geometry
circumcircle
trigonometry
IMO Shortlist
Spiral Similarity
Miquel point

"Common twist" = radical axis in this case I think.

I'm sorry I don't know the word. This is in this case for exemple the line of power of the two circles... Do you understand ?

"corde commune" in French...
"coarda comuna" in Rumania...

[hide="Here is a syntetical proof of this nice and difficult problem."]
[color=darkred][b][u]Lemma.[/u][/b] Let $ABC$ be an acute triangle. Define: the circumcircle $c=C(O)$ and the orthocentre $H$ of the triangle $ABC$; the middlepoint $M$ of the side $[BC]$; the intersection $N$ between $MH$ and the bisector of the angle $\widehat {BAC}$; $D\in AB$ and $E\in AC$ so that $H\in DE$ and $AD=AE$. Then the point $N$ belongs to the circumcircle $w=C(O_a)$ of the triangle $ADE$.
[b][u]A metrical proof.[/u][/b] $AH=2R\cos A$ and $P\in c\cap (AO_a\Longrightarrow$ $AP=2R\cos \frac{B-C}{2}\ ,$  $MP=R(1-\cos A)$. Thus,
$\frac{AN}{AP}=\frac{AH}{AH+MP}=\frac{2\cos A}{1+\cos A}\Longrightarrow \boxed {AN=\frac{2\cos A}{1+\cos A}\cdot AP}\ .$ Denote the point $N'\in AP\cap w$, i.e. $DN'\perp AB$. Thus, $\frac{AD}{\sin \left(B+\frac A2\right)}{=\frac{AH}{\cos \frac A2}}$ $\Longrightarrow$ $AD=\frac{\cos\frac{B-C}{2}}{\cos \frac A2}\cdot AH$ and $AN'=\frac{AD}{\cos \frac A2}$ $\Longrightarrow$ $AN'=\frac{\cos \frac{B-C}{2}}{\cos^2\frac A2}\cdot AH=$ $\frac{2\cos \frac{B-C}{2}}{1+\cos A}\cdot AH=$ $\frac{AP}{R}\cdot\frac{2R\cos A}{1+\cos A}=$ $\frac{2\cos A}{1+\cos A}\cdot AP$ $\Longrightarrow AN'=AN$ $\Longrightarrow$ $N\equiv N'$, i.e. the point $N$ belongs to the circumcircle of the triangle $ADE\ .$[/color]

[color=darkblue][b][u]Proof of the proposed problem.[/u][/b] Denote: $A'\in c\cap (AO$ ; the middlepoint $A_1$ of  $[AH]$ ; $N\in w\cap (AO_a$. From the above lemma results $N\in MH$. But  $AA_1=HA_1=OM$ (the point $M$ is the middlepoint of the segment $[HA']$), $OA=OA'$ and $O_aA=O_aN$. Thus the points $A_1$, $O$, $O_a$ are the middlepoints of the segments $AH$, $AA'$, $AN$ respectively and $N\in HM\equiv HA'$ $\Longrightarrow$ $O_a\in OA_1$ and $OA_1\parallel MH$. Therefore, $MH\parallel OO_a$.
[u]Remark.[/u] I denote ray $(XY$ without the point $X$.[/color][/hide] [u]See [b]PP12[/b][/u] from [url=http://www.artofproblemsolving.com/community/c1090h1163708][color=red][b]here[/b][/color][/url]

Could you explain why $AA_3$ is an angle bisector of $OAH$ and why $A', M,H$ are collinear?

[quote="Megus"]Could you explain why $AA_3$ is an angle bisector of $OAH$ and why $A',M,H$ are collinear?[/quote] [color=darkred][b][u]Lemma.[/u][/b] Let $w=C(O)$ be a circumcircle of the acute triangle $ABC$, where $AB<AC$. Denote the point $A'\in w\cap (AO$. Define the points $X\in AB$, $Y\in AC$ so that $B\in (AX)\ ,\ AX=AC$ and $Y\in (AC)\ ,\ AY=AB$. Then $m(\widehat {A'BY})=m(\widehat {A'CX})=\frac A2\ .$
[u]Standard notation.[/u] $\boxed {\ (XY\ }$ - the ray without the point $X$.[/color]

Thank you, but I don't see how to apply this lemma  :(

Another application of spiral similarity.  :)

You are a clever young because you are right, Megus ! See more above the my first reply where is the complete solution of this problem.

[quote="BogG"]Let $\triangle ABC$ be an acute-angled triangle with $AB \not= AC$. Let $H$ be the orthocenter of triangle $ABC$, and let $M$ be the midpoint of the side $BC$. Let $D$ be a point on the side $AB$ and $E$ a point on the side $AC$ such that $AE=AD$ and the points $D$, $H$, $E$ are on the same line. Prove that the line $HM$ is perpendicular to the common chord of the circumscribed circles of triangle $\triangle ABC$ and triangle $\triangle ADE$.[/quote]

First, I repost a solution given by Andre.L which was deleted for an unclear reason (to anyone who feels spoken to: according to Valentin, these things are not supposed to be deleted!).

[quote="andre.l"]The common chord of the cirmcumcircles of $ABC$ and $ADE$ is their radical axis. Thus, it's perpendicular to the line that passes through their centers. Let $O$ and $O_1$ be the circumcenters of $ABC$ and $ADE$, respectively. Then, $HM$ is perpendicular to the common chord of the circumcircles of $ABC$ and $ADE$ iff $HM$ is parallel to $OO_1$. Let $Q=OO_1\cap AH$. Since $OM\bot BC$ and $AH\bot BC$, $OM\parallel AH$. Thus, $OO_1\parallel HM \Leftrightarrow OQ \parallel HM \Leftrightarrow OMHQ$ is a parallelogram. Let $S$ be the midpoint of $OH$. Then, $OMHQ$ is a parallelogram $\Leftrightarrow MS=SQ$. But the circle with center $S$ which passes through $M$ is the nine-point circle of $ABC$. Thus, it intersects $AH$ at it's midpoint. What we will prove is that $Q$ is the midpoint of $AH$.
The point $O_1$ is on the perpendicular bisector of $DE$, which is also the angle bisector of $BAC$. Let $T$ be the intersection of $OH$ and the angle bisector of $BAC$. Now, apply Menelaus' theorem to the line $OO_1Q$ and the triangle $HTA$. We get that
$\frac{OH}{OT} \frac{O_1T}{O_1A} \frac{QA}{QH}=1$.
Thus, $QA=AH \Leftrightarrow \frac{OH}{OT} \frac{O_1T}{O_1A} =1$.
With some calculation, we can prove that the above relation is equivalent to $\cos A=2\cdot\cos^2\left(\frac{A}{2}\right)-1$, which is true.[/quote]

Let me add my solution.

Let B' and C' be the feet of the altitudes of triangle ABC from the vertices B and C, respectively. Keep in mind that these altitudes intersect at the orthocenter H of triangle ABC. Since < BB'C = 90° and < BC'C = 90°, the points B' and C' lie on the circle with diameter BC; the center of this circle is the midpoint M of the segment BC. So the circle with diameter BC is a circle with center M which passes through the points B and C and intersects the sides CA and AB again at the points B' and C', respectively. Hence, according to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=60787]IMO 1985 problem 5[/url], if R is the point of intersection of the circumcircles of triangles ABC and B'AC' (apart from A), then < MRA = 90°. Thus, the point M lies on the perpendicular to the line AR at the point R. On the other hand, since < AB'H = 90° and < AC'H = 90°, the points B' and C' lie on the circle with diameter AH. Consequently, the circumcircle of triangle B'AC' is the circle with diameter AH; therefore, since the point R lies on this circumcircle, it satisfies < ARH = 90°. Consequently, the point H lies on the perpendicular to the line AR at the point R.

Now as we know that both points M and H lie on the perpendicular to the line AR at the point R, we obtain $HM\perp AR$. Now we will prove that the point R lies on the circumcircle of triangle ADE. Once this is shown, it will follow that the points A and R are the two common points of the circumcircles of triangles ABC and ADE; this will mean that the line AR is the common chord of these circumcircles, so that the relation $HM\perp AR$ will tell us that the line HM is perpendicular to the common chord of the circumcircles of triangles ABC and ADE. This will solve the problem.

So it only remains to show that the point R lies on the circumcircle of triangle ADE.

Since AE = AD, the triangle DAE is isosceles, so that its base angle is $\measuredangle ADE=\frac{180^{\circ}-\measuredangle DAE}{2}=90^{\circ}-\frac{\measuredangle DAE}{2}=90^{\circ}-\frac{A}{2}$. In other words, $\measuredangle C^{\prime}DH=90^{\circ}-\frac{A}{2}$. Similarly, $\measuredangle B^{\prime}EH=90^{\circ}-\frac{A}{2}$. Thus, < C'DH = < B'EH. Together with < DC'H = < EB'H (both of these angles are 90°), this yields that the triangles C'DH and B'EH are similar; hence, $\frac{DC^{\prime}}{C^{\prime}H}=\frac{EB^{\prime}}{B^{\prime}H}$. On the other hand, since < BC'H = < CB'H (both of these angles are 90°) and < BHC' = < CHB', the triangles BC'H and CB'H are similar, so that $\frac{C^{\prime}H}{BC^{\prime}}=\frac{B^{\prime}H}{CB^{\prime}}$. Thus, $\frac{DC^{\prime}}{BC^{\prime}}=\frac{DC^{\prime}}{C^{\prime}H}\cdot\frac{C^{\prime}H}{BC^{\prime}}=\frac{EB^{\prime}}{B^{\prime}H}\cdot\frac{B^{\prime}H}{CB^{\prime}}=\frac{EB^{\prime}}{CB^{\prime}}$.

Now, since the point R lies on the circumcircle of triangle B'AC', we have < RC'A = < RB'A, so that < RC'B = 180° - < RC'A = 180° - < RB'A = < RB'C. On the other hand, since the point R lies on the circumcircle of triangle ABC, we have < RBA = < RCA, what becomes < RBC' = < RCB'. From < RC'B = < RB'C and < RBC' = < RCB', it follows that the triangles RBC' and RCB' are similar. The points D and E are corresponding points in these two similar triangles (since they lie on the respective sides BC' and CB' and divide them in the same ratio $\frac{DC^{\prime}}{BC^{\prime}}=\frac{EB^{\prime}}{CB^{\prime}}$); since corresponding points in similar triangles form equal angles, we thus have < RDC' = < REB', or, equivalently, < RDA = < REA. This yields that the point R lies on the circumcircle of triangle ADE, and the problem is solved.

  Darij

[quote="abdurashidjon"][quote="Virgil Nicula"][hide=Here is a syntetical proof of this nice and difficult problem.]
[b][u]A metrical proof.[/u][/b] $AH=2R\cos A$ and $P\in c\cap (AO_a\Longrightarrow$ $AP=2R\cos \frac{B-C}{2}\ ,$  $MP=R(1-\cos A)$. [/quote]
 Can someone explain how to get this ones[/quote]

Ok .We have $AH=2OM=\sqrt{4R^2-BC^2}=\sqrt{4r^2-4R^2(sinA)^2}=2RcosA$

By the low of sines $\frac{AP}{sin(B+\frac{A}{2})}=2R$  so  $AP=2Rsin(90+\frac{B-C}{2})$  so $AP=2R\cos \frac{B-C}{2}$

And $MP=OP-OM=R(1-cosA)$

BTW , mr Nicula there exist a not metrical solution to this lemma ? :?

[quote="BogG"]Let $\triangle ABC$ be an acute-angled triangle with $AB \not= AC$. Let $H$ be the orthocenter of triangle $ABC$, and let $M$ be the midpoint of the side $BC$. Let $D$ be a point on the side $AB$ and $E$ a point on the side $AC$ such that $AE=AD$ and the points $D$, $H$, $E$ are on the same line. Prove that the line $HM$ is perpendicular to the common chord of the circumscribed circles of triangle $\triangle ABC$ and triangle $\triangle ADE$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=89094][b][u]All the problems from the Swiss Imo Selection Team 2006[/u][/b][/url][/quote]

This is problem is difficult, not because it's lacking swift solutions but because it's so easy to go pursuing some typical ideas who do provide you with ways to solve it but I think, more complicated ones. Here is my proof

1) Name $O$ and $O'$ the circumcenter of $ABC$ and of the other triangle. Let $F$ be the point where $AO$ intersects again the circumcircle of $ABC$. First thing to notice is that if $MH$ is to be perpendicular to $AK$ then it will intersect it precisely at $K$. This is because $MH$ passes through $F$ and hence, if it's going to cut $AK$ perpendicularly it must do it in a point who belongs to the big circle, ergo $K$.
   1a) Notice that line $OO'$ cuts $AK$ perpendicularly at the midpoint. Hence it is parallel to line $FK$. It suffices to prove therefore that $H$ is in line $FK$ because then $M$ is in line $FH$ and they are all together beautifully.

2) With all this in mind the problem is actually half done. Triangles $BDH$ and $CEH$ are similar for $\angle HCE = \angle HBD$ and $\angle HDB = \angle HEC$. Hence
$\frac{BD}{CE} = \frac{DH}{HE}$
3) $\angle DKE = \angle BAC = \angle BKC \Rightarrow \angle BKD = \angle CKE$
4) $\angle KBD = \angle KCE$ hence, using 3, triangles $KBD$  and $KEC$ are similar hence $\frac{KD}{KE} = \frac{BD}{CE}$
5) With 3 and 4 $KH$ actually bisects $\angle DKE$ and we're set because if $G$ is the point where it touches again the small circle the $G$ and $A$ are diametrically opposite and hence $\angle HKA$ is right which means $H$ is in line $FK$.

Daniel

Here is a proof with the use of trignometry. It is quite long so I think that I had better hide it.

[hide]The sufficient and nessecery condition for $HM$ being perpendicular to the common chord is $OO' \parallel MH$. Let $OO'$ meets $AH$ at $F$, then $OO' \parallel MH \Longleftrightarrow MOFH$ being a parallelogram. Since $AH=2OM$, so it is enough to show that $AF=FH$. Since $AH=2R |\cos{A}|$, so we only need to prove that $AF= R |\cos{A}|$.

Let $AO'$ meets circle $O$ at $A'$, then since $AA'$ is the angle bisector of $\angle{BAC}$, so $O$, $M$, $A'$ are collinear. Because $AF \parallel OA'$, so we have an equality: \[\frac{AF}{AO'}=\frac{A'O}{A'O'}\] Let $AO'=r$, $A'O=R$, then we have $AA'=2R \sin{\left( B+\frac{A}{2}\right)}=2R \sin{\left( C+\frac{A}{2}\right)}$, thus: \[AA'=R \left( \sin{\left( B+\frac{A}{2}\right)}+\sin{\left( C+\frac{A}{2}\right)}\right) =2R \sin{\frac{\pi}{2}}\cos{\frac{B-C}{2}}= 2R \cos{\frac{B-C}{2}}\] Next we calculate $r$. Since $D$, $E$, $H$ are collinear, thus by theorem 4 of http://www.mathlinks.ro/Forum/viewtopic.php?t=100964,  we have: \[\frac{AB}{AD}\tan{B}+\frac{AC}{AE}\tan{C}=\tan{A}+\tan{B}+\tan{C}\] Since in any triangle $ABC$, we always have $\tan{A}+\tan{B}+\tan{C}=\tan{A}\tan{B}\tan{C}$. So: \[\frac{AB}{AD}\tan{B}+\frac{AC}{AE}\tan{C}=\tan{A}\tan{B}\tan{C}\] Hence we have \[AD=\frac{AB \tan{B}+AC \tan{C}}{\tan{A}\tan{B}\tan{C}}=2R \cdot \frac{\frac{\sin{C}\sin{B}}{\cos{B}}+\frac{\sin{C}\sin{B}}{\cos{C}}}{\tan{A}\frac{\sin{C}\sin{B}}{\cos{B}\cos{C}}}=2R\frac{\cos{B}+\cos{C}}{\tan{A}}\] Then use the fact that $AD=2r \cdot \sin{\angle{AED}}=2r \cdot \cos{\frac{A}{2}}$, we get: \[r=\frac{\cos{B}+\cos{C}}{\tan{A}\cdot \cos{\frac{A}{2}}}\cdot R = \frac{2 \cos{\frac{B+C}{2}}\cos{\frac{B-C}{2}}}{\tan{A}\cdot \cos{\frac{A}{2}}}\] Now pluge them into our original equality, we get: \[\frac{AF}{R}=\frac{r}{AA'-r}= \frac{1}{\frac{AA'}{r}-1}\] Pluge $R$, $r$ and $AA'$ into this expression, and use the fact that $\cos{\frac{B+C}{2}}=\sin{\frac{A}{2}}$ and $\sin{A}=2\sin{\frac{A}{2}}\cos{\frac{A}{2}}$, we get: \[\frac{AA'}{r}=\tan{A}\cdot \cot{\frac{A}{2}}=\frac{2\sin{\frac{A}{2}}\cos^{2}{\frac{A}{2}}}{\cos{A}\sin{\frac{A}{2}}}=\frac{2\cos^{2}{\frac{A}{2}}}{\cos{A}}\] Hence: \[\frac{R}{AF}=\frac{2\cos^{2}{\frac{A}{2}}}{\cos{A}}-1=\frac{1}{\cos{A}}\] Hence $AF=R \cos{A}$, which means $F$ is the midpoint of AH, thus HF=OM. Since $HF \parallel OM$, so $HFOM$ is a parallelogram, hence $OO' \parallel MH$. So our problem is solved[/hide].

Let $ O$ be the circumcenter, $ R$ be the radius and $ k$ be the circumcircle of $ \triangle ABC$ and let $ r$ be the radius and $ k'$ be the circumcircle of $ \triangle ADE$. Denote by $ S$ the intersection of $ AO$ with $ k$, by $ N$ the intersection of the line $ OM$ with $ k$ and let $ T$ be the intersection of the line $ AN$ with $ k'$. Moreover, let $ X$ be the intersection of $ DE$ with $ AT$. We'll prove that $ M$, $ T$, $ H$ are collinear and that $ AO$ and $ MH$ intersect in $ k$. Then it will follow that the point of intersection of the line $ MH$ with $ k$ and with $ k'$ will be a common point, let's say $ P$ and that the angle $ \angle HPA \equal{} 90$, as desired.
It's well-known that  $ AX$ and $ OM$ intersect in $ N$. Moreover, $ \angle HAX \equal{} \angle XAO$. First, we'll prove that $ S \in MH$. Let $ H_1$ and $ H_2$ be the reflections of $ H$ on $ AB$ and $ AC$, respectively. We have $ HC \Vert BS$ and thus $ BH \equal{} BH_1 \equal{} SL$ and $ HB \Vert CS$ and thus $ CH \equal{} CH_2 \equal{} BS$. Therefore, $ HBSC$ is a parallelogramm and it follows that $ HS$ intersects $ BC$ at its middle, i.e. $ M$, as desired.
It's left to prove that $ T \in MH$, i.e. $ T \in SM$. Applying Menelaos' theorem to the triangle $ \triangle NAO$, it's remains to prove that $ AN/NO \cdot OM/MN \cdot NT/TA \equal{} \minus{} 1$ or equivalently $ MN/2OM \equal{} TN/TA \equal{} AN/AT \minus{} 1$. It's well-known that $ 2OM \equal{} AH$, thus $ MN/2OM \equal{} R \minus{} OM/2OM \equal{} R/2OM \minus{} 1/2 \equal{} R/AH \minus{} 1/2$. Handling $ AN/AT$ is the more difficult part. We have $ AX \equal{} AH \cdot \cos\angle HAX$ and $ AX \equal{} AE \cdot \cos \alpha/2$. Thus, $ AE \equal{} AH \cdot \cos \angle HAR / \cos \alpha/2$. From the equality of the area of triangle $ \triangle ADE$ it follows that $ DE \equal{} AE^2 \cdot \sin \alpha / AR \Leftrightarrow DE \equal{} AH \cdot \sin \alpha \cos \angle HAX / \cos^2 \alpha/2$. But $ AT \equal{} DE/\sin \alpha$ and therefore $ AT \equal{} AH \cdot \cos \angle HAX / \cos^2 \alpha/2$. Moreover, $ AN \equal{} 2R \cdot \cos \angle XAO$. Thus, $ AN/AT \equal{} 2R \cdot \cos^2 \alpha/2 /AH$. It's left to prove that $ 1/2 \equal{} R/AH \cdot (2\cos^2 \alpha/2 \minus{} 1)$. We have $ AH/2R \equal{} OM/R \equal{} \cos \alpha$. Therefore it's left to prove that $ \cos \alpha \plus{} 1 \equal{} \cos^2 \alpha/2$, which follows from the addition theorems.

Define: $A'$ is the $A$- antipode, $H_a=AH\cap (ABC)$, Q=$HM\cap (ABC)\neq A'$, $N, L$ midpoints of arcs $BAC$, $BC$.

[b][color=#f00]Lemma[/color][/b]: $H_aN$ and $QL$ intersect on a point $P\in BC$
[b]Pf[/b]: Pascal on hexagon $NH_aA'QLL$. $\blacksquare$

Now see $Q$ is the center of the spiral similarity $\psi$ sending $\triangle BCN$ to $DEA$. From $\angle DAH=\angle BAH_a=BNP$, we get that $\psi: P\to H$.
Using Lemma 1, that means that $QH\cap AL$ is the midpoint of arc $DE$ in $(ADE)$. But $Q-H-M$ are colinear so $AQ\perp HM$. $\square$

Let $AX$, $BY$, $CZ$ be the altitudes in the $\triangle ABC$. Define point $T$ to be the intersection of $\odot(ABC)$ and $\odot(AZHY)$. We proceed in several steps.

[b]Step 1 : [/b] $\frac{ZD}{DB} =\frac{YE}{EC}$.
[b]Proof: [/b] Basically note that since $AD=AE$ we have $\angle ZDH = \angle YEH = 90^{\circ}-0.5 \angle A$. This gives us that $\angle ZHD = \angle YHE = 0.5 \angle A$. On the other hand, $\angle ZHB = \angle YHC = \angle A$; therefore
$$\angle ZHD = \angle DHB = \angle YHE = \angle EHC = 0.5 \angle A$$.
Thus line $DE$ is the angle bisector of $\angle ZHB$ and $\angle YHC$. Since $\triangle ZHB \sim \triangle YHC$ we obtain that:
$$ \frac{ZD}{DB} = \frac{ZH}{HB} = \frac{HY}{HC} = \frac{YE}{EC} $$
as desired.

[b]Step 2:[/b] Point $T$ lies on $\odot(ADE)$
[b]Proof: [/b] Observe that $\angle TBA = \angle TCA$ and $\angle TZA = \angle TYA \implies \angle TZB = \angle TYC$. This implies that $\triangle TZB \sim \triangle TYC$. Also this can be interpreted that point $T$ is centre of spiral similarity taking segment $ZB$ to $YC$. From the previous step we can conclude than then point $T$ also takes segment $ZD$ to $YE$. Thus $\triangle TDZ \sim \triangle TEY$, which implies that:
$$ \angle TDZ = \angle TDA = \angle TEY  = \angle TEA$$
This immediately implies that $T$ lies on $\odot(ADE)$ as desired.

Now we can finish the problem. Since point $T$ is $A$ Queue point, then it is well known that points $T;H,M$ are collinear. Since $\angle ATH = \angle ATM = 90^{\circ}$ problem is solved.

This problem is all about working backwards. You'll find a lot of "it suffices to prove that" in this solution.

Let $A'$ be the antipode of $A$ with respect to the circumcircle of $ABC$. We know that $H$, $M$, and $A'$ are collinear by orthocenter reflections and $\angle APA'=90^\circ$, so it suffices to prove that $\angle APH=90^\circ$. Since $\overline{PA}$ is the external angle bisector of $\angle DPE$, it suffices to prove that $\overline{PH}$ is the internal angle bisector, or $\tfrac{PD}{PE}=\tfrac{DH}{EH}$.

Since $P$ is the Miquel point of $BDEC$, $\tfrac{PD}{PE}=\tfrac{BD}{CE}$, so it suffices to prove that $\tfrac{BD}{CE}=\tfrac{DH}{EH}$. We have $\angle BDH=\angle CEH$ and $\angle ABH=\angle ACH$, so $\triangle BDH \sim \triangle CEH$. This gives $\tfrac{BD}{CE}=\tfrac{DH}{EH}$ and we are done.

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/vol19.html

then

Thaïlande ou Suisse

Sincerely
Jean-Louis

Wow there’s a sqrt(bc) inversion solution for this problem which use just angle chasing after that. Maybe I will post the solution later.

Let $L$, $N$ be feet of the altitudes from $C$ and $B$, respectively. Let $(ABC)$ and $(ALN)$ intersect at $F$. Clearly, $F$ is a spiral center taking $BL$ to $CN$. It remains to show that through the same spiral similarity, $D$ goes to $E$. Indeed, we have \[\frac{LD}{LB}=\frac{LD}{LH}\cdot \frac{LH}{LB}=\frac{EN}{HN}\cdot \frac{HN}{NC} = \frac{EN}{NC}\]
Therefore, the figure $FLDB$ is similar to the figure $FNEC$ and so what we need is proved.

Let $B', C'$ be the feet of the altitudes from $B$ and $C$, and let $P=(AB'HC')\cap (ABC)$; by well known properties of the configuration, the problem asks us to prove that $P\in (ADE)$. By the Forgotten coaxiality lemma, this is equivalent to
$$\frac{\text{Pow}_{(ABC)}(D)}{\text{Pow}_{(AB'HC')}(D)}=\frac{\text{Pow}_{(ABC)}(E)}{\text{Pow}_{(AB'HC')}(E)} \Longleftrightarrow \frac{DB}{DZ}=\frac{EC}{EY},$$
which readily follows from the similarity $\triangle HB'C\cup E \stackrel-\sim \triangle HC'B\cup D$.

Nonononoooo!! How is this possible, I can't even solve ``proper'' G1's these days. I can't believe that this problem..... (I still will keep following my rule of never calling any problem easy ever.)

[img]https://i.imgur.com/25Mpr9s.png[/img]

Let $P$ and $Q$ denote the foot of the $B$ and $C$-altitudes. Also, let $X=MH\cap\odot(ABC)$.

It is well known that $MH\perp AX$ and that $X\in\odot(APQ)$.. Note that is is enough to prove that $X$ also lies on $\odot(ADE)$. This is same as proving $\odot(ADE)$ is coaxial with $\left\{\odot(APQ),\odot(ABC)\right\}$.

Now using Forgotten Coaxiality Lemma, we get that this is equivalent to proving \[\dfrac{\operatorname{Pow}_{\odot(APQ)}(D)}{\operatorname{Pow}_{\odot(ABC)}(D)}=\dfrac{\operatorname{Pow}_{\odot(APQ)}(E)}{\operatorname{Pow}_{\odot(ABC)}(E)}\iff\dfrac{DQ\cdot DA}{DB\cdot DA}=\dfrac{EP\cdot EA}{EC\cdot EA}\iff\dfrac{DQ}{DB}=\dfrac{EP}{EC}.\]

Now note that $\angle QHD=90^\circ-\left(90^\circ-\dfrac{\angle A}{2}\right)=\dfrac{\angle A}{2}$ and similarly we also get that $\angle DHB=\angle EHP=\dfrac{\angle A}{2}$, which thus on using Angle-Bisector Theorem gives that $\dfrac{QH}{HB}=\dfrac{QD}{DB}$. We similarly also get $\dfrac{PE}{EC}=\dfrac{PH}{HC}$.

So our final condition is equivalent to, $\dfrac{DQ}{DB}=\dfrac{EP}{EC}\iff\dfrac{QH}{HB}=\dfrac{PH}{HC}\iff QH\cdot HC=PH\cdot HB$ which is evident by applying $\operatorname{Pow}_{\odot(BQPC)}(H)$ and we are done.

Let the reflection of $H$ about $M$ be $H'$. It is well known by orthocenter reflections that $H'$ lies on $(ABC)$ and $AH'$ is the diameter of $(ABC)$. Let line $HM$ intersect $(ABC)$ at $F \neq H'$. Since $\angle{AFH} = 90$, it follows that $F$ also lies on the circle with diameter $AH$. Let the circle with diameter $AH$ be $\omega_1$ and $(ABC)$ be $\omega_2$. It suffices to prove that $AFDE$ is cyclic and coaxial to $\omega_1$ and $\omega_2$ with common chord $AF$.

By the Coaxility lemma, it suffices to prove $$\frac{\text{Pow}(D,\omega_1)}{\text{Pow}(D,\omega_2)} = \frac{\text{Pow}(E,\omega_1)}{\text{Pow}(E,\omega_2)}$$

Let $\omega_1$ intersect $DE$ at $X$. It follows that since $AX \perp DE$ and $AD = AE$, we know $X$ is the midpoint of $DE$. So, our power equation above simplifies to $$\frac{DH \cdot DX}{AD \cdot BD} = \frac{EX \cdot EH}{AE \cdot CE}$$ which means that it suffices to prove $\frac{DH}{BD} = \frac{EH}{CE}$. This can be done with Law of Sines. Using the fact that $AD = AE$ and that $H$ is the orthocenter, we know that $\angle{DHB} = 180-\angle{DBH}-\angle{BDH} = 180-(90-A)-\left(90+\frac{A}{2}\right) = \frac{A}{2}$ and $\angle{EHC} = 180-\angle{ECH}-\angle{HEC} = 180-(90-A)-\left(90+\frac{A}{2}\right) = \frac{A}{2}$. Since $$\frac{DH}{AD} = \frac{\sin{DBH}}{\sin{DHB}} = \frac{\sin{(90-A)}}{\sin{\frac{A}{2}}} = \frac{\sin{ECH}}{\sin{EHC}} = \frac{EH}{CE}$$ we're done.

Solved with hints

Let the circumcircle of $\triangle ADE$ intersect $(ABC)$ at $Q$. Thus, $Q$ is the Miquel point of $DECB$. Let $B'$ and $C'$ denote the feet of the altitudes from $B$ and $C$. Consider the spiral similarity centered at $X$ that sends $DB$ to $E$.\\

We claim that this also sends $C'$ to $B'$. This is just showing that $$\frac{BD}{C'D}=\frac{CE}{B'E}.$$ Let $\angle BAC=2\alpha$. Then, $$\angle ADE=\angle AED=90-\alpha,$$ so $$\angle C'HD=\angle B'HE=\alpha.$$ Hence, we also have $$\angle BHD=\angle CHE=\alpha.$$ Thus, by Angle Bisector Theorem, $$\frac{BD}{C'D}=\frac{BH}{HC'}=\frac{CH}{HB'}=\frac{CE}{B'E},$$ where the second equality is from PoP on $BCB'C'$. Thus, this spiral similarity also sends $C'$ to $B'$.\\

Hence, $Q$ lies on $(AC'B')$, so $\angle AQH=90$. This means that $Q,H,M$ are collinear (since they all lie on the line from $Q$ to $H'$, where $H'$ is the reflection of $H$ across the midpoint of $BC,$ which we call $M$). Therefore, $(ABC),(AH),(ADE)$ are coaxial as they all have $A$ and $Q$. Hence, their centers are collinear, which means that $P$, the circumcenter of $\triangle AEF$, lies on $ON$, where $N$ is the midpoint of $AH$. Since $NO\parallel HM\perp AQ,$ we are then done.

I think slightly misplaced for a G5. From a good diagram that helps us gain intuition, by letting F be the intersection of ray MH with circle ABC, we know that the intersection of ray HM with circle ABC is the diametrically opposite point of A, whence AFM is 90 degrees, and it also lies on the circle AIHJ, where I and J are the feet of altitude from B and C, respectively. Then, we can rephrase the problem as follows: [quote]Prove that the circle ADE is coaxial with the other two circles.[/quote]

This is equivalent to proving that $$\frac{DB}{DJ}=\frac{DA*DB}{DJ*DA}=\frac{EA*EC}{EI*EA}=\frac{EC}{EI}.$$ First we chase some angles: $$JHD=90-(90-A/2)=\frac{A}{2}=\frac{JHB}{2},$$ so HD is an angle bisector. Now the problem is evident, because $$\frac{DB}{DJ}=\frac{HB}{HJ}=\frac{HC}{HI}=\frac{EC}{EI}$$ from similarity of triangles BHJ and CHI by AA (JBH=ICH by cyclic quad BCIJ), as desired. $\blacksquare$

In fact, I claim that $X$ is the $A$-queue point, which finishes by master Miquel. To do this, let the reflection of $H$ over $BC$ be $H'$ -  by [url=https://aops.com/community/p2477427]Brazil 2011/5[/url] it suffices to show that $XBH'C$ is harmonic. However, noting that $\triangle HBD \sim \triangle HCE$, we have
\[\frac{XB}{XC} = \frac{BD}{CE} = \frac{HB}{HC} = \frac{H'B}{H'C}\]
and we are done.

(a lot of the difficulty of this problem comes from the characterization of $X$ - once you successfully figure this out, the problem isnt very hard. although tbh, that might be a bit surprising to be true, which is why I took 30 min to think of that)

Let $\overline{BB_1}$, $\overline{CC_1}$ be altitudes of $\triangle ABC$ and $Q = (ABC) \cap (AB_1HC_1)$ be the $A$-Queue point of $\triangle ABC$. It's well-known that $Q$, $H$, and $M$ are collinear, and $\overline{AQ} \perp \overline{QHM}$, so it suffices to show that $AQDE$ is cyclic. By the (forgotten) Coaxiality lemma, it's equivalent to show that
$$\frac{\text{Pow}_{(ABC)}(D)}{\text{Pow}_{(AB_1HC_1)}(D)} = \frac{\text{Pow}_{(ABC)}(E)}{\text{Pow}_{(AB_1HC_1)}(E)} \iff \frac{EB_1 \cdot EA}{EA \cdot EC} = \frac{DC_1 \cdot DA}{DA \cdot DB} \iff \frac{EB_1}{DC_1} = \frac{EC}{DB}.$$
However, because $\measuredangle HDB = \measuredangle HDC_1 = - \measuredangle HEB_1 = -\measuredangle HEC,$ it's easy to see that $\triangle HDC_1 \overset{-}{\sim} \triangle HEB_1$ and $\triangle HDB \overset{-}{\sim} \triangle HEC,$ so $\tfrac{EB_1}{DC_1} = \tfrac{HE}{HD} = \tfrac{EC}{DB}$ and we are done.

Let the feet of the altitudes from $B$ and $C$ be $X$ and $Y$, respectively.

It's easy to show that $\triangle HYB \sim \triangle HXC$ with corresponding angle bisectors $HD$ and $HE$, which tells us $YD: DB = XE: EC$.

Thus there exists a spiral similarity sending $YDB$ to corresponding points $XEC$, which tells us circles $(AXY)$, $(ADE)$, and $(ABC)$ are coaxial, implying they share a common chord.

Finally we note the common chord of $(AXY)$ and $(ABC)$ is well known to be $AQ$, where $Q$ is the $A$-Queue point, which is perpendicular to $HM$. $\blacksquare$

[asy]
size(200);
pair A, B, C, H, D, E, X, Y;
A = dir(110);
B = dir(200);
C = dir(340);
H = A+B+C;
X = foot(B, A, C);
Y = foot(C, A, B);
D = extension(A, B, H, incenter(H,B,Y));
E = extension(A, C, H, incenter(H,C,X));

draw(A--B--C--cycle^^D--E^^B--X^^C--Y);
draw(circumcircle(A, D, E)^^circumcircle(A, B, C));
draw(circumcircle(A, X, Y), dashed);

dot("$A$", A, dir(90));
dot("$B$", B, dir(225));
dot("$C$", C, dir(315));
dot("$D$", D, dir(200));
dot("$E$", E, dir(30));
dot("$H$", H, dir(270));
dot("$X$", X, dir(0));
dot("$Y$", Y, dir(210));
[/asy]

[hide] Define  $(ADE) \cap (ABC)$ = $T$  , $BH \cap AC$ = $X$  and  $CH \cap AB$ = $Y$ , now  observe that  it suffices to  show   that $(AXY) $ passes through $T$   , notice that  spiral sim at  $T$ sends  $BD$ to $CE$  therefore  It is enough to show that   $\frac {BD}{DY} $ = $\frac {CE}{EX}$   which is trivial by  $\triangle HYD $ is similiar to $\triangle HXE $  and   $\triangle HYB$ is similiar to $\triangle HXC$  , hence spiral at  $T$ sends $DY$ to $ EX$  $\square $[/hide]


Let there be A={1,2,3,4,.....,2004} and B a subset of A.Knowing that B contains 1003  elements (numbers), prove that it contains at least a pair of number in which one divides the other one.... :?:

I think this topic, should be moved, because of it's proof.  Though it may be easy.

Hm, yeah.  Though this may be supplementary to mathcounts, I'll move it to Getting Started because Mathcounts doesn't expect anybody to do proofs.

Yes, that is true.

But there is proofs in National MathCounts Masters Round.

Just because it involves proofs doesn't mean it's too hard for MathCounts (master round!!!).

[hide] Suppose for the sake of contradiction that such a subset $A$ can be chosen. Write all the elements in this subset as $2^{n}y$, where y is odd. If two of the y are the same, then one of the numbers divides the other, so we need all y different. However, there are only 1002 odd numbers from 1 to 2004, so by the Pidgeonhole principle there is at least one repeat y.[/hide]

There is also a proof by induction. Try to find it.

EDIT: aw, the MathCounts police once again deprive mathletes from trying to do higher-level math...namely proofs.


Two cards are dealt at random from a standard deck of 52 cards.  What is the probability that the first card is a King and the second card is a $ \heartsuit$?

[hide="Solution"] We have two cases, where the first card is a king of hearts, and where the first card is a king that is not hearts.

[hide="CASE 1"]

We multiply the probabilities:

$ \frac 3 {52} \cdot \frac {13}{51}\equal{}\frac{39}{51 \cdot 52}$
[/hide]

[hide="CASE 2"]
Again, multiplying:

$ \frac{1}{52} \cdot \frac {12}{51}\equal{}\frac{12}{52 \cdot 51}$
[/hide]

[hide="Final Answer"]
$ \frac{39}{51 \cdot 52} \plus{} \frac{12}{51 \cdot 52}\equal{}\boxed {\frac {1}{52}}$
[/hide]
[/hide]

[quote=Eugenis]The answer of $\frac{1}{52}$ is pretty cool. The slicker way to think of this problem is the probability of pulling the King of Hearts, which is $\frac{1}{52}.$

Is this logic sound?[/quote]

Doesn't seem sound to me. Seems like the answer value is just a coincidence with the king of hearts probability.

I do not believe this is a coincidence.

[hide] Calculate the probability of getting a Queen and then a spade:

- If the first one is a spade then 1 choice. Then 12 choices for the second one. $12 \times 1=12.$
- If first one is not a spade then 3 choices. Then 13 choices for the second one. $3 \times 13=39.$

Summing we have $12+39=51.$ $51 \times 52$ total, our answer is $\frac{1}{52}.$ This is not surprising however. [/hide]

[hide] Calculate the probability of getting a spade and then a Queen.
- If the first one is a queen then 1 choice. Then 3 choices for the second one. $3.$
- If the first one is not a queen then 12 choices. Then 4 choices. $12 \times 4=48.$

Summing we have $3+48=51.$ $51 \times 52$ total, our answer is $\frac{1}{52}$ as shown before. [/hide]

[hide=Casework, I guess?]
If the King is a Club, there are 13 cases with Hearts
If the King is a Spade, there are 13 cases with Hearts
If the King is a Diamond, there are 13 cases with Hearts
If the King is a Heart, there are 12 cases with Hearts, because the King of Hearts cannot be chosen, witht there being no replacement.
Adding them all, we get a total of 51 cases. Then there are 2 cards to choose giving us $51/52*51$
When simplified, we get $1/52$[/hide]

all of the previous solutions are without replacement

[quote=ARay10]I just did $\frac1{13} \times \frac14$, and got it correct but with the wrong solution.[/quote]

WAIT LOL SAME :rotfl:


Prove that the midpoints of the altitudes of a triangle are collinear if and only if the triangle is right.

[i]Dorin Popovici[/i]

[hide="Hint"]Consider acute triangles, obtuse triangles and right triangles separately.[/hide]

The middles $ H_1,H_2,H_3$ of the altitudes of the triangle are on the sides of the median triangle.It isn't possible that 2midpoints to be the same.The only posibility to be on the same line is that two of them  to form a segment composed by the midpoints of two sides and the third one to be in the interior of this segment.This can hapen only if the triangle has a $ 90$ degree angle.

Here's the complex numbers solution from the OTIS walkthrough.

Let $(ABC)$ be the unit circle, as always, and note that the midpoints of the altitudes are $x = \frac 12\left(a+\frac 12\left(a+b+c-\frac{bc}a\right)\right)$ and cyclic permutations. So it suffices to show that $$0 = \begin{vmatrix} 3a+b+c-\frac{bc}a & \frac 3a+\frac 1b+\frac 1c - \frac a{bc} & 1 \\ 3b+a+c-\frac{ac}b & \frac 3b+\frac 1a+\frac 1c-\frac b{ac} & 1 \\ 3c+a+b-\frac{ab}c & \frac 3c+\frac 1a+\frac 1b-\frac c{ab} & 1 \end{vmatrix} = \begin{vmatrix} 2a-\frac{bc}a & \frac 2a - \frac a{bc} & 1 \\ 2b-\frac{ac}b & \frac 2b - \frac b{ac} & 1 \\ 2c-\frac{ab}c & \frac 2c - \frac c{ab} & 1 \end{vmatrix}$$ if and only if $0 \in \{a+b, b+c, c+a\}$. It turns out that this factors as $$\sum_{\mathrm{cyc}} \left(2a-\frac{bc}a\right)\left(\frac 2b-\frac b{ac}\right) - \left(\frac 2a-\frac a{bc}\right)\left(2b-\frac{ac}b\right) = -\frac{(a-b)(b-c)(c-a)(a+b)(b+c)(c+a)}{a^2b^2c^2}$$ as needed.


A prism has 15 edges. How many faces does the prism have?

Lets say the number of sides the base has is n. Then if you count the number of edges a prism has, it is 3n because there are 2 faces (2n) and the 'thickness' edge, which counts as n edges. So the base is a pentagon. Thus, there are 5+2=7 faces (thickness faces+bases)


inequalities
search
inequalities proposed
109

Prove that for a,b,c real

(a^2 + 2)(b^2 + 2)(c^2 + 2) > 9(ab + bc + ca)

The thing is,this inequallity isn't homogenuos at all  :lol:

I solved it a long time ago by expanding and then using several inequalities:
At first, we always can find to numbers $ (x,y)$ of the set $ \{a,b,c\}$ such that $ (x\minus{}1)(y\minus{}1)\ge0$. Wlog assume that these numbers are $ a,c$. Then
$ a^2b^2c^2\plus{}2\sum_{cyc} a^2b^2\plus{}4\sum_{cyc}a^2\plus{}8 \ge 9 \sum_{cyc} ab \\
\Leftrightarrow 2(\sum_{cyc} (ab\minus{}1)^2)\plus{}3(\sum_{cyc}a^2\minus{}\sum_{cyc}ab)\plus{}2b(a\minus{}1)(c\minus{}1)\plus{}(abc\minus{}1)^2\plus{}(b\minus{}1)^2\plus{}(a\minus{}c)^2 \ge 0$
By the way it's from APMO 2004

This inequality has a nice generalization : [i]Let $ a,b,c$ and $ k$ be nonnegative real numbers. Then the following inequality holds:[/i]

$ (a^2\plus{}k\plus{}1)(b^2\plus{}k\plus{}1)(c^2\plus{}k\plus{}1) \geq (k\plus{}2)^2(ab\plus{}bc\plus{}ca\plus{}k\minus{}1).$

You can see the proof here:

$ \bullet$ [url=http://www.mathlinks.ro/viewtopic.php?search_id=51447073&t=19666/]Generalization of APMO 2004[/url]

[quote="Maxim Bogdan"]This inequality has a nice generalization : [/quote][i]Let $ a,b,c$ real numbers. Then the following inequality[/i]

$ \left(3a^2 \plus{} k \plus{} 1\right)\left(3b^2 \plus{} k \plus{} 1\right)\left(3c^2 \plus{} k \plus{} 1\right) \geq (k \plus{} 4)^2\left[3(bc \plus{} ca \plus{} ab) \plus{} k \minus{} 5\right],$

[i]holds if and only if  $ k\geq 0;$ with equality if and only if $ a \equal{} b \equal{} c \equal{} \pm 1.$[/i]

[b]Proof[/b] $ \prod{\left(k \plus{} 1 \plus{} 3a^2 \right)} \minus{} (k \plus{} 4)^2\left(k \minus{} 5 \plus{} 3\sum{bc}\right)$

$ \equal{} 3\left(a \plus{} b \plus{} c \minus{} 3abc\right)^2 \plus{} 9\left(bc \plus{} ca \plus{} ab \minus{} 3\right)^2$

$ \plus{} 9k\sum{(bc \minus{} 1)^2}\plus{} \frac {3k(k \plus{} 2)}{2}\sum{(b \minus{} c)^2}\geq0.$


search
number theory unsolved
number theory

Prove that sum of digits of 4^n+25^n (in decimal)

is odd for all positive integer n.

I saw it long time ago and couldn't solve it.

I also tried to search about it but failed :blush: .

Please help me.

The sum of digits of $ 4^8 \plus{} 25^8 \equal{} 152587956161$ (in decimal) is $ 56.$


Let $a_1,a_2,\ldots, a_n > 0$ satisfying $\displaystyle \sum_{i=1}^n \frac{1}{1+a_i}=1$. Prove that
\[ \sum_{i=1}^n \frac{a_i}{(n-1)+a_i^2} \leq 1.  \]

No, It's not difficult.
$\frac{1}{1+a_i}-\frac{a_i}{n-1+a_i^2}=\frac{n-1-a_i}{(1+a_i)(n-1+a_i^2)}$
$\ge (\sum \frac{n-1-a_i}{1+a_i})\sum \frac{1}{n-a+a_i^2}=0$
(by Tchebysef).

I remember that onetime I had post this problem, general of China West MO with $n=5$.

[quote="hungkhtn"]
$\frac{1}{1+a_i}-\frac{a_i}{n-1+a_i^2}=\frac{n-1-a_i}{(1+a_i)(n-1+a_i^2)}$
$\ge (\sum \frac{n-1-a_i}{1+a_i})\sum \frac{1}{n-a+a_i^2}=0$
(by Tchebysef).[/quote]

Could you explain it ?Why this is equal to $0$?

[quote="silouan"][quote="hungkhtn"]
$\sum \frac{n-1-a_i}{1+a_i}=0$
(by Tchebysef).[/quote]

Could you explain it ?Why this is equal to $0$?[/quote]


My school doesn't offer the B exam date (we're on vacation that week), so I was hoping to find a list of some nearby schools that offered it (I'm in New York). I thought I'd email someone at the AMC organization, only I don't know who to contact  :blush: I looked at the official website, but I couldn't find anything, and the universities that offered it were all too far away (I'm in Westchester, the closest one was about an hour and a half away  :( ). Does someone know who I should contact?

Go call some nearby schools.
They probably won't turn you down (especially if they're public).

Is there anywhere there's a list for this sort of thing? I can't even find out which colleges in my area offer it.

Go to the AMC web site at [url]http://www.unl.edu/amc[/url], and click on the link that says "2007 AMC 10B/12B Higher Education Participating Sites".

Hmm... it turns out of the two of the nearby schools I checked have winter break the week the AMC B tests are offered  :(

Any AoPSer go to a school in the Westchester area that is offering the B exam and has an open spot for it?

[quote="me@home"]I thought there was a prolific AopS poster from West chester, like K8107 or something?[/quote]

Oh, thanks for point that out. Maybe his school offers the B exam  :)

[quote="Morgoth"]but I couldn't find anything, and the universities that offered it were all too far away (I'm in Westchester, the closest one was about an hour and a half away  :( ). Does someone know who I should contact?[/quote]
The school I went to write Amc A was about an hour and 40 min away. I had to wake up at 5 to get there. The ride may be boring but, you should keep it as a last resort in case all other leads die. :)

[quote="drunner2007"]None of those universities were close to me.

When is the deadline to apply for the AMC 12B?[/quote]

Actually, you can talk to schools nearby (cough Lassiter) and maybe prominent students who go to that school (cough Miles) and ask them about it.  :)

Aww darn, I emailed the professor at the college that was about an hour and a half away, but it turns out that it was already full. I'm going try to see if I can take it at a college in Connecticut (which is actually closer), but if anyone from Westchester or nearby knows of a school that has an open slot, could you PM me? Thanks.


analytic geometry
inequalities unsolved
inequalities

[quote="Zamfirmihai"]You can apply $n$ times Minkowsky[/quote]
exactly .c'est l'inegalitéé triangulaire .ok Fourrier


summer program
Mathcamp
AwesomeMath
MathPath
AMC
AIME
AoPS Books

ok so i scored around 80 on the AMC 10A this year. i would like to make that at least a 140 next year. which camp should i attend to reach that goal?

wow you have lofty goals
i said mathcamp because i don't really know much about awesomemath
i just learned that i qualified for mathcamp YAY
anyways, from my previous knowledge mathcamp is for "recreational" math and awesomemath is for "competition" math though i might be wrong (somebody has gone somewhere please say)

EDIT:  i said mathcamp and i was thinking mathpath, so i guess you can disregard this  :D

Awesomemath.

You probably won't learn anything from the actual program but there are so many smart people there that something's bound to rub off on you.

80-140? whoa
no offense but try smaller goals:
like get 120, unless you made stupid mistakes

I suggest just do some math problems from AoPS (vol 1 and 2)
that should improve your score by a lot

[quote="iamagenius"]i guess you guys are right. getting a 140 sounds much easier than it really is. my goal is just to make AIME next year.

i actually don't think aops 1 really helps me at all.[/quote]
Did you read the chapters really carefully? It helps a lot.

[quote="iamagenius"]well i mean most of the concepts are pretty easy. i can do most of the problems at the end. i feel as if i am not really learning anything. how should i make the book more useful?[/quote]
Uh no offense, but if you actually understood the concepts in the aops volume 1 thoroughly, then you wouldn't have had to post the problems from the camp qualification tests, here on this forum.
Anyways, if you know the concepts of volume 1, and still can't qualify for AIME, then I think you just need to do more problems. Do all the problems at the end of each chapter in aops 1 If  you don't know how to do a problem, don't just skip to the solutions. Skip it, and come back to it later. Struggle with it. Then I promise you'll be better.
(If you already did this, then you might want to get some of the other aops books like volume 2 or the subject books)


[color=blue] [size=150]$12{nd}$  Iberoamerican Olympiad[/size]
Guadalajara, MEXICO. [1997]

Edited by djimenez

Carlos Bravo ;)

[/color]

[quote][b]Problema 1.[/b] En un triángulo $ \triangle ABC,$ una circunferencia centrada en su incentro $ I$ corta a $ BC$ en $ D,P$ (estando D mas cerca de B)  corta a $ CA$ en $ E,Q$ (estando E mas cerca C) y corta a $ AB$ en $ F,R$ ( estando F mas cerca de A). $ S,T$ y $ U$ son las intersecciones de las diagonales de los cuadrilátero convexos $ EQFR,FRDP$ y $ DPEQ.$ Probar que las circunferencias circunscritas a $ \triangle FRT ,$ $ \triangle DPU$ y $ \triangle EQS$ concurren.[/quote]
Como $ I$ equidista de $ BC,CA,AB,$ es claro que las cuerdas interseptadas $ DP,EQ$ y $ FR$ son congruentes. Así $ EQFR ,$ $  FRDP$ y $ DPEQ$ son trapecios isosceles $ \Longrightarrow$ $ QR \equal{} EF ,$ $ FD \equal{} RP,$ $ DE \equal{} PQ.$ Ello implica que $ \triangle DEF$ y  $ \triangle PQR$ son congruentes teniendo circuncentro común $ I$ y ademas que $ SFRT,$ $TDPU$ y $ UEQS$ son cíclicos, debido a que $ \angle EFD \equal{} \angle QRP,$ $\angle FDE \equal{} \angle RPQ$ y $ \angle DEF \equal{} \angle PQR.$

$ I$ es evidentemente centro de la rotación que lleva $ \triangle DEF$ en $ \triangle PQR$ y como los puntos medios $ M$ y $ N$ de los lados $ QR, EF$ son homólogos en tal rotación, el ángulo $ \angle MIN$ vale precisamente en ángulo de rotación. Así $ \angle MIN \equal{} \angle RIF,$ pero como $ IMSN$ es cíclico, se tiene $ \angle FSR \equal{} \angle FTR \equal{} \angle FIR$ $ \Longrightarrow$ $ I \in \odot(FRT).$ Entonces por analogía se concluye que $ I \equiv \odot(FRT) \cap \odot(DPU) \cap \odot(EQS).$

[b]Problema 4.[/b] (Generalización). Sean $ P$ y $Q$ dos puntos isogonales en $ \triangle ABC.$ Los rayos $ AP,AQ$ cortan su circunferencia circunscrita en $ P',Q'.$ Se definen $ M \equiv BC \cap QP' ,$ $ N \equiv PQ' \cap BC$ y $ L \equiv PM \cap QN.$ Mostrar que $ APLQ$ es un paralelogramo.

Sea $ \triangle A'B'C'$ el triángulo pedal de $ P$ con respecto a $ \triangle ABC.$ Como $ P$ y $ Q$ son isogonales se tiene $ QA \perp B'C'$ y $ QB \perp C'A'.$ Así

$ \angle AQB \equal{} 180^{\circ} \minus{} \angle A'C'B' \equal{} 180^{\circ} \minus{} (\angle PBA' \plus{} \angle PAB') \equal{} 180^{\circ} \minus{} \angle PBP'$

$\Longrightarrow  \frac {\overline{P'M}}{\overline{MQ}} \equal{} \frac {\overline{BP'}}{\overline{BQ}} \cdot \frac {\sin \widehat{P'BC}}{\sin \widehat{CBQ}} \ , \ \frac {\overline{P'P}}{\overline{PA}} \equal{} \frac {\overline{BP'}}{\overline{BA}} \cdot  \frac {\sin \widehat{PBP'}}{\sin \widehat{ABP}}$

Por ley del seno en $ \triangle ABQ$ resulta $   \frac {\overline{BA}}{\overline{BQ}} \equal{} \frac {\sin \widehat{AQB}}{\sin \widehat{BAQ}}$

Combinando con las dos anteriores teniendo en cuenta que $ \angle AQB \equal{} 180^{\circ} \minus{} \angle PBP' ,$ $ \angle BAQ \equal{} \angle P'BC$ y  $ \angle CBQ \equal{} \angle ABP,$ se obtiene $ \frac {_{\overline{P'M}}}{^{\overline{MQ}}} \equal{} \frac {_{\overline{P'P}}}{^{\overline{PA}}}$ $  \Longrightarrow$ $ PM \parallel AQ$ y analogamente se prueba que $ QN \parallel AP.$


calculus
integration
trigonometry
analytic geometry
logarithms
LaTeX
absolute value

Find $ \int{\int_{S}{(x^2 \plus{} y^2})}dS$, where $ S \equal{} \left\{(x,y): x^4 \plus{} y^4\leq 1\right\}$.

[hide]I was trying $ x^2 \equal{} r\cos{\alpha}, y^2 \equal{} r\sin{\alpha}$, obtaining $ \int^{\frac {\pi}{2}}_{0}d\alpha{\int^{1}_{0}{\frac {r(\cos{\alpha} \plus{} \sin{\alpha})}{4\sqrt {\cos{\alpha}\sin{\alpha}}}}}dr \equal{} \int_0^{\frac {\pi}{2}}\frac {1}{8}(\sqrt {\tan{\alpha}} \plus{} \sqrt {\cot{\alpha}})d\alpha,$ which I do not know, how to integrate.[/hide]

I got a different Jacobian for this integral. Remember that it's not simply $ r$ since we're not truly using polar coordinates here, but a modified version of them.

Let's see what's different.... $ x^2 = r\cos{\theta}, y^2 = r\sin{\theta}$ implies $ |x| = \sqrt {r\cos{\theta}, }$ and $ |y| = \sqrt {r\sin{\theta}}$

For simplicity, we can knock off those absolute value bars and integrate from $ 0$ to $ \frac {\pi}{2}$  and multiply by $ 4$ since it is symmetric.

Thus, the Jacobian is
\[ %Error. "displaymode" is a bad command.
\left [\begin{array}{cc} \frac {\partial{x}}{\partial{r}} & \frac {\partial{x}}{\partial{\theta}} \\
\frac {\partial{y}}{\partial{r}} & \frac {\partial{y}}{\partial{\theta}}\end{array} \right]\]
. Using the equations relating $ x$ with $ r$ and $ \theta$, as well as $ y$ and $ r$ and $ \theta$:
\[ \left [\begin{array}{cc} \frac {2\sqrt {\cos{\theta}} \cdot}{\sqrt {r}} & - \frac {\sqrt {r} \cdot \cos{\theta} \cdot \sin{\theta}} {2} \\
\frac {2\sqrt {\sin{\theta}} \cdot}{\sqrt {r} } & \frac {\sqrt {r} \cdot \cos{\theta} \cdot \sin{\theta}} {2} \end{array} \right]\]
With simplification, assuming $ r \neq 0$ (is this valid? Can we take the integral as a limit where r approaches 0? I think so.), we get

$ \cos{\theta}\sin{\theta} \cdot (\sqrt {\cos{\theta}} + \sqrt {\sin{\theta}})$ for the Jacobian. Then, ${ \int\int f(x,y) \, dx dy = \int\int g(r,\theta) \cos{\theta}\sin{\theta} \cdot (\sqrt {\cos{\theta}} + \sqrt {\sin{\theta}})drd\theta}$. Here, $ g(r,\theta)=x^2+y^2=r(\sin{\theta}+\cos{\theta})$

Also, $ \int \sqrt {\tan{x}} \, dx$ and $ \int \sqrt {\cot{x}} \, dx$ are both elementary functions.
\[ \int \sqrt {\tan{x}} \, dx = \frac { - 2 \arctan[1 - \sqrt {2} \cdot \sqrt {\tan{x}}]}{2\sqrt {2}} + \\
\frac { 2 \arctan[1 + \sqrt {2} \cdot \sqrt {\tan{x}}] {2\sqrt {2}} + \\
\log[ - 1 + \sqrt {2} \cdot \sqrt {\tan{x}} - \tan{x}] - \\
\log[1 + \sqrt {2} \cdot \sqrt {\tan{x}} + \Tan{x}] } {2\sqrt {2}}\]
I had to seperate the numberator into two lines because it just wouldn't fit right (I'm a $ \text{\LaTeX}$ n00b)

$ \int \sqrt {\cot{x}} \, dx$ is a somewhat similar integral that is just as messy.

I'm not saying that it's easy to integrate it now, but perhaps it is [i]possible[/i]!

Thank you, rob22.

But I think, we cannot avoid $ \int\sqrt{\tan{x}}\, dx$ and $ \int\sqrt{\cot{x}}\, dx$.

[quote="rob22"]Let's see what's different.... $ x^2 = r\cos{\theta}, y^2 = r\sin{\theta}$ implies $ |x| = \sqrt {r\cos{\theta}, } and |y| = \sqrt {r\sin{\theta}}$

For simplicity, we can knock off those absolute value bars and integrate from $ 0$ to $ \frac {\pi}{2}$ and multiply by $ 4$ since it is symmetric. [/quote]

Since $ x^2\geq 0$ and $ y^2\geq 0$, we get $ \theta \in [0,\frac{\pi}{2}]$ and not $ [0,2\pi]$.

[quote="rob22"]Using the equations relating $ x$ with $ r$ and $ \theta$, as well as $ y$ and $ r$ and $ \theta$:

\[ \left [\begin{array}{cc}\frac{2\sqrt{\cos{\theta}}\cdot}{\sqrt{r}}&-\frac{\sqrt{r}\cdot\cos{\theta}\cdot\sin{\theta}}{2}\\ \frac{2\sqrt{\sin{\theta}}\cdot}{\sqrt{r}}&\frac{\sqrt{r}\cdot\cos{\theta}\cdot\sin{\theta}}{2}\end{array}\right]\] [/quote]

I think the Jacobian is $ \[ \left |\begin{array}{cc}\frac{\sqrt{\cos{\theta}}\cdot}{2\sqrt{r}}&-\frac{\sqrt{r}\cdot\sin{\theta}}{2\sqrt{\cos{\theta}}}\\ \frac{\sqrt{\sin{\theta}}\cdot}{2\sqrt{r}}&\frac{\sqrt{r}\cdot\cos{\theta}}{2\sqrt{\sin{\theta}}}\end{array}\right|=\frac{1}{4\sqrt{\sin{\theta}\cos{\theta}}}$

and we come to $ \int^{\frac{\pi}{2}}_{0}d\alpha{\int^{1}_{0}{\frac{r(\cos{\alpha}+\sin{\alpha})}{4\sqrt{\cos{\alpha}\sin{\alpha}}}}}dr$.

[quote="tmrfea"]I was trying $ x^2 = r\cos{\alpha}$, $ y^2 = r\sin{\alpha}$, obtaining $ \int^{\frac {\pi}{2}}_{0}d\alpha{\int^{1}_{0}{\frac {r(\cos{\alpha} + \sin{\alpha})}{4\sqrt {\cos{\alpha}\sin{\alpha}}}}}dr = \int_{0}^{\frac {\pi}{2}}\frac {1}{8}(\sqrt {\tan{\alpha}} + \sqrt {\cot{\alpha}})d\alpha,$ which I do not know, how to integrate.[/quote]

Now I know how to finish the task.

[b]Lemma1: [/b] For $ a > 0, b > 0$ we have $ \int_0^{\frac {\pi}{2}}\sin^{a - 1}{x}\cos^{b - 1}{x}dx = \frac {1}{2}B(\frac {a}{2},\frac {b}{2})$, where $ B(x,y)$ stands for Euler`s Beta function.

[b][i]Proof:[/i][/b]

Directly from the definition of $ B(x,y)$ using substitution $ t = sin^2{x}$.

[b]Lemma2: [/b] For $ 0 < p < 1$ we have $ \int_0^{\frac {\pi}{2}}\tan^p{x}dx = \int_0^{\frac {\pi}{2}}\cot^p{x}dx = \frac {\pi}{\cos{\frac {p\pi}{2}}}$.

[b][i]Proof:[/i][/b]

We will use Lemma1 and the two well-known identities: $ B(x,y) = \frac {\Gamma(x)\Gamma(y)}{\Gamma(x + y)}$ and $ \Gamma(z)\Gamma(1 - z) = \frac {\pi}{\sin{\pi x}}$..

\begin{eqnarray*} \int_0^{\frac {\pi}{2}}\tan^p{x}dx & = & \int_0^{\frac {\pi}{2}}\sin^{p}{x}\cos^{ - p}{x}dxdx = \frac {1}{2}B(\frac {1 + p}{2},\frac {1 - p}{2}) = \\
& = & \frac {1}{2}\frac {\Gamma(\frac {1 + p}{2})\Gamma(\frac {1 - p}{2})}{\Gamma(1)} = \frac {\pi}{\sin{\pi \frac {(1 + p)}{2}}} = \\
& = & \frac {\pi}{\cos{\pi \frac {p}{2}}} \end{eqnarray*}

Similarly also $ \int_0^{\frac {\pi}{2}}\cot^p{x}dx$.

[b]For the problem:[/b]

$ \int_{0}^{\frac {\pi}{2}}\frac {1}{8}(\sqrt {\tan{\alpha}} + \sqrt {\cot{\alpha}})d\alpha = 2\frac {1}{8}\int_{0}^{\frac {\pi}{2}}(\sqrt {\tan{\alpha}}d\alpha = \frac {1}{4}\frac {\pi}{\cos{\frac {\pi}{4}}} = \frac {\pi\sqrt {2}}{4}.$


Let ABCD is a quadrangle, AC and BD meet at O. Let I, J be the midpoints of AB and CD. Let H and K be the orthocentrics of OAD triangle and OBC triangle. Prove that the line IJ is perpendicular to the line HK.

[color=red]Thank you for the answer soon![/color]

[quote]Geometry Open Questions
An open question is a question that has no known solution up to this moment, and it is not known wheter the problem has one or not.[/quote]

See
http://www.mathlinks.ro/Forum/viewtopic.php?t=109505
http://www.mathlinks.ro/Forum/viewtopic.php?t=109289
for the same problem in a different form.

  darij


Sorry, this may come off as unclear as I'm unsure how to draw lines in Asymptote. Alternately, I drew the lines vertically and horizontally to form a square in the center. This square would have lengths of 8 so the area would be 64 for that particular square. However, we should also include four rectangles of dimensions 4 by 2. This adds 32 extra units of area for a total of 96 square units.

Edit:
[asy]draw((0,2)--(0,4)--(1,4)--(1,5)--(2,5)--(2,6)--(4,6)--(4,5)--(5,5)--(5,4)--(6,4)--(6,2)--(5,2)--(5,1)--(4,1)--(4,0)--(2,0)--(2,1)--(1,1)--(1,2)--cycle);
label("4",(0,3),W);
label("4",(3,0),N);
label("4",(6,3),E);
label("4",(3,6),N);
draw((2,5)--(4,5),blue);
draw((1,4)--(1,2),blue);
draw((5,4)--(5,2),blue);
draw((2,1)--(4,1),blue);[/asy]

I learnt how to do it from your diagram! Thank you!


Solve the following system

5^x + 9^y =86
2*(5^x) + 9^(1/y) =13

well, let f(x)= 2.9^x -9^1/x
then f'(x)>0 and so the equation has at most one solution.
but (1,2) works!


If one scores in the 97th percentile on the psat, will they definitely be named a commended scholar?

Probably.  I believe it says on the score report how many people are commended (something like 50,000?) and how many people take the test so you can do the math and try to figure out approximately what percentile gets commended.  But I believe 97th percentile should do it.

98th percentile is more sure for commended scholar status.

It's typically a scaled score of 200.

200 seems a bit low.  I scored above 200 and I was only 97th percentile.  Also, someone who scored 2 points lower than me was not commended.  Granted this was 3 years ago, so things may have changed.

last year, i recall that a friend of mine scored a 205 and made commended... if that helps

The point is, it changes from year to year.  It depends on how other people do since it's based on percentile.

is 225 definately good enough for Semifinalist?

225 will probably get you semi-finalist status, POSSIBLY, finalist, but don't count on it.

Nothing is definite.  Likely?  Yes.  But if, say 5 other people in your school got 237's, it would hurt your chances.

The score you need to get semifinalist depends on your state.  And a PSAT score of 225+ is almost definitely semifinalist, but you won't officially find out until September, but you guys should be pretty safe.  Also, finalist status has nothing to do with PSAT scores except for what you need to get semifinalist (you actually need to submit SAT scores to become a finalist).

15000/16000 of the semifinalists advance to finalist status

but getting the money is a different thing

this is the breakdown (i think)

only about 1/4 of the finalists get the 2500 from the NM foundation
another 1/4 gets money from parent's companies (these actually give a lot) or universities
and half get nothing besides the name :P

[quote="mathemonster"]well im a soph so I get abs. nothing!![/quote]

Actually me too, I was just wondering where I stand. Thanks for the info, haha

[quote="h_s_potter2002"]Also, finalist status has nothing to do with PSAT scores except for what you need to get semifinalist (you actually need to submit SAT scores to become a finalist).[/quote]

Do you know about how PSAT affects Winner status? I kinda messed up on my PSAT, but I did well enough to be a Semifinalist. (I'm pretty sure at least.)

Worthless?

It's worth [i]at least[/i] 2500 USD, which is the monetary value of the scholarship. That's what someone take home after 2.5 months of working full time at Massachusetts minimum wage (8 USD/hr).

Cutoffs for being commended range from under 210 in some states to slightly over 220 (it was 223 for MA in my year, I believe).

[b]How much does the PSAT count towards winner status, given semifinalist status?[/b]

And to answer a previous question, the highest [i]single[/i] SAT score ([i]not[/i] the highest from each section, which is what many colleges are doing) contributes more towards becoming a winner (given finalist status) than the PSAT score.

Also, those in contention for the actual scholarship have to write a short essay. I am convinced that this essay is used more to eliminate some with undesirable goals/personalities than to choose winners (except, of course, if the person is in contention for a corporate scholarship). However, I advise that you spend a lot of time on this essay, because it will likely be a helpful exercise for writing your actual college essays.

I think he might mean in terms of college applications. In which case, he's right, in a way.

[quote="le00327146"]Do you know about how PSAT affects Winner status? I kinda messed up on my PSAT, but I did well enough to be a Semifinalist. (I'm pretty sure at least.)[/quote]

PSAT scores I don't think have anything to do with moving on, except that they look at SAT scores too to make sure that your PSAT score wasn't a fluke.  But if you made semifinalist, you can't have done that bad on the PSAT.

What I heard was >220 was pretty much guarantee...I got 218 so blah. (but I'm soph so it's okay)

Yea, the score you need depends on what state you live in.  Usually, a 220 is good enough, but the scores needed in states like Maryland or Massachusetts are usually higher.

[quote="h_s_potter2002"]PSAT scores I don't think have anything to do with moving on, except that they look at SAT scores too to make sure that your PSAT score wasn't a fluke.  But if you made semifinalist, you can't have done that bad on the PSAT.[/quote]

I barely made semifinalist, based on the past couple years. Would getting a significantly better SAT score effectively replace my PSAT score?

Once you make semifinalist, your SAT scores begin to weigh in. Most people (about 15,000 out of about 16,000) who make semifinalist make finalist. However, only about 1/6 of them win the merit scholarship (another large portion win corporate scholarships).

A good friend of mine barely made semifinalist (223 PSAT) with his PSAT scores, got a 2370 SAT, and went on to become a winner. Another good friend of mine got a 240 on the PSAT's, but she got a 2270 on the SAT's and didn't win the scholarship.

Note that the SAT score used for NMS is the best single composite score, and not the sum of the best section-wise scores.


group theory
abstract algebra
superior algebra
superior algebra unsolved

4. Give an example of a group G that and an infinite subset H of G that is closed under the group operation but is not a subgroup of G.

8. Let H and K be subgroups of H. Prove that the intersection of H and K is a subgroup iff either H is a subset of  K or K is a subset of H.

4. Have you heard of the integers?   :)  Well, the nonzero ones, anyway.  Are they in any groups you recognize?

8. This is wrong, the intersection of two subgroups is always a subgroup.  You mean union.  To prove this, try taking some $ x\in H \minus{} K$.  Where is $ xy$ when $ y\in K$?


Using the numbers $ 1,2,3,4$ once each and the operations $ \plus{} , \minus{} ,\times,\div$ no more than once each, what are the minimum and maximum of the absolute value of the resulting quantity?

[quote="Jarle10"]The operations can only appear in between the numbers, so you can only use 3 of them, if you don't allow +2, or -2 for example.[/quote]

oh. oops. >_<. Lemme edit that.

EDIT: OOPS
NO MORE THAN 1 OF EACH OPERATION

obviously 0 is the minimum possible and 0 is indeed possible (4/2-3+1), so 0 is the minimum

idk how you prove the maximum other than through trial and error, but with parentheses i think it is 4*(3+2)/1=20 and without 4*3+2/1=14

Followup:

Let a,b,c,d be four positive integers such that a<b<c<d. Only by using the previously stated operations - find an expression for the largest possible value expressed by the four integers.


$ X,Y$ are 2 Independent Identically Distributed random variables of $ \mathbb{R}$

Show $ Pr(|X \minus{} Y| \le 2) \le 3Pr(|X \minus{} Y| \le 1)$

Note: From "The Probabilistic Method", chapter 1 exercises (it is starred for difficulty, but it is still driving me crazy!)

Highly non-trivial. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=153742]here.[/url]

Thank you. I don't think research problems are suitable question 3's in the exercise section to chapter 1 of any book, heh.

:P

All of math was at some point a research problem.  Also, it's not like they didn't warn you  :P


quadratics
inequalities
complex numbers
absolute value
imaginary numbers

I was wondering if a square of a number could be negative.  Maybe some sort of variable or complex number or binomial.  I'm not quite sure.

Thanks.[/code]

Yes.  This is the entire concept of [url=http://www.artofproblemsolving.com/Wiki/index.php/Imaginary_number]imaginary numbers[/url] and [url=http://www.artofproblemsolving.com/Wiki/index.php/Complex_number]complex numbers[/url].

[quote="Fly_DIA"]I was wondering if a square of a number could be negative[/quote]

All square roots of whole numbers are negative, and positive.  For example, $ \sqrt{16}\equal{} \pm4$, but the square root sign usually implies the answer being positive.  If you are looking for square roots of [i]negative[/i] numbers, $ i$ comes into play. ($ i\equal{}\sqrt{\minus{}1}$)

Usually, the [b]positive square root[/b] is implied...

However, when you are solving a quadratic definitely consider both the positive and negative roots.

The term for the positive root is [b]principal root[/b].

The $ \sqrt{  }$ sign refers to the principal root. This is why absolute value signs are needed when simplifying radical expressions containing variables.

[quote="Fly_DIA"]I was wondering if a square of a number could be negative.  code][/quote]
It has to be complex number, such as i, which means square root of -1

When solving for [i]x[/i] of an expression as such:
[i]x[/i]^2=[i]n[/i] implies [i]x[/i]=+/-√[i]n[/i]
[i]x[/i] squared equals [i]n[/i] implies [i]x[/i] equals plus or minus square root of [i]n[/i].

When solving for square root of [i]x[/i], then the answer is generally positive. This is called the [i]Principle Root[/i].

[b]Example for first choice:[/b]
Solving Quadratics
Solving Expressions with [i]x[/i]^2=[i]n[/i]
etc.

When finding the square root of a negative number, then [u]imaginary numbers[/u] or [u]complex numbers[/u] are used. The symbol for imaginary number is [i]i[/i].
[i]i[/i]=√-1
√-[i]n[/i]= [i]i[/i]√[i]n[/i]

Other imaginary facts:
i^2=-1
i^3=-i
i^4=1
i^5=i
thus, the pattern goes: [i]i[/i], -1, [i]-i[/i], 1, [i]i[/i],...

There is no real number who's square is imaginary because every real number is negative, positive, or zero. zero squared is zero, which is not negative. Obviously all positive numbers squared will be positive, and all negatives squared are positive. This is a very important inequality to know;

$ x^2 >/\equal{} 0$ for all real values of $ x$.

It's known as the trivial inequality, and you can write greater than or equal to using the \ge sign.

$ x^2 \ge 0$ for $ x \in \mathbb{R}$.

It is possible, but only possible without a radical sign.

Using the following steps, you can find a formula for a negative square.

x^2=(-n)
x=√(-n)
x=√((n)*(-1))
x=√(n)*√(-1)
[b]x=√(n)*i[/b]


inequalities
function
calculus
integration
rearrangement inequality

Sadly, I don't think I have seen it used other than in really early IMO problems... And its nice.

Anyone have problems that use rearrangement?

I mean most people know this one:

P: Prove that for all positive reals a, b, c,

a/(b+c) + b/(a+c) + c/(a+b) >= 3/2

I actually just had such a problem recently in my math class.  I am trying to remember it from the top of my head, but it was roughly:
if f and g are two continuous non-decreacing functions from [0, 1] to R, prove that
Integral[f * g] :ge: Integral[f] * Integral[g],
where all integrals are from 0 to 1.

You have to know a bunch of things other than the rearrangement inequality to do it, but it does turn out to work in the end, after much labor.  I don't know if there's a nicer way, though, but I suspect there must be.

The idea is that Chebyshef's inequality has an equivalent (bearing the same name) for integrals. On the other hand, Chebyshef's ineq can be seen as a mere application of the rearrangement inequality. I think that in order to prove that problem you posted it's enough to look at the Riemann sums (choose the dissection of [0,1] in segments [0,1/n], [1/n,2/n], .., [(n-1)/n,1]) and apply Chebyshef's inequality there.

yes the riemann sums do very much work in this case.
Though I won't post the solution since it is a straight off application of chebychev.

Joel, here's a nice way to prove your inequality.  Consider the (double) integral of (f(x) - f(y)) (g(x) - g(y)) dx dy.  Because f and g are nondecreasing, the inner expression is nonnegative, and hence so is the integral.  But if you expand the product, and then integrate each term separately, you'll get (twice) the difference of the two sides in your inequality.


A dust particle walks on the number line. Every minute, it walks either to the integer directly to its left or the integer directly to its right, with an equal probability of walking to either. After 6 minutes, what is the probablility that the dust particle returns to its starting point?

AMC problems are not Classroom Math material.

[hide="MAJORHINT"]The only way it can happen is if there are 3 forwards and 3 backwards. How can you use choose 3 forwards only? What is the probability?[/hide]

think pascal's triangle coin flipping combinatorics

well, there are 2^6 possible move choices, since he makes 6 moves and he has 2 options at each move.

and if there must 3 forward and 3 backward, then there are 6c3 ways to "choose" which moves are forwards, and after that the backwards are going to have to "fill in" the rest of the moves. 6c3=6*5*4/3/2/1=20

since probability is #successes/#of outcomes, it should be(i think) 20/2^6=5/2^4=5/16.

is 5/16 the correct answer?

i got 5/16 by saying that there are 6C3 ways to choose 3 forward. Then I multiplied by the probability of getting 3 forwards (1/8) and the probability of getting 3 backwards =

20 x 1/64 = 5/16

u can also do it as 6C3/2^6 but we maybe overcounting by saying 2^6 because it doesnt matter the order of which way the dust particle goes.

Note that this can be generalized into NC(N/2) for N turns, so N must be even. Although it does not mention block-walking, this is a good example of such problem.

[hide="sehr simple"]
it must move 3 forwards and 3 backwards. it makes 6 moves so there are $ \binom{6} {3}\equal{}20$ ways to arrange the 3 forwards and backwards. there are $ 2^6$ ways the particle can move. $ \frac{20} {64}\equal{}\boxed{\frac{5} {16}}.$[/hide]


Hi.
I've just began to learn LaTeX so please don't be angry with me if the problem is obvious:

If I want to compile the following part:
[code]
\begin{eqnarray*}
[0] &=& \{x \mid x=k\cdot m + 0,\ k \in \mathbb{Z} \} \\
[1] &=& \{x \mid x=k\cdot m + 1,\ k \in \mathbb{Z} \} \\
\end{eqnarray*}
[/code]
I get an error.

I just think it is interesting, that if I remove the third line (the one beginning with [1]) everything works, it also works if I just remove the brackets (But I have no problems with the [0] though).

I could need help...

Thanks!

Enclose the [1] in braces, like {[1]}.

LaTeX gives \\[...] a special meaning, so you have to go through this contortion to avoid it.


The family of all subsets of X containing a fixed subset, together with the empty set, is a topology for X.

1) Show this is a topology

2) under what conditions is it regular? completely regular?

For 1) we have:
i) By the assumption, X and the empty set are members of the defined family

ii) $ A \subset \bigcup_{\alpha}U_{\alpha} \iff$ there exists $ U_{\alpha'}$ containing $ A$, and the later holds.

iii) $ A \subset \bigcap_{i\equal{}1}^{n}U_{i} \iff A \subset U_{i}$ for each i, and the right hand side holds.

For 2): If A is the whole X, then we get the trivial space, which will both be regular and completely regular if X has at least two points.

I am not sure about 2) and so I would appreciate any hint on it.

Never mind about this problem. I got it figured out.


We define a fuction $ f(x)$ circulate on R if there exists $ a>0$ :
          \[ f(x+a)=f(x)\]
We define a fuction $ f(x)$ anti-circulate on R if there exists $ a>0$ :
          \[ f(x+a)= -f(x)\]
 $ a$ is a cycle of the circulate or anti-circulate function $ f(x)$
We define $ a_0$ the basic cycle of the circulate or anti-circulate function $ f(x)$ if $ a_0=min(a)$
     Prove that:
          If $ f(x)$ is anti-circulate on R with the basic cycle $ b$ then $ f(x)$ is circulate on R with the basic cycle $ 2b$

My English is not good. Maybe "periodic" better.
 And by the way, sorry for the duplication.

I think it is an easy problem !!!
  We have  f(x + b ) = - f(x)
    f(x + 2b ) = f( x+ b + b ) = -f(x+ b) = f(x)

You have just shown that $2b$ is [b]a[/b] period, but the problem asks if it's the main period, that is the [b]smallest[/b] period, and I think the answer is negative.

Sorry for that, it seems that the answer is ‘yes’ after all!

Let’s assume $b$ is a principal antiperiod for $f$ and that $a$ is a principal period of $f$ which is smaller than $2b$.
There are two possibilities: $a>b$ or $a<b$. In the first case, $f(x+a-b)=-f(x),\ \forall x\in\mathbb R,\ 0<a-b<b$, so $a-b$ is an antiperiod smaller than $b$, which is absurd. In the other case $f(x+b-a)=-f(x),\ \forall x\in\mathbb R,\ 0<b-a<b$, so $b-a$ is an antiperiod smaller than $b$ which is, again, false.


limit
trigonometry
calculus
calculus computations

Find

$ \lim_{x\to \infty} \frac{x^2[1\plus{}sin^2(x)]}{[x\plus{}sin(x)]^2}$

Thanks!

$ \lim_{ x \rightarrow \infty} \frac{x^2 (1\plus{}\sin^2(x))}{\left( x\plus{}\sin(x)\right)^2} \equal{} \lim_{ x \rightarrow \infty} \frac{1\plus{}\sin^2(x)}{\left(1\plus{}\frac{\sin(x)}{x}\right)^2}$

Now, the $ \frac{\sin(x)}{x}$ part goes to $ 0$, but the $ 1\plus{}\sin^2(x)$ part oscillates between $ 1$ and $ 2$ so the limit does not exist.


linear algebra
matrix
combinatorics
IMO Shortlist

Let $x_1,\ldots, x_n$ and $y_1,\ldots, y_n$ be real numbers.  Let $A = (a_{ij})_{1\leq i,j\leq n}$ be the matrix with entries \[a_{ij} = \begin{cases}1,&\text{if }x_i + y_j\geq 0;\\0,&\text{if }x_i + y_j < 0.\end{cases}\] Suppose that $B$ is an $n\times n$ matrix with entries $0$, $1$ such that the sum of the elements in each row and each column of $B$ is equal to the corresponding sum for the matrix $A$.  Prove that $A=B$.

Consider the matrix $A-B$. It has only $-1,0,1$ as elements, and we want to prove that it actually has only $0$. It also has the property that the sum of the elements on each line and each column is $0$. In $A$ we can't have $i\ne k,\ j\ne l$ s.t. $a_{ij},a_{kl}=1,\ a_{il},a_{kj}=0$ or vice-versa. Then in $A-B$ we can't have $i\ne k,\ j\ne l$ s.t. $a_{ij},a_{kl}=1,\ a_{il},a_{kj}\ne 1$ or vice-versa. [color=red][[b]Moderator edit:[/b] Here and in the following, $a_{uv}$ doesn't necessarily denote the $\left(u,v\right)$-th entry of the matrix $A$, but rather the $\left(u,v\right)$-th entry of whatever matrix we are currently talking about.][/color]

Consider the line in $A-B$ with the maximal number of $1$'s. If this number is non-zero, then there must be some $-1$'s as well on this line (call it line $i$). Assume $a_{ij}=-1$. In this case there is a $k$ s.t. $a_{kj}=1$ (because the sum of all numbers on column $j$ is $0$). If for all $m$ for which $a_{im}=1$ we also have $a_{km}=1$ then on the line $k$ we have more $1$'s than on the line $i$, and that's a contradiction, so there must exist an $l$ s.t. $a_{il}=1$ and $a_{kl}\ne 1$. But if we look at the four elements $a_{ij}=-1\ne 1,a_{kl}\ne 1,a_{il}=a_{kj}=1$ we see that they contradict what we said in the first paragraph.

[quote="grobber"] Then in $A-B$ we can't have $i\ne k,\ j\ne l$ s.t. $a_{ij},a_{kl}=1,\ a_{il},a_{kj}\ne 1$ or vice-versa.

[/quote]
why?

[color=red][ [b]Moderator edit:[/b] The following solution renames the matrices $A$ and $B$ as $M$ and $N$. ][/color]

We can make this really easy by generalizing to n x m matrices, and forcing the x_i (1<=i <= n), y_j (1 <= j <= m) to be nondecreasing sequences (note row/column swapping does not affect the condition if we do likewise with N).

Let a_i,j be the position in M determined by x_i , y_j.

Then consider a_(n,1). If it is 0, the entire column a_(k,1) must be 0's and therefore M, N must match in this column. Deleting this column reduces m by one.

If it is 1, the entire row a_(n,k) must be 1's and therefore M,N must match in this row. Deleting this row reduces n by one.

Therefore, by inducting on m+n we see the result.

it is easy try to prove that we have always a column with all 1 or all zero or else
a row.
then use mathematical induction.

Consider $\sum_{i=1}^{n} \sum_{j=1}^n (x_i+y_j)(a_{ij} - b_{ij})$. This sum is $0$. Now, consider all $i, j$ s.t. $a_{ij} \neq b_{ij}$. We have that is $a_{ij}=1 \implies b_{ij} =0 \implies (a_{ij} - b_{ij})(x_i + y_j)$ is positive. Similarly, $a_{ij} = 0 \implies b_{ij} = 1, x_i + y_j < 0$ so that is still positive. Summing yields a contradiction, so $A=B$.

We only consider the entries which change when $A$ switches to $B$. Let $e_{i,j}$ be $1$ if $b_{i,j}=1$ while $a_{i,j}=0$, and let $e_{i,j}$ be $-1$ if $b_{i,j}=0$ while $a_{i,j}=1$. Let $S\subseteq [n]^2$ be the subset of entries which changed. Note that for any fixed $j$,
\[
\sum_{(i,j)\in S} e_{i,j}=0
\]
and similarly for fixed $i$, varying $j$. So if $i,j$ are allowed to vary over all possible choices,
\[
\sum_{(i,j) in S} e_{i,j}(x_i+y_j) = \left(\sum_{i} x_i\sum_{(i,j) \in S} e_{i,j}\right)+\left(\sum_{j} y_j \sum_{(i,j)\in S} e_{i,j}\right)=0
\]
However for any $(i,j)\in S$, $e_{i,j}(x_i+y_j)\ge 0$, and for any $e_{i,j}=-1$, equality does not hold. Since there is at least one negative value of $e_{i,j}$, we get $0>0$, which is a contradiction.

[quote=IMO SL 2003 C4]Given $n$ real numbers $x_1$, $x_2$, ..., $x_n$, and $n$ further real numbers $y_1$, $y_2$, ..., $y_n$. The entries $a_{ij}$ (with $1\leq i \leq n$ and $1 \leq j\leq n$) of an $n\times n$ matrix $A$ are defined as follows:

$a_{ij}=\left\{ \begin{array}{c} 1\text{\ \ \ \ \ \ if\ \ \ \ \ \ }x_{i}+y_{j}\geq 0; \\ 0\text{\ \ \ \ \ \ if\ \ \ \ \ \ }x_{i}+y_{j}<0. \end{array} \right.$

Further, let $B$ be an $n\times n$ matrix whose elements are numbers from the set $\left\{0;\  1\right\}$ satisfying the following condition: The sum of all elements of each row of $B$ equals the sum of all elements of the corresponding row of $A$; the sum of all elements of each column of $B$ equals the sum of all elements of the corresponding column of $A$. Show that in this case, $A = B$.[/quote]

Firstly, shift the sequences $(x_1, \dots, x_n)$ and $(y_1, \dots, y_n)$ by a positive constant so that $x_i+y_j<0$ doesn't change for any pair $i, j$.

Note that if $A \ne B$ then the sum $$S \overset{\text{def}}{:=} \sum_{i, j} (a_{ij}-b_{ij})(x_i+y_j),$$ is positive since each of the summands is positive (in general they would be non-negative). However, we see that $$S=\sum_{i} x_i \left(\sum_{j} (a_{ij}-b_{ij}) \right)+\sum_{j} y_j \left(\sum_{i} (a_{ij}-b_{ij}) \right)=0,$$ yielding the desired contradiction. $\blacksquare$

EDIT: A few glaring errors with this proof. Thanks to Darij for providing a good fix in the post below :)

[quote=anantmudgal09]Firstly, shift the sequences $(x_1, \dots, x_n)$ and $(y_1, \dots, y_n)$ by a positive constant so that $x_i+y_j<0$ doesn't change for any pair $i, j$.[/quote]

What do you mean by "shift", and why are you doing that?

[quote=anantmudgal09]Note that if $A \ne B$ then the sum $$S \overset{\text{def}}{:=} \sum_{i, j} (a_{ij}-b_{ij})(x_i+y_j),$$ is positive since each of the summands is positive (in general they would be non-negative).[/quote]

Uhm, they're not quite all positive. First of all, $x_i + y_j$ can be $0$, which causes the term $(a_{ij}-b_{ij})(x_i+y_j)$ to vanish. Second, $a_{ij}$ can equal $b_{ij}$, with the same consequence.

The argument you probably want to make is the following: Each of the terms $(a_{ij}-b_{ij})(x_i+y_j)$ is nonnegative; thus the whole sum $\sum_{i, j} (a_{ij}-b_{ij})(x_i+y_j)$ is nonnegative. Furthermore, we claim that this sum is positive. To prove this, it suffices to find at least one positive addend. Notice that the sum of all entries of $A$ equals the sum of all entries of $B$ (since the sum of all elements of each row of $B$ equals the sum of all elements of the corresponding row of $A$); therefore, at least one entry of $A$ must be smaller than the corresponding entry of $B$ (since $A \neq B$). In other words, $a_{ij} < b_{ij}$ for [b]some[/b] $i$ and $j$. Fix such $i$ and $j$, and observe that $a_{ij}$ and $b_{ij}$ must be $0$ and $1$, respectively (since $a_{ij} < b_{ij}$). From $a_{ij} = 0$, we obtain $x_i + y_j < 0$. Thus, $\left(\underbrace{a_{ij}}_{=0}-\underbrace{b_{ij}}_{=1}\right)\left(x_i+y_j\right) = \left(0-1\right) \left(x_i+y_j\right) = - \left(x_i+y_j\right) > 0$ (since $x_i + y_j < 0$), and so we have found a positive term.

Nice argument -- just needs this little fix.

Did I do something wrong?

Permute the rows and columns of $A$ and $B$ simultaneously in the same way so that $x_1\ge x_2\ge\ldots x_m$ and $y_1\ge y_2\ge\ldots y_n$ (we generalize to $m\times n$). Now, note that our 1's form a Young diagram in $A$. A column past the 1's in the first row is all 0's so we may remove it and induct down. Otherwise, the entire first row consists of 1's, so we may remove it and induct down.

What a funny problem!

For a binary matrix $A$, we define a [i]stone move[/i] on its entries to be picking $i,j,k,\ell$ (where $i\neq j$, $k\neq \ell$) so that $a_{ik}+a_{j\ell} = 2$, $a_{i\ell}+a_{jk} = 0$, and swapping entries so that $a_{ik} = a_{j\ell} = 0$, $a_{i\ell}=a_{jk} = 1$. Now, we claim that if $A, B$ are rectangular binary matrices with the same size so that the row, column sums of $A$ and $B$ are the same, then $B$ can be reached from $A$ via a finite sequence of stone moves. Indeed, let $r_1,\ldots, r_m$ denote the row sums, and $c_1,\ldots, c_n$ denote the column sums. It suffices to show that we can reach some matrix $C$ from both $A$ and $B$ via a finite sequence of stone moves, since stone moves are undoable.

Indeed, we will demonstrate an algorithm that decreases $S = \sum\limits_{i=1}^n |a_{1i}-b_{1i}|$, which will eventually reach zero (this is essentially USAMO 2015/4). Indeed, suppose that $S \neq 0$. Then, there exist $i$ such that $a_{1i}\neq b_{1i}$, say $a_{1i} = 1, b_{1i} = 0$. There must also exist $j\neq i$ with $a_{1j} = 0$, $b_{1j} = 1$, since $\sum a_{1i}-b_{1i} = 0$. Now, we claim that there either exists $k$ so that $a_{ki} = 0$ and $a_{kj} = 1$, or there exists $k$ so that $b_{ki} = 1, b_{kj} = 0$. Indeed, suppose not, then $a_{ki}-a_{kj} \geq 0$ for $k\geq 2$, and $b_{ki}-b_{kj}\leq 0$ for $k\geq 2$, and \[
0 < \sum\limits_{k=1}^m a_{ki}-a_{kj}  = c_i - c_j = \sum\limits_{k=1}^m b_{ki}-b_{kj} < 0,
\]contradiction. So, if there exists $a_{ki} = 0$, $a_{kj} = 1$, we simply apply a stone move to $A$; else apply a stone move to $B$, and $a_{1i} = b_{1i}$, $a_{1j} = b_{1j}$, so we have decreased $S$. Eventually this forces $S = 0$, so we may discard the first row and repeat on the $m-1\times n$ matrices $A', B'$. So, by applying stone moves to $A$, $B$ we can reach a common $C$, and this means we can apply stone moves $A\to C\to B$.

However, note that if $a_{ik}+a_{j\ell} = 2$, $a_{i\ell} = a_{jk} = 0$, then we get $x_i+y_k > 0$, $x_j+y_\ell > 0$ which implies $x_i+x_j+y_k+y_\ell> 0$, but $x_i + y_\ell \leq 0$, $x_j+y_k \leq 0$, which yields $x_i+x_j +y_k+y_\ell \leq 0$, contradiction. So, we cannot apply any stone moves to $A$; since $B$ is reachable by stone move, $A=B$.

@MathStudent2002: I think you use two different definitions of "stone move", one of which has a determined direction (perhaps $i < j$ and $k < \ell$?) whereas the other doesn't.

Hm? I'm pretty sure the definition is always taking a rectangle with two opposite corners 1 and the other two 0 and swapping the 1s with the 0s?

Oops, I've misunderstood your last paragraph, sorry. Nice proof! (I've fixed a few typos.)

Notice that we can change $y_i$ to $-y_i$ and rephrase the problem:

[quote = rephrased problem]
Let $x_1,\ldots, x_n$ and $y_1,\ldots, y_n$ be real numbers. Let $A = (a_{ij})_{1\leq i,j\leq n}$ be the matrix with entries \[a_{ij} = \begin{cases}1,&\text{if }x_i \geq y_j;\\0,&\text{if }x_i < y_j.\end{cases}\] Suppose we are given sum of each row and sum of each column, then show that we can find all values of $A$.
[/quote]

[b]Solution.[/b] Arrange $n$ numbers in increasing order (not equal). Now sum of each row say the $i^{th}$, means the number of $y_j$'s greater than $x_i$. Now place the $x_i$'s according the $n$ numbers assuming the $n$ numbers were $y_j$'s in increasing order. Now sum of each column say the $j^{th}$ mean the number of $x_i$'s greater than $y_j$, so label the $n$ numbers now.

Therefore we can define the entire set $A$.

[quote=darij grinberg]What $n$ numbers are you arranging, @TheDarkPrince?[/quote]

Some $n$ numbers. Notice that $y_i$'s are reals and only their relative position matter, so we choose any $n$ numbers and accordingly fix $x_i$ and $y_i$.

[quote=darij grinberg]Sorry, I really don't follow.[/quote]

Sorry, I'm not able to express it too well :(

We need to find the values of matrix $A$. We have defined the elements of $A$ in such a way that $x_i$'s and $y_i$'s value doesn't matter only their relative value matters, i.e which is greater.

Sum of each row means number of values of $y_i$'s greater than that particular $x_j$.
-----
We'll take an example:

Suppose $n = 3$, sum of row$_1 = 2$, sum of row$_2 = 3$, sum of row$_3 = 3$, sum of column$_1 = 3$, sum of column$_2 = 3$, sum of column$_3= 2$.

We'll try to find the values of matrix $A$ with this information.

[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(4cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -5.664651162790702, xmax = 17.033023255813966, ymin = -6.063255813953488, ymax = 6.141395348837212;  /* image dimensions */

 /* draw figures */
 /* dots and labels */
dot((0.,0.),linewidth(4.pt) + dotstyle);
dot((3.,0.),dotstyle);
dot((5.,0.),dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */
[/asy]

Sum of row$_1 = 2$ means $x_1$ is less than one ($3-2 = 1$) $y_i$.

[asy]
import graph; size(4cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -5.664651162790702, xmax = 17.033023255813966, ymin = -6.063255813953488, ymax = 6.141395348837212;  /* image dimensions */

pair A = (0.,0.), B = (3.,0.), C = (5.,0.), x_1 = (3.7493023255813975,0.);
 /* draw figures */
 /* dots and labels */
dot(A,linewidth(4.pt) + dotstyle);
dot(B,dotstyle);
dot(C,dotstyle);
dot(x_1,dotstyle);
label("$x_1$", (3.8237209302325605,0.1879069767441865), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */
[/asy]

Sum of row$_2 = 3$ means $x_2$ is less than zero ($3-3 = 0$) $y_i$.

[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(4cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -5.664651162790702, xmax = 17.033023255813966, ymin = -6.063255813953488, ymax = 6.141395348837212;  /* image dimensions */

pair A = (0.,0.), B = (3.,0.), C = (5.,0.), x_1 = (3.7493023255813975,0.), x_2 = (5.795813953488376,0.);
 /* draw figures */
 /* dots and labels */
dot(A,linewidth(4.pt) + dotstyle);
dot(B,dotstyle);
dot(C,dotstyle);
dot(x_1,dotstyle);
label("$x_1$", (3.8237209302325605,0.1879069767441865), NE * labelscalefactor);
dot(x_2,dotstyle);
label("$x_2$", (5.870232558139539,0.1879069767441865), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */
[/asy]

Repeating this for columns gives:

[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(4cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -5.664651162790702, xmax = 17.033023255813966, ymin = -6.063255813953488, ymax = 6.141395348837212;  /* image dimensions */

pair y_1 = (0.,0.), y_2 = (3.,0.), y_3 = (5.,0.), x_1 = (3.7493023255813975,0.), x_2 = (5.795813953488376,0.), x_3 = (6.744651162790702,0.);
 /* draw figures */
 /* dots and labels */
dot(y_1,linewidth(4.pt) + dotstyle);
label("$y_1$", (0.06558139534883684,0.1506976744186051), NE * labelscalefactor);
dot(y_2,dotstyle);
label("$y_2$", (3.079534883720932,0.1879069767441865), NE * labelscalefactor);
dot(y_3,dotstyle);
label("$y_3$", (5.070232558139538,0.1879069767441865), NE * labelscalefactor);
dot(x_1,dotstyle);
label("$x_1$", (3.8237209302325605,0.1879069767441865), NE * labelscalefactor);
dot(x_2,dotstyle);
label("$x_2$", (5.870232558139539,0.1879069767441865), NE * labelscalefactor);
dot(x_3,dotstyle);
label("$x_3$", (6.819069767441865,0.1879069767441865), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */
[/asy]

This gives the matrix \[\begin{bmatrix}
1 & 1 & 0\\
1 & 1 & 1\\
1& 1 &  1
\end{bmatrix}\]

Was this kinda clear?

No, I still don't follow. You're saying nothing about $B$.

We will prove the result in general for an $m\times n$ matrix. Strong induct on $m+n$, the base case of $m+n=2$ being trvial.

We first apply the permutation $\pi_1$ to $\{x_i\}$ and $\pi_2$ to $\{y_i\}$ to sort them into decreasing order, and also permute $A$ and $B$ into $A'$ and $B'$ by applying $\pi_1$ to the rows and $\pi_2$ to the columns. To retrieve $A$ and $B$, we apply $\pi_1^{-1}$ to the rows and $\pi_2^{-1}$ to the columns. So if we can show $A'=B'$, we are done.

Since $x$ and $y$ are sorted, $A$ will form a Young Tableux. (This is where each row has a number of 1's, and then 0's, with the constraint that the number of 1's in the rows is non-increasing.) If there are any rows of 0's at the bottom of $A$, $B$ must also have these rows of 0's, since the row sum is 0. Then we can delete these rows, and since $m+n$ decreased, we finish by induction. Similarly for columns of 0's at the right. If there are no rows or columns of 0's, there must be a row of 1's in $A$. Then $B$ must have this same row of 1's, since the row sum is maximum. So we can delete it and finish by induction.

WLOG $\{x_i\}$ and $\{y_i\}$ are both sorted. Then, the $1$s in $A$ form a Young's Tableau. If we consider the first column of $B$, the amount of $1$s it has is equal to the number of rows with at least one $1$, so it is fixed. In a similar manner, the second column is now fixed, as its sum is equal to the number of rows which still have another $1$, and so forth. So, $B$ is fixed, and thus must be $A$, as desired.

We will prove the general result for $m \times n$
Induction on $m+n$
The case $m+n=3$ is trivial. Assume for $m+n=k-1$
Now let $m+n=k$ ,
Consider the row in $A$ with maximum number of $1's$
Case 1- This row contains only $1's$ , then the corresponding row in $B$ must also contain all ones, hence we may delete this row and complete the problem by induction hypothesis.
Case 2 - This row contains a $0$, now we claim that the column of this $0$ ,say $i$, contains only $0's$ and then complete the proof as before.
If possible let this column $i$ contain a $1$ . We show that the row of this $1$ ,say $l$, contains more $1's$ than our chosen row, say $k$,m contradicting maximality of chosen row. Let there be a $1$ in $j'th$ column of row $k$. We show that there must be a $1$ in $j'th$ column of row $l$.Whereas we already have an extra $1$ in row $l$, hence proof will be complete.
If possible let $a_{k,j}=1$ but $a_{l,j}=0$, we already know $a_{k,i}=0$ and $a_{l,i}=1$ .
 Hence $x_l+x_j<0, x_k+x_i<0 \implies x_i+x_j+x_l+x_k<0$
And $x_k+x_j\geq0 x_l+x_i\geq0 \implies  x_i+x_j+x_l+x_k\geq0$
Contradiction.
Hence proved.

Nice problem! Here's my solution: Rearrange the rows of $A$ so that number of ones in a row is a non-increasing sequence as we go down. Do the same row transformations with $B$. Note that this doesn't affect the problem, since only the order of the rows is affected. This change is equivalent to ordering the $x_i$'s as $x_1 \geq x_2 \geq \dots \geq x_n$. Then, for a fixed $j \in [n]$, we have $$x_1+y_j \geq x_2+y_j \geq \dots \geq x_n+y_j$$
In particular, this means that if we have a $1$ in some cell, then all cells above in the same column must also contain $1$. Do an analogous transformation of the columns, which corresponds to arranging the $y_i$'s as $y_1 \geq y_2 \geq \dots \geq y_n$. This gives that if there is a $1$ in some cell, then all cells to its left (in the same row) must also have $1$'s. Let $a_i$ denote the number of $1$'s in the $i^{\text{th}}$ row (in this new matrix). Then the above transformation forces $a_1 \geq a_2 \geq \dots \geq a_n$. If $a_n \neq 0$, then the first $a_n$ columns must contain all $1$'s. So the first $a_n$ columns of $B$ must also contain $1$'s. But, since the last row of $B$ also has $a_n$ $1$'s, which are already filled in the first $a_n$ columns, so the remaining elements of the last row of $B$ must be zero (i.e. same as $A$). In that case, a simple induction on $n \geq 1$ finishes the argument. Else, if $a_n=0$, then the last row of $A$ has only zeros. So the last row of $B$ must also be zero. In that case, consider the largest $i$ such that $a_i \neq 0$. Then all rows of $A$ below the $i^{\text{th}}$ row consists of only $0$'s, so the same must be true for $B$. But, the first column contains exactly $i$ $1$'s in both $A$ and $B$, and there are already $(n-i)$ $0$'s in the first column of $B$, so the first column's initial $i$ cells must all have $1$. So in this situation again the inductive step suffices. Hence, done. $\blacksquare$

[b]REMARKS:[/b] Tbh I didn't like my writeup much, coz there are just too many words :P, but I guess the basic idea is pretty simple. The resulting matrix after the row/column transformations is what I had used in my solution to the first problem [url=https://artofproblemsolving.com/community/c260h2021934p14452440]here[/url].

We gerenalize the problem for an $m \times n$ board. Suppose WLOG that $x_1 \leq x_2 \leq ... \leq x_m$ and $y_1 \leq y_2 \leq ... \leq y_n$.

Now, we induct on $m+n$ (the base cases are easy).

If the first row  of $A$ has all its entries $0$, we can erase it and induct, since the correspondent row in $B$ must have all its entries $0$ as well. Otherwise, there exists a $k \leq n$ such that $x_1+y_k \geq 0$. Thus, since $$x_1+y_n \geq x_1+y_{n-1} \geq ... \geq x_1+y_k \geq 0$$ the cell $(1,n)$ is equal to $1$.

Furthermore, $$x_m+y_n \geq x_{m-1}+y_n \geq ... \geq x_1+y_n \geq 0$$ $\implies$ all cells above $(1,n)$ are equal to $1$. Then, we can erase this column and induct, since the correspondent column in $B$ must contain only $1$'s in it.

Hence, we are done.
$\blacksquare$

We consider the general $m\times n$ case, we begin by induction. The case when $m=1$ is easy to check. Now we consider the case $(m,n)$ with the assumption that all cases $(m',n')$ with $m'\leq m$,$n'\leq n$ and $(m',n')\neq (m,n)$ holds. Now WLOG assume $|x_1|=\max\{|x_1|,|x_2|,...,|x_n|,|y_1|,...,|y_n|\}$. Therefore, the sum of the column corresponding to $x_1$ must have entries either all equal to $1$ or $0$. Therefore the entries of $B$ must also be the same, now we an apply the inductive hypothesis to the remaining matrix and we are done.

Nice problem!
Generalize for $x_1> x_2> \cdots> x_n; y_1>y_2> \cdots >y_m$ (WLOG the numbers are sorted, otherwise interchange some and rows and columns) and we will induct. Obviously the original matrix $A$ is an Young Tableux now (i.e. the rows and columns have decreasing number of 1's from left to right and from uppermost edge to the lowermost edge). If there is an all-zero row or column in $A$, we are done (this row or column will appear also in $B$, so we can remove it and induct down). Otherwise there is an all-one row (the uppermost one), so this row must be also in $B$, so we can remove it and induct down.

Catloe walk is OP!

Assume FTSOC that the two matrices are not equal. Consider where the two matrices differ. If $A$ has a $1$ in some spot that $B$ has a $0$ in, call it a Ritwino cell. If it is the other way around, call it a Jatloeo cell.

Now, pick any Ritwino cell in the grid and let Catloe be on it. Note that if a row or column has one changed cell, it must have at least two changed cells. So, select Jatloeo cell in the row, and let Catloe walk to it in a straight line. Now, select another Ritwino cell in the column Catloe is in, and let Catloe walk there. This repeats until Catloe reaches a cell he was already in. Catloe has completed a loop that only turns on Ritwino and Jatloeo cells. By parity, the turns are also alternating Ritwino and Jatloeo.

Now, consider the Ritwino and Jatloeo turns. Clearly, if we focus on any row or column, it has the same number of Ritwino and Jatloeo turns. Then the sum of $x_i+y_j$(all nonnegative) on the Ritwino turns is equal to the sum of $x_i+y_j$(all negative) on the Jatloeo turns(from matrix $A$), contradiction.

Consider
\[S=\sum_{i=1}^{n}{\sum_{j=1}^{n}{(x_i+y_j)(a_{ij}-b_{ij})}}=\sum_{i=1}^{n}{\left(x_i\sum_{j=1}^{n}{y_j(a_{ij}-b_{ij})}\right)}+\sum_{j=1}^{n}{\left(y_j\sum_{j=1}^{n}{x_i(a_{ij}-b_{ij})}\right)}=0\]
If $x_i+y_i\ge 0$ then $a_{ij}-b_{ij}=1-b_{ij}\ge 0$ and if $x_i+y_i<0$ then $a_{ij}-b_{ij}=-b_{ij}\le 0$. Either way, $(x_i+y_i)(a_{ij}-b_{ij})\ge 0$ so $S\ge 0$. Equality holds if and only if $x_i+y_i=0$ or $a_{ij}=b_{ij}$ for each $i,j$. In both cases, $a_{ij}\ge b_{ij}$ so if the sum of row and column equality were to hold, $a_{ij}=b_{ij}$ as desired.

WLOG, let $x_1 \leq x_2 \leq \dots \leq x_n$ and same for $y$.
Then, $a_{i1} \leq \dots \leq a_{in}$. And same for the columns.
Now, FTSOC, let there exist $B\neq A$ with those property.
Now we create $C = A-B$ then the entries of $C$ will be $-1,0,1$. And notice that if $c_{ip} = -1$ and $c_{iq} = 1$ then, $a_{ip} =0, a_{iq} =1 \implies p < q$. And same for entries in the same column. $\qquad(1)$

Now, we can have a sequence of $2k$ cells $z_1, \dots z_{2k}$ of $C$ s.t. $z_{2i}, z_{2i+1}$ are in the same row and $z_{2i-1}, z_{2i}$ are on the same column. And $z_{2i} =1, z_{2i+1}=-1$.
But this cycle contradicts because of $(1)$ so we are done. $\blacksquare$

Used with hint on OTIS 20\%

    We can generalize to mxn by induction on a fixed n with varying m, and vice versa, since the process is the same, with the key being that we can remove a row/column if it has only 1s or only 0s since B MUST match up there. The base case m=n=1 is obvious. Now, assume there is a row (resp. column) with at least 1 1, we will take this with indices 1n, where we order indices s.t. $x_1\le x_2\le ...\le x_m,y_1\le y_2\le ...\le y_n$. Then since $x_m+y_n\ge x_{m-1}+y_n\ge\dots\ge x_1+y_n$, this entire row has 1s; in particular, we can remove it and it is reduced to inductive hypothesis. (If 1n was 0 then the entire row with indice x=1 is all 0s, and remove that).

We generalize to all $n$ by $m$ rectangular matrices. The problem is essentially saying that there is only one possible matrix given the row and column sums.\\

Call a row or column boring if it consists of only 0s or only 1s.\\

The main claim of the problem is the following:

\begin{claim}
    There exists a boring row or column.
\end{claim}

\begin{proof}
Suppose not. Then, arrange $x_1\dots x_n$ and $y_1\dots y_m$ in nondecreasing order in both ways (this only rearranges the rows and columns so it does not affect the claim). Note that each row and column is nondecreasing. Therefore, if there are no boring rows, each row must end with 1. Similarly, if there are no boring columns, each column must start with 0. However, this is a contradiction as the top right corner is both 0 and 1.
\end{proof}

We want to show that we can uniquely reconstruct the original matrix given the row and column sums. We will use induction. This is clearly true if either $n=1$ or $m=1$. We induct on $m+n$.\\

We know that there is a boring row or column. Thus, we can select a boring row or column (we can do this as one exists and we are given the row and column sums so we can pinpoint which one it is and we also know whether it consists of all 0s or all 1s) and consider the matrix excluding that boring row or column. We know the row and column sums of this remaining matrix (by taking the original row and column sums and subtracting away the boring row or column), and since this is a smaller value of $m+n$, by induction we are able to uniquely reconstruct this remaining matrix. Since we also know the contents of the boring row or column, we can reconstruct the entire matrix, hence done.

Note that in any such $m \times n$ matrix, there must be at least one row or column with only 0's or 1's, as the number with highest absolute value surpresses the other entries. Thus this row or column is fixed, so we can simply remove it.

The resulting matrix has the same property, so we conclude by downwards induction. $\blacksquare$

[hide="Storage:"]

We prove this result by induction on $n$. The base case is trivial for $n=1$. Now let $n \ge 2$ and consider the matrix $A$ defined in the problem. Take $x_i$ to be maximal and $y_j$ to be minimal. Consider the $i$'th row. If it contains a zero, then by minimality of $y_j$, so does the cell $(i,j)$. Thus, by maximality of $x_i$, the $j$'th column is full of zeroes. Now if $x_w$ is minimal, then consider the $w$'th row. If cell $(w,z)$ contains a $1$, then by minimality the column $z$ is full of ones. We deduce that the matrix $B$ coincides with $A$ on the column $j$, and the $1$'s of row $w$. Since the sum of the $w$'th row in both $B$ and $A$ are equal, it follows that $B$ coincides with $A$ in both the row $w$ and the column $j$, so we can reduce $A$ and $B$ to two matrices $A'$ and $B'$ of size $(n-1) \times (n-1)$ by removing $x_w,y_j$ (row/column), which are equal by induction $\implies A=B$.

Now if the $i$'th row is full of $1$'s take $y_p$ maximal. Repeating the same argument for $y_p$ leaves us with the case of row $i$ full of $1$'s and column $p$ full of $1$'s ; In this case, they coincide with $B$, so we can once again remove them and finish by induction.

$$\mathbb{Q.E.D.}$$

[/hide]