m\left(\widehat{POR}\right)=2\phi$ , where $0\le\phi \le\frac {\pi}{2}$ . Prove easily that $[PQR]=\frac 12\cdot PR\cdot PQ\cdot\sin\phi =$ $\frac 12\cdot 2r\sin\phi\cdot 2r\sin\phi\cos\phi\cdot\sin\phi\iff$

$\boxed{[PQR]=2r^2\sin^3\phi\cos\phi}$ . Thus, $[PQR]$ is maximum $\iff$ $\sin^3\phi\cos\phi$ is maximum on $\left[0,\frac {\pi}{2}\right]$ $\iff$ $\sin^6\phi\cos^2\phi$ is maximum $\iff$

$\frac {\sin^2\phi}{3}\cdot\frac {\sin^2\phi}{3}\cdot\frac {\sin^2\phi}{3}\cdot\cos^2\phi$ is maximum. Observe that $\frac {\sin^2\phi}{3}+\frac {\sin^2\phi}{3}+\frac {\sin^2\phi}{3}+\cos^2\phi =1$ (constant). Therefore, $[PQR]$ is maximum $\iff$

$\frac {\sin^2\phi}{3}=\cos^2\phi =\frac 14\iff$ $\boxed{\ \phi =\frac {\pi}{3}\ }$ and in this case the maximum value of the area $[PQR]$ is $\frac {3r^2\sqrt 3}{8}$ , i.e. $\boxed{\ [PQR]\le\frac {3r^2\sqrt 3}{8}\ }$ .

Proof 2. Denote $\{P,T\}=PO\cap w$ , $S\in PO$ for which $RS\perp PO$ and $SP=u$ , $ST=v$ , where $\boxed{u+v=2r}$ . Observe that

$[PQR]=$ $[PSR]=$ $\frac 12\cdot SP\cdot SR$ , i.e. $\boxed{[PQR]=\frac 12\cdot u\sqrt {uv}}$ . In conclusion, $[PQR]$ is maximum $\iff$ $u^3v$ is maximum $\iff$ $\frac u3\cdot\frac u3\cdot\frac u3\cdot v$

is maximum. Since $\frac u3+\frac u3+\frac u3+v=2r$ (constant) obtain that $[PQR]$ is maximum $\iff$ $\frac u3=v=\frac r2$ , i.e. $[PQR]\le \frac {3r^2\sqrt 3}{8}$ .

Remark. $3=\frac uv=\left(\frac {RP}{RT}\right)^2=\tan^2\widehat{PTR}=\tan^2\widehat {QPR}\implies$ $\tan\widehat{QPR}=\sqrt 3\iff$ $\phi =\frac {\pi}{3}$ (see the first proof).


PP20. $\triangle ABC\Longrightarrow \frac{1}{b+c} +\frac{1}{c+a} +\frac{1}{a+b} \le \frac{3(R+r)}{4S}$ .

Proof. From $(b+c)^2\ge 4bc$ obtain that $\sum\frac {1}{b+c}\le $ $\sum\frac {b+c}{4bc}=$ $\frac 14\cdot \sum\left(\frac 1b+\frac 1c\right)=$ $\frac 12\cdot\sum \frac 1a=$ $\frac{ab+bc+ca}{2abc}$ . Let's prove

that $ \frac{ab+bc+ca}{2abc}\le \frac{3(R+r)}{4S}$ . This is equivalent with $ \frac{s^2+r^2+4Rr}{8RS}\le $ $\frac{3(R+r)}{4S}$ $\Longleftrightarrow $ $s^2+r^2+4Rr\le$ $ 6R^2+6Rr$ $\Longleftrightarrow$

$ s^2+r^2\le 6R^2+2Rr$ . Using that $ s^2\le 4R^2+4Rr+3r^2$ $\implies$ $ s^2+r^2\le $ $4R^2+4Rr+4r^2$ . So we must prove that

$ 4R^2+4Rr+4r^2\le $ $6R^2+2Rr$ $\Longleftrightarrow $ $0\le R^2-Rr-2r^2$ $\Longleftrightarrow $ $0\le (R+r)(R-2r)$ which is true .


PP21. Let $ABCD$ be a parallelogram with $AD=a$ and let $E\in (BD)$ be a point for which denote

$\left\{\begin{array}{c} F\in AE\cap BC\\\ G\in AE\cap CD\end{array}\right\|$ . Suppose that $\left\{\begin{array}{c} AE=2\\\ CF=3\end{array}\right\|$ . Find the ratio $\frac {[ABE]}{[DEG]}$ .

Proof. Denote $EG=x$ . Observe that $\frac {EB}{ED}=\frac {EA}{EG}=\frac 2x$ and $\frac {GD}{GC}=\frac {AD}{CF}=\frac a3$ . Apply the Menelaus'

theorem to the transversal $\overline{FGE}$ and $\triangle BCD\ :$ $\frac {FC}{FB}\cdot\frac {EB}{ED}\cdot\frac {GD}{GC}=1$ $\iff$ $\frac {3}{3+a}\cdot \frac 2x\cdot\frac a3=1\iff$

$x(a+3)=2a\iff$ $\boxed{x=\frac {2a}{a+3}}$ . In conclusion, $\frac {[ABE]}{[DEG]}=\left(\frac 2x\right)^2\implies$ $\frac {[ABE]}{[DEG]}=\left(\frac {a+3}{a}\right)^2$ .


PP22. If the triangles $BCM\ ,\ CDN$ are equilateral and are outside of the parallelogram $ABCD$ , then the triangle $AMN$ is equilateral.

Proof 1 (complex numbers). Denote $w=\cos\frac {\pi}{3}+i\cdot\sin\frac {\pi}{3}$ . Observe that $\left\{\begin{array}{c} w^3=-1\ ,\ w^2-w+1=0\\\\ \overline w=1-w=\frac 1w=-w^2\end{array}\right|$ .

Thus, $A(a)$ , $B(b)$ , $C(c)$ , $D(d)$ and $M(m)$ , $N(n)$ so that $a+c=b+d$ and $\left\{\begin{array}{c} m-c=w(b-c)\\\\ n-d=w(c-d)\end{array}\right|\ \implies$

$\left\{\begin{array}{c} m=(1-w)c+wb\\\\ n=(1-w)d+wc\end{array}\right|\ (*)$ . Therefore, $\triangle MAN$ is equilateral $\iff$ $a-m=w(m-n)\iff$

$a=(1-w)m+wn\iff$ $b+d=c+m+w(n-m)\stackrel{(*)}{\iff}$ $\left(w^2-w+1\right)(b-2c+d)=0$ , what is truly.

Remark. $b-2c+d=0\iff$ the point $C$ is the midpoint of the diagonal $[BD]$ , what is falsely. In conclusion, $b+d\ne 2c$ .

Proof 2 (synthetic). Denote $\phi =m\left(\widehat{ABC}\right)$ . Observe that $\left\{\begin{array}{c} AD=MB=MC\ ;\ DN=BA=CN\\\\ m\left(\widehat{ADN}\right)=m\left(\widehat{MBA}\right)=m\left(\widehat{MCN}\right)=\phi +60^{\circ}\end{array}\right|$ $\implies$

$\triangle ADN$ $\cong\triangle MBA\cong$ $\triangle MCN$ $\implies\ AN=MA=MN$ . In conclusion, the triangle $AMN$ is equilateral.


PP23. Solve the equation $(\sin x - 2\cos x)^4 + (\sin x - 2\cos x)(5 - 7\sin 2x + 4\cos ^2x)\cos x - \cos ^4x = 0$ .

Proof. $(\sin x - 2\cos x)^4 + (\sin x - 2\cos x)(5 - 7\sin 2x + 4\cos ^2x)\cos x - \cos ^4x = 0\iff$

$(\tan x - 2)^4+ (\tan x - 2)\cdot\frac {5 - 7\sin 2x + 4\cos ^2x}{\cos^2x} - 1 = 0\ (*)$ Observe that $\frac {5 - 7\sin 2x + 4\cos ^2x}{\cos^2x}=$

$\frac {5\left(\sin^2x+\cos^2x\right) - 14\sin x\cos x + 4\cos ^2x}{\cos^2x}=$ $\frac {5\sin^2x - 14\sin x\cos x + 9\cos ^2x}{\cos^2x}$ . Therefore,

$\frac {5 - 7\sin 2x + 4\cos ^2x}{\cos^2x}=5\tan^2x - 14\tan x + 9$ . Denote $\boxed{\tan x-2=t}$ . Thus, the equation $(*)$ becomes

$t^4+t\left[5(t+2)^2-14(t+2)+9\right]-1=0$ , i.e. $t^4+5t^3+6t^2+t-1=0\iff$ $(t+1)^2(t^2+3t-1)=0$ a.s.o.


PP24. Solve ${{\left( \dfrac{1+i}{\sqrt{2}} \right)}^{x}}+{{\left( \dfrac{1-i}{\sqrt{2}} \right)}^{x}}=\sqrt{2}\ ,\ x\in\mathbb Z\ .$

Proof. Observe that ${{\left( \dfrac{1+i}{\sqrt{2}} \right)}^{x}}\cdot {{\left( \dfrac{1-i}{\sqrt{2}} \right)}^{x}}=1$ and $\left( \dfrac{1\pm i}{\sqrt{2}} \right)^8=1$ . Thus, with the substitution ${{\left( \dfrac{1+i}{\sqrt{2}} \right)}^{x}}=y$ our equation becomes

$y+\frac 1y=\sqrt 2\implies $ $y^2-y\sqrt 2+1=0\implies$ $y\in \left\{\frac {1\pm i}{\sqrt 2}\right\}\implies$ ${{\left( \dfrac{1+i}{\sqrt{2}} \right)}^{x}}=\left\{\frac {1\pm i}{\sqrt 2}\right\}$ $\implies x\in \left|\left| 8k\pm 1\right| k\in\mathbb Z\right|$ .


PP25 (2012 Philippine Math Olympiad National Finals Oral Round). Solve the system

$(x+y)(z+1)=(x+z)(y+1)=(y+z)(x+1)=2a(a+1)$ .

Proof 1. The system is symmetrically. From the first equation get $x-z=y(x-z)$ , i.e. $x=z\ \vee\ y=1$ .

$\blacktriangleright$ Case $1$ . If $x=z$ the system becomes $(x+y)(x+1)=2x(y+1)=2a(a+1)$ ,

i.e. $x^2+y=xy+x\ \vee\ x(y+1)=a(a+1)\iff$ $x(x-y)=x-y\ \vee\ x(y+1)=a(a+1)$ . Thus,

$x=z=1$ and $y=a^2+a-1$ or $x=z=y$ and $x^2+x-a(a+1)=0\begin{array}{cc} \nearrow & a\\\\ \searrow & -a-1\end{array}$ .

Therefore, in this case the solutions are $\left\|\begin{array}{ccc} 1 & a^2+a-1 & 1\\\\ a & a & a\\\\ -a-1 & -a-1 & -a-1\end{array}\right\|\ \ (*)$ .

$\blacktriangleright$ Case $2$ . If $y=1$ the system becomes $(x+1)(z+1)=2(x+z)=2a(a+1)$ , i.e. $\left\{\begin{array}{c} x+z=a(a+1)\\\\ 1+xz=x+z\end{array}\right\| \iff$

$z=a(a+1)-x$ and $x^2-a(a+1)x+a^2+a-1=0\begin{array}{cc} \nearrow & a^2+a-1\\\\ \searrow & 1\end{array}$ .

So get $\left\|\begin{array}{ccc} a^2+a-1 & 1 & 1\\\\ 1 & 1 & a^2+a-1\end{array}\right\|$ . In conclusion, all solutions are $\left\|\begin{array}{ccc} a^2+a-1 & 1 & 1\\\\ 1 & a^2+a-1 & 1\\\\ 1 & 1 & a^2+a-1\\\\ a & a & a\\\\ -a-1 & -a-1 & -a-1\end{array}\right\|$ .

Proof 2. Using the substitution $\left\{\begin{array}{c} x+yz=u\\\ y+zx=v\\\ z+xy=w\end{array}\right\|$ obtain the system $u+v=v+w=w+u=2a(a+1)$ what has the solution

$u=v=w=a(a+1)$ , i.e. the initial system becomes $x+yz=y+zx=z+xy=a(a+1)$ which is equivalently with the

system $\left\{\begin{array}{c} (x-y)=z(x-y)\\\ (y-z)=x(y-z)\\\ x+xz=a(a+1)\end{array}\right\|\iff$ $\left\|\begin{array}{c} x=y\ \vee\ z=1\\\ y=z\ \vee\ x=1\\\ x+yz=a(a+1)\end{array}\right\|$ . In conclusion,

$\left\{\begin{array}{cc} x=y & \begin{array}{cccccc} \nearrow & y=z & \implies & x=y=z & \in & \{a,-(a+1)\}\\\\ \searrow & x=1 & \implies & x=y=1 & ; & z=a^2+a-1\end{array}\\\\ z=1 & \begin{array}{cccccc} \nearrow & y=z & \implies & y=z=1 & ; & x=a^2+a-1\\\\ \searrow & x=1 & \implies & x=z=1 & ; & y=a^2+a-1\end{array}\end{array}\right\|$ $\implies\left\|\begin{array}{ccc} a & a & a\\\\ -a-1 & -a-1 & -a-1\\\\ 1 & 1 & a^2+a-1\\\\ a^2+a-1 & 1 & 1\\\\ 1 & a^2+a-1 & 1\end{array}\right\|$ .


PP26. Let $x^2-ax+b=0$ be a equation with the roots $x_1$ and $x_2$ . Ascertain the relation $\mathrm R(a,b)$ so that $x_1^2=x_2+1\ \ \vee\ \ x_2^2=x_1+1$ . For $a=5$ find $b$ .

Proof 1 (directly). Observe that $x_1+x_2=a$ and $x_1x_2=b$ . Thus, $x_1^2=x_2+1\ \vee\ x_2^2=x_1+1\iff$ $x_1^2-x_2-1=0\ \vee\ x_2^2-x_1-1=0\iff$ $\left(x_1^2-x_2-1\right)\left(x_2^2-x_1-1\right)=0\iff$

$x_1^2x_2^2-\left(x_1^3+x_2^3\right)-\left(x_1^2+x_2^2\right)+x_1x_2+\left(x_1+x_2\right)+1=0\iff$ $b^2-\left(a^3-3ab\right)-\left(a^2-2b\right)+b+a+1=0\iff$ $\boxed{a^3+a^2-(1+3b)a=b^2+3b+1}$ .

For $a=5$ obtain that $150-5(3b+1)=b^2+3b+1\iff$ $b^2+18b-144=0\iff$ $b\in\{6,-24\}$ .

Proof 2 (common root). Observe that $x_1+x_2=a$ and $x_1x_2=b$ . Thus, $x_1^2=x_2+1\ \vee\ x_2^2=x_1+1\iff$ $x_1^2=a-x_1+1\ \vee\ x_2^2=a-x_2+1\iff$ $x_1^2+x_1-(a+1)=0$ or

$x_2^2+x_2-(a+1)=0\iff$ the equations $\left\{\begin{array}{c} x^2-ax+b=0\\\\ x^2+x-(a+1)=0\end{array}\right|$ have at least a common root $\iff$ $\left|\begin{array}{cc} 1 & b\\\\ 1 & -(a+1)\end{array}\right|^2=$ $\left|\begin{array}{cc} 1 & -a\\\\ 1 & 1\end{array}\right|\cdot\left|\begin{array}{cc} -a & b\\\\ 1 & -(a+1)\end{array}\right|\iff$

$(a+b+1)^2=(a+1)\left(a^2+a-b\right)\iff$ $a^2+2a(b+1)+(b+1)^2=a(a+1)^2-b(a+1)\iff$ $\boxed{a^3+a^2-(1+3b)a=b^2+3b+1}$ .

Proof 3 ("brute force"). Eliminate $x_1$ and $x_2$ between the relations $\left\{\begin{array}{c} x_1+x_2=a\\\\ x_1x_2=b\\\\ x_1^2=ax_1-b=x_2+1\end{array}\right|\implies$ $\left\{\begin{array}{ccc} x_1x_2 & = & b\\\\ x_1+x_2 & = & a\\\\ ax_1-x_2 & = & b+1\end{array}\right|\implies$ $\left\{\begin{array}{ccc} x_1x_2 & = & b\\\\ x_1 & = & \frac {a+b+1}{a+1}\\\\ x_2 & = & \frac {a^2-(b+1)}{a+1}\end{array}\right|\implies$

$\left[a+(b+1)\right]\cdot\left[a^2-(b+1)\right]=b(a+1)^2\iff$ $a^3-a(b+1)+a^2(b+1)-(b+1)^2=ba^2+2ba+b\iff$ $\boxed{a^3+a^2-(1+3b)a=b^2+3b+1}$ .
This post has been edited 145 times. Last edited by Virgil Nicula, Nov 22, 2015, 2:33 PM

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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 ryanbear, May 9, 2023, 6:10 AM

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    by tigerzhang, Aug 1, 2021, 12:02 AM

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    by the_mathmagician, Jan 17, 2021, 7:43 PM

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    by OlympusHero, Dec 30, 2020, 6:08 PM

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    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 :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

72 shouts
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  • Total entries: 456
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