geometry
Triangle
Circumcenter
IMO
IMO 2001
hojoo lee
geometric inequality

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Like usual, we let : $\angle A = \alpha$, $\angle B = \beta$ and $\angle C = \gamma$.
Then, $\angle BCO = \frac 12(\pi - \angle BOC) = \frac{\pi}2 - \frac 12(2 \alpha)$.
Since $\triangle ABC$ is acute-angled, $O$ is inside the triangle, and $P$ is between $B$ and $C$. So, we just have to prove that $\angle COP = \angle POC$, or, equivalently : $PO > PC$. Let $OC = R$, then $b = 2R \sin{\beta}$ (thanks to the sine law).
Since $PC = b \cos{\gamma}$, we have :
\begin{eqnarray*}PC^2 < PO^2 &\iff& PC^2 < R^2 + PC^2 - 2Rb \cos{\gamma} \cos{(\frac{\pi}2 - \alpha)} \\ &\iff& 0 < \frac{b^2}{4 \sin^2{\beta}} - \frac{b^2 \cos{\gamma} \sin{\alpha}}{\sin{\beta}} \\ &\iff& 4 \sin{\alpha} \sin{\beta} \cos{\gamma} < 1 \\ &\iff& 2 \sin{\alpha}(\sin{(\beta + \gamma)} - \sin{(\gamma - \beta)}) <1 \\ &\Longleftarrow& 2 \sin{\alpha}(\sin{\alpha} - \sin{\frac{\pi}6}) \le 1 \\ &\Longleftarrow& 2 \sin{\alpha}\left(1 - \frac 12 \right) \le 1 \\ &\Longleftarrow& 2 - 2 \left(\frac 12 \right) \le 1 \end{eqnarray*}
which is true.

The problem condition $C\ge B+30^\circ$ is equivalent to

$\sin(C-B) \ge \sin 30^\circ = \frac{1}{2}\ \Longleftrightarrow\ \sin C\cos B-\sin B\cos C\ge \frac{1}{2}\ \Longleftrightarrow$

$\frac{c}{2R}\cdot \frac{c^{2}+a^{2}-b^{2}}{2ca}-\frac{b}{2R}\cdot \frac{a^{2}+b^{2}-c^{2}}{2ab}\ge \frac{1}{2}\ \Longleftrightarrow\ c^{2}-b^{2}\ge aR$

The angles $A+\angle COP$ add to $90^\circ$ in an acute $\triangle ABC,$ iff the triangle $\triangle COP$ is isosceles with $\angle COP = \angle OCP,$ because it is always trus that $\angle OCP \equiv \angle OCB = 90^\circ-A$. Thus $A+\angle COP < 90^\circ$ is equivalent to $PO > PC.$ Let M be the midpoint of BC. By Pythagorean theorem for the right triangle $\triangle OMP,$

$PO^{2}= OM^{2}+MP^{2}$

$MP = CM-CP,\ \ CM = \frac{a}{2},\ \ CP = b \cos C = \frac{a^{2}+b^{2}-c^{2}}{2a}$

$PO > PC\ \Longleftrightarrow\ OM^{2}+(CM-CP)^{2}> CP^{2}\ \Longleftrightarrow$

$OM^{2}+CM^{2}-2CM \cdot CP > 0\ \Longleftrightarrow OM^{2}+\frac{a^{2}}{4}-\frac{a^{2}+b^{2}-c^{2}}{2}> 0\ \Longleftrightarrow$

$OM^{2}+\frac{c^{2}-b^{2}}{2}> \frac{a^{2}}{4}$

The consequence $c^{2}-b^{2}\ge aR$ of the prolem condition and the trivial facts $OM > 0,$ $R > \frac{a}{2}$ for an acute triangle imply

$OM^{2}+\frac{c^{2}-b^{2}}{2}\ge OM^{2}+\frac{aR}{2}> OM^{2}+\frac{a^{2}}{4}> \frac{a^{2}}{4}$

hence $A+\angle COP < 90^\circ$.

Can anybody post solution without trigonometry, if there is one. Please

$ AP, AO$ and parallel to $ BC$ through $ A$ cut $ (O)$ again at $ X, Y, Z,$ respectively. $ YZ$ cuts $ BC$ at $ Q.$ $ XY \parallel BC,$ $ YX \perp AX$ $ \Longrightarrow$ $ AXYZ$ is rectangle, $ XY \equal{} PQ \equal{} AZ$ and $ M$ is common midpoint of $ BC, PQ.$ For any acute $ \triangle ABC,$ $ OM > 0,$ $ R > \frac {BC}{2}$ and

$ \angle COP < \angle OCP \equal{} 90^\circ \minus{} \angle A$ $ \Longleftrightarrow$ $ CP^2 < OP^2$ $ \Longleftrightarrow$ $ \frac {(BC \minus{} PQ)^2}{4} < OM^2 \plus{} \frac {PQ^2}{4}$ $ \Longleftrightarrow$ $ OM^2 \plus{} \frac {BC \cdot PQ}{2} > \frac {BC^2}{4}$

From $ \angle XAY \equal{} \angle C \minus{} \angle B \ge 30^\circ$ $ \Longrightarrow$ $ PQ \equal{} XY \ge \frac {AY}{2} \equal{} R.$ Therefore,

$ OM^2 \plus{} \frac {BC \cdot PQ}{2} \ge OM^2 \plus{} \frac {BC \cdot R}{2} > \frac {BC^2}{4}$ $ \Longrightarrow$ $ CP < OP$ $ \Longrightarrow$ $ \angle COP \plus{} \angle A < 90^\circ.$

Blah here's my silly way to show $OP>CP$.

From the power of point $P$ we see that $R^2-OP^2=BP\cdot CP$, so it is equivalent to show that
\[OP>\frac{R^2-OP^2}{BP}=CP,\]or by the Law of Cosines on $\angle{BPO}$, that
\begin{align*}
OP\cdot BP+OP^2 &> R^2 = BO^2 = BP^2+OP^2-2BP\cdot OP\cos\angle{BPO}\\
\Longleftrightarrow OP+OP(2\cos\angle{BPO}) &> BP.
\end{align*}But by the Law of Sines on $\triangle{APO}$, we have
\[2\cos\angle{BPO} = 2\sin\angle{APO} = 2\frac{AO\sin(C-B)}{OP} \ge \frac{AO}{OP}.\]Hence
\[OP+OP(2\cos\angle{BPO}) \ge OP+AO = OP+BO > BP\]by the Triangle Inequality, so we're done.

Draw $OK\perp AP$ and $OD\perp BC$.Obviously $KOPD$ is a rectangle.
$AO=CO$ means $\triangle ABC$ is isosceles. So $\angle CAO=\frac{1}{2}(\pi-\angle AOC)=\frac{\pi}{2}-\angle B$

[asy]
/* File unicodetex not found. */

 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(11.74cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -4.3, xmax = 7.44, ymin = -2.64, ymax = 6.3;  /* image dimensions */

 /* draw figures */
draw(circle((1.82,1.42), 3.76));
draw((0.29,4.86)--(-1.59,-0.16));
draw((-1.59,-0.16)--(5.21,-0.21));
draw((5.21,-0.21)--(0.29,4.86));
draw((0.29,4.86)--(0.27,-0.17));
draw((1.82,1.42)--(0.28,1.44));
draw((1.82,1.42)--(-1.59,-0.16));
draw((1.82,1.42)--(0.27,-0.17));
draw((1.82,1.42)--(1.8,-0.19));
 /* dots and labels */
dot((1.82,1.42),dotstyle);
label("$O$", (2,1.46), NE * labelscalefactor);
dot((0.29,4.86),dotstyle);
label("$A$", (0.38,4.98), NE * labelscalefactor);
dot((-1.59,-0.16),dotstyle);
label("$C$", (-1.42,-0.38), NE * labelscalefactor);
dot((5.21,-0.21),dotstyle);
label("$B$", (4.78,-0.52), NE * labelscalefactor);
dot((0.27,-0.17),dotstyle);
label("$P$", (0.24,-0.52), NE * labelscalefactor);
dot((0.28,1.44),dotstyle);
label("$H$", (0.36,1.56), NE * labelscalefactor);
dot((1.8,-0.19),dotstyle);
label("$D$", (1.74,-0.48), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
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[/asy]

$\angle CAP=\frac{\pi}{2}-\angle C$, so angle $PAO=\angle CAO-\angle CAP=\angle C-\angle B\ge 30^\circ$.

 So we have $KO\ge \frac{1}{2}AO$(You may verify this using sine rule as well).As $KOPD$ is a rectangle $KO=PD>\frac{1}{2}AO=\frac{1}{2}CO>\frac{1}{2}CD$. So $PC<PD<OP\implies \angle COP<\angle PCO=\angle OCD \dots (*)$.

We now have $\angle COP+ \angle A=\angle COP+\angle COD<\angle OCD+\angle COD=90^\circ$.

Another approach that utilises power of a point:

As in all the other solutions here, we establish that we need to prove $OP>CP$. This in turn is equivalent, obviously, to:

$OP^2>CP^2$
$\Leftrightarrow R^2-PC \cdot PB>CP^2$

This is because of Power of a Point. We have that this is in turn equivalent to:

$R^2>CP^2+PC \cdot PB=CP(CP+PB)=CP \cdot CB$

Now note that $PB=AB\cos B=2R\sin C\cos B$, and similarly $PC=2R\sin B\cos C$. It follows that $PB-PC=2R\sin(C-B) \geq R$, where the last step follows from the problem's condition. Since $PB+PC=BC$, we deduce that $PB \geq \frac{R+BC}{2}$, and hence $PC \leq \frac{BC-R}{2}$. So we have:

$CP \cdot CB \leq \frac{BC(BC-R)}{2}$, so it suffices to show that:

$\frac{BC(BC-R)}{2}<R^2$
$\Leftrightarrow \frac{a(a-R)}{2}<R^2$
where $BC=a$. Bearing in mind that $a<2R$, we have:

$a-R<R$
$\Leftrightarrow (a-R)(a+R)<R(a+R)$
$\Leftrightarrow a^2-R^2<aR+R^2$
$\Leftrightarrow a^2-aR<2R^2$
$\Leftrightarrow a(a-R)<2R^2$
$\Leftrightarrow \frac{a(a-R)}{2}<R^2$

as desired.

Extend $CO$ to meet the circumcircle of $ABC$ at $D$.  Then $\angle DAB+\angle A=90^\circ$ and we want to prove that
\[\angle A+\angle COP<90^\circ=\angle DAB+\angle A\Longleftrightarrow \angle COP<\angle DAB.\]
We also have $\angle DAB=\angle BCD$ so what we want to prove is equivalent to proving
\[\angle COP<\angle BCD=\angle PCO\]
which is also equivalent to proving $OP>CP$.

We have
\begin{align*}
OP&>CP\\
\Longleftrightarrow OP^2&>CP^2 \\
\Longleftrightarrow CP^2+OC^2-2(CO)(CP)\cos\angle PCO&>CP^2\\
\Longleftrightarrow OC^2&>2(OC)(CP)\cos\angle PCO \\
\Longleftrightarrow OC&>2(CP)\cos \angle PCO
\end{align*}
Now let $\angle B=\beta $ and $\angle C=\gamma$, then $\angle PCO=\beta+\gamma-90$.  Also, $PC=b\cos \gamma$ and $OC=\frac{b}{2\sin \beta}$.  Now what we want to prove is
\[\frac {b}{2\sin \beta}>2b\cos \gamma\cos(\beta+\gamma-90)=2b\cos\gamma\sin(\beta+\gamma)\]
or
\[1>4\sin\beta\cos\gamma\sin(\beta+\gamma).\]
By the condition, $\gamma\geq \beta+30$ and the triangle is acute so $\cos\gamma<\cos(\beta+30)$.  Then
\[4\sin\beta\cos\gamma\sin(\beta+\gamma)<4\sin\beta\cos(\beta+30)=2\sin\beta(\cos\beta\sqrt3-\sin\beta).\]
Taking the derivative of the RHS we get
\[2(\cos \beta(\cos\beta\sqrt3-\sin\beta)+\sin\beta(-\sin\beta\sqrt3-\cos\beta)) \\ =2(\sqrt{3}(\cos^2\beta-\sin^2\beta-2\sin\beta\cos\beta))=2(\sqrt3\cos2\beta-\sin2\beta)\]
We find the critical points:
\begin{align*}
2(\sqrt3\cos2\beta-\sin2\beta)&=0 \\
\sqrt{3}\cos 2\beta=\sin2\beta
\end{align*}
Since $\beta\neq 45^\circ$, the LHS is not equal to 0 so we divide
\[\tan 2\beta=\sqrt{3}\]
and the only solution to this is such that $0^\circ<\beta<60^\circ$ $2\beta=60\circ\implies \beta=30^\circ$.  Thus the critical points are $\beta=\{0^\circ,30^\circ,60^circ\}$.  Plugging the two critical points in, we have
\[2\sin\beta(\cos\beta\sqrt3-\sin\beta)=0\]
for $\beta=0^\circ$ and $\beta=60^\circ$ and
\[2\sin\beta(\cos\beta\sqrt3-\sin\beta)=1\]
when $\beta=30^\circ$ therefore the maximum is achieved when $\beta=30^\circ$ however that means $\gamma =90^\circ$ which is impossible since the triangle is acute so therefore the maximum is not achievable and thus
\[4\sin\beta\cos\gamma\sin(\beta+\gamma)<1\]
as desired. $\blacksquare$

[u][b]Outlines of my proof[/b][/u]

Let $\angle{COP}=\theta$.Easy to show that its acute.
Sine rule in $\triangle{COP}$ gives
 $\frac{R}{CP}=\frac{cos(A-\theta)}{sin\theta} \Rightarrow cosA cot\theta+sinA=\frac{1}{2sinBcosC}=\frac{1}{sinA-sin(C-B)} \ge \frac{1}{sinA-\frac{1}{2}}=\frac{2}{2sinA-1}$.

From here we get the upper bound of $\tan\theta$.Now it is easy(using some elementary calcuations) to show that $tanA tan\theta <1$ so that $tan{A+\theta}>0$ and the result follows...

[quote="orl"]Consider an acute-angled triangle $ABC$. Let $P$ be the foot of the altitude of triangle $ABC$ issuing from the vertex $A$, and let $O$ be the circumcenter of triangle $ABC$. Assume that $\angle C \geq \angle B+30^{\circ}$. Prove that $\angle A+\angle COP < 90^{\circ}$.[/quote]
My solution:
[hide][img]http://www.artofproblemsolving.com/Forum/download/file.php?mode=view&id=49282&[/img][/hide]
Let $D$ be the foot of the altitude of $\triangle ABC$ issuing from $B.$ It is known that $DP \perp OC$ and $\triangle CPD \sim \triangle CAP.$

Let $E=OC \cap DP,$ we have \[\cos A=\cos \angle CPE=\frac{PE}{PC}\] and \[\sin \angle COP=\frac{PE}{OP}.\] Therefore, \[\angle A+\angle COP<90^{\circ} \Leftrightarrow \sin \angle COP<\sin (90^{\circ}-\angle A)=\cos A\Leftrightarrow OP>CP.\] Now, let us compute the lengths of $OP$ and $CP.$ Let $M$ be the midpoint of $BD$ and $R$ be the radius of $(O),$ we have \[CP=AC\cdot \cos \angle ACP=2R\cdot \sin B\cos C\] and \[CM=OC\cdot \sin COM=R\cdot \sin A,\quad OM=OC\cdot \cos \angle COM=R\cdot \cos A.\] Therefore, \[\begin{aligned} OP^2&=OM^2+PM^2=OM^2+(CM-CP)^2\\ &=R^2\big[ \cos^2A+(\sin A- 2\sin B\cos C)^2\big].\end{aligned}\] It suffices to prove that \[4\sin^2B\cos^2C<\cos^2A+(\sin A-2\sin B\cos C)^2,\] or \[1>4\sin A\sin B\cos C.\] We have \[\begin{aligned} 4\sin A\sin B\cos C&=2\sin A \big[ \sin A-\sin (C-B)\big] \le 2\sin A(\sin A-\sin 30^{\circ})\\ &=2\sin^2A-\sin A =1+(\sin A-1)(2\sin A+1)<1.\end{aligned}\] The proof is done.

I hope this solution would not be duplicate with any other presented above.

Since $\angle PCO = 90 - \angle A$ it is well noted that to prove $PO > PC $ is enough.

It is also well-known that $\angle OAP = \angle C - \angle B \ge 30$

Let point $M$ to be the midpoint of $BC$ then

$MP = R\sin{\angle OAP} \>= R \sin{30} = \frac{R}{2}$

since $ 30\le \angle OAP < \angle A \le 90$

$ R < a = 2R \sin{A} < 2R $

therefore,

$ PC = MC - MP \le \frac{a-R}{2} < \frac{R}{2} \le MP < PO $ $Q.E.D$

the last inequality hold since $PO$ is the hypotenuse of the right triangle $\triangle PMO$

My solution:

Let $ A' $ be the reflection of $ A $ in the perpendicular bisector of $ BC $ .
Let $ M, N $ be the midpoint of $ BC, AA' $, respectively .

Since $ \angle AOA'=2(\angle ACB-\angle CBA) \geq 60^{\circ} $ ,
so we get $ AA' \geq OA \Longrightarrow MP=AN \geq \tfrac{1}{2} OA $ ,
hence $ CP=MC-MP<OC-MP \leq \tfrac{1}{2} OA \leq MP \Longrightarrow \angle COP<\angle OCB $
$ \Longrightarrow \angle BAC+\angle COP=\angle MOC+\angle COP<\angle MOC+\angle OCB=90^{\circ} $

Q.E.D

Here's a solution, from working with [b]swimmer[/b]:

We just have to show $PO > PC$. But, using the following diagram:

[img]https://hostr.co/file/J105WwHiuEme/IMOGEO.png[/img]

we can see that $PO > FO > EF > CP$, so we are done.

It suffices to show $OP>CP$. If $M$ is the midpoint of $BC$, this will follow from $MP>CP$. Now let $B_1OC_1$ be a diameter of $(ABC)$ parallel to $ABC$ with $B_1$ on the same side as $B$, and let $AP$ meet $B_1C_1$ at $P_1$. Since $\angle C-\angle B\ge 30$, we know $\angle C_1B_1A\le 30$, so
\[
CP\le P_1C_1\le P_1O=MP<OP
\]
which is what we want.

Consider the following

[b]Lemma.[/b] $CM>2CP$ where $M$ is the midpoint of $BC$.

(Proof) Rewrite the given as $$\tfrac{a^2}{2}>a^2+b^2-c^2 \iff a^2<2(c^2-b^2).$$ Apply sine law to get the equivalent claim $$\sin^2 A < 2(\sin^2 C-\sin^2 B)=2\sin(C-B)\sin A \implies \sin A <1,$$ which is clearly true. $\blacksquare$

Apply the lemma to get $CP<PM<PO$, so $\angle COP<\angle OCP=90^{\circ}-\angle A$ as desired. $\blacksquare$

It is equivalent to prove $OP>CP$.
Let $R$ denote circumradius of $\Delta ABC$.

WLOG assume $R=\frac{1}{2}$. Some length chase gives $CP=\sin B \cos C$.

Using Law of Cosines on $\Delta POC$,
\begin{align*}
OP^2-CP^2&=R^2-2\cdot R \cdot CP\cos(90^{\circ}-A)\\
&=\frac{1}{4}-\sin A \sin B \cos C\\
\end{align*}
It is now equivalent to prove $$LHS>0 \implies 4\sin A \sin B \cos C < 1$$

Note-(\star)=reason (written below)

\begin{align*}
4\sin A \sin B \cos C &= 2\sin A \cdot (2\sin B \cos C)\\
&=2\sin A \cdot (\sin A - \sin (C-B))\\
&<2\sin A \cdot (1-\frac{1}{2})[(\star) \sin A<1, (-\sin(C-B))<\frac{1}{2}]\\
&<\sin A\\
&<1\\
\end{align*}
As desired.

The conclusion rewrites as  \begin{align*} 	\angle COP &< 90^{\circ} - \angle A = \angle OCP \\ 	\iff PC &< PO \\ 	\iff PC^2 &< PO^2 \\ 	\iff PC^2 &< R^2 - PB \cdot PC \\ 	\iff PC \cdot BC &< R^2 \\ 	\iff ab \cos C  &< R^2 \\ 	\iff \sin A \sin B \cos C & < \frac14. \end{align*} Now  \[ \cos C \sin B  	= \frac{1}{2} \left( \sin(C+B)-\sin(C-B) \right) 	\le \frac{1}{2} \left( 1 - \frac{1}{2} \right) = \frac 14 \] which finishes when combined with $\sin A < 1$.

[b]Remark[/b]:  	If we allow $ABC$ to be right then 	equality holds when $\angle A = 90^{\circ}$, 	$\angle C = 60^{\circ}$, $\angle B = 30^{\circ}$. 	This motivates the calculation above.

[quote=orl]Consider an acute-angled triangle $ABC$. Let $P$ be the foot of the altitude of triangle $ABC$ issuing from the vertex $A$, and let $O$ be the circumcenter of triangle $ABC$. Assume that $\angle C \geq \angle B+30^{\circ}$. Prove that $\angle A+\angle COP < 90^{\circ}$.[/quote]
[asy]
import olympiad;
import geometry;
size(7cm);
pair O=origin;
pair A,B,C,T,P,M;
A=dir(60);
B=dir(210);
C=dir(330);
P=foot(A,B,C);
M=(B+C)/2;
draw(A--B--C--A,springgreen);
T=extension(A,B,O,M);
draw(segment(A,P),green);
draw(segment(O,C),green);
draw(segment(O,P),green);
draw(segment(M,T),green);
dot(A^^B^^C^^M^^O^^T^^P);
draw(segment(T,C),green);
label("$A$",A,N);
label("$B$",B,SW);
label("$C$",C,SE);
label("$T$",T,NW);
label("$O$",O,NE);
label("$M$",M,SW);
label("$P$",P,SW);
[/asy]
\[\angle A+\angle COP < 90^{\circ} \Longleftrightarrow \angle COP < 90^{\circ}-\angle A \Longleftrightarrow \angle COP < \angle OCP \Longleftrightarrow PC < OP \Longleftrightarrow PC^2 < OP^2\].Note that $OP^2=R^2-BP\cdot PC$ by Power of a point.Hence \[R^2-BP\cdot PC > PC^2 \Longleftrightarrow R^2 > BP\cdot a \Longleftrightarrow \tfrac{a^2+b^2-c^2}{2} < R^2\]
 ---------
Now  we exploit the condtion $\angle C\geq \angle B+30^{\circ}$.Since $\angle TCB=\angle TBC\implies \angle ACT\geq 30^{\circ} \implies \sin{ACT}\geq \tfrac{1}{2}$.Hence we have  \[\dfrac{c^2-b^2}{2aR}=\sin{ACT}\geq \tfrac{1}{2} \implies (b^2-c^2)\leq aR\].Thus \[\dfrac{a^2+b^2-c^2}{2}\leq \dfrac{a^2+aR}{2}=\dfrac{4R^2\sin{A}^2 + 2R^2\sin{A}}{2} < R^2 \] \[\implies 2\sin{A}^2+\sin{A}<1 \implies (\sin{A}+1)(2\sin{A}-1)<0 \implies \sin{A}>\tfrac{1}{2}\implies \angle{A}>30^{\circ}\]. So we just have to show $\angle{A}>30^{\circ}$.Assume not then  \[2\angle{C}\geq 30^{\circ}+\angle{B}+\angle{C} \geq \angle{A}+\angle{B}+\angle{C}=180^{\circ} \implies \angle{C}\geq 90^{\circ} \] a contradiction since $ABC$ is acute.Hence $\angle{A}<30^{\circ}$ and we are done.$\blacksquare$

WLOG $R=1$, and let $\angle{COP}=x$. Since $\sin$ is increasing over $[0, 180]$, we just need to prove $\sin(x)<\sin(90-A)=\cos A$. We have $\sin x=\cos A\cdot \frac{CP}{OP}$, so it suffices to prove $OP>CP$, or $OP^2>CP^2$. But $OP^2>CP^2\iff OM^2+CP^2+CM^2-2(CP)(CM)>CP^2\iff OC^2>2(CP)(CM)\iff (CP)(CM)< 1/2$. Since $R=1$ we have $BC=2\sin A$ so $CM=\sin A$; also, $CP=AC\cos C=2\sin B\cos C=\sin A-\sin(C-B)$. Thus, we desire to prove $\sin A(\sin A-\sin(C-B))<1/2$; since $\sin(C-B)\ge \sin 30=1/2$, it suffices to prove $\sin A(\sin A-1/2)<1/2$. This expression is clearly increasing in $A\in (0, 90]$, but $A<90$, so $\sin A(\sin A-1/2)<\sin 90(\sin 90 -1/2)=1/2$, and we are done.

As per the usual, let $a = BC, b = AC, c = AB, \angle BAC = a, \angle ABC = b, \angle ACB = c,$ and $R$ be the circumradius of $ABC$. It’s clear that $$\angle C \ge \angle B + 30 \Longleftrightarrow C -  B \ge 30 \Longleftrightarrow \sin( C - B )\ge \frac{1}{2}$$
This is the same as saying:
$$\sin C \cos B - \sin B \cos C \ge \frac{1}{2} \Longleftrightarrow \frac{c}{2R} \left( \frac{c^{2}+a^{2}-b^{2}}{2ca} \right)- \frac{b}{2R} \left( \frac{a^{2}+b^{2}-c^{2}}{2ab} \right)\ge \frac{1}{2}$$
After some cleaning, this simplifies to:
$$\frac{2c^2-2b^2}{4R} \ge \frac{a}{2} \Longleftrightarrow c^2-b^2 \ge aR$$
Now, we should work a bit on the condition $A + \angle COP < 90$. Note that $\angle OCB = 90-a$, so we want $\angle OCP > \angle COP \Longleftrightarrow OP > CP$.
Denote $M$ the midpoint of side $BC$. Notice that $OM^2 = CO^2 - CM^2 = R^2 - \frac{a^2}{4}$. Let $CP$ have side length $x$; then $PB = a-x$. Therefore, since $AP^2 = AC^2-CP^2 = AB^2 - PB^2$, we know that:
$$b^2 - x^2 = c^2 - (a-x)^2 \Longleftrightarrow b^2=c^2-a^2-2ax \Longleftrightarrow x = \frac{b^2-c^2+a^2}{2a}$$
It follows that $PM = \frac{c^2-b^2}{2a}$. Therefore,
$$OP^2 = PM^2 + OM^2 = \frac{(c^2-b^2)^2}{(2a)^2} + R^2 - \frac{a^2}{4}$$
$$CP^2 = \left( \frac{c^2-b^2}{2a} \right)^2$$
and
$$OP^2 - CP^2 = \left( \frac{c^2-b^2}{2a} \right)^2 -  \left( \frac{b^2-c^2+a^2}{2a}\right)^2 + R^2 - \frac{a^2}{4} = R^2 - \frac{a^2}{4} + \frac{a^2}{2a}\left(\frac{2c^2-2b^2-a^2}{2a} \right) = R^2 - \frac{a^2}{2} + \frac{c^2-b^2}{2}$$
Using what we found earlier: namely that $c^2-b^2 \ge aR$, we see that:
$$OP^2 - CP^2 = R^2 - \frac{a^2}{2} + \frac{c^2-b^2}{2} \ge R^2 - \frac{a^2}{2} + \frac{aR}{2} = \left( R + \frac{A}{4}\right)^2 - \left(\frac{3A}{4} \right)^2 = \left( R - \frac{a}{2}\right) \left(R + a \right)$$
Clearly, $R + a \ge 0$, so all that is left to do is show that $2R \ge A$, which is a trivial corollary of the Law of Sines. Therefore, $OP^2 - CP^2 \ge 0 \Longleftrightarrow OP \ge CP$, which we showed earlier is equivalent to $\angle A + \angle COP < 90$.

Let $M$ the midpoint of $BC$,
then it's not difficult to see that $MP \geq R/2$ (this is because $MP=Rsin(\angle C-\angle B).$
After this, $OP \geq R/2 \geq CP $ , this is what we wanted to show.

Note that
\[\angle COP+\angle CAB=180^\circ - \angle OCP - \angle OPC + \angle CAB = 180^\circ - (90^\circ - \angle CAB) - \angle OPC+\angle CAB=\]
\[90^\circ+2\angle CAB-\angle OPC.\]
Thus, it is equivalent to show that
\[\angle OPC>2\angle CAB = \angle COB.\]
Now, note that because $\angle ACB>\angle ABC$, we have that $P$ lies closer to $C$ than to $B$ along line $BC$.

Note that if $PC<PO$, then $\angle COP<\angle PCO$ by Law of Sines, so
\[\angle OPC = 180^\circ - \angle COP - \angle PCO > 180^\circ - 2\angle PCO = 180^\circ - 2\angle BCO = \angle COB.\]
Thus, it suffices to show $PC < PO$. By the Law of Cosines,
\[PO^2 = PC^2+OC^2-2OC\cdot PC \cos \angle OCP,\]
so it suffices to show
\[OC > 2PC\cdot \cos \angle OCP = 2PC \cdot \cos \angle OCB = 2PC \cdot \sin \angle CAB = 2AC \cdot \cos \angle ACB \cdot \sin \angle CAB=\]
\[AC \cdot [\sin (\angle ACB+\sin \angle CAB)-\sin (\angle ACB - \angle CAB)].\]
Note that if $R$ is the circumradius, then $OC = R = \frac{AC}{2\sin \angle ABC}$. Thus, we can divide through by $R$ to see that we wish to show
\[1 > 2\sin \angle ABC \cdot \left(\sin \angle ABC - \sin (\angle ACB - \angle CAB)\right) = 1-\cos 2\angle ABC -\]
\[\cos (\angle ACB - \angle CAB - \angle ABC)+\cos (\angle ACB - \angle CAB +\angle ABC)=\]
\[1-\cos 2\angle ABC -\cos (2\angle ACB - 180^\circ)+\cos (2\angle CAB - 180^\circ) = \]
\[1-\cos 2\angle ABC + \cos 2\angle ACB - \cos 2\angle CAB.\]
Finally, we see that it suffices to show
\[\cos 2\angle ACB < \cos 2\angle ABC + \cos 2\angle CAB.\]
Take $2\angle ABC=r,\angle ACB - \angle ABC = x$. We want to show
\[\cos (r+2x) < \cos r + \cos (360^\circ - 2r-2x) = \cos r + \cos (2r+2x).\]
Equivalently, we wish to show
\[0< \cos (2r+2x)+[\cos r-\cos (r+2x)] = \cos (2r+2x) + 2\sin (r+x)\sin x.\]
Remark that $\sin x \ge 1/2$, then it suffices to show that for all $r+x$ that
\[0 < \cos (2r+2x)+\sin (r+x) = 1 - 2\sin^2(r+x)+\sin (r+x).\]
If $\sin (r+x)=k$, we want to show that $2k^2-k-1 < 0$. By rearranging, this is equivalent to showing that $(4k-1)^2 < 9$; that is, showing that $|4k-1| < 3$ or that $k<1$ (as it is clear that $0\le k$). However, it is clear that $\sin (r+x) \ne 1$; otherwise we would get $\cos (2r+2x) = -1 \implies \cos 2\angle CAB = -1 \implies \angle CAB = 90^\circ$, which is untrue. Thus, we are done.

Let $M$ be the midpoint of $BC$. Note that $\angle C\geq \angle B$, so $\angle PAO=\angle C-\angle B\geq 30^{\circ}$. Seeing that the perpendicular from $O$ to $AP$ is equal to $PM$, we have the relation \[PM=AO\sin{\angle PAO}\geq \frac{1}{2}AO=\frac{1}{2}R\] where $R$ is the circumradius. Thus, \[PM\geq {1}{2}R\iff 2(MC-PC)=BC-2PC\geq R\] thus $PC<\frac{1}{2}(BC-R)$.

I claim that $\angle A+\angle COP<90^{\circ}$ is equivalent to showing $OP>CP$. Indeed, note that $\angle OCP=90^{\circ}-\angle A$ so \[OP>CP\iff \angle OCP>\angle COP\iff 90^{\circ}-\angle A>\angle COP\iff \angle A+\angle COP<90^{\circ}\] From right triangle $OPM$, $OP>PM$, so \[2OP>PM\geq R \iff OP>\frac{1}{2}R\] Since $2R\geq BC$, we have $PC<\frac{1}{2}(BC-R)\leq\frac{1}{2}(2R-R)=\frac{1}{2}R<OP$ as desired.

[quote=orl]Consider an acute-angled triangle $ABC$. Let $P$ be the foot of the altitude of triangle $ABC$ issuing from the vertex $A$, and let $O$ be the circumcenter of triangle $ABC$. Assume that $\angle C \geq \angle B+30^{\circ}$. Prove that $\angle A+\angle COP < 90^{\circ}$.[/quote]
[b][color=#f00]Solution[/color]:[/b] Observe that $$\angle A+\angle COP < 90^{\circ} \iff \angle COP < 90^{\circ} - \angle A = \angle OCB = \angle OCP \iff PC < OP \iff PC^2 < OP^2-(\star)$$ Now, let $\Omega$ denote the circumcircle of $\triangle ABC$ and $R$ be the circumradius. So, we have $|pow_{\Omega}(P)|=OP^2 - R^2$
$=PB \cdot PC$. Hence from $(\star)$ we have
$$PC^2 - R^2 < OP^2 - R^2 = PB \cdot PC \iff PC^2 - PB \cdot PC < R^2 \iff PC(PC-PB)<R^2 \iff \underbrace{PC(PC+BP)}_\text{(using directed lengths)} < R^2$$
$$\iff PC \cdot BC < R^2 \iff ab\cos C < R^2 \iff \underbrace{4\sin A\sin B\cos C}_\text{(using extended sine rule)} < 1-(\bigstar)$$
Now note that $2\sin B\cos C = \sin (B+C) - \sin (C-B)=\sin A - \sin (C-B)$. We also had $B \geqslant C + 30^{\circ} \iff (B-C) \geqslant 30^{\circ}$ now since the $\triangle ABC$ is an acute-angled, we have $\sin (B-C) \geqslant \frac12$ and $\sin A < 1$. On combining all these we get:
$$4\sin A\sin B\cos C = (2\sin A) (2\sin B \cos C) = (2\sin A)(\sin A - \sin (C-B)) < 2 \left(1 - \frac12 \right)=1 \quad \blacksquare$$

Clearly we have to prove that $OP>CP\iff OP^2>CP^2$.
We introduce Cartesian coordinates. $O=(0,0), B=(-c,y), C=(c,y), A=(p,q), P=(p,y)$. The point $A_1=(-p,q)$ is on $(ABC)$. $$2p=AA_1= 2R\sin \gamma - \beta \geq 2R \sin 30=R= OC=\sqrt{c^2+y^2} > c$$
$CP^2=(c-p)^2<p^2<p^2+y^2=OP^2$

Same synthetic solution as above posts.
[asy]
unitsize(48);
pair A, B, C, M, O, P, Q;
A = dir(60); B = dir(200); C = dir(340);
O = (0, 0); M = (B+C)/2; P = foot(A, B, C); Q = foot(O, A, P);

draw(unitcircle);
draw(A--B--C--cycle);
draw(A--O^^B--O^^C--O);
draw(A--P);
draw(O--P);
draw(O--M);
draw(O--Q);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$O$", O, dir(130));
dot("$M$", M, dir(270));
dot("$P$", P, dir(270));
dot("$Q$", Q, dir(0));
//label("$\ge30^\circ$", A, 10*dir(dir(B-A)+dir(C-A)), black+fontsize(6pt));
[/asy]
Let $M$ be the midpoint of $\overline{BC}$ and let $Q$ be the foot from $O$ to $\overline{AP}$, so $OMPQ$ is a rectangle. Let $r$ be the circumradius of $\triangle ABC$ (lowercase $r$ is used to avoid confusion with point names).

Note $\angle OAQ = \angle C - \angle B \ge 30^\circ$, so $OQ \ge \tfrac12OA = \tfrac12r$ and thus $MP \ge \tfrac12r$. Thus
\begin{align*}
OP & > MP = \frac12r\\
PC & = MC - MP < OC - MP
\le r - \frac12r = \frac12r.
\end{align*}
Thus $\boxed{OP > PC}$, which is enough to finish:
\[ \angle A + \angle COP < \angle A + \angle OCP = 90^\circ. \]

Let CO meet circle at D. we have ∠CAD = 90 so we need to prove ∠COP < ∠BAD or ∠COP < ∠PCO or PC < PO. Let OH be perpendicular to BC.
PC^2 = (b.cos ∠C)^2 , PO^2 = OH^2 + PH^2 = OH^2 + (a/2 - (b.cos ∠C))^2 so PO^2 - PC^2 = OH^2 + a^2/4 - (a^2 + b^2 - c^2)/2
so we need to prove OH^2 + a^2/4 - (a^2 + b^2 - c^2)/2 > 0 or in fact OH^2 + (c^2 - b^2)/2 > a^2/4.
sin ∠(C-B) ≥ sin(30) = 1/2 so sin ∠C. cos ∠B - sin ∠B. cos ∠C ≥ 1/2 so c^2 - b^2 ≥ aR which R is radius of circle.
we needed to prove OH^2 + (c^2 - b^2)/2 > a^2/4 and we have OH^2 + aR/2 > a^2/4 so we need to prove R ≥ a/2 which is true.
we're Done.

Interesting.

Let $M$ and $X$ be the feet of the perpendiculars from $O$ to $BC$ and $AP$, respectively. Easy angle chasing gives $\angle PAO = \angle B - \angle C\ge 30^\circ$, so $MP = OX\ge \frac{1}{2}OA = \frac{1}{2}OC$.

Let $P_1$ be the point on $MC$ such that $MP_1 = \frac{1}{2}OC$. We know that $P_1$ lies on segment $MC$ because $\angle COM = \angle A > 30^\circ$ from the angle condition and the fact that $ABC$ is acute. Since $\angle COP\le \angle COP_1$, clearly it suffices to show that $\angle A + \angle COP_1 < 90^\circ$.

Now let $N$ be the midpoint of $OC$ and let $P'$ be the intersection of the perpendicular bisector of $OC$ with $CM$. Note that $MN = MP_1$. If $\angle COM\le 45^\circ$ we are done, because then $\angle COP_1 + \angle A < 2\angle A\le 90^\circ$. Otherwise, $P'$ lies on segment $CM$. Then easy angle chasing gives $\angle P_1NC = \frac{3}{2}\angle A - 45^\circ < 90^\circ = \angle P'NC$, where the second to last inequality follows from the fact that $ABC$ is acute. Finally, we have $CP_1 < CP' = OP' < OP_1$, so

$\angle COP_1 + \angle A < \angle OCP_1 + \angle A = 90^\circ$

as desired.

Imagine bashing such a beautiful problem.

[img width=50]https://media.discordapp.net/attachments/925784397469331477/931973262831915058/Screen_Shot_2022-01-15_at_11.08.56_AM.png?width=1220&height=1169[/img]

Let $D$ be on $AP$ and $E$ be on $BC$ such that $ODPE$ is a rectangle. $\angle ADO=90^\circ$ and \[\angle DAO=\angle PAB-\angle OAB=90^\circ-\angle B-90^\circ -\angle C=\angle C-\angle B\ge 30^\circ\] so it follows that $DO\ge\frac{1}{2}AO,$ so $EP\ge\frac{1}{2}OC>\frac{1}{2}EC.$

Thus, $OP>EP>PC,$ so $\angle COP<\angle OCP=90^\circ-\angle A,$ as desired.

Imagine not bashing

[asy] import geometry; size(8cm); point A = dir(45), B = dir(205), C = dir(335), O = (0,0), Cp = dir(155), M = (B+C)/2; point P = projection(line(B,C))*A;
draw(A--B--C--A); draw(circle(A,B,C)); draw(C--Cp--A); draw(M--O--P--A, dashed);
markrightangle(Cp,A,C); markrightangle(O,M,C); markangleradiusfactor=-10; markangle(line(C,O), line(C,P)); markangleradiusfactor=-5; markangle(line(A,Cp),line(A,B));
dot("$A$",A,dir(45)); dot("$B$",B,dir(205)); dot("$C$",C,dir(335)); dot("$O$",O,dir(90)); dot("$M$",M,dir(270)); dot("$P$",P,dir(270)); dot("$C'$",Cp,dir(155)); [/asy]
We first start with some basic synthetic observations.
[b][color=red]Claim:[/color][/b]  We are done if and only if $OP > PC$.
[i]Proof.[/i]  Rearrange to show that we want to prove that $\angle COP < 90 - \angle A$. Furthermore, define $C'$ to be the intersection of $CO$ and the circumcircle of $(ABC)$ ($\neq C$, of course). Then, note that  \[ 90 - \angle A = \angle C'AB = \angle C'CB = \angle OCB \] which we want to show is $> \angle COP$, i.e., that $OP > PC$. $\blacksquare$
Note trivially that $OP > PM$, where $M$ is the midpoint of $BC$. We therefore are done if we prove that $MP \geq PC$, or equivalently if $PC \leq \frac{1}{3} PB$. We now perform a bash.
[b][color=red]Claim:[/color][/b]  Given the problem conditions, $PC \leq \frac{1}{3}PB$.
[i]Proof.[/i]  Write: \[ \frac{PC}{PB} = \frac{\frac{PC}{PA}}{\frac{PB}{PA}} = \frac{\tan \angle B}{\tan \angle C} \] and furthermore note that since $\angle C < 90$, $\tan \angle C \geq \tan (\angle B + 30)$ (tan is an increasing function from $0$ to $90$). Furthermore, note that $30 < \angle B < 60$, or else $\angle A$ or $\angle C$ is obtuse. Thus, let $x = \tan B$, and we bash. We have: \[ \frac{\tan \angle B}{\tan \angle C} \leq \frac{\tan \angle B}{\tan(\angle B+30)} = \frac{x}{\frac{x+\tan 30}{1- x \tan 30}} =  \frac{x - \frac{x^2}{\sqrt{3}}}{x+\frac{1}{\sqrt{3}}} = \frac{x\sqrt{3} - x^2}{x\sqrt{3}+1} \] which we want to show is $\leq \frac{1}{3}$. Evaluating this expression at $x = \tan 30$ gives us exactly $\frac{1}{3}$, so we only need to show that this is decreasing between $x = \tan 30$ and $x = \tan 60$. To do this, we take the derivative: \[ \left(\frac{x\sqrt{3} - x^2}{x\sqrt{3}+1}\right)' = \frac{(\sqrt{3}-2x)(x\sqrt{3}+1) - \sqrt{3}(x\sqrt{3}-x^2)}{(x\sqrt{3}+1)^2} = \frac{-\sqrt{3}x^2-2x+\sqrt{3}}{(x\sqrt{3}+1)^2} \] which has a clearly positive denominator and negative numerator when $x \geq \frac{1}{\sqrt{3}}$, and is therefore negative, which finishes. $\blacksquare$

Let $A_1$ be the point on $(ABC)$ such that $AA_1 \parallel BC$, the midpoint of $BC$ be $M$, and $Q$ lie on $BC$ such that $QC = QO$.

Observe that $$\angle ACA_1 = \angle ACB - \angle BCA_1 = \angle C - \angle AA_1C = \angle C - \angle ABC \ge 30^{\circ}$$ so $$\angle AOA_1 = 2 \angle ACA_1 \ge 60^{\circ}.$$ Thus, since $OA = OA_1$, we have $AA_1 \ge OA = R$, implying $$MP = \frac{AA_1}{2} \ge \frac{R}{2}.$$ Now, because $ABC$ is acute, we know $CB < 2R$, which also means $CM < R$. It follows that $$CB \cdot CP < 2R \cdot (CM - MP) < 2R \cdot (R - MP)$$ $$\le 2R \cdot \left( R - \frac{R}{2} \right) = R^2 = CO^2 = CB \cdot CQ$$ so $CP < CQ$. Now, we know $$OP > OQ = CQ > CP$$ so $$\angle A + \angle COP < \angle A + \angle OCP = 90^{\circ}$$ as required. $\blacksquare$

[b]Remarks:[/b] To show $OP > OQ$, just notice that $MP > MQ$ follows from $CP < CQ$, yielding $$OP^2 - OQ^2 = MP^2 - MQ^2 > 0$$ as required.

Also, I actually discovered the key idea and solved this question while watching a movie :P.

[quote=fuzimiao2013]
apparently this is a "gem" according to channing421

Let $M$ be the midpoint of $BC$ and $Q$ be the foot of the altitude issuing from the vertex $C$.  It suffices to prove that $\color{blue}OP>PC$, since then $\angle COP < \angle PCO = 90^{\circ} - \angle MOC = 90^{\circ} - A$.

By LoC on $\triangle OPC$,\begin{align*}OP^2 &= PC^2 + CO^2 - 2\cdot PC \cdot CO \cdot \cos \angle BCO \\ &= PC^2 + R^2 - 2R\cdot PC \cdot \cos \angle BCO \\ &> PC^2.\end{align*}Substituting $b\cos C$ for $PC$, $R$ for $OC$, and simplifying with $\angle BCO = \angle QCA$, \begin{align*}\Leftrightarrow R &> 2\cdot PC\cdot \cos \angle BCO \\ \Leftrightarrow \frac{b}{2\sin B} &> 2\cdot b\cos C \cdot \cos (90^{\circ} - A) \\ \Leftrightarrow \frac 14 &< \sin A\sin B\cos C = \sin(B+C)\sin B\cos C.\end{align*}This final proof is left as an exercise to the reader. $\square$

[hide=Remark]aka i dont rly wanna write that last janky trig ineq up[/hide]
[/quote]

[hide=Storage]
Note that $$\angle A + \angle COP < 90 \iff \angle COP < 90 - \angle A \iff PC < PO$$ Then, \begin{align*}
    PC < PO &\iff \\
    PC^2 < R^2 + PC^2 - 2 \cdot R \cdot PC \cos\angle OCP &\iff \\
    2 \cdot PC \cos\angle OCP < R &\iff \\
    \sin B \cdot \cos C \cdot \sin A < \frac{1}{4} \\
\end{align*}

Since $\angle C \ge \angle B + 30$, The final inequality follows as $$\sin B \cdot \cos C \cdot \sin A < \sin B \cdot \cos (B + 30) \cdot \sin (A)$$ and \begin{align*}&\sin B \cos (B + 30) = \sin B \sin (60  - B) \\ &= \frac{1}{2}(\sin(B + (B + 30)) - \sin(B - (B + 30))) \le \frac{1}{4}\end{align*}
[/hide]

[b]Claim:[/b] $R^2 > ab \cos C$.
[i]Proof:[/i] By the extended law of sines, $R^2 = \frac{a}{2 \sin A} \cdot \frac{b}{2\sin B}$. We can divide out $ab \cos C$ to get $\frac{R^2}{ab \cos C} = \frac{1}{4\sin A \sin B \cos C}$. But $\sin B \cos C = \frac 12 ( \sin(B+C) + \sin(B-C)) = \frac 12(\sin A - \sin (C-B))$, so $$\frac{R^2}{ab \cos C} = \frac{1}{2 \sin A (\sin A - \sin(C-B))}.$$ But notice that $2\sin A (\sin A - \sin (C-B))$ is maximized when $\sin A$ is maximized and $\sin(C-B)$ is minimized, and since $A<90$, $\sin A < 1$ and since $C-B \geq 30, $ $\sin(C-B) \geq 1/2$, so $2\sin A ( \sin A - \sin (C-B)) < 1$, or $\frac{R^2}{ab \cos C} > 1$, implying $R^2 > ab\cos C$, as desired.

[b]Claim:[/b] $PO>CP$.
[i]Proof:[/i] We find that $CP = b \cos C$ and $PO = \sqrt{(\frac a2 - b \cos C)^2 + R^2 - (\frac a2)^2} = \sqrt{b^2 \cos^2 C + R^2 - ab \cos C}.$ Since $R^2 - ab\cos C$, $\sqrt{b^2 \cos^2 C + R^2 - ab \cos C} > \sqrt{b^2 \cos^2 C} = b \cos C$, so $PO>CP$.

Since $PO>CP$, then $90 - \angle A =\angle PCO > \angle COP$, but then $\angle A + \angle COP < \angle A + \angle PCO = 90$, so we are done.

Though similar to others, I find this trig bash very less bashy
The key claim is as follows,
[color=#f00][b]Claim :[/b][/color] $R^2 \geq ab cos{c}$
Proof : We first prove a lemma
[b]Lemma:[/b] $$sin(a)sin(b)cos(c) \leq \frac{1}{4}$$
equivalently $$sin(a)(sin(b+c)-sin(c-b)) \leq \frac{1}{2}$$
                      $$sin(a)(2sin(a)+1) \leq 1$$
$$(sin{a}-1)(2sin{a}+1) \leq 0$$
which is true
By sine rule the claim immediately follows, and by cosine rule in $\triangle OCP$ the problem gets finished.
EDIT: just noticed this is identical to mathmanman's soln lol

Note that $A + \angle COP < 90 \equiv \angle COP < 90 - A = \angle OCP \equiv CP<OP$. We now bash. Note that $OP>CP$ is equivalent to \[ R^2\cos^2a + \left(\frac{a}{2}-b \cos C\right)^2 > b^2\cos^2C \equiv ab \cos C < R^2\cos^2A+\frac{a^2}{4} = R^2 = \frac{(abc)^2}{16 \cdot [ABC]^2} = \frac{a^2b^2c^2}{4 \cdot bc \sin A \cdot ac \sin B} \equiv \sin A \sin B \cos C < \frac{1}{4}. \] Switch to radians. If $B \le \frac{\pi}{6}$, we need to prove $\sin B \sin \left(B+\frac{\pi}{6}\right) = f(B) \le 1/4$. We use calculus to see $f'(B) = \sin\left( 2B+\frac{\pi}{6} \right) > 0$. Thus we only need to check $B=\frac{\pi}{6}$, which is obvious. We now note that if we fix $B$ and vary $C$ from $B+\frac{\pi}{6}$, $\cos C$ will decrease and $\sin A$ will decrease. Thus we want $C=B+\frac{\pi}{6}$. Thus we need to prove $g(x) = \sin\left(2x+\frac{\pi}{6}\right) \sin x \cos \left(x+\frac{\pi}{6}\right) < \frac{1}{4}$ for $\frac{\pi}{6} < x < \frac{\pi}{3}$. We use calculus again. Note that $g'(x) = \cos\left(2x+\frac{\pi}{6}\right) \sin x \cos \left(x+\frac{\pi}{6}\right) + \sin\left(2x+\frac{\pi}{6}\right) \cos \left(2x+\frac{\pi}{6}\right) = \cos\left(2x+\frac{\pi}{6}\right) \left( \sin x \cos \left( x + \frac{\pi}{6} \right) + \sin \left( 2x +\frac{\pi}{6} \right) \right) < 0$, so we just need to check $x = \frac{\pi}{6}$ which is obvious and we're done.

[asy]
unitsize(2.8cm);
pair A,B,C,O,P;
A=(0.4,0.9165151389912);
B=(-0.8,-0.6);
C=(0.8,-0.6);
O=(0,0);
P=(0.4,-0.6);
draw(circle((0,0),1));
draw(A--B--C--cycle);
draw(O--C);
draw(B--O--P--A);
label("$A$",A,NE);
label("$B$",B,SW);
label("$C$",C,SE);
label("$O$",O,N);
label("$P$",P,S);
[/asy]
Note that $\angle BOC=2\angle A$ due to the Inscribed Angle Theorem. Since $\triangle BOC$ is isosceles with $BO=OC$, it follows that
\begin{align*}
\angle OBC&=\angle OCB\\
&=\frac{180^{\circ}-\angle BOC}2\\
&=\frac{180^{\circ}-2\angle A}2\\
&=90^{\circ}-\angle A.
\end{align*}
Thus, proving that $\angle A+\angle COP<90^{\circ}$ is equivalent to proving that $\angle COP<90^{\circ}-\angle A=\angle OCP.$ Applying the Law of Sines, we find that $\frac{OP}{\angle OCP}=\frac{PC}{\angle COP}$, so that $$\frac{OP}{PC}=\frac{\sin\angle OCP}{\sin\angle COP}.$$ If $OP<PC$, then $\sin\angle OCP<\sin \angle COP$ and $\angle COP<\angle OCP$, implying that $\angle OCP+\angle COP>180^{\circ}$, a contradiction. It therefore follows that having $\angle A+\angle COP<90^{\circ}$ is equivalent to $\angle COP<\angle OCP$, which in turn is the same as having $PC<OP$, as proven above. Thus, it suffices to show that $PC<OP.$

Defining the side lengths $(BC,CA,AB)=(a,b,c)$, as usual, we note that $$PC=b\cos\angle C$$
by analysis on $\triangle ACP.$ Moreover, applying the Law of Cosines on $\triangle OCP$ gives
\begin{align*}
OP^2&=OC^2+CP^2\\
&=OC^2+CP^2-2OC\cdot CP\cos\angle OCP\\
&=PC^2+R^2-2R(b\cos\angle C)\cos(90^{\circ}-\angle A)\\
&=PC^2+R^2-2Rb\cos\angle C\sin\angle A,
\end{align*}
where $R=OA=OB=OC$ is the circumradius of $\angle ABC.$

Now, note that $R=\frac{b}{2\sin\angle B}$ by the Law of Sines, it follows that
\begin{align*}
\frac{2(OP^2-PC^2)\sin\angle B}{bR}&=\frac{2R\sin\angle B-4b\cos\angle C\sin\angle B\sin\angle A}b\\
&=\frac{b-4b\cos\angle C\sin\angle B\sin\angle A}b\\
&=1-4b\cos\angle C\sin\angle B\sin\angle A.
\end{align*}
Since $\frac{2\sin\angle B}{bR}$ is clearly positive, we see that $\frac{2(OP^2-PC^2)\sin\angle B}{bR}=1-4b\cos\angle C\sin\angle B\sin\angle A$ and $OP^2-PC^2$ share the same sign. Thus, to prove that $PC<OP$, it suffices to show that $\cos\angle C\sin\angle B\sin\angle A<\frac14.$
\paragraph{}Note that $\sin\angle A<1$ due to the acuity of $\angle ABC$; in addition, the fact that $\angle C\ge\angle B+30^{\circ}$ implies that $\cos\angle C\le\cos(\angle B+30^{\circ})$ as the cosine function is decreasing on the range $[0^{\circ},180^{\circ}].$ Therefore
\begin{align*}
\cos\angle C\sin\angle B\sin\angle A&<\cos\angle C\sin\angle B\\
&\le\cos(\angle B+30^{\circ})\sin\angle B\\
&=\sin(90^{\circ}-(\angle B+30^{\circ})\sin\angle B\\
&=\sin(60^{\circ}-\angle B)\sin\angle B.
\end{align*}
Applying the Sum-to-Product Identity for products of sines, we find that
\begin{align*}
\cos\angle C\sin\angle B\sin\angle A&<\sin(60^{\circ}-\angle B)\sin\angle B\\
&=\frac{\cos((60^{\circ}-\angle B)-\angle B)-\cos((60^{\circ}-\angle B)+\angle B)}2\\
&=\frac{\cos(60^{\circ}-2\angle B)-\cos60^\circ}2\\
&\le\frac{1-\frac12}2\\
&=\frac14,
\end{align*}
as desired. $\blacksquare$

[b]Remark: [/b]Reasonably easy problem, if you ask me. In any diagram, it looks to be the case that COP+BAC<OCP+BAC=90, finishing, so it suffices to prove CP<OP. There's also a sort of degree of "equality" even though the thing we want is strict, and even though this is geometry: When I saw this, I thought of "What if BAC approaches 90 degrees? What becomes degenerate?" Well, O lies on CP, and any right angles dropped from O would lie on BCP, so weaker inequalities like leg<hypotenuse are viable.

Indeed, construct Q=foot from O on BC, R=foot from O on AP; $$ARO=90,RAO=BAP-BAO=90-B-(90-C)=C-B\ge 30\implies PQ=OR\ge\frac12AO=\frac12CO>\frac12CQ\implies OP>PQ>CP,$$ as desired. $\blacksquare$

Denote the projections from $O$ onto $BC$ and $AP$ as $M$ and $N$, respectively. Then
\[\angle OAN = C-B \ge 30 \implies \sin OAN \ge \frac 12.\]

Thus we get that
\[\frac 12 \leq \frac{NO}{AO} = \frac{PM}{CO} < \frac{PM}{CM} \implies PC < PM < PO.\]

As a result, we have $\angle COP < \angle PCO = 90-A$, giving the desired. $\blacksquare$

Let $D$ be the second intersection of $CO$ and $(ABC)$. Now, notice that
\begin{align*}
    \angle CAB + \angle COP &< 90^\circ\\
    \angle COP &< 90-\angle CAB\\
    \angle COP &< \angle DAB \\
    \angle COP &< \angle DCB = \angle OCP
\end{align*}
Thus, the required conclusion rewrites to showing $\angle COP < \angle OCP$ or in turn $PC < OP$.\\
Now, let $X = \overline{OC} \cap \overline{AP}$. First, notice that
\[PC < XC \text{ ($\angle XPC= 90^\circ$ and thus $\angle PXC < 90$)}\]
Next by Ratio Lemma,
\begin{align*}
    \frac{OX}{XC} &= \frac{AO}{AC} \cdot \frac{\sin{\angle OAP}}{\sin{\angle PAC}}\\
    OX &= \left( \frac{AO}{AC}\cdot \frac{\sin{\angle OAP}}{\sin{\angle PAC}} \right) XC.
\end{align*}
Now,
\begin{align*}
    \frac{AO}{AC}\cdot \frac{\sin{\angle OAP}}{\sin{\angle PAC}} &= \frac{\sin{\angle OCA}}{\sin{\angle AOC}}\cdot \frac{\sin{\angle OAP}}{\sin{\angle PAC}} \\
    &= \frac{\sin{90-\angle B}}{\sin{\angle B}} \cdot \frac{\sin{\angle B + \angle C}}{\sin{90-\angle C}}\\
    &= \left(\frac{BP}{AB}\cdot\frac{AC}{PC}\right) \left(\frac{\frac{AP}{AB}\cdot\frac{PC}{AC}+\frac{AP}{AC}\cdot\frac{BP}{AB}}{2\cdot \frac{AP}{AB}\cdot\frac{BP}{AB}}\right)\\
    &= \frac{PC+BP}{2PC} > 1
\end{align*}
since $A$ is closer to $C$ than $B$ ($\angle BCA > \angle ABC$) and $P$ is the foot of the altitude. Further, $$\angle OXP = 90 + \angle OCP > \angle OPA \text{ (this angle is strictly less than $90^\circ$)}$$
So, $OX< OP$ as well. Thus, we have that
\[PC < XC < OX < OP\]
and thus, $PC < OP$ as needed.

Notice that because $\angle OCP = 90^\circ - \angle A$, it suffices to show that $\angle COP < \angle OCP$, or $CP < OP$. Using the Law of Cosines, $$CP^2 < OP^2 = CP^2 + OC^2 - 2  (OC) (CP)( \sin A)  \iff 1 < 4 \sin B \cos C \sin A$$ as $CP = 2R \sin B \cos C$. On the other hand, note that $$ 2\sin B \cos C \sin A < 2\sin B \cos C = \sin(B+C) - \sin(C-B) \leq \sin(B+C) -\frac 12 \leq \frac 12$$ because $C-B\geq 30^\circ$, as needed.


The sides of a triangle have lengths of 15, 20, and 25. Find the length of the shortest altitude. \[ \text {(A) } 6 \qquad \text {(B) } 12 \qquad \text {(C) } 12.5 \qquad \text {(D) } 13 \qquad \text {(E) } 15  \]

[hide="bleh"]
Notice that 15, 20, 25 is a Pythagorean triple.  Thus, the area of the triangle is 150.

Then, the altitude to the side with length 25 is $\frac{150}{25}\cdot 2=12$.  This is clearly less than the other altitudes (15 & 20).

$\boxed{B}$
[/hide]

[hide="answer"]This is a right triangle (since 15,20, and 25 are multiples of 3,4, and 5), so two of the altitudes have lengths 15 and 20. Their product is 300. Using 25 as a base, we can find the length of the third altitude x by 25x=300, and x=12. [b]B [/b]is our answer.[/hide]

[hide]
Sine this is a right triangle, we use the formula Altitude=ab/c, thus getting 15*20/25=12, so B.
[/hide]


[quote="maokid7"]For $x+y+z=1$ and $x, y, z >0$ prove that

$0\leq xy+yz+xz-2xyz\leq \frac{7}{27}$[/quote]
$xy+yz+xz-2xyz\leq \frac{7}{27}\Leftrightarrow$
$\Leftrightarrow27(x+y+z)(xy+xz+yz)-54xyz\leq7(x+y+z)^3\Leftrightarrow$
$\Leftrightarrow7\cdot\sum_{cyc}(x^3-x^2y-x^2z+xyz)+\sum_{cyc}(x^2y+x^2z-2xyz)\geq0,$
which true by Schur and Muirhead.


Let $ ABCD$ be a convex quadrilateral (not necessarily cyclic). Let lines $ AB$ and $ CD$ meet at $ E$, $ AD$ and $ BC$ meet at $ F$. Suppose $ X$ is a point inside the quadrilateral such that $ \angle AXE \equal{} \angle CXF$.
Prove that $ \angle AXB \plus{} \angle CXD \equal{} \pi$.

can anyone post solution nice of problem?
I know,this problem is Bulgari TST 2005.
But I haven't official solution  .
Can anyone post them?


superior algebra
superior algebra unsolved

Let $ x\in G$ where $ G$ is a group and $ x,x^{2},x^{4},x^{5}\neq e$ as well as $ x^{6}\equal{}e$, where $ e$ is the identity element of $ G$.  Is this enough information to say what the order of $ x$ is?  I'm having trouble trying to find out if $ x^{3}\equal{}e$ or not using this data.

Of course not.  For example, $ \text{ord}_{7}(2) \equal{} 3$.

[quote="amcavoy"]Let $ x\in G$ where $ G$ is a group and $ x,x^{2},x^{4},x^{5}\neq e$ as well as $ x^{6}\equal{} e$, where $ e$ is the identity element of $ G$.  Is this enough information to say what the order of $ x$ is?  I'm having trouble trying to find out if $ x^{3}\equal{} e$ or not using this data.[/quote]
Consider for example $ \left<\mathbb{Z}_{7},\plus{}\right>$. And let $ x\equal{}1$. Then $ x,x^{2},x^{4},x^{5},x^{6}\not \equal{} 0$ but yet $ x^{3}\not 0$ instead $ x^{7}\equal{}0$. (Here $ e\equal{}0$). So this shows what you are trying to show is wrong.

The problem is actually stated as follows:

Let $ x$ belong to a group.  If $ x^{2}\neq e$ and $ x^{6}\equal{} e$, prove that $ x^{4}\neq e$ and $ x^{5}\neq e$.  What can we say about the order of $ x$?

Can we prove anything about $ x^{3}$?

[quote="amcavoy"]The problem is actually stated as follows:

Let $ x$ belong to a group.  If $ x^{2}\neq e$ and $ x^{6}\equal{} e$, prove that $ x^{4}\neq e$ and $ x^{5}\neq e$.[/quote]
This means that $ x\not \equal{} e$ because otherwise surly $ x^{2}\equal{} e$ a contradiction. Now, $ e \equal{} x^{6}\equal{} (x^{2})^{3}$ so the order of $ x^{2}$ is at most three; it cant be $ 1$ because $ x^{2}\not \equal{} e$ and it cant be $ 2$ because otherwise it must be the case that $ 2|3$ which is an impossibility, the only only possibility is that the order of $ x^{2}$ is $ 3$. Okay, that means $ (x^{2})^{2}$ cant be the order, i.e. $ x^{4}\not \equal{} e$. Finally, $ x^{5}\not \equal{} e$ because if $ x^{5}\equal{} e$ then since $ e \equal{} (x^{2})^{3}\equal{} x^{6}$ we have $ x^{5}\equal{} x^{6}$, by group cancellation we got $ x \equal{} e$ which is too an impossibility.

[quote]  What can we say about the order of $ x$?[/quote]
I guess we dont know, it can either be $ 3$ or it can be $ 6$.


Solve in $ \mathbb{R}$ the equation $ \lvert\ x\minus{}\frac{\pi}{6}\rvert \plus{}\lvert\ x\plus{}\frac{\pi}{3}\rvert\equal{}\arcsin \frac {x^3\minus{}x\plus{}2}{2}$  :?:

_______________________________
$ \textbf{Azerbaijan Land of Fire }$

[quote="Pirkuliyev Rovsen"]Solve in $ \mathbb{R}$ the equation $ \lvert\ x \minus{} \frac {\pi}{6}\rvert \plus{} \lvert\ x \plus{} \frac {\pi}{3}\rvert \equal{} \arcsin \frac {x^3 \minus{} x \plus{} 2}{2}$  :?: [/quote]

For $ x\le \minus{} \frac {\pi}3$, the equation is $ \minus{} 2x \minus{} \frac {\pi}6 \equal{} \arcsin\frac {x^3 \minus{} x \plus{} 2}2$ : in this equation, $ LHS\ge\frac {\pi}2$ while $ RHS\le\frac {\pi}2$ and so equality may only occur for $ x \equal{} \minus{} \frac {\pi}3$, which does not fit.

For $ \minus{} \frac {\pi}3\le x\le\frac {\pi}6$, the equation is $ \frac {\pi}2 \equal{} \arcsin\frac {x^3 \minus{} x \plus{} 2}2$, and so $ x^3 \minus{} x \plus{} 2 \equal{} 2$ and so $ x \equal{} 0$ (solution $ x \equal{} 1\notin [ \minus{} \frac {\pi}3,\frac {\pi}6]$) or $ x \equal{} \minus{} 1$ ([u][b]edited due to Prester's remark[/b][/u])

For $ \frac {\pi}6\le x$, the equation is $ 2x \plus{} \frac {\pi}6 \equal{} \arcsin\frac {x^3 \minus{} x \plus{} 2}2$ : in this equation, $ LHS\ge\frac {\pi}2$ while $ RHS\le\frac {\pi}2$ and so equality may only occur for $ x \equal{} \frac {\pi}6$, which does not fit.

Hence the two solutions $ x \in\{ \minus{} 1, 0\}$  ([u][b]edited due to Prester's remark[/b][/u])

[quote="pco"][u]xref4search [/u]: solve in R | x - pi/6 | + | x + pi/3 | = arcsin( x^3 - x + 2 ) / 2 ); abs ; arcsin [quote="Pirkuliyev Rovsen"]Solve in $ \mathbb{R}$ the equation $ \lvert\ x \minus{} \frac {\pi}{6}\rvert \plus{} \lvert\ x \plus{} \frac {\pi}{3}\rvert \equal{} \arcsin \frac {x^3 \minus{} x \plus{} 2}{2}$  :?: [/quote]

For $ x\le \minus{} \frac {\pi}3$, the equation is $ \minus{} 2x \minus{} \frac {\pi}6 \equal{} \arcsin\frac {x^3 \minus{} x \plus{} 2}2$ : in this equation, $ LHS\ge\frac {\pi}2$ while $ RHS\le\frac {\pi}2$ and so equality may only occur for $ x \equal{} \minus{} \frac {\pi}3$, which does not fit.

For $ \minus{} \frac {\pi}3\le x\le\frac {\pi}6$, the equation is $ \frac {\pi}2 \equal{} \arcsin\frac {x^3 \minus{} x \plus{} 2}2$, and so $ x^3 \minus{} x \plus{} 2 \equal{} 2$ and so $ x \equal{} 0$ (solution $ x \equal{} 1\notin [ \minus{} \frac {\pi}3,\frac {\pi}6]$)

For $ \frac {\pi}6\le x$, the equation is $ 2x \plus{} \frac {\pi}6 \equal{} \arcsin\frac {x^3 \minus{} x \plus{} 2}2$ : in this equation, $ LHS\ge\frac {\pi}2$ while $ RHS\le\frac {\pi}2$ and so equality may only occur for $ x \equal{} \frac {\pi}6$, which does not fit.

Hence the unique solution $ x \equal{} 0$[/quote]

I found also $ x\equal{}\minus{}1$,  because $ \minus{}1 > \minus{}\frac {\pi}3$

[quote="prester"] I found also $ x \equal{} \minus{} 1$,  because $ \minus{} 1 > \minus{} \frac {\pi}3$[/quote]

Quite exact !  :oops:

So set of solutions is $ \{\minus{}1,0\}$

Thanks for your correction.


probability
inequalities
function
combinatorics unsolved
combinatorics

You have to organize a fair procedure to randomly select someone  from $ n$ people  so that every one of them would be chosen with the probability $ \frac{1}{n}$. You are allowed to choose two real numbers $ 0<p_1<1$ and $ 0<p_2<1$ and order two coins which satisfy the following requirement: the probability of tossing "heads" on the first coin $ p_1$ and the probability of tossing "heads" on the second coin is $ p_2$. Before starting the procedure, you are supposed to announce an upper bound on the total number of times that the two coins are going to be flipped altogether. Describe a procedure that achieves this goal under the given conditions.

[quote="rem"]You have to organize a fair procedure to randomly select someone  from $ n$ people  so that every one of them would be chosen with the probability $ \frac {1}{n}$.
You are allowed to choose two real numbers $ 0 < p_1 < 1$ and $ 0 < p_2 < 1$ and order two coins which satisfy the following requirement: the probability of tossing "heads" on the first coin $ p_1$ and the probability of tossing "heads" on the second coin is $ p_2$.
Before starting the procedure, you are supposed to announce an upper bound on the total number of times that the two coins are going to be flipped altogether.
Describe a procedure that achieves this goal under the given conditions.

Taken from http://www.artofproblemsolving.com/Forum/viewtopic.php?p=967228#967228.[/quote]

My general approach to this--I'm not going to rigorously fill in all the details--is to pick a sufficiently large power of $ n \minus{} 1$--let's say $ (n \minus{} 1)^k$--such that "most" of the $ (n \minus{} 1)^k \plus{} 1$ integers $ \binom {(n \minus{} 1)^k}{0}$, $ \binom {(n \minus{} 1)^k}{1}$, $ \binom {(n \minus{} 1)^k}{2}$, ..., $ \binom {(n \minus{} 1)^k}{(n \minus{} 1)^k}$ are divisible by $ n \minus{} 1$.  (If $ n \minus{} 1$ is prime this is very easy.)  We then toss a coin--its heads probability not yet determined--exactly $ (n \minus{} 1)^k$ times.  Our strategy is that if it comes up heads $ i$ times, and $ (n \minus{} 1)\not|\binom {(n \minus{} 1)^k}{i}$, to pick the first person--and if $ (n \minus{} 1)|\binom {(n \minus{} 1)^k}{i}$ to pick one of the remaining $ n \minus{} 1$ people.  Because there are $ \binom {(n \minus{} 1)^k}{i}$ outcomes where there are $ i$ heads, all equally likely, these can be fairly assigned to the $ n \minus{} 1$ remaining people when $ (n \minus{} 1)|\binom {(n \minus{} 1)^k}{i}$.  It remains only to pick the heads probability, $ p$, such that the probability of the number of heads satisfying $ (n \minus{} 1)|\binom {(n \minus{} 1)^k}{i}$ is exactly $ \frac {n \minus{} 1}{n}$.  I claim that this is possible as long as "most" values of $ i$ satisfy $ (n \minus{} 1)|\binom {(n \minus{} 1)^k}{i}$.  Specifically we require that $ \sum_{0\le i\le (n \minus{} 1)^k: (n \minus{} 1)|\binom {(n \minus{} 1)^k}{i}} \frac {1}{2^{(n\minus{}1)^k}} \binom {(n \minus{} 1)^k}{i} > \frac {n \minus{} 1}{n}$ and it should always be possible to satisfy this inequality if we pick $ k$ to be large enough.

My question, though, is this approach actually requires only [i]one[/i] coin, not two coins as contemplated by the problem.  Although there are many individual values of $ n$ for which I can find a better solution--requiring fewer coin tosses--if we are allowed two coins, I cannot get any improvement in the general case of all $ n$--either in terms of elegance of solution or number of coin tosses required--by allowing two coins.  The "worst case" which is attained for some values of $ n$ is pretty much as described above.

So I'm wondering if others who solved this problem found the same thing or if having the second coin really helps us at all in the general case.

In a town of $n$ people, a president is to be elected. The people have decided that the best way to do this is randomly: they want to construct a random process that elects exactly one of the $n$ people, all with equal probability of being elected president.

However, these are ancient times and the only way these people can obtain random results is by flipping a coin. However, they know that the process is probably impossible with a fair coin, so they have asked their blacksmith to build a skewed coin, which comes up heads with probability $p$ and tails with probability $1-p$, for some $0<p<1$.

Is it always possible to choose an adequate $p$, such that the fair election of a president is possible?

[hide="Solution"]

Yes, it is always possible. Number the people from 1 to $n$, where the $n$-th person is the blacksmith.

Take any $m$, and take any sequence of $m$ binary digits. Let ${{m}\choose{i}} = (n-1)x_i+y_i$ be the number of possible sequences that contain $i$ digits "1", where $x_i \ge 0$ and $0 \le y_i \le n-2$. Arrange these sequences in any order. Assign the first $x_i$ sequences to the first person, the second $x_i$ sequences (from $x_i+1$ to $2x_i$) to the second person, and so on. "Burn" the last $y_i$ sequences. The first $n-1$ people have sequences assigned to them, and the last person doesn't.

Now, toss a coin (with $p$ to be determined) $m$ times. If a sequence that is not burnt is the result, then the corresponding person is elected president. The first $n-1$ people all have the same probability of being elected. If a sequence that is burnt is the result, then the blacksmith is elected president. If we can guarantee that the probability of a burnt sequence being the result is $1/n$, then we will be done. Note that the probability of a burnt sequence being the result is

$Q(p) = \displaystyle\sum_{k=0}^m p^k(1-p)^{m-k}y_k$

where $Q$ is a polynomial. Notice that $y_0=1$. Then, $Q(0) \ge 1^my_0 = 1$. If we can prove that $Q(0.5) \le 1/n$, then, since $Q$ is continuous, there must exist a $0.5 \le p \le 0$ such that $Q(p)=1/n$, and choosing this $p$ we will be done.

However,

$Q(0.5) = \displaystyle\sum_{k=0}^m \displaystyle\frac{y_k}{2^m} \le \displaystyle\frac{(m+1)(n-2)}{2^m}$,

which grows as a linear function in $m$ divided by an exponential function in $m$. Therefore, for sufficiently big $m$, $Q(0.5) \le 1/n$, and we're done.
[/hide]

I'm rooting for the blacksmith.

(This problem is a generalization of the original problem, using only 1 coin).


[quote="bgbgbgbg"]find the exact value of sin(pi/11),cos(pi/11),tan(pi/11) please show the way[/quote]

see http://www.literka.addr.com/mathcountry/algebra/trigvalues/values11.htm


Find all $ a,b,c\in \mathbb{Z}$ such that

$ a^2\plus{}b^2\plus{}c^2\equal{}a^2b^2$.

[hide]We need only consider nonnegative $ a,b,c$, as we could simply replace any of them with their negative and not change the equation.

Note that $ (0,0,0)$ is a solution; we will show it is the only one.

Consider the equation mod 4. If 2 or 3 of $ a,b,c$ are odd, then the LHS is 2 or 3 mod 4, but they equal $ (ab)^2$; contradiction. So, at most 1 of $ a,b,c$ is odd. If 1 of them is odd, then the LHS is odd. But by Pigeonhole, one of $ a,b$ must be even, so the RHS is even; contradiction. So they're all even.

Let $ 2a_1\equal{}a,2b_1\equal{}b,2c_1\equal{}c$.

The equation becomes

\[ a_1^2\plus{}b_1^2\plus{}c_1^2\equal{}4a_1^2b_1^2 \hspace{3cm} (*)\]

Since the RHS is odd, there are an even number of odd terms on the LHS.

If 2 of $ a,b,c$ are odd, consider the equation $ (*)$ mod 4. Then the LHS is 2 mod 4, but the RHS is 0; contradiction.

So they're all even again. Generally, we see that if $ a_i,b_i,c_i$ (where $ a\equal{}2^ia_i$, etc.) satisfy

\[ a_i^2\plus{}b_i^2\plus{}c_i^2\equal{}2^ia_i^2b_i^2\]

then they are all even. So, $ 2^i|a$ for all positive integers $ i$, that is, $ a\equal{}0$. This holds similarly for the other two, and we are done.

P.S. I think this was in Engel.[/hide]


if $ABC$ is a triangle and $m_a , m_b , m_c$ are its medians, $h_a , h_b , h_c$ are the altitudes
 show that
 $\frac{h_a}{m_b}+\frac{h_b}{m_c}+\frac{h_c}{m_a} \le 3$

[quote="11235813"]Very Easy.

Step1) Cauchy

step2) expand  will get <(a-b)(b-c)(c-a)>^2>0[/quote]
Please, explain step 1 ;)

[quote="campos"]not cauchy nor expansion are needed in this problem :?

it follows immediately from $h_a\leq m_a$[/quote]
Very good and easy proof!   :D  :rotfl:

[quote="campos"]not cauchy nor expansion are needed in this problem :?

it follows immediately from $h_a\leq m_a$[/quote]
You are wrong. So easyly you can't do it. We have inequality $\frac{h_a}{m_b}+\frac{h_b}{m_c}+\frac{h_c}{m_a}\leq 3$, but not $\frac{h_a}{m_a}+\frac{h_b}{m_b}+\frac{h_c}{m_c}\leq 3$ ;)

[quote="11235813"]Very Easy.

Step1) Cauchy

step2) expand  will get <(a-b)(b-c)(c-a)>^2>0[/quote]

Can you explain because I don't think cauchy works.

:oops: i misread the problem, i'll give it a try  ;)


calculus
derivative
logarithms
calculus computations

Does anyone know how to work out the differential of the following, could you tell me the identities you used in your solution??

[b]diff[/b](1/[b]ln[/b]2){(1-x)[b]ln[/b]((1-x)/(1-y)) + x[b]ln[/b](x/y)} [b]w.r.t y[/b]

Thankyou

i have moved your previous post into a more appropriate subforum: http://www.mathlinks.ro/Forum/viewtopic.php?t=82769 . before posting something again, try thinking why it disappeared...

Hint to your question: $\ln\frac{a}{b}=\ln a-\ln b$ helps simplifying the term into a form easily differentiable. Since you differentiate with respect to y, you can treat any terms which don't contain y as constants. I won't say more since this is probably homework.

Hi there,

this isnt homework....i am reading through some 'neural networks' notes....could you just tell me the identities for differentiating '[b]ln[/b]((1-x)/(1-y))' w.r.t. y or [b]ln[/b](1/(1-y) w.r.t y

Can anyone help me?

Thankyou

Okay,

$\frac{d}{dy}\ln\frac{1-x}{1-y}=\frac{d}{dy}(\underbrace{\ln(1-x)}_{=\text{ constant}}-\ln(1-y)))=-\frac{d}{dy}\ln(1-y)=-(-\frac{1}{1-y})$
$=\frac{1}{1-y}$

(in order to differentiate $\ln(1-y)$, we used the $(f(g(x)))^{\prime}=g^{\prime}(x)\cdot f^{\prime}(g(x))$ rule for $f(x)=\ln x$ and $g(x)=1-x$).

The second one is simpler, do it yourself.

  darij

Thankyou very much for your help!

Kind regards

Mandip


What is the difficulty level of this book? Easy AIME probs?

lol I don't think they're easy AIME problems. These are two random problems, take a look:

> Prove that the Simson lines of diametrically opposite points on the circumcircle of a triangle are perpendicular to each other and meet on the nine-point circle.

> If squares are erected on two sides of a triangle, prove that their circumcircles intersect on the circle whose diamenter is the third side, and the centers of these three circles are the vertices of an isoceles right angled triangle.

Yeah, probably ranges from medium-hard AIME, easy Olympiad (maybe beyond, I'm not sure). It was too hard for me a year ago, so I put it down. I need to go find it again soon.


function
inequalities proposed
inequalities

[quote="TRAN THAI HUNG"]Is this right? :)
$ a_1^3 \plus{} a_2^3 \plus{} ... \plus{} a_{100}^3 \ge \frac {{(a_1 \plus{} a_2 \plus{} ... \plus{} a_n )^3 }}{{100^2 }} \equal{} \frac {{2009^3 }}{{100^2 }}$[/quote]
Of course not! Consider that all of the variables are positive integers but your equality shows that a_1=a_2=...=a_100=2009/100, which is obviously not an integer.

Thanks.I haven't noted that they are integer. :oops:

We have
Step1 $ a_i^3  \plus{} a_j^3  \ge m_i^3  \plus{} m_j^3$
 $ m_i  \plus{} m_j  \equal{} a_i  \plus{} a_j$ $ m_i  \minus{} m_j$ is as smallest as
 possible. $ m_i , m_j$ are integer.
So $ a_1^3 \plus{} a_2^3 \plus{} ... \plus{} a_{100}^3$ achieve min when there are 91 number is equal to 20 and 9 number equal to 21.
The min is 811349 :) .Is that right now? :(

[quote="Algadin"]Let $ a_1,a_2,...,a_{100}\in N^*$ and $ a_1 \plus{} a_2 \plus{} .... \plus{} a_{100} \equal{} 2009$
Find min of $ M \equal{} a_1^3 \plus{} a_2^3 \plus{} .... \plus{} a_{100}^3$[/quote]
[hide]Hint:
Considering that function f(x)=x^3 is a strictly convex function in the closed interval [1,1910], what can we infer about the maximum function of M?? [/hide]

[quote="Algadin"]I don't understand[/quote]
It means when we have a sequence
$ a_1,a_2,...,a_{100}$ with M=k
we can make k is smaller by change $ a_i,a_j$ into $ m_i,m_j$
$ m_i \plus{} m_j \equal{} a_i \plus{} a_j$, $ m_i \minus{} m_j$ is as smallest as
 possible(equal to 0 or 1). $ m_i , m_j$ are integer. :lol:  :)  :roll:


I need to show using the definition, that weak-* topology is weaker than weak topology(that is every open set in weak-* topology is open set in weak topology). Help me please.

Let $ V$ be a normed space, $ V^{*}$ its dual and $ V^{**}$ its double dual. Then:
- the weak topology on $ V^{*}$ is the weakest topology in which all functionals from $ V^{**}$ are continuous
- the weak-* topology on $ V^{*}$ is the weakest topology in which all point-evaluation are continuous.

Every element $ v \in V$ gives you a functional $ J(v) \in V^{**}$, given by: $ J(v)(\phi) \equal{} \phi(v)$, i.e. point evaluation. Because every evaluation is a functional on $ V^{*}$ (but not necessarily vice versa), the weak-* topology is weaker than the weak topology. Informally, weak-*topology requires less functions to be continuous, so it has fewer open sets.

In other words, the basic open sets of the weak topology on $ V^{*}$ are $ \bigcap_{v^{**} \in \Psi}N(\phi_{0},\Psi_{0},\epsilon) \equal{} \{\phi \in V^{*} : |v^{**}(\phi) \minus{} v^{**}(\phi_{0})| < \epsilon$ (prove this!), where $ \Psi$ is any finite set of functionals from $ V^{**}$. For the weak-* topology, you can write a similar base of open sets, only now you take $ \Psi$ to be a subset of $ J(V) \subseteq V^{**}$, not a subset of the whole $ V^{**}$. Of course the basis for the weak topology contains all basic sets that form the basis of weak-* topology, so the latter is weaker.

Hope that clarifies the matter.

Exercise: he weak-* topology is equivalent to pointwise convergence of functionals, i.e. if $ \phi_{n} \in V^{*}$, then $ \phi_{n} \rightarrow \phi$ in weak-*topology if and only if $ \phi_{n}(v) \rightarrow \phi(v)$ for all $ v \in V$.


1) (i) Find values of $ t$ from $ t^3 \minus{} 3t^2 \minus{} 3t \plus{} 1 \equal{} 0$
 Given $ \tan3\theta \equal{} \frac {3\tan\theta \minus{} \tan^3\theta}{1 \minus{} 3\tan^2\theta}$, find the exact value of $ \tan15$ (15 degrees) using (i)

2) How do you solve an equation like

                            $ x \equal{} \frac {4}{3}\sin x$
edit : there was a mistake in the first question, fixed

problem b:
We have 3/4x=sinx
Both sides of this equation are functions the variable x, so we can draw two graphs of two functions on the plane. Determine the intersections, and we have the roots of the equation.

hello, the given equation $ t^3\minus{}3t^2\minus{}3t\plus{}1\equal{}0$ is equivalent to $ (t\plus{}1)(t^2\minus{}4t\plus{}1)\equal{}0$, from here we get the solutions $ t_1\equal{}\minus{}1,t_2\equal{}2\minus{}\sqrt{3},t_3\equal{}2\plus{}\sqrt{3}$. We want to calculate $ \tan(15^\circ)$. Using your
formula we get $ \tan(45^\circ)\equal{}\frac{3\tan(15^\circ)\minus{}\tan^3(15^\circ)}{1\minus{}3\tan^2(15^\circ)}$. Let
$ \tan(15^\circ)\equal{}x$ so we get our equation above and we have $ \tan(15^\circ)\equal{}2\minus{}\sqrt{3}$.

i should have written the full question, i'm trying to edit it right now but im not being allowed to edit my original post

The original question stated to find tan(15) and tan(45) , so yeah Dr Sonnhard i dont think one can use tan45 = 1 since we have to find it. I did not write this before because i thought once we find tan(15) it would not be hard to find tan(45).

for $ \tan 45$, just consider a $ 45\minus{}45\minus{}90$ triangle.  The two angles are congruent, so so are the opposite sides.  Thus the tangent is $ 1$.


PenguinIntegral made an IRC channel for aops but it's kind of i don't complicated so I think this will better.
We can do math contests in real time and other stuff. Check out the freewebs site I made. [url]http://www.freewebs.com/aopsmath[/url]

PenguinIntegral wants me to add: [size=150][b]Live Math help 24/7[/b][/size]

My idea was way too complicated. Check nonies  out, its great.

[quote="anirudh"]we already have an AoPS Classroom.[/quote]
But this one is on 24/7 :D

[quote="nonie"][quote="anirudh"]but no latex[/quote]
unfortunately...[/quote]
Just use the texer and host the picture.

[quote="solafidefarms"]It's a program that converts text into Tex.

Your chat is quite popular at the current time; you have 2 people in it :([/quote]
We need to hold events, maybe one this afternoon? Post about what events you would like and appropriate times.

If you cant see the chat room, then click the 'help with web chats' and make sure you install the latest version of java.


If the non negative real numbers $ x , y, z$ add to $ 2$ ( $ x \plus{} y \plus{} z \equal{} 2$), prove that :

  $ x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2 \plus{}xyz\leq 1$

 Babis

[color=red][b]Sorry, edited[/b][/color]

[quote="stergiu"]If the non negative real numbers $ x , y, z$ add to $ 2$ ( $ x \plus{} y \plus{} z \equal{} 2$), prove that :

  $ x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2 \leq 1$

 Babis[/quote]
Mr Babi i think it is $ x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2 \plus{}xyz \leq 1$

[quote="Dimitris X"][quote="stergiu"]If the non negative real numbers $ x , y, z$ add to $ 2$ ( $ x \plus{} y \plus{} z \equal{} 2$), prove that :

  $ x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2 \leq 1$

 Babis[/quote]
Mr Babi i think it is $ x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2 \plus{} xyz \leq 1$[/quote]

 Yes, it was a typo. I apologize :(  !

 Babis

[quote="stergiu"]If the non negative real numbers $ x , y, z$ add to $ 2$ ( $ x \plus{} y \plus{} z \equal{} 2$), prove that :

  $ x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2 \plus{} xyz\leq 1$

 Babis

[color=red][b]Sorry, edited[/b][/color][/quote]
Let $ 3u \equal{} a \plus{} b \plus{} c, 3v^2 \equal{} ab \plus{} bc \plus{} ca, w^3 \equal{} abc, u^2 \equal{} tv^2$, so $ t \ge 1$.
Then we have to prove after homogenizing: $ 8uw^3 \plus{} 9u^4 \ge 16v^4$. If $ t \ge \frac {4}{3}$ we have $ 8uw^3 \plus{} 9u^4 \ge 9u^4 \ge \frac {4^2}{3^2}9v^4 \equal{} 16v^4$.
So assume $ t < \frac {4}{3}$.
From $ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \ge 0$ we get $ w^3 \ge 3uv^2 \minus{} 2u^3 \minus{} 2\sqrt {(u^2 \minus{} v^2)^3}$.
Thus we just have to proof:
$ 24u^2v^2 \minus{} 16u^4 \minus{} 16 u \sqrt {(u^2 \minus{} v^2)^3} \plus{} 9u^4 \ge 16v^4 \iff$
$ ( \minus{} 16t^2 \plus{} 24t \minus{} 16\sqrt {t}\sqrt {(t \minus{} 1)^3}) \plus{} (9t^2 \minus{} 16) \ge 0 \iff$
$ (t \minus{} \frac {4}{3})( \minus{} 16t \plus{} \frac {8}{3}) \plus{} 16(\frac {2}{9} \minus{} \sqrt {t}\sqrt {(t \minus{} 1)^3}) \plus{} 3(t \minus{} \frac {4}{3})(3t \plus{} 4) \ge 0 \iff$
$ (\frac {4}{3} \minus{} t)(7t \minus{} \frac {19}{3}) \plus{} 16(\frac {2}{9} \minus{} \sqrt {t}\sqrt {(t \minus{} 1)^3}) \ge 0 \iff$
But since $ 1 \le t < \frac {4}{3}$ we get $ (\frac {4}{3} \minus{} t)(7t \minus{} \frac {19}{3}) > 0$ and $ 16(\frac {2}{9} \minus{} \sqrt {t}\sqrt {(t \minus{} 1)^3}) > 0$. And we are done. Equlaity iff $ t \equal{} \frac {4}{3}$ and $ w \equal{} 0$. That is $ a \equal{} b \equal{} 1, c \equal{} 0$ or permutations.

[quote="stergiu"]If the non negative real numbers $ x , y, z$ add to $ 2$ ( $ x \plus{} y \plus{} z \equal{} 2$), prove that :

  $ x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2 \plus{} xyz\leq 1$

 [/quote]
After homogenization we need to prove that $ \sum_{cyc}(x^4 \plus{} 4x^3y \plus{} 4x^3z \minus{} 10x^2y^2 \plus{} 4x^2yz)\geq0,$ which is obvious.


modular arithmetic
LaTeX
analytic geometry
geometry
rectangle

Whenever I use the \pmod{some number here} command in TeXnic Center and compile the document, there is a large space between the number that precedes the parantheses mod. Is there anyway to fix this? Also, I am trying to incorporate the attached gif image to my LaTeX document, and I know that the \includegraphics* command is the best way to do this. However, I have some questions about attaching the picture itself. For example, do I have to use Virtual PDF Printer to create a .pdf file of the gif image, and then put picture1.pdf in place of myimage, or can I directly place picture1.gif where myimage is? Can TeXnic Center handle .gif files, of are only .jpeg/.pdf files allowed? What am I supposed to put in for the numbers after viewport? The file I am trying to include has a length of 500 pixels and has a height of 102 pixels.

The LaTeX guide had the following example, but I dont know what the numbers after viewport stand for ;)

[code]\includegraphics*[viewport=100 200 500 650]{myimage}[/code]

Finally, how would one create alligned bullets in LaTeX? For example,

\bullet text text text
\bullet text text text

First, use \bmod instead of \pmod: $3\equiv8\bmod5$.

I don't know if you can include gif's. I always just use the virtual printer and put in pdf's with \includegraphics* like this website instructs.

The numbers in the viewport are the coordinates of the lower left corner and then the upper right corner. E.g., [viewport=100 200 500 650] says to take the part of the diagram from the lower corner with coordinates of (100,200) to the upper right corner of (500,650). In other words, it takes the rectangle where x ranges from 100 to 500 and y ranges from 200 to 650. You'll have to play with the numbers and compile a few times to find what fits.

For a list with aligned bullets, use itemize:
[code]\begin{itemize}
\item First thing goes here.
\item Second thing goes here.
\item Use as many items as you want.
\end{itemize}[/code]

Put \usepackage{graphicx} in your header. Then use the command \includegraphics{figure.png} to include any .png file. You can easily create PNG-s with any program including Paint. ;)


Six numbers are randomly selected from integers 1 to 45 inclusive. Let p be the probability that at least 3 of the numbers are consecutive. Find the greatest integer less than or equal to 1000p.

Not sure if this is right.

[hide="Solution"]We can use PIE here. We first count the cases with 3 consecutive digits. There are 43 triples (i.e., 1,2,3 ; 2,3,4 ; 3,4,5 ; 4,5,6 ; etc.). And once a triple is picked there are 45-3 = 42 numbers left. We pick three of them in $ \binom{42}{3}$ ways. So far, our running total is $ 43\binom{42}{3}.$ But if we have something like 1,2,3,4,7,9 or something like that, then using the method above, we have counted this once for 1,2,3 and once for 2,3,4. Therefore, such configurations are counted twice, and must be subtracted once. There are 42 4-tuples (i.e., 1,2,3,4 ; 2,3,4,5 ; 3,4,5,6 ; etc.). And once they are picked, there are 45-4 = 41 numbers left, 2 of which can be picked in $ \binom{41}{2}$ ways. Therefore, we have $ 42\binom{41}{2}.$ Subtracting the latter count from the former yields
\[ 43\binom{42}{3} \minus{} 42\binom{41}{2} \equal{} 459200.\]
The total number of ways to pick 6 numbers out of 45 is $ \binom{45}{6} \equal{} 8145060.$ So our probability is $ \dfrac{459200}{8145060} \equal{} \dfrac{80}{1419} \equal{} 0.05637773080.$ Therefore, $ \lfloor 1000p \rfloor \equal{} \lfloor 56.37773080 \rfloor \equal{} \boxed{56}.$ [/hide]

I dont know how to do this problem, but I'm pretty sure you need to also subtract if 5 or 6 numbers are consecutive
ex: 1,2,3,4,5,6 would be repeated 4 times 1,2,3   2,3,4   3,4,5   4,5,6
so assuming you method is correct
$ \[ 43\binom{42}{3}\minus{}42\binom{41}{2}\minus{}41\binom{40}{1}\minus{}40\binom{39}{0}$
which gets you something, but I can't type now because my mom is kicking me off the computer  :blush:

Don't forget that {a,a+1,a+2,k,k+1,k+2} its same thing as {k,k+1,k+2,a,a+1,a+2} ,so u have to substract those to... also dont forget that you substract twice some pairs so you have to add them back ,and calculate as the inclusion-exclusion formula says....it's pretty hard....

[quote="RobRoobiks"]I dont know how to do this problem, but I'm pretty sure you need to also subtract if 5 or 6 numbers are consecutive
ex: 1,2,3,4,5,6 would be repeated 4 times 1,2,3   2,3,4   3,4,5   4,5,6
so assuming you method is correct
$ \[ 43\binom{42}{3} \minus{} 42\binom{41}{2} \minus{} 41\binom{40}{1} \minus{} 40\binom{39}{0}$
which gets you something, but I can't type now because my mom is kicking me off the computer  :blush:[/quote]

No, you do not need to subtract further because the ones that have 5 consecutive digits are counted 3 times when we counted the 3 consecutive digits, and counted 2 times when we counted the 4 consecutive digits, for a total count of +3 - 2 = 1 times. The same applies to the 6 consecutive digits.


Fie $a,b$ reale pozitive. Demonstrati inegalitatea:

$\frac{1}{\sqrt{1+a^{2}}}+\frac{1}{\sqrt{1+b^{2}}}\geq \frac{2}{\sqrt{1+ab}}$

Daca a=2 si $b=\frac{1}{2}$ atunci
  $\frac{1}{\sqrt{1+a^{2}}}+\frac{1}{\sqrt{1+b^{2}}}=\frac{1}{\sqrt{5}}+\frac{2}{\sqrt{5}}< \frac{2}{\sqrt{2}}=\sqrt{2}$
Deci inegalitatea este falsa.

Imi pare cunoscuta problema. Nu cumva trebuie si conditia $a+b=1$?

Daca imi amintesc corect, s-a dat in Sankt Petersburg pentru $a,b \in [0,1]$.

da, greseala mea, avem $a,b \in (0,1)$ sau $ab\geq 3$

Voi rezolva pentru a,b din (0;1).Avem ca $a=\tan x,b=\tan y$.Inegalitatea devine $\cos x+\cos y\geq \frac{2\sqrt{\cos x \cos y}}{\sqrt{\cos{(x-y)}}}$.Asadar avem de demonstrat $\frac{(\cos x+\cos y)^{2}}{4\cos x \cos y}\geq \frac{1}{\cos{(x-y)}}$.Scadem 1 din ambele parti si devine $\frac{(\cos x-\cos y)^{2}}{4\cos x \cos y}\geq\frac{1-\cos{(x-y)}}{\cos(x-y)}$.Acuma avem ca $1-\cos(x-y)=2\sin^{2}{ \frac{x-y}{2}}$ si de asemenea $\cos x-\cos y=-2\sin{\frac{x+y}{2}}\sin{\frac{x-y}{2}}$.Ramanem cu inegalitatea $\sin^{2}{\frac{x+y}{2}}\geq\frac{2\cos x \cos y}{\cos{(x-y)}}$

[color=darkblue][b]Cazul 1.[/b] $\boxed{\ \{a,b\}\subset [0,1]\ }\ .$ In acest caz vo arata ca ${\frac{1}{\sqrt 1+a^{2}}}+\frac{1}{\sqrt{1+b^{2}}}\le \frac{2}{\sqrt{1+ab}}\ \ (*)\ .$

Intr-adevar, presupunem fara a restrange generalitatea ca $0\le a< b\le 1\ .$

Asadar, $(*)\Longleftrightarrow$ $\frac{1}{\sqrt{1+a^{2}}}-\frac{1}{\sqrt{1+ab}}\le \frac{1}{\sqrt{1+ab}}-\frac{1}{\sqrt{1+b^{2}}}$ $\Longleftrightarrow$

$\frac{\sqrt{1+ab}-\sqrt{1+a^{2}}}{\sqrt{1+a^{2}}}\le\frac{\sqrt{1+b^{2}}-\sqrt{1+ab}}{\sqrt{1+b^{2}}}$ $\Longleftrightarrow$ $\frac{a(b-a)}{1+a^{2}+\sqrt{(1+a^{2})(1+ab)}}\le\frac{b(b-a)}{1+b^{2}+\sqrt{(1+b^{2})(1+ab)}}$ $\Longleftrightarrow$

$a(1+b^{2})-b(1+a^{2})\le \sqrt{1+ab}\cdot\left(b\sqrt{1+a^{2}}-a\sqrt{1+b^{2}}\right)$ $\Longleftrightarrow$

$(b-a)(ab-1)\le\frac{(b^{2}-a^{2})\sqrt{1+ab}}{b\sqrt{1+a^{2}}+a\sqrt{1+b^{2}}}$ $\Longleftrightarrow$

$ab-1\le 0\le\frac{(a+b)\sqrt{1+ab}}{b\sqrt{1+a^{2}}+a\sqrt{1+b^{2}}}\ ,$

ceea ce este evident deoarece $ab\le 1\ .$ In concluzie, [u]enuntul este eronat ![/u]

[b]Cazul 2.[/b] $\boxed{\ ab\ge 3\ }\ .$ In acest caz vom arata ca $\frac{1}{\sqrt{1+a^{2}}}+\frac{1}{\sqrt{1+b^{2}}}\ge \frac{2}{\sqrt{1+ab}}\ \ (**)\ .$

Intr-adevar, $ab\ge 3$ $\Longleftrightarrow$ $\frac{4(1+ab)}{1+ab)^{2}}\le 1\Longrightarrow$ $\frac{4(1+ab)}{(1+ab)^{2}}\le \frac{4(1+ab)+(a-b)^{2}}{(1+ab)^{2}+(a-b)^{2}}$ $\Longrightarrow$

$\frac{4}{1+ab}\le \frac{(1+a^{2})+(1+b^{2})+2(1+ab)}{(1+a^{2})(1+b^{2})}\le$

$\frac{(1+a^{2})+(1+b^{2})+2\sqrt{(1+a^{2})(1+b^{2})}}{(1+a^{2})(1+b^{2})}=$ $\left(\frac{\sqrt{1+a^{2}}+\sqrt{1+b^{2}}}{\sqrt{1+a^{2}}\cdot\sqrt{1+b^{2}}}\right)^{2}$ $\Longrightarrow$

$\frac{4}{1+ab}\le\left(\frac{1}{\sqrt{1+a^{2}}}+\frac{1}{\sqrt{1+b^{2}}}\right)^{2}$ $\Longrightarrow$ $\boxed{\ \frac{2}{\sqrt{1+ab}}\le \frac{1}{\sqrt{1+a^{2}}}+\frac{1}{\sqrt{1+b^{2}}}\ }\ .$

[b]Observatie.[/b] Avem egalitate in acest caz daca si numai daca $a=b\ ,$ adica inegalitatea $(**)\ .$

Am folosit urmatoarele inegalitati simple :

$1\blacktriangleright\ x>0\ ,\ y>0\ ,\ z\ge 0\ : \ 1\ge \frac{x}{y}\Longrightarrow\frac{x+z}{y+z}\ge \frac{x}{y}\ .$

$2\blacktriangleright\ 1+ab\le \sqrt{1+a^{2}}\cdot\sqrt{1+b^{2}}\ .$[/color]

Asa cum se observa si din solutia mea trigonometrica de sus aveam $ sin^{2}{\frac{x+y}{2}}\geq 1$ ceea ce era fals.Pentru cealalta parte frumoasa solutie.

[color=darkred][b][u]Un exercitiu similar :[/u][/b] $|ab|\le 1\Longrightarrow (1+ab)^{3}\le (3-ab)(1+a^{2})(1+b^{2})\ .$[/color]

[color=darkred][b][u]Lema.[/u][/b] $\left\{\begin{array}{c}f\in C^{2}(I)\ ,\ f'\le 0\ ,\ f''\le 0\\\\ n\in N^{*}\ ,\ \{x_{1},x_{2},\ldots ,x_{n}\}\subset I\end{array}\right\|$ $\Longrightarrow$ $\sum_{k=1}^{n}f(x_{k})\le n\cdot f\left(\sqrt [n]{x_{1}x_{2}\ldots x_{n}}\right)\ .$[/color]

[color=darkblue][b]Dem.[/b] $\left\{\begin{array}{c}f'\le 0\ ,\ \frac{1}{n}\cdot \sum x_{k}\ge \sqrt [n]{x_{1}x_{2}\ldots x_{n}}\Longrightarrow f\left(\frac{1}{n}\cdot\sum x_{k}\right)\le f\left(\sqrt [n]{x_{1}x_{2}\ldots x_{n}}\right)\\\\ f''(x)\le 0\Longrightarrow \mathrm{(Jensen)\ }\sum f(x_{k})\le n\cdot f\left(\frac{1}{n}\cdot\sum x_{k}\right)\end{array}\right\|\ .$

[u]Exemplu (generalizarea inegalitatii propuse).[/u]

$n\in N\ ,\ n\ge 2\ ;\ \{x_{1},x_{2},\ldots ,x_{n}\}\subset [0,1]$ $\Longrightarrow$

$\boxed{\ \sum\frac{1}{\sqrt [n]{1+x^{n}}}\le \frac{n}{\sqrt [n]{1+x_{1}x_{2}\ldots x_{n}}}\ }\ .$

Intra-devar, pentru $n=2$ am aratat intr-un reply precedent.

Fie $n\ge 3\ .$ Notam $f: [0,1]\rightarrow \mathbb R\ ,\ f(x)=\frac{1}{\sqrt [n]{1+x^{n}}}\ .$

$\left(1+x^{n}\right)f^{n}(x)=1\Longrightarrow$ $(1+x^{n})f'(x)+x^{n-1}f(x)=0\Longrightarrow \boxed{\ f'(x)\le 0\ }$ $\Longrightarrow$

$(1+x^{n})f''(x)=$ $x^{n-2}f(x)\left[(n+1)x^{n}-(n-1)(1+x^{n})\right]=$

$x^{n-2}f(x)\left[2x^{n}-(n-1)\right]\le x^{n-2}f(x)[2-(n-1)]=$

$x^{n-2}f(x)(3-n)\le 0$ $\Longrightarrow$ $\boxed{\ f''\le 0\ }\ .$[/color]

[url]https://artofproblemsolving.com/community/c6h590467p13312183[/url]
Let $a, b $ be positive real numbers such that $ab\ge 3$ , prove that\[\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}\ge \frac{2}{\sqrt{1+ab}}.\]Prove the inequality :$$\frac{1}{\sqrt{1+ab}}<\frac{1}{\sqrt{1+a^{2}}}+\frac{1}{\sqrt{1+b^{2}}}\leq \frac{2}{\sqrt{1+ab}}$$for $0\leq a,b\leq 1$.
[url]https://artofproblemsolving.com/community/c6h546551p3163690[/url]
[url]https://artofproblemsolving.com/community/c6h1975798p13710540[/url]


Time for this forum to end its state of hibernation...

So, who's going?  I'll be there with Westchester/Putnam A, hopefully we'll be the first-placed non-NYC team like two years ago.

[quote="TimReyn"]Time for this forum to end its state of hibernation...[/quote]

Agreed.

I will be going with AAMC A. We just might have three whole teams this year.  :!:

This time I will [i]not[/i] forget anyone I am meeting. :)

[quote="Mgccl"]I wonder is Long Island part of Upstate new york... xDD
if not... then there is no place for Long Island people in the entire forum :([/quote]

NYC is close enough, right?

http://www.artofproblemsolving.com/Forum/index.php?f=233

On topic, who else is going?


Solve for $ x$: $ 5x^2 \plus{} 6x \plus{} 1 \equiv 0 (mod 23)$

[hide="Solution"]23 is prime.

$ 23|(5x\plus{}1)(x\plus{}1) \implies 23|(x\plus{}1) \; \text{or} \; 23|(5x\plus{}1)$

The first possibility occurs when $ x\equal{}23k\minus{}1$.

The second possibility occurs when $ 23h$ ends in 1 or 6. This occurs when $ h \equiv 2 \pmod {5} \iff h \equal{} 5j\plus{}2$.
then, $ 5x\plus{}1 \equal{} 23(5j\plus{}2) \equal{} 5(23j)\plus{}46 \iff x \equal{} 23h\plus{}9$

Thus, we require $ x \equiv \minus{}1, 9 \pmod{23}$
[/hide]


there are three integer a, b, c where a+b+c=13 and $ \frac{b}{a}\equal{}\frac{c}{b}$. find the maximum and minimum of a.

[quote="bjh7790"]there are three integer a, b, c where a+b+c=13 and $ \frac {b}{a} \equal{} \frac {c}{b}$. find the maximum and minimum of a.[/quote]

It's easy to see that $ a,b,c \neq 0$ and $ a,c$ are either both positive or both negative.

So the condition gives $ |b|\equal{}\sqrt{ac} \le \frac{|a|\plus{}|c|}{2}\equal{}\frac{|a\plus{}c|}{2}$

As $ |a\plus{}c| \ge 2|b|>|b|$, $ a,c$ must be both positive.

Now $ 13\equal{}(a\plus{}c)\plus{}b \ge |b|$, so $ 26 \ge a\plus{}c \ge 2$

Just try $ a\equal{}1$ we get $ (a,b,c)\equal{}(1,3,9)$ is a solution. So min. value of $ a$ is $ 1$.

Now if $ a$ is not a perfect square, then from $ b^2\equal{}ac$ we get there must be a prime divisor divides all of $ a,b,c$, i.e. divides $ 13$ also. In this case the prime divisor must be $ 13$ and it easily leads to the solution $ (a,b,c)\equal{}(13,\minus{}13,13)$.

If $ a$ is a perfect square, then we first try $ a\equal{}5^2,c\equal{}1$, impossible.
Then we try $ a\equal{}4^2$, which leads to $ (a,b,c)\equal{}(16,\minus{}4,1)$.

So max. value of $ a$ is $ 16$


Consider the Newton equation $ x''\equal{}x^2$.

Look for solutions of this equation in the form $ x(t)\equal{}c_1(t\plus{}c_2)^a$. Using this find all solutions. Check whether they are defined on the whole real axis, and if not, explain why it does not contradict the uniqueness theorem.

[quote="Countdownmath"]Consider the Newton equation[/quote]

Why is this a Newton equation?

if $ x$ is considered as a position, then $ x'$ is the velocity and $ x''$ is acceleration


pms finally sent out!

[b][size=200]Mafia Game 40!!![/size][/b]

[size=150]Nightmare in Dream Land Part II[/size]

Kirby was minding his own business one day when a vortex appeared in the middle of his backyard.
"OMGS NOOOOO!!!!" he cried out, "I JUST RAKED THOSE LEAVES!! MR. DARK EVIL MAGICIAN PERSON WHOS MAKING THIS VORTEX YOU OWE ME $ \$1000$ FOR LABOR ISSUES AND EMOTIONAL TRAUMA!"

But what would come out of the vortex would make Kirby even more scared and depressed.

The first figure to step out had a tornado for a body.

"GASP! NIGHTMARE!! I THOUGHTS I KILLED HIM BACK IN THE 2ND KIRBY NINTENDO DS GAME!"

He was followed by some more "friends". Dark Matter being one of them.

"OH NOES!!!" Kirby cried out. How could he stop these monsters? No doubt they were stronger than the last time he beat them on easy multiplayer level 1. But he had a plan. He snuck into his basement, found the secret cloner in the back of his basement (because doesn't everyone have one of those?) and cloned himself. Many, many times.

Meanwhile....

"Hey where are we? Near that brat Kirby?" Dark Matter asked
"hmm....the mailbox says here '1 Kirby Place. Note: This is not Kirby's house, no matter what the address says'. Stupid magician. He dropped us in the wrong spot! Looks like we'll hafta go into town to find him!" Nightmare answered.
"Like this? We'll be slaughtered!" Dark Matter cried out
"Fine, we'll call upon 5 of our highest lieutenant Waddle Dees to help us accomplish this job, and i will place a spell on everyone so they all look the ssame...including us!" Nightmare said
"Ugh. and i just got my nails done. whatever, if it'll help us get rid of kirby, fine then"

[quote="Normal, nothing special townie PM"]You are [b][useless power] Kirby[/b]! You are [b]pro-Dream Land [/b] Unfortunately, there is a seeming lack of [item] here, so this power is all but useless. You win when all evil is vanquished from Dream Land [/quote]

[u][b]Players:[/b][/u]

1. 1=2
2. AIME15
3. johnnybeard
4. Dojo
5. cf249
6. 142857
7. Wickedestjr
8. Mewto55555
9. hedgehog12
10. rt_08
11. alexhhmun
12. pythag011
13. iphailedusamo
14. eggylv999
15. james4l
16. chenhsi
17. aleph0
18. CCM
19. Brut3Forc3
20. zapi2007
21. Hamster1800
22. xoangieexo
23. mathblitz
24. erabel
25. andersonw
26. Peter
27. isabella2296
28. infinitypi
29. sup3rcrah3r
30. BOGTRO

[u][b]Deaths:[/b][/u]
None

Let the game begin. or you can standby for the rules (newbies choose the latter)

[color=violet][b][size=200]Rules: (stolen from pacman who stole it from perfect who stole it from a person[/size][/b][/color] READ THEM

Votes must be in bold. If you do not bold your vote, it will not be counted.

Please be attentive and unvote, if necessary, before casting a new vote. This is not required, but I’d appreciate it.

Executions will require a simple majority of votes. Once a player has reached the necessary majority, his pleas are useless and any attempts to unvote will be unheeded. You may vote: no lynch – majority votes of this kind are necessary to end the day without a death.

The game is not to be discussed outside of the thread unless your role specifically states that you may do so.

Once your death scene has been posted, you’re dead. Stop typing. The only thing you can type is a "bah" post.

Don’t edit/delete previously submitted posts

To be sure you are reading this any newbies your first post after this post must have the word "fail" in it

Don’t quote any PMs from me. Neither true, nor false. Seriously.

If you have a night choice to make, it is due by the posted deadline. I will not wait for you – if you do not submit a choice to me, tough toenails! No choice will be made.

If you anticipate being unavailable for more than a 48-hour period, please post a notice to that effect in the thread. Treat this game as a commitment. Be considerate – don’t leave us hanging.

Remember, it’s a game. It’s supposed to played. It’s supposed to be fun. Do your part.

I'm here

Don't forget to attribute your rules to the person who took it off another person who took it off another person who stole it from another person who copied and pasted from another person.

sorry. i edited. looks like i gotta modkill myself  :wink:

zapi read the rules theres something in there you missed  :wink:

Zapi fails, ggnore

[b]Vote: zapi2007[/b] as punishment.

Alright so basically, this is pretty nice how much info we've been given.

So basically we've got a tornado guy (it looks like his name is nightmare) probably gf.

Dark matter and 5 waddle dees. So we've got 7 scum in all.

TY for the role pm btw. Mewto5555 is an imitation mew.

and btw zapi is not a nooby.

[b]Fos:Hamster[/b] although random votes all we got.

just to say, hamster is no newbie neither

(always use alliteration!)

[quote="zapi2007"]that was edited
trust me...
anyone who has been on 4 minutes ago[/quote]

the only edited part was the purple parentheses

[u][i][b]The "zapi fails" Votecount[/b][/i][/u] (otherwise known as the first votecount)
[b]zapi2007 (1)[/b]: Hamster1800

[i]Not voting(29): everyone else[/i]

Can Peter delete all of the beginning confirm posts, so that 1. Official intro post is the second post and 2. Rules is the third post?

[b]vote: zapi[/b] for betraying me in that mafia game, and being lazy/suicidal in another.

OMG can i actually post? Every time I try to post, someone intercept-posts me.

[quote="1=2"]I'm light kirby. I have powers. I can cancel all night actions once (used night 1), and I can revive one person once (not used).[/quote]

Revive isabella2296 might be a good idea.

I'm UFO Kirby. Townie.

[quote="pythag011"]
Revive isabella2296 might be a good idea.
[/quote]

why?

Since izzy was KIRBY, who I'm assuming is the most powerful town role? (I don't actually know though, but from the name, unless mz94 made THE MAIN KIRBY not the main kirby.)

and the fact that its in light blue (nothing we've seen before)

and the fact that i believe he said you guys suck or something to the tune of that

I'm bomb kirby, if I'm killed at night, I kill the person who killed me. (At least, that's what it essentially said)...

[b]vote: pythag011[/b]

Right now we don't have much inclination to finish so how about we kill random people and not let this be player abandoned, hm?

I think that we should [b]vote:Brut3Forc3[/b].

The reason is that he has been extremely active, yet hasn't really posted that much. I remember in one of the mafia games I read about here, that this "seemingly inactive" guy was mafia, and nobody voted him, because they thought he was inactive. Then he sprang up and killed the townies and won. Brut3Forc3 seems scummy to me.

well, is the mod here?

@SMB:  most people have been active, but we still lack posts
but i guess the point is just to target someone who hasn't roleclaimed

im here

if u wanna continue post

otehrwise this will be abandoned

I'd rather not have this abandoned... Anyway's I think Brut3Forc3 hasn't posted in here like for four months!!! Definitely scummy.

well, without enough active players, we'll never get a lynch to get this game moving...
deadline, maybe?

ok, i agree with you guys
if we have to take out someone...
let's just lynch someone who has seemed suspicious, without enough evidence to be totally sure

but that's what this game is all about

[b]Vote: Brut3forc3[/b]

ive just decided to kill brut3forc3 and start night seeing as theres not much activity

he was a waddle dee!

one more to go

[b][u]Dead: [/u]
[color=darkred]hedgehog12 (King Dedede), part-anti-Dream Land, killed Day 1 of complete overkill [/color]
[color=green]Dojo (Hammer Kirby), pro-Dream Land, killed Day 2 of omnipotent being's boredom
iphailedusamo (Cook Kirby), pro-Dream Land, killed Night 2 of kitchen knife wounds
142857 (Throw Kirby), pro-Dream Land, killed Night 2 of an unknown big explosion
andersonw (Tornado Kirby), pro-Dream Land, killed Day 3 of pacman2812's anger at his inactivity [/color]
sup3rcrash3r (Unknown), unknown alignment, modkilled Day 3 of infinitypi's happiness and clumsiness
infinitypi (Unknown), unknown alignment, modkilled Day 3 of mewto55555's iron fist of pushing
[color=violet]mathblitz (Hi-Jump Kirby), pro-Dream Land, killed Night 3 from a flame-infused glass cutter (pretty sad if you ask me ;P) [/color]
[color=red]aleph0(pacman2812), (Waddle Dee), anti-Dream Land, killed Night 3 from a medieval arrow (you know with the feathers and all) [/color]
erabel (Unknown), unknown alignment, modkilled Day 4 of National Treasure allusion-type speech and mod's forgetfulness
[color=cyan]isabella2296 (Kirby), pro-Dream Land, lynched Day 4. FAIL[/color]
[color=green]cf249 (Cutter Kirby), pro-Dream Land, killed night 4 from a series of 3rd-degree burns
crazypianist1116 (Spark Kirby), pro-Dream Land, killed night 4 from a single stab wound [/color]
[color=red]Peter (Waddle Dee), anti-Dream Land, killed night 4 from a sword to the heart [/color]
[color=green]Wickedestjr (Flame Kirby), pro-Dream Land, lynched day 5 from over-pie-ification [/color]
[color=red]ernie (Waddle Dee), anti-Dream Land, killed night 5 of a sword to the heart [/color]
BOGTRO (Meta Knight), pro-himself, killed night 5 of a burst of flame and a knife to the head
[color=green]eggylv999 (Mike Kirby), pro-Dream Land, killed day 6 of stuff
Hamster1800 (Ice Kirby), pro-Dream Land, killed night 6 by an arsonist [/color]
[color=red]alexhhmun (Nightmare), anti-Dream Land, killed night 6 by a (or the dark knight)
Brut3Forc3 (Waddle Dee), anti-Dream Land, lynched day 7 with slight suspicion[/color]
crazychinesemaniac (ERR: SYSTEM OVERLOAD), ERR: SYNTAX, modkilled day 7 cause, well, has anyone even seen him/her lately?[/b]

its night

alive:
1=2
AIME15
Bugi
SMB
pythag011
hongsquare
zapi2007
xoangieexo

lastpost (this is used to preserve order of locked topic)


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $A$ be a square real matrix whose characteristic polynomial has real roots and assume that $e^A$ is diagonalisable. Prove that $A$ is diagonalisable. What if you do not impose the condition on the roots of the polynomial?

A counterexample for the second question would be $\begin{pmatrix}0&1\\-\pi^2&0\end{pmatrix}$, I suppose :). Its exponential matrix is $-I_2$.

As for the first question, $A$ can be brought to its Jordan form through real similarity, due to the given condition, so we may as well assume it was in Jordan form to begin with. The fact that the exponential matrix of a non-diagonalizable Jordan matrix is not diagonalizable is obvious: it's easy to prove that such an exponential matrix is equal to a lower-triangular matrix having the same Jordan block structure as the initial matrix, except that its eigenvalues are $e^{\lambda_i}$, where $\lambda_i$ are the eigenvalues of the initial matrix $A$, by writing $A$ as the sum between its diagonal and its strictly lower triangular part, which commute (what alekk keeps calling Dunford decomposition :)).

[quote="grobber"]A counterexample for the second question would be $\begin{pmatrix}0&1\\-\pi^2&0\end{pmatrix}$, I suppose :). Its exponential matrix is $-I_2$.
[/quote]

But this matrix has eigenvalues $\pm \pi I$, so it is diagonalizable (over $\mathbb{C}$, of course, but real matrix with not-real eigenvalues can be hardly diagonalizable over $\mathbb{R}$, and no conditions for exponent may help apriori.

I know it's diagonalizable; maybe I misunderstood the question :). I thought harazi asked whether it was possible for $e^A$ to be diagonalizable in $\mathcal M_n(\mathbb R)$ without $A$ having the same property.

I don't see what else the second question could mean, since if we work in $\mathcal M_n(\mathbb C)$, we can do exactly what I did above to show that $A$ is diagonalizable in $\mathcal M_n(\mathbb C)$ whenever $e^A$ is.


can we calculate x in "sinx = a" manullay ??????without using calculater or trignometric tables can we calculate x in "sin x= a" where "a" is a fixed constant.
plz help and give authentic arguements.

I dont think so but if $x$ is very small , then we have $sin x\approx x$

Maybe he's talking about that : http://mathworld.wolfram.com/NewtonsMethod.html


Let's inscribe a circle in a triangle. Let the tangent points of the circle to the sides of the triangle divide a triangle (1) into 4 smaller triangles. Let's prove that the incenters of incircles in 3 smaller triangles (those which have a common side with the big one) lay on the biggest inscribed circle (the first one).
Hope it's understandable because I needed to translate this problem.

Moderator: please delete pelao_malo's reply and ban Withy1. It's another cheating attempt (see http://www.mathlinks.ro/viewtopic.php?t=223905 problem 3). Also look at Withy1's other posts - it's his third attempt of cheating.