IMO Shortlist
geometry
homothety
Triangle
inequalities
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]  :)

Sorry if it's a bit murky. I remember posting this one too, and giving a solution I was proud of (can't remember if I actually posted it). However, I can't find that solution.. [Moderator edit: This was at http://www.mathlinks.ro/Forum/viewtopic.php?t=220 .]

Let $D',E'$ be the images of $D,E$ through the homothety of center $F$ and ratio $4$. We have to show that $D'E'\le AB+AC$, so it would be enough to show $AE'+AD'\le AB+AC$. Again, we notice that it's enough to show $AD'\le AC\ (*)$. Let $X$ be the vertex of the equilateral triangle $CAX$, lying on the opposite side of $CA$ as $B$. Clearly, $AX=AC$, so $(*)$ is equivalent to $FD'\le FX=FA+FC$ (the last equality is well-known, and it follows from Ptolemy's equality applied to the cyclic quadrilateral $AFCX$) or, in other words, $4FD\le FA+FC$. In terms of areas, this means $4S(FAC)\le S(AFCX)\iff 3S(FAC)\le S(XAC)$, and this is clear since for fixed $XAC$, the area $FAC$ reaches its maximum when $FA=FC$, and in this case we have equality in the above inequality.

I think this pretty much ends the proof: we have shown that $4FD\le FX$, which is, as we have shown, equivalent to the initial problem.

I saw this problem these days and I was pretty sure it was an ISL problem.

Lets take the equilateral triangles $ ACP$ and $ ABQ$ on the exterior of the triangle $ ABC$.

We have that  $ \angle{APC} \plus{} \angle{AFC} \equal{} 180$, therefore the points $ A,P,F,C$ are concyclic.

But $ \angle{AFP} \equal{} \angle{ACP} \equal{} 60 \equal{} \angle{AFD}$, so $ D \in (FP)$.
Analoguosly we have that $ E \in (FQ)$.

Now observe that $ \frac {FP}{FD} \equal{} 1 \plus{} \frac {DP}{FD} \equal{} 1 \plus{} \frac {[APC]}{[AFC]}\geq 4$, and the equality occurs when $ F$ is the midpoint of $ \widehat{AC}$.

Therefore $ FD \leq \frac {1}{4}FP$, and $ FE \leq \frac {1}{4}FQ$.

So, by taking it metrical, we have that:
$ DE \equal{} \sqrt {FD^{2} \plus{} FE^{2} \plus{} FD \cdot FE}\leq \frac {1}{4}\cdot \sqrt {FP^{2} \plus{} FQ^{2} \plus{} FP \cdot FQ} \equal{} \frac {1}{4}PQ$

But $ PQ \leq AP \plus{} AQ \equal{} AB \plus{} AC$, and thus the problem is solved.

This post was also a spam and as I am unable to delete this post,i am writing the proof of $\frac{[APC]}{[AFC]} \ge 3$.

Note that
$(AF-CF)^2 \ge 0 \Rightarrow AF^2+CF^2+AF*CF \ge 3AF*CF \Rightarrow AC^2 \ge 3AF*CF \Rightarrow AP*CP\sin60^{\circ} \ge 3AF*CF\sin120^{\circ} \Rightarrow \frac{[APC]}{[AFC]} \ge 3$.

This is a really nice problem! Thanks to [b]sayantanchakraborty[/b] for giving some crucial hints leading to the following solution.

Since $\angle BFC, \angle CFA, \angle AFB$ are all equal and sum up to $360^{\circ},$ they must each be equal to $120^{\circ}.$ Construct a point $B'$ outside $\triangle ABC$ such that $\triangle ABB'$ is equilateral. Define point $C'$ analogously. Since $\angle AB'B + \angle AFB = 60^{\circ} + 120^{\circ} = 180^{\circ},$ points $A,B',B,P$ are concyclic. Furthermore, since $\angle B'FB = \angle B'AB = 60^{\circ} = 180^{\circ} - \angle BFC,$ points $C,F,D,B'$ are collinear.

I claim that $FB' \geq 4FD.$ This is equivalent to
\begin{align*}
[\triangle AB'C] & \geq 3[\triangle APC] \\
\iff AB' \cdot B'C \cdot  \sin (60^{\circ}) & \geq 3 \cdot AF \cdot CF \cdot \sin (120^{\circ}) \\
\iff AB' \cdot B'C & \geq 3 \cdot AF \cdot CF .\end{align*}
By cosine rule on $\triangle AB'C,$
\begin{align*}
AC^2 & = AB'^2 + B'C^2 - 2 \cdot AB' \cdot B'C \cdot \cos (60^{\circ})  \\
& = AB'^2 + B'C^2 - AB' \cdot B'C \\
& \geq AB' \cdot B'C , \end{align*}
where we have used the trivial inequality $AB'^2 + B'C^2 \geq 2 \cdot AB' \cdot B'C.$ Hence, it suffices to show that
\begin{align*}
AC^2 & \geq 3 \cdot AF \cdot CF \\
\iff AF^2 + CF^2 - 2 \cdot AF \cdot CF \cos (120^{\circ}) & \geq 3 \cdot AF \cdot CF \\
\iff AF^2 + CF^2 & \geq 2 \cdot AF \cdot CF,\end{align*}
which is true. Similar arguments show that $FC' \geq 4FE.$

The rest is obvious. Both the dilations centered at $F$ which map to $B$ to $B'$ and $C$ to $C'$ have ratio at least 4, so $B'C' \geq 4DE.$ By triangle inequality, we have that
\[AB'+A'C \geq B'C' \implies AB+AC \geq 4DE. \quad \blacksquare\]

For equality to hold, we need $AF=BF=CF,$ that is, the Fermat point must be the circumcenter of $\triangle ABC.$ This is possible iff $\triangle ABC$ is equilateral, because $ \angle AFB = 2 \angle ACB \implies \angle ACB = 60^{\circ}$ and similarly $\angle ABC= 60^{\circ}.$

Erect equilateral triangles $AMC$ and $ANC$ outwardly on the sides of $\triangle ABC$. It is well known that $F \in BM$ and $D \in CN$.

$\blacksquare\boxed{\text{Lemma 1}}$ $\frac{1}{FD}=\frac{1}{FA}+\frac{1}{FC}$ and  $\frac{1}{FE}=\frac{1}{FA}+\frac{1}{FB}$ .
[u][b]Proof[/b][/u]
Taking $\angle FAC= \alpha$ and $\angle FCA= \beta$ and using $\alpha + \beta =60$ ,$$\frac{FD}{FA}+\frac{FD}{FC}=\frac{sin(\alpha)}{sin(60+\beta)}+\frac{sin(\beta)}{sin(60+\beta)}=1$$ and other part is analogously proved.

$\blacksquare\boxed{\text{Lemma 2}}$ $FA+FC \geq 4FD$ and $FA+FB \geq 4FE$
[u][b]Proof[/b][/u]
By lemma 1 , $\frac{FA+FC}{FD}= \left(\frac{1}{FA} + \frac{1}{FC} \right) \cdot ( FA+FC ) \geq 4 \implies FA+FC \geq 4FD$
The other part follows analogously.
Using lemma 2 $$ AB+AC =
 AN+AM
\geq MN = \sqrt{FN^2+FM^2+FN \cdot FM }
= \sqrt{ (FA+FB)^2+(FA+FC)^2+(FA+FB)\cdot(FA+FC)}
\geq 4 \cdot \sqrt{FC^2+FD^2+FC \cdot FD}=4DE$$

Construct equilateral triangles $ACX$ and $ABY$ outside of $ABC$, so it's well known that $BFDX$ and $CFEY$ are lines. $YAFB$ is cyclic, so consider the tangent at the point $T$, the antipode of $Y$, labeled as line $\ell$. Note that $d(Y,AB):d(Y,\ell)=3:4$, so then $\dfrac{FE}{FY}\le \dfrac 14$ with equality only when $F=T$, and similarly $\dfrac{FD}{FX}\le \dfrac 14$. Let $M$ and $N$ be the points on segments $FY$ and $FX$, respectively, such that $\dfrac{FM}{FY}=\dfrac 14$ and $\dfrac{FN}{FX}=\dfrac 14$. Then since $FE\le FM$ and $FD\le FN$, $DE=\sqrt{FD^2+FE^2+FD\cdot FE}\le \sqrt{FN^2+FM^2+FN\cdot FM}=MN=\dfrac{XY}{4}\le \dfrac{AB+AC}{4}$, as desired.

cute

[hide="solution"]
Lemma 1: Given equilateral triangle $\triangle XYZ$, an $X$-cevian intersects $YZ$ and minor arc $YZ$ at $U, V$ respectively. Then $XU \ge 4UV$.
Proof: Just note that equality occurs when we take the median, and rotating the cevian increases $XU$ but decreases $XV$ and thus $UV$ as well.
Lemma 2: Given two points $P, Q$ on an obtuse triangle, $|PQ|$ is maximized when it is the longest side.
Proof: obvious.
Back to the problem:
Append equilateral triangles $\triangle ABP, \triangle ACQ$ outside $\triangle ABC$; it's well-known that $F = BQ\cap CP$. Apply a homothety of factor $4$ w.r.t $F$ sending $D\mapsto D', E\mapsto E'$; by Lemma 1, $D'E'$ is a segment inside obtuse triangle $\triangle FPQ$, so by Lemma 2, $4DE \le D'E' \le PQ$. But by triangle inequality, $PQ \le AP+AQ = AB+AC$, so done.[/hide]

Trigonometry is the best weapon.
[quote=orl]Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \][/quote]
Clearly $\angle AFB=\angle BFC=\angle CFA=120^o.$ Now, erect equilateral triangles $ABC', BCA', CAB'$ on the sides, externally. Then $AFBC'. AFCB'$ are cyclic. Hence, $\angle C'FA+\angle AFC=\angle C'BA+\angle AFC=60^o+120^o=180^o,$ and so $C, F, C'$ are collinear. We get two more symmetric results and so $F$ is teh Fermat point given by $AA' \cap BB' \cap CC'.$

[b]Claim:[/b] $FE: FC' \le 1:4.$
[i]Proof:[/i] Ptolemy yields $FC'=FA+FB.$ Hence, it suffices to show
$$FA+FB \ge 4FE$$
Let $\angle FAB=x.$ Then it suffices to show
$$\frac{FA}{FB}+1 \ge \frac{4FE}{FB} \Leftrightarrow \frac{sin(60^o-x)}{sinx}+1 \ge \frac{4sin(60^o-x)}{sin(60^o+x)}$$
$$\Leftrightarrow \left(cosx-\sqrt{3}sinx \right)^2 \ge 0$$
which is true. $\square$

Similarly we get $FD:FB' \le 1:4$ and so we get $4DE \le B'C' \le AC'+AB'=AB+AC,$ as desired. $\blacksquare$

[quote=orl]Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \][/quote]

The problem is a masterpiece.

[hide=Solution (bash bash bash)]
Let $AF = a, BF = b, CF = c$. By Law of Cosines, $AB = \sqrt{a^2 + b^2 + ab}, AC = \sqrt{a^2 + c^2 + ac}$. Now observe that by Angle Bisector Theorem, $AE = \frac{a}{a + b} \cdot AB = \frac{a}{a + b} \cdot \sqrt{a^2 + b^2 + ab}$. Next, see that by Law of Cosines on $\triangle AEF$,
$$
\frac{a^2 + EF^2 - \left ( \frac{a}{a + b} \right )^2 \cdot (a^2 + b^2 + ab)}{2 \cdot a \cdot EF} = \frac{1}{2},
$$and solving for $EF$ via quadratic formula yields $EF = \frac{ab}{a + b}$. Similarly, $DE = \frac{ac}{a + c}$. Then once again by Law of Cosines,
\[
DE^2 = DF^2 + EF^2 + DF \cdot EF = \left ( \frac{ab}{a + b} \right)^2 + \left (\frac{ac}{a + c} \right)^2 + \frac{ab}{a + b} \cdot \frac{ac}{a + c}.
\]Thus it remains to show the inequality, for reals $a, b, c$,
$$
\sqrt{a^2 + b^2 + ab} + \sqrt{a^2 + c^2 + ac} \ge 4 \cdot \sqrt{ \left ( \frac{ab}{a + b} \right)^2 + \left (\frac{ac}{a + c} \right)^2 + \frac{ab}{a + b} \cdot \frac{ac}{a + c}}.$$As this inequality is homogenous, we set $a = 1$. Now observe that the above inequality is equivalent to
$$
\frac{\sqrt{a^2 + b^2 + ab} + \sqrt{a^2 + c^2 + ac}}{2} \ge \sqrt{ \left ( \frac{2}{\frac{1}{a} + \frac{1}{b}} \right)^2 + \left (\frac{2}{\frac{1}{a} + \frac{1}{c}} \right)^2 + \frac{2}{\frac{1}{a} + \frac{1}{b}} \cdot \frac{2}{\frac{1}{a} + \frac{1}{c}}},$$and by GM-HM, this inequality is weaker than
$$\frac{\sqrt{a^2 + b^2 + ab} + \sqrt{a^2 + c^2 + ac}}{2} \ge \sqrt{ab + ac + a\sqrt{bc}}.$$By AM-GM, this inequality is weaker than
$$\sqrt[4]{(a^2 + b^2 + ab)(a^2 + c^2 + ac)} \ge \sqrt[4]{(ab + ac + a\sqrt{bc})^2}.$$Take both sides to the fourth power, and divide both sides by $a^2bc$ to get
$$\left (b + 1 + \frac{1}{b} \right ) \left (c + 1 + \frac{1}{c}\right) \ge \left (\sqrt{\frac{b}{c}} + 1 + \sqrt{\frac{c}{b}}\right)^2,$$recalling that $a = 1$. Now we expand both sides and subtract the right hand side to get
$$bc + b + c + 1 + \frac{1}{b} + \frac{1}{c} + \frac{1}{bc} - \frac{b}{c} - \frac{c}{b} - 3 - 2 \sqrt{\frac{b}{c}} - 2 \sqrt{\frac{c}{b}} \ge 0,$$which after rearranging is equivalent to
$$bc + \frac{1}{bc} + \left(\sqrt{b} - \sqrt{\frac{1}{c}}\right)^2 + \left(\sqrt{c} - \sqrt{\frac{1}{b}}\right)^2 \ge 2,$$which is obviously true. $\square$[/hide]

Construct points $X, Y, Z$ forming equilateral triangles $\triangle BCX$, $\triangle CAY$, $\triangle ABZ$ sticking out from the triangle. Evidently, $F$ is the intersection of the three circumcircles of these equilateral triangles.

Observe additionally that $F \in \overline{AX}$, and in particular $F$ is the concurrence point of $\overline{AX}$, $\overline{BY}$, and $\overline{CZ}$. Note now that $\angle DFE = \angle BFC = 120^\circ$.

Thus, we can calculate \begin{align*}
    DE^2 = DF^2 + FE^2 - 2 \cdot DF \cdot FE \cdot \cos{120^\circ} = DF^2 + FE^2 + DF \cdot FE.
\end{align*} As $F$ varies along $(ABZ)$ with length $AB$ fixed, note that the maximum length of $FE$ occurs when $\overline{FZ} \perp \overline{AB}$, and this is also the case for the minimum length of $FZ$. Thus $\frac{FE}{FZ} \leq \frac{1}{4}$.

As a result, we know \begin{align*}
    DE^2 &= DF^2 + FE^2 + DF \cdot FE \\
    &\leq \frac{1}{16} (FZ^2 + FY^2 + FZ \cdot FY) \\
    &= \frac{1}{16} YZ^2 \\
    &\leq \frac{1}{16} (AZ + AY)^2 \\
    &= \frac{1}{16} (AB + AC)^2,
\end{align*} as desired.

Solved with Alex Zhao, Elliott Liu, Connie Jiang, Groovy (\help), Jeffrey Chen, Nicole Shen, and Raymond Feng.

Externally construct equilateral triangles \(ACY\) and \(ABZ\), so that \(B\), \(F\), \(D\), \(Y\) are collinear  and \(C\), \(F\), \(E\), \(Z\) are collinear.

[asy]         size(7cm); defaultpen(fontsize(10pt));         pair A,B,C,Y,Z,F,D,EE;         A=dir(110);         B=dir(220);         C=dir(320);         Y=A+(C-A)*dir(60);         Z=B+(A-B)*dir(60);         F=extension(B,Y,C,Z);         D=extension(B,Y,A,C);         EE=extension(C,Z,A,B);

draw(D--EE);         draw(B--Y,gray);         draw(C--Z,gray);         draw(circumcircle(A,F,C),linewidth(.3));         draw(circumcircle(A,F,B),linewidth(.3));         draw(C--Y--A--Z--B);         draw(A--B--C--cycle,linewidth(.7));

dot("\(A\)",A,dir(105));         dot("\(B\)",B,S);         dot("\(C\)",C,S);         dot("\(F\)",F,dir(265));         dot("\(Y\)",Y,dir(30));         dot("\(Z\)",Z,dir(150));         dot("\(D\)",D,dir(-5));         dot("\(E\)",EE,dir(210));     [/asy]

Observe that \(FY/FD\ge4\) and \(FZ/FE\ge4\). It follows that \begin{align*}     AB+AC=AY+AZ\ge YZ&=\sqrt{FY^2+FZ^2+FY\cdot FZ}\\     &\ge4\sqrt{FD^2+FE^2+FD\cdot FE}=4DE, \end{align*} as needed.

[quote=orl]Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \][/quote]

See also here [url]https://artofproblemsolving.com/community/c6t243f6h2624066_a_refinement_of_imo_shl_2002[/url]

First ISL solution in a while. This is the type of problem where if you don't know this property of the Fermat point it's hard to solve (I certainly couldn't do it) and if you know it it's very quick (after I got a hint with the property I solved it within fifteen minutes).

It is well-known that $F$ is the Toricelli/1st Fermat point of $\triangle ABC$. It is a well-known property of $F$ that if $ABG$ and $ACH$ are equilateral triangles erected outward from $AB$ and $AC$, respectively, $C,F,G$ are collinear and $AGBF$ is cyclic (similarly $B,F,H$ are collinear and $AHCF$ is cyclic).

Notice that as $F$ is on minor arc $AB$, the minimum possible value of $\frac{GE}{EF}$ occurs when $F$ is on the midpoint of the arc, as this maximizes $EF$ and minimizes $EG$. In that case, it is easy to show that $\frac{EG}{EF}=3$ and thus $\frac{FG}{FE}=4$, hence it must be true that $\frac{FG}{FE}\ge 4$ and similarly $\frac{FH}{FD}\ge 4$.

Let $FG=\alpha FE$ and $FH=\beta FD$, where $\alpha, \beta \ge 4$. Since $\angle EFD = 120^{\circ}$, by LoC on $\triangle FED$ we get $$ED=\sqrt{FE^2+FD^2-2FE\cdot FD \cos{120^{\circ}}}=\sqrt{FE^2+FD^2+FE\cdot FD}.$$ Using LoC on $\triangle FGH$, we get

\begin{align*}
HG &= \sqrt{FG^2+FH^2-2FG\cdot FH\cos{120^{\circ}}} \\ &= \sqrt{FG^2+FH^2+FG\cdot FH} \\ &= \sqrt{(\alpha FE)^2+(\beta FD)^2+(\alpha FE)(\beta FD)} \\ &= \sqrt{\alpha^2 FE^2+\beta^2 FD^2+\alpha\beta FE\cdot FD} \\ & \ge \sqrt{16FE^2+16FD^2+16FE\cdot FD} \\ &= 4\sqrt{FE^2+FD^2+FE\cdot FD} \\ &= 4ED.
\end{align*}

By the Triangle Inequality, $AG+AH\ge GH \ge 4ED$, finishing the problem.

Let $B', C'$ be such that $ACB', ABC'$ are equilateral. We have that $C,F,C'$ and $B,F,B'$ are collinear.

[color=#f00][b]Claim:[/b][/color] $EC' \ge 3EF$

[color=#00f][b]Proof:[/b][/color] Note that $\frac{FE}{C'E} = \frac{AF}{AC'}  \frac{\sin \angle FAB}{\sin \angle C'AB} = \frac{AF}{c} \frac{2\sin \angle FAB}{\sqrt{3}}$. Let $AF = x$, $BF = y$. Note that $R$, the circumradius of $(AFBC')$, is equal to $\frac{c}{\sqrt{3}}$.

We have $2R = \frac{y}{\sin \angle FAB} \implies \sin \angle FAB = \frac{y}{2R} = \frac{\sqrt{3}y}{2c}$.

So $\frac{FE}{C'E} = \frac{x}{c} \frac{y}{c} = \frac{xy}{c^2}$.

Observe that $c^2 = x^2 + y^2 + xy \ge 3xy \implies \frac{xy}{c^2} \le \frac{1}{3}$.

So we have $\frac{FE}{C'E} \le \frac{1}{3} \implies C'E \ge 3FE$, as claimed. $\square$.

From the claim, we have $DE \le \frac{B'C'}{4} \le \frac{AB' + AC'}{4} = \frac{AB+AC}{4} \implies AB + AC \ge 4DE$, as desired. $\blacksquare$

Note that ∠AFB = ∠BFC = ∠CFA = 120 so making regular triangles with bases AB and AC is a good move.
Let S and K be outside ABC such that ABS and ACK are regular triangles. Note that AFBS and AFCK are cyclic. Let O1,O2 be reflections of F across AB and AC. FE/ES = [AFB]/[ABS] = [AO1B]/[ASB] ≤ 1/3 so FS ≥ 4FE. Same way we can prove FK ≥ 4FD. so SK ≥ 4DE and SK ≤ AS + AK = AB + AC.
we're Done.

Try the refinement
[url]https://artofproblemsolving.com/community/c6t243f6h2624066_a_refinement_of_imo_shl_2002[/url]

What.

[img width=50]https://media.discordapp.net/attachments/925784397469331477/952399059321245766/Screen_Shot_2022-03-12_at_7.53.43_PM.png?width=864&height=1170[/img]

Let $G$ be on $FD$ extended such that $\angle AGC=60^\circ.$ Let $H$ be on $FE$ extended such that $\angle AHB=60^\circ.$ Note that $AFCG$ is cyclic. Also, $\angle AFD=60^\circ$ and $\angle CFD=60^\circ$ so $\angle CAG=\angle ACG=60^\circ.$ Thus, $ACG$ is equilateral. Similarly, $AHB$ is equilateral. Now, $AB+AC\ge HG.$ Since $\angle HFG$ is obtuse, it suffices to show $HF\ge 4EF$ and $GF\ge 4DF$ to prove that $HG\ge 4ED.$

Note that $\triangle HFA\sim \triangle HAE$ by AA so $\frac{HE}{HF}=\left(\frac{HE}{HA}\right)^2\ge \left(\frac{\sqrt{3}}{2}\right)\ge \frac{3}{4},$ which implies the result that $HF\ge 4EF$. Similarly, $GD\ge 4DF.$ Now, WLOG suppose the parallel line through $E$ parallel to $HG$ lies outside of $\triangle EDF.$ Then this line intersects $FG$ at $J.$ $HG\ge EJ\ge ED$ as desired.

Construct equilateral triangles $\triangle ABX$ and $\triangle ACY$ outside of $\triangle ABC$ and note that $AXBF$ and $AYCF$ are cyclic. It's easy to see that $FX\ge 4FE$ and $FY\ge 4FD$ so by the Law of Cosines we easily obtain $XY\ge 4DE$ so that
\[AB+AC=AX+AY\ge XY\ge 4DE.\]
We are done. $\blacksquare$

Really cute :)
[quote=orl]Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \][/quote]

Draw points $K, L$ such that $AKB$ and $ALC$ are equilateral triangles. Clearly, $AFCL, AFBK$ are cyclic quads, and $\angle AFL=\angle ACL=180^{\circ}-\angle AFB=60^{\circ}$ implies $B, F, D, L$ are collinear. Similarly, $C, F, E,$ and $K,$ are collinear. Now $AB+AC=AK+AL \ge KL$ so it suffices to show that $DE \leq \tfrac{1}{4} KL$.

We will show that $FE \leq \tfrac{1}{4}FK$ and $FD \leq \tfrac{1}{4}FL$. It suffices to prove the following two lemmas to finish:

[b]Lemma 1.[/b] Point $W$ lies on arc $\widehat{YZ}$ of the circumcircle of equilateral triangle $XYZ$ not containing $X$ and line $XW$ meets $YZ$ at point $T$. Then $WX \geq 4WT$.

[i]Proof:[/i] Indeed, it is enough to show $XT \ge 3WT$. Now $XT$ is larger than the $X$-median of $\triangle XYZ$ and $WT$ is smaller than the length it achieves when $W$ is antipodal to $X$. For rigour, this follows as $XW \cdot XT$ is fixed by shooting lemma. When $W$ is antipodal, equality is achieved, proving the lemma.

[b]Lemma 2.[/b] In obtuse triangle $XYZ$ with obtuse angle at $X$, points $Y_1, Z_1$ lie on rays $XY, XZ$ such that $XY \geq 4XY_1$ and $XZ \geq 4XZ_1$. Then $YZ \geq 4Y_1Z_1$.

[i]Proof:[/i] Scale by a factor of $4$ to assume $XY_1 \leq XY$ and $XZ_1 \le XZ$. Now $Y_1Z_1<Y_1Z$ as $\angle Y_1Z_1Z>\angle Y_1XZ>90^{\circ}$ and $Y_1Z<YZ$ as $\angle YY_1Z>\angle YXZ>90^{\circ}$, so $Y_1Z_1<YZ$ unless $Y_1=Y$ and $Z_1=Z$, proving the claim.

Finally, by combining Lemma 1 and Lemma 2 in triangle $FKL$ for points $D$ and $E$, the conclusion follows.

r poblem

Construct equilateral triangles $ABX$ and $ACY$ so that both are not in the of $ABC$. Then it is well known that $A$, $F$, $X$ and $B$, $F$, $Y$ are collinear, and moreover $AXBF$, $AYCF$ are cyclic. Now we can obtain $FX\ge 4FE$, $FY\ge 4FD$ and hence by LOC $XY \ge 4DE$. To finish,
\[AB+AC=AX+AY\ge XY\ge 4DE\]
as desired. $\square$

Point $F$ is $\textbf{First Fermat Point}$ and construction can easily be found from the theorem thonk:/


calculus integration conics hyperbola rotation trigonometry SFFT

How do you solve this problem, and problems like these in general?

Find all integral points (x,y) that satisfy:

$\frac{1}{x}+\frac{1}{y}= \frac{1}{4}$

For example, $(2,-4)$ is one such integral point.

Take a look at [url=http://www.artofproblemsolving.com/Movies/simon1.php]this[/url] and [url=http://www.artofproblemsolving.com/Movies/simon2.php]this[/url] and see if you can figure out how to apply them to the problem.

What kind of curve is this when plotted on the plane? A hyperbola? :maybe:

I know you can rotate it to rid the equation of the xy term, and I know it's overkill, I just want a solution for the general case. I just don't want to do the trig right now.

EDIT: I know you can do the general case with SFFT.

[hide]Multiplying through by $xy$, we get $xy-4x-4y=0$. Factoring using SFFT and rearranging, we get $(x-4)(y-4)=16$. So our solutions are $(5,20)$, $(6,12)$, $(8,8)$, $(12,6)$, $(20,5)$, $(3,-12)$, $(2,-4)$, $(0,0)$, $(-4,2)$, $(-12,3)$. Obviously, $(0,0)$ is extraneous since this results in a zero denominator. So we have a total of 9 solutions, the ones above excluding $(0,0)$.[/hide]

Can we solve $\frac{1}{x}+\frac{1}{y}= \frac{1}{z}$ like this?
(Meaning find the number of integer pairs (x,y) that satisfy the equation)
$\frac{1}{x}+\frac{1}{y}= \frac{1}{z}$
$x+y = \frac{xy}{z}$
$zx+zy-xy = 0$
$xy-zx-zy = 0$
$xy-zx-zy+z^{2}= z^{2}$
$(x-z)(y-z) = z^{2}$

So then we find the number of factors of $z^{2}$, multiply by 2 to account for negatives, and subtract one for the extraneous $(0,0)$ solution.  Would that work all the time?

P.S.

Those were really good videos!  The best part was when he was like "READ THE QUESTION."


can someone please tell me what is the fundamental law governing principle of equivalence...i really didn't get it

also what is the equivalent weight of an elemnt say O because it gives both H_2 O_2 and H_2O

or law of equivalence...i equivalent combines with another equivalent


({2, 4, 6, 9, 12, 18, 27, 36, 48, 60, 72}, |).
Explain all the following as I am unable to follow all these?

a. Find all maximal elements.
b. Find all minimal elements.
c. Is there a greatest element?
d. Is there a least element?
e. Draw the Hasse diagram.


geometry
Pythagorean Theorem
cyclic quadrilateral
geometry unsolved

Well, is it possible to prove Pythagorean theorem with inversion? If I'm not mistaken, I have seen it somewhere on this forum, but I couldn't find it.

Further, we can prove Ptolemy's theorem by using inversion, which is basically applying an inversion with center in one of the vertices of the quadrilateral that carries the other three into a straight line. Using the formula that gives the inverse distances in terms of the primitive distances, we obtain the Ptolemy's relation. The Pythagorean theorem is just  a particular case when the cyclic quadrilateral is a rectangle.

Thanks for the tip. I am quite familiar with the proof of Ptolemy's theorem with inversion.

Edit:
But still, is there a proof with inversion without application of Ptolemy's theorem?

I am not good at all at geometry but what Luis said, is that you can prove the Pythagorean Theorem by using the same method as the one you use to prove Ptolemey's Theorem. Instead of any quadrilateral, just pick a rectangle. You do not actually apply Ptolemey's Theorem anywhere... :wink:


Let $ f: \mathbb N \to \mathbb Z$ be a function such that for all positive integers $n$, \[(f(n\plus{}1)\minus{}f(n))(f(n\plus{}1)\plus{}f(n)\plus{}4)\leq0.\] Prove that $f$ is not injective.

[quote="posabogdan"]Let $ f: N \to Z$ s.t $ (f(n \plus{} 1) \minus{} f(n))(f(n \plus{} 1) \plus{} f(n) \plus{} 4)\leq0,\forall n\in N$.Prove that f is not injective.[/quote]

Suppose $ f(x)$ is injective. So $ f(n\plus{}1)\neq f(n)$. Then :

$ f(n\plus{}1)>f(n)$ $ \implies$ $ f(n\plus{}1)\plus{}f(n)\leq \minus{}4$ $ \implies$ $ \minus{}4\minus{}f(n)\geq f(n\plus{}1)>f(n)$ $ \implies$ $ \minus{}2\minus{}f(n)\geq 2\plus{}f(n\plus{}1)>2\plus{}f(n)$ $ \implies$ $ |f(n\plus{}1)\plus{}2|\leq |f(n)\plus{}2|$

$ f(n\plus{}1)<f(n)$ $ \implies$ $ f(n\plus{}1)\plus{}f(n)\geq \minus{}4$ $ \implies$ $ \minus{}4\minus{}f(n)\leq f(n\plus{}1)<f(n)$ $ \implies$ $ \minus{}2\minus{}f(n)\leq 2\plus{}f(n\plus{}1)<2\plus{}f(n)$ $ \implies$ $ |f(n\plus{}1)\plus{}2|\leq |f(n)\plus{}2|$

And so the function $ |2\plus{}f(n)|$ is a non negative non increasing function and so has a limit
So for $ n$ great enough, $ |f(n\plus{}1)\plus{}2|\equal{}|f(n)\plus{}2|$ and, since $ f(n\plus{}1)\neq f(n)$ : $ f(n\plus{}1)\plus{}2\equal{}\minus{}2\minus{}f(n)$ for all $ n$ great enough.

But this implies $ f(n\plus{}1)\equal{}\minus{}4\minus{}f(n)$ and so $ f(n\plus{}2)\equal{}\minus{}4\minus{}f(n\plus{}1)\equal{}f(n)$, so contradiction.

And no injective function fits the requirement.
Q.E.D.


Let $ ABC$ be a triangle whose vertexs, in cartesian coordinates, are $ A \equal{} (1, 0), B \equal{} (3, 0)$ and $ C \equal{} (2, 1)$.
Calculate the inclination $ m$ of the line that passes by the point $ (0, 0)$ and divides this triangle in two regions of equal areas.

[hide="Answer"]

$ m\equal{}\dfrac{4\minus{}\sqrt{7}}{9}$[/hide]

[hide="approach"]Use the formula that uses triangle coordinates in a determinant, and then set top triangle's area to 1/2 the total[/hide]


It is very hard to define what mathematics is. Today's mathematics is so huge. If someone would compare calculus and pure geometry, he/she won't find much in common. My dictionary says that math is study of numbers, but we all know that this is not a good definition.
There is other point. Different people seems to draw line differently. For example, some people think statistics is a branch of math, and the others(like me) do not think so.
What is your definition of mathematics?

[quote="lightrhee"]
For example, some people think statistics is a branch of math, and the others(like me) do not think so.
What is your definition of mathematics?[/quote]

Why would you not consider statistics a branch of mathematics? :huh:

To me, mathematics is the study of the deductive consequences of what are commonly agreed upon mathematical "truths" - or axioms. How those axioms are chosen is a bit flexible, but for the most part they are based on abstractions of observations of natural phenomena.

So basically, mathematics = some set of axioms (based on natural laws) + the study of their deductive consequences.

In my book, mathematics = number + shape.

For those of you opting for single-word definitions, define "language," "number," and "shape."

I had to do this project on math and art this year.  For my definition, I had math is a science dealing with the logic of quantity, shape, and arrangement.  I think I got that using google.

Language is just a way of communicating, so I think it is appropriate to call mathematics a method of communication.
I saw another definition somewhere:
Mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we are saying is true.

I thought that is pretty clever  :lol:

Hey, stupidked, nice username  :) .

Of course statistics is a branch of mathematics.

[quote="4everwise"]Of course statistics is a branch of mathematics.[/quote]
Why? I cannot defend my position since the term math is not defined, but I think statistics is more like science that adopts tons of mathematics.

As to what mathematics is, I really can't say. The reading I have done on this question and its history has shown me how complicated it really is. Maybe after I've done some real university math I'll have a better perspective and I'll be able to take a position, but now I'm just confused about what math really is (I even read an essay called "The Crisis in Modern Mathematics" - this is no laughing matter!). So, basically I have nothing to add  :roll:

As to the question of statistics' place in math, I agree with lightree that it is a science that uses mathematical methods. Like theoretical physics, "pure" statistics is highly mathematical but is still a science. But really I barely know anything about statistics so I'll leave that question to you guys  ;)

I want to hear some answer from more people. What is definition of mathematics?

What's the definition of green?

The English language is a giant circular reference with over 600,000 words.

I personally like this one.

"Mathematics is a man-made universe" - anonymous(or I just dont know who said it)

[quote="Geehoon"]"Mathematics is a man-made universe"[/quote]

That's pretty good, actually -- a simple formulation that is neither limiting nor overly broad.

[quote="bubala"] What's the definition of green? [/quote]

Green is the perception of light with wavelengths between 490 and 540nm, with some equivocating at the boundaries.

Math is a set of definitions/ axioms along with all the theorems that can be proved using them.
statistics has too much formulas, it doesn't feel pure enough to be considered math :D
[quote=JBL]Green is the perception of light with wavelengths between 490 and 540nm, with some equivocating at the boundaries.[/quote]
great definition, no why can't everything be so simple :lol:

[quote="Mariag"]Math is a set of definitions/ axioms along with all the theorems that can be proved using them.[/quote]That's a little simplistic I think. What is this set of definitions/axioms? I mean.. define a number, for example.

I think her definition is perfect!

The set of definitions/axioms can be anything you want - though we usually choose them to model fundamental properties of measurement and structure in the physical world.

For example, Euclid's Five Axioms of Geometry (which were later refined and expanded to 21 Axioms by Hilbert) were based on observations of physical "truths" of measurement. But it turns out he could just as easily have chosen a different set of axiom which lead to other interesting "truths" of geometry (non-Euclidean geometry).

I would say that Mathematics is more of a philosophy... One could argue that going to the next step from the current would be using logic, as there are so many different paths to take. Lots of mathematicians, for example, Descartes, were also successful philosophers  :)

I'm not sure about a definition of mathematics, but I think statistics more so a branch of math than science (however, I do agree that it is more of a science than many others). It deals with many mathematical concepts (probability being one of them) and uses a lot of math, though I have to say there are parts of statistics that actually require no math at all, which I guess is some justification of statistics not being part of math. In any case, I would still categorize statistics as math over science.

[quote="JBL"][quote="bubala"] What's the definition of green? [/quote]

Green is the perception of light with wavelengths between 490 and 540nm, with some equivocating at the boundaries.[/quote]

Ok, define 490--better yet, go out into the world and find 490.  Not 490 apples or trees--just find 490.

[quote="manuel"]mathematics is art[/quote]
It is impossible to be a mathematician without being a poet in soul.
- Sophia Kovalevskaya

[quote="lightrhee"] My dictionary says that math is study of numbers, but we all know that this is not a good definition.[/quote]

Well, unless your name is Nicholas Bourbaki  :lol:
Alternately Principia Mathematica + Goedel coding

Seriously, I believe math to be inherently undefinable, and given any definition, there will always be an exception. (confer with Wittgenstein's discussion of "game" in [u]Philisophical Invesigations[/u].).

modern philosphy is not concerned with definitions anymore.
since the internal ambiguity behind language we can not rely in definitions. so it is worthless to keep defining stuff.

Keith Devlin once defined mathematics something like this:
"The abstract study of pattern and structure"
I like the presence of the 2nd, 5th and 7th words.  These seem to get to the heart of what doing math is all about.


Hey everyone.  I just joined here and I'm sure I'll be posting lots of questions since I am not very good at Math.. :(

I am doing an online class with Intermediate Algebra (or Algebra 2) and I have a question about rationalizing the denominator.

I have a problem here that is quite simple looking but I cannot for the life of me figure this out.

It is
[u]10[/u]
fifth root of 16

I hope that makes sense..it looks like a fraction with a sqaure root of 16, but in the "dip" of the square root symbol it has a small 5.

Any help is appreciated...Thanks!

hello there this is a bit easy for this forum but i will answer without latex:
[hide]x^1/y
power is the yth root of x
so you mean
(10)/(16^(1/5))
which to rationalize we would multiply by 1(16^(4/5)/(16^(4/5)
to get
(10)(16^(4/5))/(16)
so basically we didn't do anything to our equation because y/y=1 and multiplying anything by 1=itself
[/hide]


[quote="shadysaysurspammed"]1)$1,2,\frac{1}{2} , 3, 1 , \frac{1}{3}, 4, \frac{3}{2}, \frac{2}{3} , \frac{1}{4} , 5$

In the above series,find the $2005th$ term.

2)Evaluate the sum

$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\cdots++\frac{1}{22350}$[/quote]

for $1)$, we can divide the sequence into nth groups, where the first term of the group is n. and there are n number of terms in the nth group. so we need to find which group the 2005th term is located.

consider the arithmetic sequence where $a =1, d = 1$
so we need to find the smallest value of $n$ where
$\to  \frac{(1 + n)n}{2} > 2005$
i got $n = 63$. so the 2005th term lies in the $63th$ group. $\frac{(1 + n)n}{2} = 2016$, so in the group, the 2005th term is the $63 - (2016 - 2005) = 52$th term
note that the denominator of each term in the group is the postition of the term in the group.
so,the answer should be $\frac{63-52+1}{52} = \frac{12}{52}$, IF there are no computations mistakes  :D


probability
conditional probability
combinatorics unsolved
combinatorics

Conventionally, we take the definition of conditional probability as $P(A|B) = \frac{P(A \cap B)}{P(B)}$(1)

This implies that $P(A \cap B ) = P(A|B)P(B) = P(B|A)P(A)$.

I have some doubts with these. It does not seem 'obvious' that $P(A|B)P(B) = P(B|A)P(A)$. Let us suppose we do not know the formula above for conditional probability (for a moment, suppose it is untrue). How do we prove for the general case that $P(A|B)P(B) = P(B|A)P(A)$?? Please keep in mind that we just understand the 'intuitive' meaning of conditional probability and NOT the formula above.

My question basically pertains to 'why' the above formula for conditional probability is correct? Why is (in the general case) $P(A|B)P(B) = P(B|A)P(A) = P(A \cap B)$ ?? Why do we define conditional probability through the formula above? Is the above formula (1) a definition of conditional probability or is it a consequence of some other definition? Intuitively, shouldn't $P(A \cap B ) = P(A|B)P(B) + P(B|A)P(A)$ ( at least my 'logic' seems to suggest that!) ?

[hide]Mod: Please move this if you find this forum is not appropriate to post my question[/hide]


Compute the number of ordered pairs of positive integers (x,y) such that :
x^2+y^2[u]<[/u]50.

33 yes!  Just make sure it isn't over 50, and dont repeat, and you are sure to get it. I tried it be4.

This level question should be in high school basics for future reference.

Anyways,

[hide]
x[u]<[/u]7
y[u]<[/u]7

Starting from the "edges,"

1+3+5+4+3+2+1 = 19

19(2)-5 = 33.[/hide]


A 25 foot ladder is placed against a vertical wall of a building.  The foot of the ladder is 7 feet from the base of the building.  If the top of the ladder slips 4 feet, then the foot of the ladder will slide:
\[ \text{(A)}\ 9\ \text{ft.} \qquad \text{(B)}\ 15\ \text{ft.} \qquad \text{(C)}\ 5\ \text{ft.} \qquad \text{(D)}\ 8\ \text{ft.} \qquad \text{(E)}\ 4\ \text{ft.} \]

[hide][b]D[/b]

Use right triangles.[/hide]

Yup its D. Since the ladder is connected to the wall, the corresponding change in the height, will always be equal to the change in the distance from the base since the length of the ladder is constant.


function
ratio
real analysis
real analysis unsolved

hi all,

this problem seems hard. Any idea ?
find all functions from R to R such that: fof=f '

bye

Here's a starting point.

Assume there is an $x_0$ s.t. $f'(x_0)>0$. Then there is an $\alpha>0$ s.t. $f'(x)>0,\ \forall x\in I=(x_0-\alpha,x_0+\alpha)$, and $f\circ f>0$ on $I$. Now it's easy to show that, in fact, $f$ is increasing on $(x_0,\infty)$, and $f^{(n)},\ n\ge 1$ are all positive on $(x_0,\infty)$, and this is why from now on it's enough to work with $x$ which are large enough (so assume I mention that $x$ is large enough after each statement :)).

Show now that $f(x)$ is large compared to $e^x$ (their ratio goes to $\infty$ for $x\to\infty$). We then have $f'=f\circ f>e^{f}$. Let's show that this is impossible.

We have $[e^{-f}]'=-f'\cdot e^{-f}$, which has modulus $\le -1$, and this is clearly impossibl because $e^{-f}>0$ is bounded from below by $0$.

This is a contradiction, so $f'\le 0$, so $f$ is decreasing (or non-increasing, if you prefer :)), and $f\circ f=f'$ is increasing, and $f\to \ell\le 0$ as $x\to\infty$. This means that $f$ is convex and decreasing. $f'$ tends to a limit $\ell\le 0$ as $x\to\infty$, and if $f$ is unbounded from below then $f'$ eventually becomes $>0$ because a convex function is unbounded from above, so $f\to\infty$ as $x\to -\infty$. This means that $f$ is bounded from below, so $f\to a<0, f'\to 0$ as $x\to\infty$. Moreover, $f'\to a$ as $x\to -\infty$.

I'm depleted for now, but I'll keep thinking on it.


geometry
inequalities
abstract algebra
geometry proposed

Hi :icecream

I've got a problem:

Given $\triangle ABC$ and a point $P$, let $x=PA$, $y=PB$, $z=PC$. Show that $ayz+bzx+cxy \geq abc$.

A question... What is it about Paracevian and precevian???

Thanks

[quote="gemath"]we can use complex number for solution
using the fact
$(b-c)(m-b)(m-c)+(c-a)(m-c)(m-a)+(a-b)(m-a)(m-b)=(a-b)(b-c)(c-a)$ with any complex number $A(a),B(b),C(c),M(m)$ now using inequality for module of complex number:$|x|+|y|\ge|x+y|$ we will have your problem![/quote]

Nice proof but whena does the inequality holds?

There is a nice elementary solution to this problem, but only when $P$ is inside of the triangle. Equality holds when  $P$ is an orthocentar of the triangle.

Idea of the solution is: Take point $X$ such that $\angle BPX=\angle PCA$ and $\angle XBP=\angle PAC$, and also point $Y$ s.t. $\angle YCP=\angle BAP$ and $\angle YPC=\angle ABP$. Then triangles $(YCP,BPA)$, $(BPX,APC)$ and $(XYP,ABC)$ are similar (in given pairs). So compute $BX,XY,YC$ in terms of $x,y,z,a,b,c$ and take that $BX+XY+YC\geq BC$ and you get the wanted inequality...

Bye


I took my State Competition on March 3rd, and after the test, I thought, WOW! that was easy.  As it turned out, I got 3rd in the state of WV, but I don't know my score.  I also thought that this year's Chapter Competition was one of the hardset ever.  I got a 27, and 3rd Place in the Chapter.  Scores across the board were down from last year, my teacher said.  With a 27, I lead my school in score.  One of the kids on my team (my team took 3rd in the state as well) got like - an 18.  So, now I leave you with this question - Was this year's Chapter hadrer than this year's State?  Or was it just a difficult Chapter?

The ninja thinks:

They were about the same.  Like obviously, relative to other years, the chapter was very hard, and the state somewhat easy.

Then again, 3 out of the top 5 from our chapter got higher scores at state than at chapter.

Me on the other hand, i had a HUGE off day, and dropped 15 points off my chapter score.

Definitely not. My score for School vs. Chapter vs. State is
42 vs. 41 vs. 36

Even though I got 2nd on all of those, these were hard.

School was hard, Chapter was Hard, State was HARD!!!!

one comment:

a state test would not be at state if it was easier than chapter.
state was noticeably harder.

i think that it just depends on the type of problems that u r good at.  i found the state test to be a lot easier on the brain than the chapter.  i got 6 of the target questions right, and around 20-23 of the sprints.  That should have given me a score of around 32-35, which was about 10 points higher than my chapter score

My scores
School: 44 (28/16)
Chapter: 39 (23/16)
State: 30 (18/12)

Maybe I was more prepared for school (the easiest school I've seen in years except for #30. but I still got that one right. I think I missed two of the first 10 actually)

I though Chapter was harder than last year
but that had to be true
cause last year's Chapter was really easy

I might have had a off day at state (idk)

i just got my scores from my teacher.
school - 33
chapter - 27
state - 32

the cutoff for wv was 30

States were easier than chapters.  Some old Nationals such as 1990 (I got 29/16 on a practice) were easier than chapters.

My Scores
School-We don't do a school round
Chapter-21 lol  :rotfl:
State-32

Chapter-40
State-36

so a little bit easier for chapter...but I thought they were very comparable compared to most years

ok now new topic, did the target for the states seem easier than the target for the chapter, or am i just imagining it?

School        Chapter      State
Target (/16)   14               10             4 (not accurate :mad: )
Sprint (/30)    27               16             19
Total (/46)     41               26             23

just look at the sprints

jorian

that right there proves it, well, except for ur target score.  either i really messed up the chapter, or this state test was extremely easier compared to other years.  just the fact that i scored 5-7 points higher on the state should be an indicator.  but maybe it was just luck.  where's that is mathcounts all about luck forum? :P (jk) :P

haha
i think machack would like to explain that one
(and how he got owned by fanatic)

what r u talkin about?? :huh:  :huh:

whos machack
how did he get owned by Fanatic

did u read the topic?

u should know if you've seen it

For some strange reason the state Sprint seemed harder than the Target... at least the last 10-15 problems of it did.

I agree completely about how the sprint was harder than the target.

I got a 12 on target and a 9 :blush: on sprint. I silly-mistaked so many questions omg

Chapter: 16/22

Our state's average dropped so much I think it was by something like 12 points...then again...it's South Carolina

We made 5th (team) because some stupid guy made a 4 on the TEAM. OMFG. We would've had at least 3rd if not for this... :furious:

I think this year's state was pretty hard. Normally, when I did practice ones, I got around 38, but I got a 35 on the state.

well since i can't release my state score yet, i can say this: i scored somewhere from 0-46 at chapter and scored less at state. so state was harder in that sense but the funny thing is that our team got a higher score at state than chapter because andrew did better on the state sprint than at chapter

what do you mean you can't release it yet???

the Indiana coach (mr fischer) doesn't want other states knowing Indiana scores.

i thought that state and chapter were about the same. The chapter target was maybe a little harder...even though I got a higher score on it. :rotfl:

they were apparently equally hard.  45=45.  hmm... wonder if nats is gonna be harder?

@choi: did you get 45 on school?

-jorian

State was harder!

School:Chapter:State
36:37:28 (<-bad day, could have gotten 33)

my school doesnt do school round.  i got 45 state, chapter

perhaps ill download it and do it

State was harder than Chapter, but 2007 Chapter definitely was harder than usual.

Chapter,State:
41,41  :|

I think chapter was a lot harder than the state. I scored higher on the state.

ummmm.... spam is that fake ham that comes in little pouches.  dont dis spam. besides everyone knows that chapter had to have been harder than states.  i even thought that this year's school was harder than states.

lmao. state wouldn't be state if it was easier than chapter.


$ 1\blacktriangleright\{x,y,z\}\subset \mathcal R^*_ \plus{} \implies \sum\frac {x}{2x \plus{} y \plus{} z}\le\frac {3 \plus{} (x \plus{} y \plus{} z)^2}{(1 \plus{} x \plus{} y \plus{} z)^2}\ .$

$ 2\blacktriangleright\{x,y,z\}\subset \mathcal R^*_ \plus{} \implies \sum\frac {1}{2x \plus{} y \plus{} z}\ge 2\cdot\frac {3 \plus{} (x \plus{} y \plus{} z)}{(1 \plus{} x \plus{} y \plus{} z)^2}\ .$

[b]Remark.[/b] The upper inequalities are equivalently !

[quote="Virgil Nicula"]$ 1\blacktriangleright\{x,y,z\}\subset \mathcal R^*_ \plus{} \implies \sum\frac {x}{2x \plus{} y \plus{} z}\le\frac {3 \plus{} (x \plus{} y \plus{} z)^2}{(1 \plus{} x \plus{} y \plus{} z)^2}\ .$
[/quote]
$ \sum_{cyc}\frac {x}{2x \plus{} y \plus{} z}\leq\frac {3}{4}\leq\frac {3 \plus{} (x \plus{} y \plus{} z)^2}{(1 \plus{} x \plus{} y \plus{} z)^2}.$
[quote="Virgil Nicula"]
$ 2\blacktriangleright\{x,y,z\}\subset \mathcal R^*_ \plus{} \implies \sum\frac {1}{2x \plus{} y \plus{} z}\ge 2\cdot\frac {3 \plus{} (x \plus{} y \plus{} z)}{(1 \plus{} x \plus{} y \plus{} z)^2}\ .$
[/quote]
$ \sum_{cyc}\frac {1}{2x \plus{} y \plus{} z}\geq\frac{9}{4(x\plus{}y\plus{}z)}\geq2\cdot\frac {3 \plus{} (x \plus{} y \plus{} z)}{(1 \plus{} x \plus{} y \plus{} z)^2}.$


Let (a,b,c) represent $\sum_{sym} x^a y^b z^c$ where a,b,c are constants and x,y,z are variables.
Prove $(4,1,0)+(2,2,1) \geq (3,2,0)+(3,1,1)$

For example, $(4,1,0) \geq (3,2,0)$ by Muirhead

WTS:$\sum_{sym}x^{4}y+\sum_{sym}x^{2}y^{2}z\geq\sum_{sym}x^{3}y^{2}+\sum_{sym}x^{3}yz$
This is equivalent to
$\sum_{sym}x^{4}y-\sum_{sym}x^{3}y^{2}\geq\sum_{sym}x^{3}yz-\sum_{cyc}x^{2}y^{2}z$.
$(LHS)=\sum_{cyc}x^{4}y+xy^{4}-x^{3}y^{2}-x^{2}y^{3}=\sum_{cyc}xy(x+y)(x-y)^{2}$
$(RHS)=2xyz\sum_{cyc}x^{2}-xy=xyz\sum_{cyc}(x-y)^{2}$
Therefore $(LHS)\geq(RHS)$ is equivalent to
$\sum_{cyc}(x-y)^{2}xy(x+y-z)\geq0$.
If $x,y,z$ are sides of a triangle, then this is valid. Otherwise, wlog we can assume $x\geq y\geq z$, then $x\geq y+z$.
Then $\sum_{cyc}(x-y)^{2}xy(x+y-z)\geq(x-y)^{2}xy(x+y-z)+(y-z)^{2}yz(y+z-x)$
$\geq(x-y)^{2}(y-z)xy-(y-z)^{2}(x-y)yz=y(x-y)(y-z)(x^{2}-xy-yz+z^{2})$
$=y(x-y)(y-z)(x(x-y)-yz+z^{2})\geq y(x-y)(y-z)((y+z)z-yz+z^{2})$
$=y(x-y)(y-z)(2z^{2})\geq 0$, and this completes the proof.

note. If the original inequality is changed to
$\sum_{sym}x^{4}y+\sum_{cyc}x^{2}y^{2}z\geq\sum_{sym}x^{3}y^{2}+\sum_{cyc}x^{3}yz$,
then the solution becomes far easier, as this is equivalent to $(x^{3}+y^{3}+z^{3}+3xyz-\sum_{sym}x^{2}y )(xy+yz+zx)\geq0$.

Thnx for the nice soln.

Next challenge:
Prove $2(4,4,0)+3(3,3,2) \geq 5(4,2,2)$

[quote="spanferkel"][quote="dominicleejun"]Next challenge:
Prove 2(4,4,0)+3(3,3,2) >= 5(4,2,2)[/quote]
false. Put $y=z=1, x\to\infty$ :huh:[/quote]

As both hands would be going to infinity, this is not proper counterexample. But $x=y=\frac{1}{n},z=n$ can be the counterexample when $n$ is going to infinity.


If you want a formula for $y$ in terms of $x$...
[hide]$y=2^{x-1}$[/hide]

[quote="jeez123"]x,y
-5,0.015625
-4, 0.03125
-30, 0.0625
-2, 0.125
-1, 0.25
0, 0.5
1, 1
2, 2
3, 4
4, 8
5, 16

the differences of the y is just the multiple of 2... I don't know where to go after that
thanks![/quote]
[hide] When a number is multiplied by a constant (2, in this case) to obtain the next number, the function is exponential. So the function is in the form $f(x)=a\cdot b^{x}$, so we plug in 2 values to get:
$1=a\cdot b^{1}\Longrightarrow ab=1$
$2=a\cdot b^{2}\Longrightarrow ab^{2}=2$
Divide to get $b=2$, so $a=\frac{1}{2}$.
Thus, $f(x)=\frac{2^{x}}{2}$

Also, since we know it is multiplied by 2 each time, we know that $b=2$, and can easily solve for $a$ from there.

Of course it's much easier to just be able to recognize it as $f(x)=2^{x-1}$. :D [/hide]


[quote="RHSLilSweetie07"]A flight attendant pulls her 64.8 N flight bag a distance of 305 m along a level airport floor at a constant velocity.  The force she exerts is 33.8 N at an angle of 54.6 degrees above the horizontal.  Find the coefficient of kinetic friction between the flight bag and the floor.

I have no idea how to work this problem.  Please help.[/quote]

if you take horizontal components of the forces acting on the bag, you'll find that the force against motion (normal force times the coefficient of kinetic friction, which equals the weight of the bag) equals the force in the direction of motion (equal to the force she exerts times the cosine of the angle to the horizontal). Draw a free body diagram if you can't see why...

The from this equation you'll be able to find the coefficient of kinetic friction.

[color=blue]I think that it's unnecessary to give the distance that the flight attendant pulls her flight bag. The answer is $\large \mu=\frac{F\cos\theta}{P-F\sin\theta}$
with $F$ is the force of the girl and $P$ is the weight of her bag[/color]


linear algebra
matrix
linear algebra unsolved

I have a matrix $ A \in \mathbb{C}^{n\times n}$. I want to know whether it is possible to write the following summation as a matrix product form ? \[ \sum_{i\equal{}1}^{n}\sum_{j\equal{}1}^{n}\left[(a_{i,j}\minus{}a_{i,j\minus{}1})^2\plus{}(a_{i,j}\minus{}a_{i,j\plus{}1})^2\plus{}(a_{i,j}\minus{}a_{i\minus{}1,j})^2\plus{}(a_{i,j}\minus{}a_{i\plus{}1,j})^2\right]\] Waiting for a reply!

Bhupala!

Your expression is homogeneous of degree 2 over the reals, hence there is a $ \mathbb R$-bilinear mapping $ \beta: M_n(\mathbb C)\times M_n(\mathbb C)\to \mathbb C$ such that $ \beta(A,A)$ is equal to your expression. This approach does not deal only with your particular matrix, but with all complex $ n\times n$ matrices.

bhupala,
In case you were looking for an explicit wait to express your sum, here is one way to proceed.
Let $ J$ the permutation matrix who $ i,j$ coefficient equals 1 iff $ j\equal{}i\plus{}1$ (index considered modul $ n$, of course).
Then look at $ AJ$, hence you get $ \mathrm{tr}\bigl( (I_n\minus{}J)^TA^TA(I_n\minus{}J)\bigr) \equal{}\sum_{i \equal{} 1}^{n}\sum_{j \equal{} 1}^{n} (a_{i,j} \minus{} a_{i,j \minus{} 1})^2$.
You should be able to conclude easily.


Find all functions $ f: \mathbb{R}^\plus{}\rightarrow\mathbb{R}^\plus{}$ such that

$ f(x)f(yf(x))\equal{}f(x\plus{}y)$ for all $ x,y>0$

Very nice problem. :)  It was one of the few func eqns for which I didn't guess the answer before proving. :lol:

For some $ x$, if $ f(x) > 1$ then take $ y: \equal{} \frac {x}{f(x) \minus{} 1}$.
This $ y$ satisfies $ yf(x) \equal{} x \plus{} y$ hence $ f(yf(x)) \equal{} f(x \plus{} y) > 0$ from the main equation $ f(x) \equal{} 1$. Contradiction.
Thus $ f(x)\le 1$

$ f(x \plus{} y) \equal{} f(x)f(yf(x))\le f(x)$
Hence $ f$ is decreasing.

For some $ x,z > 0$ put $ y: \equal{} \frac {z}{f(x)}$. It gives $ f(x)f(z) \equal{} f(x \plus{} \frac {z}{f(x)})\forall x,z > 0\cdots (*)$

Assume for some $ m>0$, $ f(m) \equal{} 1$
Putting $ x \equal{} z \equal{} m$ in $ (*)$ we have $ f(2m) \equal{} 1$
Thus $ f(2^km)\equal{}1$ for all $ k\in \mathbb N$

Thus there exists arbitrarily large $ x$ satisfying $ f(x) \equal{} 1$

$ f$ decreasing implies $ f(x) \equal{} 1$ for all $ x$

If we keep aside that solution:
there is no $ m$ satisfying $ f(m) \equal{} 1$. (1)

If $ f(x \plus{} y) \equal{} f(x)$ for some positive $ x,y$ we have $ f(yf(x)) \equal{} 1$ from the main equation.
But this contradicts property (1).
Hence $ f$ is injective.

Since the LHS in $ (*)$ is symmetric, so must the RHS. Implying $ f(x \plus{} \frac {z}{f(x)}) \equal{} f(z \plus{} \frac {x}{f(z)})$
Thus $ x \plus{} \frac {z}{f(x)} \equal{} z \plus{} \frac {x}{f(z)}\implies \frac {x}{z} \equal{} \frac {1 \minus{} \frac {1}{f(x)}}{1 \minus{} \frac {1}{f(z)}}$
If $ g(x) \equal{} \frac {1}{f(x)} \minus{} 1$ then $ \frac {x}{z} \equal{} \frac {g(x)}{g(z)}\forall x,z$
Thus $ g(x) \equal{} cx$
Since $ f(x) \equal{} \frac {1}{g(x) \plus{} 1}$ we have $ f(x) \equal{} \frac {1}{cx \plus{} 1}$.
The only way to have $ f(x) > 0$ is $ c\ge 0$

Putting back in the main equation we see that it is readily satisfied.

$ \boxed{f(x) \equal{} \frac {1}{cx \plus{} 1}\ \forall x > 0\text{ where }c\ge 0\text{ is a constant }}$

[quote="Akashnil"]Assume $ m$ is the maximum $ x$ for which $ f(x) \equal{} 1$
Putting $ x \equal{} z \equal{} m$ in $ (*)$ we have $ f(2m) \equal{} 1$
Hence the assumption of existence of $ m$ is contradicted.

There can be two conclusions:
1) there is no $ x$ satisfying $ f(x) \equal{} 1$
2)There exists arbitrarily large $ x$ satisfying $ f(x) \equal{} 1$[/quote]

Your forgot:
3)There exists an infinite, increasing sequence of $ x$ satisfying $ f(x) \equal{} 1$,
and converging to (but not reaching) some bound $ y$.

For instance, $ f(x) \equal{} 1$ might hold iff $ 0<x<1$.

Thank you. That can be easily corrected.

I edited:

[hide="from this"][quote="I"]Assume $ m$ is the maximum $ x$ for which $ f(x) \equal{} 1$
Putting $ x \equal{} z \equal{} m$ in $ (*)$ we have $ f(2m) \equal{} 1$
Hence the assumption of existence of $ m$ is contradicted.

There can be two conclusions:
1) there is no $ x$ satisfying $ f(x) \equal{} 1$
2)There exists arbitrarily large $ x$ satisfying $ f(x) \equal{} 1$

If the later case is true then $ f$ decreasing implies $ f(x) \equal{} 1$ for all $ x$

If we keep aside that solution, we have property (1).[/quote][/hide]

[hide="to this"][quote="I"]Assume for some $ m > 0$, $ f(m) \equal{} 1$
Putting $ x \equal{} z \equal{} m$ in $ (*)$ we have $ f(2m) \equal{} 1$
Thus $ f(2^km) \equal{} 1$ for all $ k\in \mathbb N$

Thus there exists arbitrarily large $ x$ satisfying $ f(x) \equal{} 1$

$ f$ decreasing implies $ f(x) \equal{} 1$ for all $ x$

If we keep aside that solution:
there is no $ m$ satisfying $ f(m) \equal{} 1$. (1)[/quote][/hide]

P.S. Thank you for checking my solution. Tell me if it is OK now.


geometry
analytic geometry
algebra
polynomial
algorithm
function
trigonometry

[quote="maersklandro"]Anyone, by chance, now what is the expression for the area of intersection of whatever 2 triangles based (obviously) on the coordinates of all 6 tips?

No, of course not. Stupid question. But does anyone know of any research in this area? At least what approaches, if any, have been applied to this problem?

I truly hope that there's no solution to this merely because too few people bothered with it not because one could not be found. I have a sneaking suspicion of the latter, though altough there is very little material on the matter to be found on the web.

Any help (or links) would be much appreciated. 10q[/quote]
If you know the coordinates of all six tips, then some simple analytic geometry could get you a formula.

It wouldnt be a pretty formula, but it would be a formula.

This is how you come up with the formula:

You draw a sketch as to where the intersection points will be. At worst they will form a hexagon - for simplicity assume that this is the case.

You find the equations of the lines of the first triangle and the equations of the lines of the second triangle. Set some equalities and you will have the coordinates of the six points of intersection.

Now divide this hexagon into 4 triangles. Obviously the area of the hexagon is the sum of the areas of the triangles.

Find the lengths of the sides of each triangle using the distance formula (since you know the coordinates.) Then use Heron's formula to find their areas, and add them together to find the area of the hexagon.

Finding a closed formula for this would take about 20 pages of algebra and I am WAY too lazy to do that.  ;)

There's even a formula for the area of a hexagon (or general n-gon) that looks kind of like jmerry's (I think jmerry's formula is a slightly modified special case of it).  I believe it is:

$Area=\frac{1}{2}|(x_1y_2+x_2y_3+...+x_{n-1}y_n+x_ny_1)-(y_1x_2+y_2x_3+...+y_nx_1)|$

[quote="jmerry"]There are much better ways than Heron's formula to find an area if you know coordinates: the are of a triangle with vertices at $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is $\frac12\left|(x_2-x_1)(y_3-y_1)-(x_3-x_1)(y_2-y_1)\right|$.[/quote]

Really? Wow, that's quite interesting.

A closed formula is exactly what I'm after. And believe you me, it's a lot harder than you might think. At least for me it's proved so :( .

I need a "one size fits all" formula. I need it to work for any 2 triangles and intersection case. It should even work if the triangles do not overlap, in which case is should return 0.

I'm planning to use it for a 3D software renderer I'm working on. Instead of sampling at the pixel centres or in several specific points of a pixel (as all the current 3D renderers do) this one will sample 2 triangles/pixel (2 pixel halfs). Calculating the overlap area between the projected triangle and each half of the current screen pixel is what I want. Thus I can EXTREMELY accurately determine what influence that particular projected triangle will have on that particular screen pixel as far as determining it's final colour. This, for now, excludes texture mapping which is even a more touchy issue with this "sampling" method. Combined with Binary Space Partitioning and hidden surface removal, this will a be a revolution in 3D graphics :D . If i can get it to work :P. Except for practically not being able to texture map :oops: ...

However, I am more & more inclined to believe that coming up with such a formula would be more of an achievement than the renderer it's intended for. Pls. help :( .

No matter what you do, you're going to need to split into many cases.

Here's a procedure: For each vertex of the original triangle, test whether it is inside or outside the other triangle. Take all of the inside vertices, and all intersections of the sides of the two triangles. These form a unique convex polygon, which we can calculate the area of using the formula given by Sly Si. (If this polygon has fewer than three vertices, it has area zero)

If there are only three vertices, any order makes the intersection triangle convex.
If there are four vertices, either all four are on the boundary of one triangle and the order is easy to determine or the four include one vertex of each triangle. These should be opposite vertices of the quadrilateral.
If there are five or six vertices, all will be on the boundary of the same triangle and the order is easy to determine.

One thing's for sure. When (more like if, actually) I find the "1 for all" expression, I'm keepin' quiet about it :D :first: !!!

You should be aware that there can't possibly be a "nice" formula, for example a simple polynomial in the coordinates.  You're writing a computer program -- all you really need is an algorithm, and jmerry effectively gives it to you.

Actually, I was kind of hoping it'd be a "nice", simplifiable little formula :D . A polinomial was exactly what i had in mind :yup: . Seems there's nothing but grief reserved 4 me on this issue :ewpu: .

It certainly can't be polynomial: fix one of the triangles and two vertices of the other triangle so that the triangle and segment we determine don't intersect, and consider just moving the remaining vertex.  The resulting function will be zero everywhere except for a triangular region (the interior of the first triangle).  No polynomial can behave like that.

Since you're writing a computer program, wouldn't you just want a function instead of a "closed-form" formula? What language are you programming this in? (C++/OpenGL I'm assuming)

If i understand your question right, you want the area of two triangles in which one of their vertices touch..., if you know the coordinates of everything, you could just use Shoelace thoerem...

No. I need a formula that works for any (whatever) 2 triangles and returns the correct overlap area. Even if there is none at all. WHATEVER 2 triangles. And it needs to be independent of clockwise/counterclockwise enumeration of tips.

It is only practical if i can get it to a 'closed formula' form. Not only that, i need the most compact iteration possible. I have to do this for every single visible triangle in the scene, twice for every pixel to be rendered.

An algorithm is only a poor substitute for an expression/formula you simply do not know.

It's precisely because I do not want to resort to classic real-time rendering, such as is the case with all OpenGL and DirectX renderers, that I am trying to develop new math specifically for better rendering tehniques.

These engines 'solve' through brute force (of the GPU) what, I believe, can be achieved better, more elegantly (far less waste of processing power) by applying more advanced math to the problem.

There is absolutely no attempt to fundamentally improve rendering tehniques through quality refinement of the software but merely ever increasing hardware processing power to cope with insolently inefficient rendering tehniques. SoftWare developers have become complacent. Or, perhaps, they just don't see there are other ways.

With all the development in the field of 3D graphics and the sacrifices in wasted processing power, there is NO 3D RENDERER today to offer PERFECT SUB PIXEL ACCURACY.

This is because they all work by sampling a finite amount of points in the scenes by calculating the intersection between imaginary lines strewn through the pixel centers. The colour of the first thing these rays hit in the 3D scene is sampled at the specific point of intersection through the use of texture mapping and shading tehniques. It gets a bit more complicated with reflective or translucent surfaces, but that is besides the point.

By using some sort of formula to calculate the exact area of overlap between each of the 2 halfs of the pixel and a given triangle in the scene you can extremely accurately determine the exact amount of influence the colour of that triangle is to have on the final colour of that pixel. Unlike current rendering which just shoots 1 RAY through the centre of the pixel (or, more recently a larger, yet still finite, number of rays, see Full Scene Anti Aliasing). It is obvious to anyone that you can never possibly shoot enough rays through the pixel to reach anywhere near the precision of the alternative method, with triangle overlap area calculation. This, for the time being, excludes classic renderring tehniques.

Until I find a PRACTICAL, EFFICIENT (read another version of the formula that handles repeating patterns of coloured squares - textures -) way to use the triangle overlap method on the very texture itself. But seing as how I can't yet even apply it to triangles that need to be textured, it's a long ways off.

This is what I am working on. An alternative to all current rendering tehniques as employed in such api-s like OpenGL or DirectX. And, as this being the case, why would i want to use any of these 2?

Of course this method is a lot harder than classic, ray rendering. But if achieved, it is infinitely superior.

I applied to forum(s) as a last ditch, desperate measure. I suspect i shall either find the answer myself, or never find it at all. I can't foresee finding it myself. But i am trying for help... I do not want to risc not having tried every opportunity.

Sorry for the long post.

Sorry, but a method that fits all your criteria does not exist :D. But casework isn't too bad for programming... it doesn't actually affect the runtime significantly I don't think...

Wow, one jump assembler instruction is not going to ruin your renderer.

Also, what is the deal with sub-pixel accuracy. Anything sub pixel in rendering is a waste of time, because you can't see it.

No, it is not a waste of time. The fact that it can't be done using ray rendering is quite another thing.

In real life, what a camera sees, that is smaller than a single pixel on the display, has no influence on the colour of that pixel? Just because it is not dead centre pixel? That's just plain wrong.

I mean, cameras don't use antialiasing or texture filtering and yet, while working at lower resolutions than computer displays usually do, you get no aliasing.

Isn't realistic 3D CGI the fundamental goal of all 3D researchers?

An alternative way of seing things might be using a triangle / circle overlap area formula, the triangle being the polygon and the circle being the analogue to a light sensitive cell in the retina. One circle per each pixel, dead centre, 1 pixel in diametre.

I haven't tried using a formula for this. Not even sure there is one. I went for perfect coverage from the beginning but instead of using a triangle / square overlap method i thought it more sensible to try triangle / 1st. triangle + triangle / 2nd. triangle overlapping.
The other method can be deduced from this one, so this one must be simpler, and easier to find the math for, right? A square is the sum of 2 triangles, right? That was my reasoning.

And it's not just rendering applications such a formula would have (had?) applications in, but also digital image resizing, for example.

Anyway, about using an algorithm instead of a closed formula on this... The answer is that it all adds up. In processor time, of course.
No argument can convince me that an algorithm could possibly be as efficient as an expression parsed at compile time, and already simplified by me pre-compile. And what about applying this rendering approach to texture mapping? How could texture mapping ever be implemented? Or will we be using mip maps and ray rendering till the end of time?

You see, using a ray rendering engine you have to apply a lot of patches to the basic rendering process. One for each limitation and failing of ray rendering. What lacks in the maths of the process we must try and compensate through piles upon piles of additional software.

For example: you need to use mip mapping and anisotropic filtering to artificially improve how textures look (basic mip mapping uses 3 times the memory normal texturing does) and also antialiasing to improve how the polygon edges look. All eat up processing power in huge amounts, needlessly. It's just such a huge waste. And none of it is actually correct with real life physics. They're just the typical answer that programmers turned to, not being able to get their heads around the maths. They will always do that. Try and make the computer do their maths for them.

Off topic:

Some of you may have heard about scanline buffering. And perhaps you have asking yourself  why it never really took off. One reason is that nobody (none of the programmers that worked on/with it) could ever figure out what was the maximum possible number of visible projected segments they would need to handle. So they used dynamic memory allocation. I just can see no other reason why anyone in their right mind would ever use dynamic memory allocation in a real-time renderer. And the lengths they would go to to get a clock cicle or two of their memory allocation routines. While at the same time wasting so many more by using dynamic memory itself. It's more than comical. It's awful. I mean, in the worst case scenario, the renderer would have to deal with the same number of segments regardless whether it used dynamic or static allocation. But what a difference in performance doing away with the memory routine would have made. I never followed up on the subject so I don't really know whether any of those enthusiastic programmers ever took the time, paper, and pen to determine the maximum possible number of segments. I did. And I'll only tell you that it's a linear increase with the number of polygons in the scene. Not exponential. Linear. I won't say exactly what the number is. The only important, and real, letdown of scanline buffering and the second main reason it never took off, in my opinion, is the processor time cost of handling and keeping track of the scanline to scanline intersections.

On topic:

And not just that, but an algorithm will increase the size of errors due to the floating point precision limit.

All development seems centred on superficial software and hardware development. Nobody tries anything to upgrade the theory.

Some of you said that there is no formula for this, others that it's simple using analytic geometry, yet others that I should try a case by case approach or to use an algorithm. What am I supposed to do? I don't want to believe there is no expression for the overlap area. I just can't get my mind around it. I say: how can there not be?!!!

Again, sorry for the extremely long post.

Almost any formula you can write down is smooth. Absolute values cheat this a little, but they're really piecewise defined functions. The area of the intersection you care about is not smooth- every time a vertex crosses a side of the other triangle, there's a corner or related discontinuity in some derivative. In order to find what you want, you [b]must[/b] break into cases based on these crossings. In each case, we could boil it down to a rational function in the coordinates of the vertices. The problem is, there are a lot of cases, and each uses a different rational function, some of which are rather complicated.

Algorithms [u]are[/u] often faster than a formula (and the floating point thing is nonsense, a formula is also evaluated piece per piece, where an algorithm could even use lower operations)...
Just a silly example: evaluate $sin(|x|)-|x|^3$. This is given as formula, thus requires your property. But to distinguish between $x\geq 0$ and $x<0$ first end then evaluating is more efficient.

And don't defame software developers as complacent or anything else as long as you don't even know if your method is better (I'm sure if it is, someone already uses it).
But I'm also not convinced why your's should be perfectly exact. How do you e.g simulate bowl just by triangles in a perfect way.

At least just evaluate $n^2$ subpixel for every pixel and to take their medium value works well, is probably faster and more or less near to what a camara does (indirectly).

So, just cast a bunch of rays...

You aren't limited to one ray per pixel...

[quote="maersklandro"]I am quite sure there is only ray rendering. Whether it be real-time or not is besides the issue. It still is ray rendering. There just is NOTHING ELSE.

On the odd chance there is someone, somwhere using something else, it's certainly a secret to me. All 3D game engines, ray tracers and 3D visualisation & design software use only ray rendering. That means shooting a finite (and actually quite small) number of rays through each pixel, or even rendering the scene at a higher resolution (larger image) and subsequently resizing the resulting 2D screenshot inside the GPU rendering pipeline to the final, smaller resolution which is sent to the display. Whether a Z-buffer, an S-buffer or some other method of culling hidden surfaces (different to backface culling which i think all renderers use or can use, whatever they are ray renderers or not) is irrelevant. Ray rendering is the basis of the pipelines of every existing GPU.

And about this method being better, there is no doubt about that. In my mind at least. However, for it to achieve its full applicative potential it would need half the interest, attention and research ray rendering is getting. It gets none but my own. You just can't expect to reach the same state of sofistication ray rendering has reached over many years of continuos and intensive research over night. And with so few people researching the topic too.

"My method" is better because it is far closer to modelling real physics than ray rendering is. Ray rendering offers finite precision whereas triangle overlap rendering can potentially offer PRACTICALLY (actually floating point precision limited) infinte precision.

Off topic, why has there been no research into such a branch of mathematics? Is it of no interest or use to anyone? Having formulas for the area of intersection of various shapes, that is. Or at least the most efficient algorithms possible. Isn't there a formula for the area of intersection between whatever 2 circles? I'll look up the subject.

Anyway, I won't give up, even if i might eventually have to resort to an algorithm to get the overlap area calculation done. And run it the number of visible poligons in the scene * twice the number of pixels to be rendered and at least 25 times a second. Certainly not an elegant solution. Any constructive suggestions?[/quote]
No, we can't get ray tracing to go fast enough. Right now we use raserization.

[quote="maersklandro"]Isn't there a formula for the area of intersection between whatever 2 circles? I'll look up the subject.[/quote]That should be rather easy to compute given the radii of the two circles, and either the coordinates of the centers, or the distance between them.

No, wait....I think you are reinventing the wheel.

[url=http://www.devmaster.net/articles/software-rendering/part2.php]Read this.[/url]

[url=http://www.devmaster.net/articles/software-rendering/part3.php]Then this.[/url]

[quote="maersklandro"]What do you mean, "reinventing the wheel"? What has how i project to do with the finished, projected vertex coordinates?

And besides, in addition to the types of projection listed there there's another one. Here, have a look at my tutorial on the subject. It's pretty old, however. This type of projection is very difficult to implement accurately. And it isn't really being used. At least not in real-time renderers. Ray tracers can achieve the same effect, I believe. But I think it's through the use of much more complex maths. Not really sure.

I want to calculate overlap areas between projected triangles and pixel halfs (also triangles). It's kind of irrelevenat what type the of projection was used, as long as as the transformation is linear. I came upon the circle-circle overlap area formula by chance, while looking for triangle-triangle overlap area calculation methods. I haven't ever needed it. I would probably have reinvented it before searching on the net. Anyway, it seems it's only what I need and can't come up with myself that I can't find on the net.[/quote]
Ok, give me a complete overview of what you want to do. Do you want to rasterize with pixel halfs?
Why do you need to areas?

[quote="maersklandro"]It's 4:30 AM.  Well, I want an easy, compact, simple and FAST method of calculating the area of intersection between whatever 2 triangles :D . This would naturally be a part of the rasterization process, and more.

PLEASE, PLEASE read my attachement!!!

The process steps are these:

Projection (any type, provided that a straight line in world space is projected to a straight line onscreen; so no fisheye lens projection). The vertexes that are the tips of the triangles are projected from world space coordinates to screen coordinates and stored in a separate array.

For each and every screen pixel cycle through all the VISIBLE triangles (at that pixel, determined using a variation on the Scanline buffer, perhaps) that have been projected and calculate the overlap area between the projected triangles and each half of that particular screen pixel. On a triangle to triangle basis add the two areas together (the result will be <= 1) and multiply this with each of the Red, Green, Blue, Red & Alpha Colour components of the triangle. Add these to a colour components buffer, respectively. After cycling through all the triangles visible at the current pixel, round (or truncate) and combine the colour components in the buffer and you have the final colour for that pixel. An effective hidden surface removal method is essential, obviously. Otherwise triangles hidden behind other triangles at that pixel will wrongly influence the final colour of the pixel.[/quote]
So it is a really fancy way of determining which triangle gets the edge pixels?

[quote="maersklandro"]No. Look. You are looking at this wrong.

Using triangle overlap, a pixel can have more than a single triangle contributing to its final colour. It can have more than 16 triangles (actually, all of the triangles in the scene can converge on 1 pixel, for example, and all would be accounted for) contributing to its final colour, which is the maximum currently possible with multi sampling rendering. And what's more, each triangle that covers the pixel (even partially) contributes to its final colour via a WEIGHED AVERAGE of the colours of all the triangles according to how much of the pixel each one covers.

Unlike ray rendering which only renders the luckiest triangle(s), those that happen to cover the centre or 2, 4, 6, 8 or 16 various arbitrary points on the pixel, while covering very little of the pixel itself in terms of surface. This obviously doesn't take into acccount how much of the pixel each sampled triangle covers. Am I clear enough? See the attachement.

So, instead of sampling a few points to determine the final colour of the pixel, this process is far more accurate. It's practically perfect. Theoretically, you could even use this method to correctly render virtual starfields accurately. Far beyond the capabilities of ray rendering, where whole star systems might get lost "in space" between sample points. But heck, what do they matter anyway?[/quote]
Yes, very nice, but I don't see why you call opengl ray rendering. Ray rendering doesn't use triangles, it can used any mathematically defined object. For example, your method is just better triangle rendering, and until I see some screenshots, I am going to assume the increase in quality is very minor. In ray tracing, I can do intersection tests with mathematically defined curves and circles, not triangle approximations. That is why the quality is so much better.

To answer your original question in screen space, do a test each pair of triangles to determine the coords of intersection, then use the formulae given on the previous page.

Whatever you do, all of these tests are going to be extremely slow.

Yes, I know. It's exactly what you said. Better triangle rendering. I'm not after anything else right now. And the quality will be very much better. At least for the same expense in processing power. After I'll have the triangle overlapping issue solved I might try to tackle texturing. Depens on how and IF i manage to get over this first hurdle.

I call OpenGL ray rendering because that is exactly what it is. I understand what you mean. You think i don't realize there's a difference between ray-tracers and 3D engines. I know there's a difference. But just because 3D engines are real-time renderers does not mean they do not use ray rendering. They use hardware implemented math. procedures to calculate the intersection of imaginary rays that spring from the camera space origin and go through each pixel centre of the virtual camera, which is in fact the computer screen.

The first triangle (or other primitive in case of ray tracers altough i was under the impression Quake III also was the first game to use real-time rendered splines for curved surfaces) these rays hit (for the sake of argument, let's assume it is fully opaque and not mirror grade reflective) is taken to be what is seen at that pixel. This is exactly what a Z-buffer is built upon. Even if you do not PHYSICALLY MODEL the imaginary ray it doesn't mean that your rendering pipeline is not based on it. I believe ray tracers differ from most 3D engines specifically in that they do physically model a huge amount of rays (along with many environmental interactions that alter the "light" properties and direction of the ray) as to try to closer model the reality. All, however, use rays that originate with the camera, not the light sources. How do you suppose texturing is performed? The "samples" are the points where the imaginary rays of each pixel hit the triangles in the scene. Even if you don't physically model the rays to get the triangle "sample" points, it's pretty clear how you get these points, isn't it? Ray tracers do, however, phisically model the rays.

Altough OpenGL is ray rendering, I never said ray rendering is the same as ray tracing. Ray tracing is more complex ray rendering. However, aren't the recursive mirrors in Quake III ray tracing?

I want and NEED a formula for the overlap area in order to make this work. I need something that has no "if"s, no "case"s, no "<"s or ">"s. I need something, simple, easy to sistematise. I want an implementation that is fast. Those tests are not. There must be some other way.

[quote="maersklandro"]I want and NEED a formula for the overlap area in order to make this work. I need something that has no "if"s, no "case"s, no "<"s or ">"s. I need something, simple, easy to sistematise. I want an implementation that is fast. Those tests are not. There must be some other way.[/quote]

OK, again (I feel like the $101$th person telling that):
there is no such function in terms of rational functions and most other standard functions like $\sin$.

There is indeed one by using $|x|$, but that is indentical to using cases and makes computation longer when used than when using cases.

So now stop that asking for a closed formula, there is none :!:

Sorry for bugging you so much... :blush:

10q anyways. :D


find all functions f on positive integers such that

                                  (x+y) f(f(x)y) = x^2 (f(f(x)+f(y)))

Your friend has already posted it before  :lol:
It's here [url]http://www.mathlinks.ro/Forum/topic-35009.html[/url]


Any matrix of the form:

$ X=\left[ \begin{array}{cccc}a&b&c&d\\\noalign{\medskip}0&a&b&c\\\noalign{\medskip}0&0&a&b\\\noalign{\medskip}0&0&0&a\end{array}\right]$

where $ a,b,c,d \in \mathbb{R}$ will work.

(For lack of a trick, I solved this in a straightforward way...)

We can replace $ A$ with $ 2I-A$ without changing the problem. The matrix $ S=2I-A$ represents the shift operator: $ Se_{i}=e_{i-1}$ and $ Se_{1}=0$. Using this observation, it is not hard to prove that $ X$ must have a diagonal-constant form (i.e. a Toeplitz matrix).


Find $ lim_{n \rightarrow \infty}( 1 \minus{} \frac {1}{1!} \plus{} \frac {1}{2!} \minus{} \frac {1}{3!} \plus{} ... \plus{} ( \minus{} 1)^n \frac {1}{n!})$

Try to plug -1 into Teylor series for $ e^x$.

Well first of all it is $ \cosh{1}\minus{}\sinh{1}$ and second of all that equals $ \frac{1}{e}$.  :wink:

Oops! I was in a rush, heading to a class.


If a circle of radius 2 is made to roll along the x-axis, what is an equation for the path of the center of the circle?

[hide="hint"] the radius of the circle is always the same length. this means that the center is always the same length from the x-axis.[/hide]

[hide="that was too easy"]
$y=2$ because the center of the circle makes no vertical movement
or you could say $y=-2$ if the circle was on the other side of the x-axis
[/hide]

I got it.

How about if the question was tristed in terms of the y-axis?

If a circle of radius 2 is made to roll down the y-axis, what is an equation for the path of the center of the circle?

by similar reasons its:

$x=\pm2$ depending on which side of the y-axis the circle is rolling down on

better way to 'twist' the question: what if we have a square 'rolling' along some flat surface? let's say the side length is 2 and it's on the x-axis. what path does the 'center' of the square trace out?


Prove that if $a,b,c,d > 0$ and $a \le 1$, $a+b \le 5$, $a+b+c \le 14$, $a+b+c+d \le 30$, then

$\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d} \le 10$.

[hide="hint"] Karamata = one-line proof [/hide]

[quote="Elemennop"]I know of a proof using substitution $w = \sqrt{a} + 1$, $x = \sqrt{b} + 2$, etc.

What is Karamata?[/quote]
It's an inequality theorem like Jensen's Inequality.
Look this link: http://www.artofproblemsolving.com/Forum/topic-14975.html

For the problem, u use concavity of the function, right?

[quote="paladin8"]Yeah, since $(1,4,9,16) \succ (a,b,c,d+e)$ with $e = 30-a-b-c-d$ and $f(x) = \sqrt{x}$ is concave, we have $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d} \le \sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d+e} \le 1+2+3+4 = 10$.[/quote]

That was TWO lines! HA! Just kidding :P I've never even heard of Karamata though...

[quote="Arne"]Cauchy works nicely:

\[ \begin{aligned} \left(\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}\right)^2 & \leq \left(\frac{a}{1} + \frac{b}{2} + \frac{c}{3} + \frac{d}{4}\right)\left(1 + 2 + 3 + 4\right) \\ & = 10\left(\frac{1}{4}(a + b + c + d) + \frac{1}{12}(a + b + c) + \frac{1}{6}(a + b) + \frac{1}{2}a\right) \\ & \leq 10\left(\frac{30}{4} + \frac{14}{12} + \frac{5}{6} + \frac{1}{2}\right) \\ & = 100 \end{aligned}  \]

so we're done :)[/quote]

Yes, it's very nice but I don't understand this part :?
\[ \begin{aligned} \left(\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}\right)^2 & \leq \left(\frac{a}{1} + \frac{b}{2} + \frac{c}{3} + \frac{d}{4}\right)\left(1 + 2 + 3 + 4\right)\end{aligned}  \]

Perfect. As you can see, there are very few inequalities which can NOT be proved by Cauchy ;)

Thank you all. I known this inequality but it was named Schwarz inequality and Cauchy was AM-GM :) Never mind.


[quote="nguoivn"]Given $ a, b, c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$.
Prove that: $ \sqrt [3]{1 \plus{} 3a^2} \plus{} \sqrt [3]{1 \plus{} 3b^2} \plus{} \sqrt [3]{1 \plus{} 3c^2} \geq\ 2(\sqrt [3]{\frac {a}{b \plus{} c}} \plus{} \sqrt [3]{\frac {b}{c \plus{} a}} \plus{} \sqrt [3]{\frac {c}{a \plus{} b}})$
 :)[/quote]

We have:

$ ab\plus{}bc\plus{}ca\equal{}3\Leftrightarrow a\equal{}\frac{3\minus{}bc}{b\plus{}c}$

$ \equal{}> 1\plus{}3a^2\minus{}\frac{8a}{b\plus{}c}\equal{}\frac{(b\minus{}c)^2\plus{}3(bc\minus{}1)^2}{(b\plus{}c)^2}\ge 0$

$ \Leftrightarrow \sqrt[3]{1\plus{}3a^2}\ge 2\sqrt[3]{\frac{a}{b\plus{}c}}$


give me cheats you know on topics in math yu know,plz

There are no "cheats." It's all based on conceptual understanding of topics and applications in problems.


AMC AIME FTW geometry rectangle USA(J)MO USAMO

This would be an awesome tournament spanning over 10 months!!! This is the [b][size=200]CTC CHAMPIONSHIP!!![/b][/size]

Here's my idea:

9 Qualification tournaments: (One a month from October-June)

Qualification tournaments are as such:

Everyone who signs up plays 4 completely random opponents. These games against opponents are 5 countdowns (not first to 5,not best out of 5, but 5) meaning the final result will be 3-2,4-1,or 5-0.
Your total number of cds won is totaled up. The top 4 scores will go into a single elimination playoff that is a best of 9 cds (so first to 5) so that 1st,2nd,3rd,and 4th are decided (so the two first-round losers play to decide 3rd place). After the tournament, everyone's tournament score is calculated.

You get:
10 point participation bonus
5 points for every cd you win(during the first 4 rounds. CDs in the final do not count)
Bonus: 10 pts for 4th, 20 for 3rd,30 for 2nd, and 40 for 1st.

Everyone may participate in as many qualification tournaments as they want. Then, come July, everyone's top 5 scores are added together. The top 16 scorers will be invited to participate in the CTC Championship, which has different rules.

The CTC Club Championship:
Everyone will be divided into 4 groups of 4 based on seeding (1,8,9,16),(2,7,10,15),(3,6,11,14), and (4,5,12,13).

A round-robin tournament will be played. Again it will be 5 cds. The scores will be totaled. The top 2 players in each group will automatically advance to round 2. The other 8 will be seeded according to the score. They will then play one best 5 out of 9 cd series against their opponent. Losers are out. Now in round 2, the final 12 will participate in a triple-elimination tournament, best 5 out of 9 cds. Game continues until the finals are left, which will be a FIRST TO 10 CDS!!!

It is now time to signup for the first of these (the October) qualification rounds.

[hide="People eligible"]
Skaldjfhrulez
Kl2836
Xoangieexo
FantasyLover
Vallon22
Dojo
Pacman2812
Myyellowducky82
Vamathletes
Guineapiggies
Pinkpiglet
Superllamel
Jain_harshi
Splatyango
Mathblitz
Weird888
Dragonlaird7
Dragon96
Dannyhamtx
Kookamunga
Pi guy
27r
Yclarinet
Chenhsi
Textangle
Aime_is_hard
Pi-dimension
Mind Storm
Bisbmard9
Emilylambkin
Johnnybeard
Discrete_math
AsianPascal
Bugi
Irsmart22
Bogtro
Michaelwong256

Note:If you are not on this list, pm me and ask to be in CTC[/hide]

[hide="People joined"]
1. Mewto55555
2. RomanianGenius
3. Jjx1
4. Peregrinefalcon88
5. AIME15
6. Cherry_pi
7. Violin321
8. James4l
9. Yoh2241
10. Ernie
11. Mathnerd711
12. Sup3rsolv3r
13. Jongao
14. Isabella2296
15. Lotsofmath
16. Nikeballa96[/hide]

Signups will end Sunday September 28th. All games will be pmed, and you will have over a month to play your four tournament games.

this is weird. and i bet lots of peopel will complain. but i'll join. just tell me, what matches i have. i'll try to come on after school each day. cna't come on weekends.

i would like to join to this is my first tournament though just FYI but it sounds interesting the biggest aops tourny ever

[quote="james4l"]sure i'll join if you give us PM notifications[/quote]

Silly jamesy-wamesy that is something that mewmew will always do. That and call you by nicknames that make you think I'm your grandmother :rotfl:

Guys, if you have nothing else to say, private message mewto.

Okay, since I said that, I join.  :)

hm i see you signed me up w/o asking me. although good job predicting i would join  :lol:

[hide="People eligible"]
Skaldjfhrulez
Kl2836
Xoangieexo
FantasyLover
Vallon22
Dojo
Pacman2812
Myyellowducky82
Vamathletes
Guineapiggies
Pinkpiglet
Superllamel
Jain_harshi
Splatyango
Mathblitz
Weird888
Dragonlaird7
Dragon96
Dannyhamtx
Kookamunga
Pi guy
27r
Yclarinet
Chenhsi
Textangle
Aime_is_hard
Pi-dimension
Mind Storm
Bisbmard9
Emilylambkin
Johnnybeard
Discrete_math
AsianPascal
Irsmart22
Bogtro
Michaelwong256

Note:If you are not on this list, pm me and ask to be in CTC[/hide]

[hide="People joined"]
1. Mewto55555
2. RomanianGenius
3. Jjx1
4. Peregrinefalcon88
5. AIME15
6. Cherry_pi
7. Violin321
8. James4l
9. Yoh2241
10. Ernie
11. Mathnerd711
12. Sup3rsolv3r
13. Jongao
14. Isabella2296
15. Lotsofmath
16. Nikeballa96
17. CA Math
18. Bugi
19. Program4
20. Itex89[/hide]

Expected participation in this tournament is in the neighborhood of 30 people.

Signups will end Sunday September 28th. All games will be pmed, and you will have over a month to play your four tournament games.

The thingis I haven't seen mathnerd on FTW ever. I am gonna pm him.

You can PM me or mewto55555 on that, instead of forum posting. :wink:

JONGAO DEFEATS ITEX
4-3,4-2,4-1,4-0,4-1.
xD mewto :lol:

A reminder: You will play everyone in your group once. You play a [b]total[/b] of 5 cds against your opponent, meaning the final score will be 3-2, 4-1, or 5-0.

\begin{tabular}{|l\parallel{}c|c|c|c|c| |c|}\hline & Against Mewto & Against Violin321 & Against Jongao & Against CA Math & Against Itex89 & Total \\
\hline Mewto55555 & X & 5 & 5 & ? & 4 & ? \\
\hline Violin321 & 0 & X & 0 & ? & ? & ? \\
\hline Jongao & 0 & 5 & X & ? & 5 & ? \\
\hline CA Math & ? & ? & ? & X & ? & ? \\
\hline Itex89 & 1 & ? & 0 & ? & X & ? \\
\hline \end{tabular}

\begin{tabular}{|l\parallel{}c|c|c|c|c| |c|}\hline & Against RG & Against AIME15 & Against Yoh & Against Lotsofmath & Against Program4 & Total \\
\hline RomanianGenius & X & 3 & 5 & 5 & ? & ? \\
\hline AIME15 & 2 & X & 5 & 3 & ? & ? \\
\hline Yoh2241 & 0 & 0 & X & 0 & 0 & 0 \\
\hline Lotsofmath & 0 & 2 & 5 & X & ? & ? \\
\hline Program4 & ? & ? & 5 & ? & X & ? \\
\hline \end{tabular}

\begin{tabular}{|l\parallel{}c|c|c|c|c| |c|}\hline & Against Jjx1 & Against James4l & Against Mathnerd711 & Against Isabella2296 & Against Bugi & Total \\
\hline Jjx1 & X & ? & ? & 0 & 3 & ? \\
\hline James4l & ? & X & ? & ? & ? & ? \\
\hline Mathnerd711 & ? & ? & X & ? & ? & ? \\
\hline Isabella2296 & 5 & ? & ? & X & 4 & ? \\
\hline Bugi & 2 & ? & ? & 1 & X & ? \\
\hline \end{tabular}

\begin{tabular}{|l\parallel{}c|c|c|c|c| |c|}\hline & Against Falcon & Against Cherry & Against Ernie & Against Sup3r & Against Nikeballa & Total \\
\hline Peregrinefalcon88 & X & ? & ? & ? & ? & ? \\
\hline Cherry_pi & ? & X & 0 & ? & ? & ? \\
\hline Ernie & ? & 5 & X & 5 & ? & ? \\
\hline Sup3rsolv3r & ? & ? & 0 & X & ? & ? \\
\hline Nikeballa96 & ? & ? & ? & ? & X & ? \\
\hline \end{tabular}[/quote]

[quote="mewto55555"]A reminder: You will play everyone in your group once. You play a [b]total[/b] of 5 cds against your opponent, meaning the final score will be 3-2, 4-1, or 5-0.

\begin{tabular}{|l\parallel{}c|c|c|c|c| |c|}\hline & Against Mewto & Against Violin321 & Against Jongao & Against CA Math & Against Itex89 & Total \\
\hline Mewto55555 & X & 5 & 5 & 2 & 4 & 16 \\
\hline Violin321 & 0 & X & 0 & ? & ? & ? \\
\hline Jongao & 0 & 5 & X & ? & 5 & ? \\
\hline CA Math & 3 & ? & ? & X & ? & ? \\
\hline Itex89 & 1 & ? & 0 & ? & X & ? \\
\hline \end{tabular}

\begin{tabular}{|l\parallel{}c|c|c|c|c| |c|}\hline & Against RG & Against AIME15 & Against Yoh & Against Lotsofmath & Against Program4 & Total \\
\hline RomanianGenius & X & 3 & 5 & 5 & ? & ? \\
\hline AIME15 & 2 & X & 5 & 3 & ? & ? \\
\hline Yoh2241 & 0 & 0 & X & 0 & 0 & 0 \\
\hline Lotsofmath & 0 & 2 & 5 & X & ? & ? \\
\hline Program4 & ? & ? & 5 & ? & X & ? \\
\hline \end{tabular}

\begin{tabular}{|l\parallel{}c|c|c|c|c| |c|}\hline & Against Jjx1 & Against James4l & Against Mathnerd711 & Against Isabella2296 & Against Bugi & Total \\
\hline Jjx1 & X & ? & ? & 0 & 3 & ? \\
\hline James4l & ? & X & ? & ? & ? & ? \\
\hline Mathnerd711 & ? & ? & X & ? & ? & ? \\
\hline Isabella2296 & 5 & ? & ? & X & 4 & ? \\
\hline Bugi & 2 & ? & ? & 1 & X & ? \\
\hline \end{tabular}

\begin{tabular}{|l\parallel{}c|c|c|c|c| |c|}\hline & Against Falcon & Against Cherry & Against Ernie & Against Sup3r & Against Nikeballa & Total \\
\hline Peregrinefalcon88 & X & ? & ? & ? & ? & ? \\
\hline Cherry_pi & ? & X & 0 & ? & ? & ? \\
\hline Ernie & ? & 5 & X & 5 & ? & ? \\
\hline Sup3rsolv3r & ? & ? & 0 & X & ? & ? \\
\hline Nikeballa96 & ? & ? & ? & ? & X & ? \\
\hline \end{tabular}[/quote][/quote]

How about for ratings...

if the score is 3-2, then weighing factor = 35

if the score is 4-1, then weighing factor = 40

if the score is 5-0, then weighing factor = 50

Even though the first month has already started, can I join for the second month?

I would also like to join CTC.

Sure, I don't think we have started singups yet though, but welcome to CTC!

For more info, go to the "Introducing CTC!" thread.

[quote="mewto55555"][quote="mewto55555"]A reminder: You will play everyone in your group once. You play a [b]total[/b] of 5 cds against your opponent, meaning the final score will be 3-2, 4-1, or 5-0.

\begin{tabular}{|l\parallel{}c|c|c|c|c| |c|}\hline & Against Mewto & Against Violin321 & Against Jongao & Against CA Math & Against Itex89 & Total \\
\hline Mewto55555 & X & 5 & 5 & 2 & 4 & 16 \\
\hline Violin321 & 0 & X & 0 & ? & ? & ? \\
\hline Jongao & 0 & 5 & X & ? & 5 & ? \\
\hline CA Math & 3 & ? & ? & X & ? & ? \\
\hline Itex89 & 1 & ? & 0 & ? & X & ? \\
\hline \end{tabular}

\begin{tabular}{|l\parallel{}c|c|c|c|c| |c|}\hline & Against RG & Against AIME15 & Against Yoh & Against Lotsofmath & Against Program4 & Total \\
\hline RomanianGenius & X & 3 & 5 & 5 & ? & ? \\
\hline AIME15 & 2 & X & 5 & 3 & 4 & 14 \\
\hline Yoh2241 & 0 & 0 & X & 0 & 0 & 0 \\
\hline Lotsofmath & 0 & 2 & 5 & X & ? & ? \\
\hline Program4 & ? & 1 & 5 & ? & X & ? \\
\hline \end{tabular}

\begin{tabular}{|l\parallel{}c|c|c|c|c| |c|}\hline & Against Jjx1 & Against James4l & Against Mathnerd711 & Against Isabella2296 & Against Bugi & Total \\
\hline Jjx1 & X & ? & ? & 0 & 3 & ? \\
\hline James4l & ? & X & ? & ? & ? & ? \\
\hline Mathnerd711 & ? & ? & X & ? & ? & ? \\
\hline Isabella2296 & 5 & ? & ? & X & 4 & ? \\
\hline Bugi & 2 & ? & ? & 1 & X & ? \\
\hline \end{tabular}

\begin{tabular}{|l\parallel{}c|c|c|c|c| |c|}\hline & Against Falcon & Against Cherry & Against Ernie & Against Sup3r & Against Nikeballa & Total \\
\hline Peregrinefalcon88 & X & ? & ? & ? & ? & ? \\
\hline Cherry_pi & ? & X & 0 & ? & ? & ? \\
\hline Ernie & ? & 5 & X & 5 & ? & ? \\
\hline Sup3rsolv3r & ? & ? & 0 & X & ? & ? \\
\hline Nikeballa96 & ? & ? & ? & ? & X & ? \\
\hline \end{tabular}

[quote="mewto55555"][quote="mewto55555"][quote="mewto55555"]A reminder: You will play everyone in your group once. You play a [b]total[/b] of 5 cds against your opponent, meaning the final score will be 3-2, 4-1, or 5-0.

\begin{tabular}{|l\parallel{}c|c|c|c|c| |c|}\hline & Against Mewto & Against Violin321 & Against Jongao & Against CA Math & Against Itex89 & Total \\
\hline Mewto55555 & X & 5 & 5 & 2 & 4 & 16 \\
\hline Violin321 & 0 & X & 0 & ? & ? & ? \\
\hline Jongao & 0 & 5 & X & ? & 5 & ? \\
\hline CA Math & 3 & ? & ? & X & ? & ? \\
\hline Itex89 & 1 & ? & 0 & ? & X & ? \\
\hline \end{tabular}

\begin{tabular}{|l\parallel{}c|c|c|c|c| |c|}\hline & Against RG & Against AIME15 & Against Yoh & Against Lotsofmath & Against Program4 & Total \\
\hline RomanianGenius & X & 3 & 5 & 5 & ? & ? \\
\hline AIME15 & 2 & X & 5 & 3 & 4 & 14 \\
\hline Yoh2241 & 0 & 0 & X & 0 & 0 & 0 \\
\hline Lotsofmath & 0 & 2 & 5 & X & ? & ? \\
\hline Program4 & ? & 1 & 5 & ? & X & ? \\
\hline \end{tabular}

\begin{tabular}{|l\parallel{}c|c|c|c|c| |c|}\hline & Against Jjx1 & Against James4l & Against Mathnerd711 & Against Isabella2296 & Against Bugi & Total \\
\hline Jjx1 & X & 0 & 0 & 0 & 3 & 3 \\
\hline James4l & 5 & X & ? & ? & ? & ? \\
\hline Mathnerd711 & 5 & ? & X & ? & ? & ? \\
\hline Isabella2296 & 5 & ? & ? & X & 4 & ? \\
\hline Bugi & 2 & ? & ? & 1 & X & ? \\
\hline \end{tabular}

\begin{tabular}{|l\parallel{}c|c|c|c|c| |c|}\hline & Against Falcon & Against Cherry & Against Ernie & Against Sup3r & Against Nikeballa & Total \\
\hline Peregrinefalcon88 & X & ? & ? & ? & ? & ? \\
\hline Cherry_pi & ? & X & 0 & ? & ? & ? \\
\hline Ernie & ? & 5 & X & 5 & ? & ? \\
\hline Sup3rsolv3r & ? & ? & 0 & X & ? & ? \\
\hline Nikeballa96 & ? & ? & ? & ? & X & ? \\
\hline \end{tabular}

Bah Jjjx1 drops out.

I seriously doubt that this tournament will ever end.

Ok due to the obscene lack of participation I will cancel this tournament (sorry to thoise of you who have been playing) and create a new format. Same rules, but only people who have played will play eachother (i.e. i will not match you up with people who dont play games except maybe in the first round).

Can someone lock this?

I suggest we halt all tournaments until summer, when people have more time.

Locked.