geometry circumcircle trigonometry vector

[quote="N.T.TUAN"]Let $ABC$ be a triangle. Triangles $PAB$ and $QAC$ are constructed outside of triangle $ABC$
such that $AP = AB$ and $AQ = AC$ and $\angle{BAP}= \angle{CAQ}$. Segments $BQ$ and $CP$ meet at $R$. Let $O$ be the circumcenter of triangle $BCR$. Prove that $AO \perp PQ.$[/quote]

Well, let's solve this one with complex numbers.

We denote $\angle{BAP}= \angle{CAQ}=\phi$ then because the triangles $PAC$ and $QAB$ coincide after rotation with center $A$ and angle $\phi$, we have that the angle formed by the segments $CP$ and $QB$ is equal to $\phi$. Then we get that $\angle{BOC}=2\phi$ and $BO=CO$. Now we take $A$ as the origin of the complex plane. Then if $z=\cos{\phi}+i\sin{\phi}$ we have that $p=\frac{b}{z}$ and $q=cz$ and also $\frac{b-o}{c-o}=z^{2}$ thus $o=\frac{z^{2}c-b}{z^{2}-1}=\frac{q-p}{z-\frac{1}{z}}$. Hence $\frac{o}{\bar{o}}=-\frac{q-p}{\bar{q}-\bar{p}}$ which means that $OA$ is perpendicular to $PQ$, as desired. :wink:

Here is my synthetic solution

There is no new philosophy in this problem. The idea is how to relate the circumcenter $O$ with $PQ$. Since we have so many pairs of equal angles here, probably it would help if we can take advantage of some hidden cyclic quadrilaterals and some spirality.

Let $\measuredangle{PAB}= \measuredangle{CAQ}= \alpha$. It is not hard to see that $APBR$ and $AQCR$ are cyclic. Indeed, the congruence of two triangles $APC$ and $ABQ$ gives us $\measuredangle{ABR}= \measuredangle{APR}$, $\measuredangle{AQR}= \measuredangle{ACR}$. Let's draw the circles $(ARBP)$ and $(ARCQ)$. Denote $U, V$ respectively the intersections of $AO$ with these two circles.

Notice that $\measuredangle{BOC}= 2 \alpha$. Hence, let 's bisect this angle by calling $T$ the midpoint of the arc $BC$ of the circumcircle of $BRC$. Then we have some spiralities here: $BPA$ and $BTO$, $CQA$ and $CTO$ $(*)$. Thus $BUTO$ is cyclic and $P, T, U$ are collinear. Similarly, $COVT$ is cyclic and $C, T, V$ are collinear.

We need now to prove $\measuredangle{PUA}+\measuredangle{UPQ}= 90^{0}$. This is equivalent to proving $\measuredangle{TPQ}= \frac{\alpha}{2}$. Hence, it is sufficient for us to prove that actually $PTQ$ is isosceles at $T$ and $\measuredangle{PTQ}= 180^{0}-\alpha$. The latter is very easy by some angle-chasing. To prve $TP = TQ$, use $(*)$.

although it's a little longer than the proofs above, the lemma $AB\perp CD \Leftrightarrow AC^{2}+BD^{2}=AD^{2}+BC^{2}$ and some trigonometry solve the problem  :D

What's surprise! the main idea of my first solution is similar with yours, treegoner. The idea is how to relate the circumcenter O with PQ
This is my first solution
Triangles $APC$ and $ABQ$ are congruent so $\angle APR=\angle ABR$. then $A,P,B,R$ are cyclic.
So $\angle BRP=\angle PAB$. But $\angle PRB= \angle BOC/2$ so $\angle PAB=\angle BOC/2$
Let the perpendicular line of $BC,AP,AQ$ through $O,B,C$ meet $BC,AP,AQ$ at $D,E,F$.
triangle $CFA$ and $BEA$ are similar so $AE/AF=AB/AC=AP/AQ$ so $EF//PQ$ $(1)$
We have triangle $CDO$ and $CFA$ are similar. Then triangle $CFD$ and $CAO$ are similar
thus $FD/AO=CD/OC$ and $\angle FDC=\angle AOC$ $(2)$
Similarly $ED/AO=BD/BO$ and $\angle BDE=\angle BOA$ $(3)$
from (2) and (3) $DE=DF$ and $\angle EDF=180-\angle BOC$
then  $\angle EFD=\angle BOC/2=\angle CAQ$ $(4)$
$FD$ meets $AO$ at $K$. we have $\angle CFK=\angle CAK$ so $A,K,C,F$  are cyclic.
then $\angle CAQ=\angle CKF$ $(5)$
from (4) and (5) $EF//CK$ $(6)$
from (1) and (6) $PQ//CK$
but $\angle CKA=\angle CFA=90$ so $PQ\bot AO$ $q.e.d$

May be this soluiton is shorter
Triangle $APC$ and $ABQ$ are congruent so $\angle APR=\angle ABR$.
Then $A,P,B,R$ are on circle $(O1)$. Similarly $A,R,C,Q$ are on circle $(O2)$
$(O1), (O2)$ meet $AQ,AP$ at $E,F$
Let $(I)$ be the circumcircle of triangle $EAF$
$O1I,O2I$ meet $AE,AF$ at $M,N$.
we have $\angle AEP=\angle ARC=\angle AFQ$ so triangles $AEP, AFQ$ are similar.
$M,N$ are midpoint of $AE,AF$ so triangles $PAM$ and $QAN$ are similar
so $\angle MPN=\angle MQN$. Then $P,M,N,Q$ are cyclic
And with $I,M,A,N$ are cyclic we have $\angle AQP=\angle MNA=\angle MIA$.
So $I,M,H,Q$ are cyclic. ($IA$ meets $PQ$ at $H$) Thus $IA\bot PQ$
But $PQ$ is the radical axis of $(I)$ and $(O)$ so $IO\bot PQ$. (because $PA*PF=PR*PC$ and $QA*QE=QB*QR$)
This means $AO\bot PQ$ $q.e.d$

[color=darkblue][u][b]May be this metrical proof is longest ![/b][/u] I"ll show that $OP^{2}-OQ^{2}=c^{2}-b^{2}$, i.e. $AO\perp PQ$.

Denote $2x=m(\widehat{BAP})$, the circumcircle $C(O,\rho )$ of $\triangle BRC$ and $S\in AR\cap BC$.  Thus, $\{\begin{array}{c}AC=AQ\\\ AP=AB\\\ \widehat{CAP}\equiv\widehat{QAB}\end{array}\|$ $\implies$ $\triangle ACP\sim\triangle AQB$ $\implies$

$\{\begin{array}{c}CP=QB\\\ \widehat{APC}\equiv\widehat{ABQ}\\\ \widehat{ACP}\equiv\widehat{AQB}\end{array}\|$ $\implies$ $\{\begin{array}{c}\widehat{APR}\equiv\widehat{ABR}\\\ \widehat{ACR}\equiv\widehat{AQR}\end{array}\|$ $\implies$ the quadrilaterals $APBR$, $AQCR$ are cyclically with the radical axis $AR$.

Observe that $\{\begin{array}{c}m(\widehat{BRS})=m(\widehat{BPA})=90^{\circ}-x\\\ m(\widehat{CRS})=m(\widehat{CQA})=90^{\circ}-x\\\ m(\widehat{BRP})=m(\widehat{BAP})=2x\end{array}\|$ $\implies$ the ray $[RS$ is the bisector of the angle $\widehat{BRC}$.

Prove easily that $\{\begin{array}{c}m(\widehat{BOC})=4x\\\ a=2\rho\sin 2x\end{array}$. Thus, $\{\begin{array}{c}PA=c\ ,\ QA=b\\\ PB=2c\cdot\sin x\\\ QC=2b\cdot \sin x\end{array}$ and $\{\begin{array}{c}m(\widehat{CBO})=m(\widehat{BCO})=90^{\circ}-2x\\\ m(\widehat{PBO})=180^{\circ}+3x-B\\\ m(\widehat{QCO})=180^{\circ}+3x-C\end{array}$.

Apply the generalized Pythagoras' theorem in the triangles $POB$ , $QOC$ to the sides $OP$, $OQ$ respectively :

$\{\begin{array}{c}OP^{2}=BP^{2}+\rho^{2}-2\rho\cdot BP\cos \widehat{PBO}\\\ OQ^{2}=CQ^{2}+\rho^{2}-2\rho\cdot CQ\cos \widehat{QCO}\end{array}$ $\implies$ $OP^{2}-OQ^{2}=4(c^{2}-b^{2})\sin^{2}x+4c\rho\sin x\cos (B-3x)-4b\rho \sin x\cos (C-3x)=$

$4(c^{2}-b^{2})\sin^{2}x+4c\rho \sin x(\cos B\cos 3x+\sin B\sin 3x)$ $-4b\rho \sin x(\cos C\cos 3x+\sin C\sin 3x)=$

$4(c^{2}-b^{2})\sin^{2}x+4\rho\sin x\cos 3x$ $(c\cos B-b\cos C)+4\rho \sin x\sin 3x(c\sin B-b\sin C)=$

$4(c^{2}-b^{2})\sin^{2}x+4\rho\sin x\cos 3x(\frac{a^{2}+c^{2}-b^{2}}{2a}-\frac{a^{2}+b^{2}-c^{2}}{2a})=$ $4(c^{2}-b^{2})\sin^{2}x+\frac{4\rho (c^{2}-b^{2})}{a}\sin x\cos 3x=$

$(c^{2}-b^{2})(4\sin^{2}x+\frac{2\sin x\cos 3x}{\sin 2x})=c^{2}-b^{2}$. I used the simple relations $c\sin B=b\sin C$ and $4\sin^{2}x+\frac{2\sin x\cos 3x}{\sin 2x}=1$.[/color]

This problem was in Brazilian Training Lists!
I solved this using Spiral Similarity, some Trigonometry and the lemma mentioned by campos!

I found a really nice proof:

[hide]Consider this sequence of three rotations:

(1) about A of angle $ \minus{}\theta$
(2) about O of angle $ 2\theta$
(3) about A of angle $ \minus{}\theta$

1. $ APBR$ is cyclic

We have the spiral similarity (congruence, actually)
$ \triangle PAB \sim \triangle CAQ$, so $ \triangle PAC \sim \triangle BAQ$, implying that $ \angle APC \equal{} \angle ABQ$, so $ \angle APR \equal{} \angle ABR$ so $ APBR$ is cyclic

2. $ \angle PRB \equal{} \theta$

follows from (1) after noting that $ \angle PRB \equal{} \angle PAB \equal{} \theta$

3. $ \angle BOC \equal{} 2\theta$

follows after angle chasing using (2): $ \angle BRC \equal{} 180 \minus{} \angle PRB \equal{} 180 \minus{} \theta$ and so $ \angle BOC \equal{} 2\theta$

4. rotation (2) takes B to C

follows from (3) and the fact that $ OB \equal{} OC$ (since O is the circumcenter of BRC)

5. rotation (1) takes P to B and rotation (3) takes C to Q

trivial

6. the sequence of rotations (1), (2), (3) take P to Q

trivial from (4) and (5)

7. the triple rotation is actually a translation by vector $ \overrightarrow{\text{PQ}}$

follows from properties of rotations (we have a net angle of 0, so its a translation)

Consider the image of A in this triple rotation/translation. Thinking of this as a translation, it goes to a point A' such that $ AA' \parallel{} PQ$ and $ AA' \equal{} PQ$ (actually, we only need the former).

Now we think of it as a triple rotation:
(1) The first rotation does nothing.
(2) The second rotation takes it to a point $ A_1$ such that $ \angle AOA_1 \equal{} 2\theta$ and $ AO \equal{} OA_1$. This implies that $ \angle OAA_1 \equal{} 90 \minus{} \theta$.
(3) The third rotation takes $ A_1$ to a point $ A'$ such that $ \angle A_1AA' \equal{} \theta$.

Combining the angles from (2) and (3), we get that $ \angle OAA' \equal{} 90$, so that $ OA \perp AA'$. Since $ AA' \parallel{} PQ$, $ OA \perp PQ$, as desired.[/hide]

Let the perpendicular bisectors of $AB$ and $AC$ intersect $AP$ and $AQ$ at $O_1$ and $O_2$, respectively. Since quadrilaterals $ABPO_1$ and $ACQO_2$ are similar, it follows that $AO_1/AP=AO_2/AQ$ and hence that $PQ$ and $O_1 O_2$ are parallel. Now let $\mathcal{O}_1$ and $\mathcal{O}_2$ be the circles centered at $O_1$ and $O_2$ with radii $O_1 A$ and $O_2 A$, respectively. Let the second intersection of $\mathcal{O}_1$ and $\mathcal{O}_2$ be $K$. By the inscribed angle theorem, it follows that $\angle{(KB,KA)}=\angle{(KA,KC)}=90^\circ - \angle{(AP, AB)}$ which implies that $\angle{CK, BK}=2\angle{(AP, AB)}$. Now note that since $A$ is the center of spiral similarity taking $PB$ to $CQ$, it follows that $APBR$ and $AQCR$ are cyclic which implies that $\angle{(CR, BR)}=\angle{(AP, AB)}$ and hence that $\angle{(CO, BO)}=2\angle{(AP, AB)}=\angle{(CK, BK)}$. Therefore $BOCK$ is cyclic and since $AK$ bisects $\angle{BKC}$ and $OB=OC$, it follows that $A$, $K$ and $O$ are collinear. Hence $AO$ is the radical axis of $\mathcal{O}_1$ and $\mathcal{O}_2$, which implies that $AO$ is perpendicular to $O_1 O_2$ and therefore perpendicular to $PQ$ since $PQ$ is parallel to $O_1 O_2$.

Hello, I have a very funny solution to this problem.

[hide="Solution"]
Let $AB$ and $AC$ hit the circumcircle of $\triangle BCR$ at $B', C'$, and let $AO$ intersect $BC', B'C$ at $X, Y$ and let $BC'$ interset $B'C$ at $Z$ and let $BC$ intersect $B'C'$ at $P$.

Clearly we have $\angle BB'C=\angle BC'C=\angle PAB=\angle QAC$.  Ok so we really just need to prove $\dfrac{\cos \angle OAP}{\cos \angle OAQ}=\dfrac{AQ}{AP}$; RHS is obviously $\dfrac{AC}{AB}$.

Now HMMM darn cosines wat?  well we have $\angle OAP=\angle XYZ$ and $\angle OAQ=\angle ZXY$...now wait Brokard says $PZ \perp XY$ - is this
really going to work - !

Wel we have $\LHS =\dfrac{\cos \angle XYZ}{\cos \angle YXZ}=\dfrac{\sin \angle PZC}{\sin \angle PZC'}=\dfrac{PC}{PC'}=\dfrac{AC}{AB}$ - it works!

QED
[/hide]

Yes ! But the formulation of the problem screams for a complex bash

Dear Mathlinkers,
you can also see :
http://perso.orange.fr/jl.ayme  vol. 16 Deux triangles adjacents... p. 62 which indicate p. 60-61 .

Sincerely
Jean-Louis

My solution:

Let $ O' $ be the circumcenter of $ \triangle ACQ $ .
Let $ M, N $ be the midpoint of $ CQ, CR $, respectively .

Easy to see $ R \in (O') $ .

Since $ O', M, N, C $ are concyclic ,
so we get $ \angle AO'O=\angle QCP  $ . ... $ (1) $
Since $ \angle RO'O=\angle BQC, \angle O'OR=\angle CBQ $ ,
so we get $ \triangle ORO' \sim \triangle BCQ $ ,

hence $ \frac{O'A}{CQ}=\frac{O'R}{CQ}=\frac{O'O}{QB}=\frac{O'O}{CP} $ . ... $ (2) $

From $ (1) $ and $ (2) $ we get $ \triangle AOO' \sim \triangle QPC $ ,
so from $ OO' \perp PC $ and $ AO' \perp QC \Longrightarrow AO \perp QP $ .

Q.E.D

My solution:

[u][b]Lemma:[/b][/u] $AO \perp PQ$ if only if $AP^2-AQ^2=OP^2-OQ^2$ (It's well known)
 It is easy too see that: $\triangle APC\cong \triangle ABQ$  $\Longrightarrow $ $BQ=CP=1$ and $RP=d,RQ=c$
Let $AP=AB=a,CA=CQ=b$ and $AR=x$ $\Longrightarrow $ By power of point in $\odot (RBC)$ we get :
$AP^2-AQ^2=OP^2-OQ^2$ if only if  $a^2-b^2=d-c$ since $OP^2-OQ^2=d-c$ (By power of point in $\odot (RBC)$
By Stewart's theorem in $\triangle ACP$ we get: $a^2(1-d)+b^2d=x^2+d(1-d)...(1)$
By Stewart's theorem in $\triangle ABQ$ we get: $b^2(1-c)+a^2c=x^2+c(1-c)...(2)$
By $(1)-(2)$ we get:$a^2-b^2=d-c$ which it is true

$\Longrightarrow $ $a^2-b^2=d-c$ $\Longrightarrow $ $AO \perp PQ$... :P

Needing help for a complex number solution, I'm trying to learn that technique at the moment and working through an article by Robin Park. And well, there is a step in the calculation that I can't seem to be able to comprehend.

Noticing those rotations, we can find $p = a+e^{-i \theta}(b-a)$, $q=a+e^{i \theta}(c-a)$ and $o= \tfrac{b-ce^{2i \theta}}{1-e^{2i \theta}}$, where $\theta = \angle{BAP}$. It remains to verify $\tfrac{a-o}{\overline{a}-\overline{o}}=\tfrac{p-q}{\overline{p}-\overline{q}}$. To ease the amount of calculation we also set $A$ to be the origin. Also, we let $z = e^{i \theta}$.Then we get \[ \frac{o}{\overline{o}} = \frac{p-q}{\overline{p}-\overline{q}} \iff \frac{\frac{b-cz^2}{1-z^2}}{\frac{\frac{1}{b}-\frac{1}{cz^2}}{1-\frac{1}{z^2}}} = \frac{\frac{b}{z}-cz}{\frac{z}{b}-\frac{1}{cz}} \iff \dots \]
That above is how the author of the article presented the solution; I didn't formulate it like him.
Now to my problem: how did he get to that equivalent form? Especially, I assume he got $\overline{p}-\overline{q}=\tfrac{z}{b}-\tfrac{1}{cz}$. Why is that true?

Also, admittingly, I'm still very weak at manipulating complex numbers as I rarely use those.

Here's a solution similar to Robin Park's. Whoops, I'm bad at this complex bashing stuff.

We use complex numbers.

Let the complex number corresponding to a point be its lowercase letter. Set $a$ as the origin, $c=1$, and $\angle PAB = \angle QAC = \theta$. By standard formulas, we have $p=e^{-i \theta}b$ and $q=e^{i \theta}c$.

Also, I claim $APBR$ and $AQCR$ are both cyclic. Observe $\triangle PAB \sim \triangle CAQ$, with spiral center at $A$, so the standard Yufei spiral similarity configuration tells us $R \in \odot(ABP)$ and $R \in \odot(ACQ)$. Now angle chase to get $\angle BOC = 2 \theta$.

By the rotation formula, you can get $o = \frac{b-ce^{i \theta}}{1-e^{i 2 \theta}}$. Now we want $\frac{o-a}{p-q}$ as pure imaginary, that is
\[ \frac{(b-e^{i 2\theta})}{(1-e^{i 2\theta})(e^{-i \theta}b-e^{i \theta})} = \frac{-(\overline{b}-\frac{1}{e^{i 2\theta}})}{(1-\frac{1}{e^{i 2\theta}})(e^{i \theta} \overline{b}-\frac{1}{e^{i \theta}})} \]
After some simplifying, one can see that this is clearly true, so $OA \perp PQ$ and we are done.

Finally got around to actually doing this.  Looks like my solution bears resemblance to [b]Huyền Vũ[/b]'s in post 5, though the last part is a bit strange because I didn't know how to construct the argument elegantly.

Construct point $T$ outside $\triangle ABC$ such that $\triangle TBC\sim\triangle ABP\sim\triangle ACQ$.  By considering the rotation sending $\triangle APC$ to $\triangle ABQ$ (which are congruent by SAS) we see that the angle between lines $PC$ and $QB$ is equal to $\angle BAP = \angle BTC$.  Thus $B$, $C$, $T$, and $R$ lie on a common circle, so $O$ is the circumcenter of $\triangle TBC$.  We will subsequently dispose of point $R$.

Denote by $D_1$ and $D_2$ the projections of $B$ onto $AP$ and $C$ onto $AQ$ respectively, and let $M$ be the midpoint of $\overline{BC}$. I claim that $D_1M = D_2M$.  To prove this, remark that by simple angle chasing $\angle BOC = \angle ABD_1$, so $\triangle AD_1B$ is directly similar to $\triangle CMB$.  By the duality of spiral similarity, $\triangle BD_1M\sim\triangle BAO$ with similarity ratio $r = \tfrac{BM}{BO}$.  Analogously, $\triangle CD_2M\sim\triangle CAO$ with the same similarity ratio $r = \tfrac{CM}{CO}$.  Since $AO$ is taken to $D_1M$ and $D_2M$ under these spiral similarities, we must have $D_1M = D_2M$.

Now let $\psi$ denote the measure of angle $\angle ABD_1 = \angle ACD_2$, and remark that \[\angle D_1MD_2 = 180^\circ - \angle BMD_1 - \angle CMD_2 = 180^\circ - \angle BOC = 2\psi.\]  It follows that, upon letting the perpendicular from $M$ to $D_1D_2$ intersect $AC$ at $X$, we have $\angle XMD_1 = \psi$, so quadrilateral $CD_1XM$ is cyclic.  Thus the angle between $MX$ and $AC$ is equal to $\angle MD_1C = \angle OAC$, which implies $AO\perp D_1D_2$.  But $D_1D_2\parallel PQ$ by similarity, and so we obtain the desired perpendicularity. $\blacksquare$

Here is a synthetic solution constructing only one point. I would appreciate if someone simplifies my final directed angle chasing.

Let $\theta = \angle BAP$. By Spiral Similarity, $ \triangle APC\stackrel{+}{\sim} \triangle ABQ$ so $\angle APR = \angle ABR\implies \angle BRP = \theta$. Thus $\angle BOC = 2\theta$. Now for the key step, construct point $X$ outside $ \triangle BOC$ such that $\triangle OXB\sim \triangle ABC$. Then $\angle XOC = \angle BAC + 2\theta = \angle APQ$ thus,
$$\frac{OX}{OC} = \frac{OX}{OB} = \frac{AB}{AC} = \frac{AP}{AQ}\implies \triangle OXC\stackrel{-}{\sim} \triangle APQ.$$
Furthermore, $\angle XBC = \angle C + (90^{\circ} - \theta) = \angle OCA$ so
$$\frac{BC}{BX} = \frac{AC}{OB} = \frac{AC}{OC}\implies \triangle BCX\stackrel{-}{\sim} \triangle CAO.$$
Hence, using directed angle, we find
\begin{align*}
\measuredangle(AO, PQ) &= \measuredangle(AO, AC) + \measuredangle(AC, AQ) + \measuredangle(AQ, PQ)\\
&=-\measuredangle(CX, BC) + \measuredangle(AC, AQ) - \measuredangle(OC, CX) \\
&= \measuredangle(AC, AQ) + \measuredangle(BC, OC) \\
&= 90^{\circ}
\end{align*}
as desired.

Here is a quick and easy solution. Clearly $ABPR$ and $ACQR$ are inscribed; next, meet $AP$ and $(ACQR)$ at $X$, and $AQ$ and $(ABPR)$ at $Y$. Since $\measuredangle AXQ = \measuredangle ARQ = \measuredangle ARB = \measuredangle AYP$, $PQXY$ is inscribed. So, if $\rho$ is the radius of $(BCR)$, \[
PO^2-PA^2 = PR\cdot PC + \rho^2 - PA^2 = PA\cdot PX +\rho^2 - PA^2 = PA\cdot AX + \rho^2 = \mathcal P(A,(PQXY)),
\]where $\mathcal P$ denotes power. Since the last expression is the same for $P,Q$, this means $PO^2-PA^2 = QO^2-QA^2$, done!

Denote $\omega_b \equiv \odot(ABP)$ and $\omega_c \equiv \odot(ACQ).$ Let $AP$ meet $\omega_c$ for a second time at $X.$ Let $AQ$ meet $\omega_b$ for a second time at $Y.$

Because $A$ is the center of spiral similarity sending $\overline{PB} \mapsto \overline{CQ}$, we know that $R$ is the second intersection point of $\omega_b$ and $\omega_c.$ By Reim's theorem for $\omega_b, \omega_c$ cut by $PX, BQ,$ it follows that $PB \parallel XQ.$ Because $\triangle APB$ is isoceles, $PB$ is parallel to the tangent to $\omega_b$ at $A.$ Therefore, by the converse of Reim's theorem for $\omega_b$ cut by $PX, YQ,$ it follows that $PQXY$ is cyclic. Thus, there exists $r \in \mathbb{C}$ such that \[ r^2 = \text{pow}(A, \odot(PQXY)) = AP \cdot AX = AQ \cdot AY. \]
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dot((1007.5572715124796,1032.9105790739395),linewidth(3.pt) + dotstyle);
label("$X$", (1007.0982038590661,1033.2893737659476), N * labelscalefactor);
dot((997.599658737153,1029.7899089804253),linewidth(3.pt) + dotstyle);
label("$Y$", (997.316957584563,1030.3814356843386), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
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[/asy]

Define a new circle $\gamma \equiv \odot(A, r).$ Because \[ PR \cdot PC = PA \cdot PX = PA(PA + AX) = PA^2 - r^2, \] we see that $P$ lies on the radical axis of $\odot(BCR)$ and $\gamma.$ Similarly, $Q$ lies on the radical axis of $\odot(BCR)$ and $\gamma.$ Therefore $PQ \perp AO.$

Skimming through (not very carefully), I haven't seen this solution yet.

Note that $A$ is the spiral similarity that sends $PB$ to $CQ$, so it rotates $PC$ to $BQ$ implying $PC=BQ$. Also, $APBR, AQCR$ are cyclic.

Let $N$ be the intersection of $(PRQ), (BRC)$. Then $N$ sends $BQ$ to $CP$, but $BQ=CP$, so $N$ is the midpoints of arcs $PQ$ and $BC$ in the circles.

Note that $\measuredangle BRA = \measuredangle BPA = \measuredangle AQC = \measuredangle ARC$ so $AR$ bisects $BRC$ and $AR$, $NO$ concur on $(BRC)$. Therefore, $\measuredangle BPA = \measuredangle BRA = \measuredangle BNO$, so (since $AP=AB, ON=OB$), by (a working SSA similarity), $\triangle BON\sim \triangle BAP$.

Therefore, a spiral similarity sends $NO$ to $PA$, and $AO$ to $PN$. Thus, the angle between $AO$ and $PQ$ is \[\measuredangle (AO, PN)+\measuredangle (PN, PQ)=\measuredangle ABP+\measuredangle NPQ = 90^{\circ}\]

Oh jk this is the second solution in post 3

Much shorter solution.

Let $\theta = \angle BAP$. Note that $\triangle ABQ\stackrel{+}{\sim}\triangle ACP$ thus $\angle(BQ, CP) = \theta)$. Therefore $\angle BRC = 2\theta$. Let $C'$ be the reflection of $C$ across $AQ$.

Note that $\triangle ABP\stackrel{+}{\sim}\triangle AC'Q$ thus $\triangle ABC'\stackrel{+}{\sim}\triangle APQ$ which means $\angle(PQ,BC') = \theta\hdots (\spadesuit)$. Moreover, $\triangle CAC'\stackrel{+}{\sim}\triangle COB$ thus $\triangle CAO\stackrel{+}{\sim}\triangle CC'B$. Thus $\angle(BC', AO) = \angle(CA,CC') = 90^{\circ}-\theta$. Combining this with $(\spadesuit)$ gives the conclusion.

its easiest p6 i seen ! solution: its obvious $APBR , BRCQ$ are cyclic let $T , S$ be intersection of this circles with lines $AP$ and $AQ$ respectively.
its obvious $PQST$ is cyclic quadrilateral and with this observations according power of point theorem we have: $QR.QB-PR.PC=QO^2-PO^2=AQ^2-AP^2$
problem solved.

We have $\angle BOC = 2(\pi - \angle BRC) = 2\angle BAP$. Let $(ABC)$ be the unit circle, and let $q = a + (c-a)z$ and $p = a + \frac{b-a}{z}$ for some complex number $z$ on the unit circle. The point $O$ is given by
\[\frac{b-o}{c-o} = z^2 \implies o = \frac{b-z^2c}{1-z^2}.\]
Then
\[\frac{a-o}{p-q} = \frac{a - \frac{b-z^2c}{1-z^2}}{\frac{b-a}{z} - z(c-a)} = \frac{-z}{1-z^2}\]
which is purely imaginary.

[hide=Diagram]
[img]https://lh3.googleusercontent.com/-DqvQ8QGvY7E/X6j-48YxtlI/AAAAAAAAG_g/qHhq5dEGOWUqf_GPG7TdcAmfEeD46Y82wCLcBGAsYHQ/s0/tst2006_6.PNG[/img]
[/hide]

Let $\theta:=\angle PAB=\angle CAQ$. Note that $A$ is the center of spiral similarity $PB\mapsto CQ$, so it also maps $PC\mapsto BQ$. Thus $\angle (\overline{PC},\overline{BQ})=\theta$, in particular $\angle BRC =\pi-\theta$. Hence $\angle BOC = 2(\pi-\angle BRC) = 2\theta$. This eliminates $R$ entirely! Indeed, $O$ is now located on the perpendicular bisector of $\overline{BC}$ with $\angle BOC=2\theta$.

Now use complex numbers with $A=0$. Let $t=e^{i\theta}$. Then
\begin{align*}
    p=b/t,\quad q=tc.
\end{align*}
We know $\angle BOC=2\theta$ and $BO=CO$. Hence $|\tfrac{b-o}{c-o}| = 1 = |t^2|$, so actually
\begin{align*}
    \frac{b-o}{c-o}\div t^2 = 1 \implies o=\frac{t^2c-b}{t^2-1}.
\end{align*}
Finally, confirm
\begin{align*}
    \frac{o-a}{p-q} = \frac{t^2c-b}{(t^2-1)(b/t-tc)} = \frac{t}{1-t^2} \in i\mathbb{R}.
\end{align*}
[hide=Remark]
One thing that had me confused for a bit was that when you locate $O$ as the point on the perp. bisector of $\overline{BC}$ with $\angle BOC=2\theta$, there should be \textit{two} points which satisfy this, but the equation
\[ \frac{b-o}{c-o} = t^2\]
gives one. The reason is that the second value of $O$ is the reflection over $\overline{BC}$, and then the RHS above would be $1/t^2$, not $t^2$, even though both are ``$\angle BOC$''.
[/hide]

See my solution to this problem on my Youtube channel here:
[url]https://www.youtube.com/watch?v=Hz8kyLKP0_k[/url]

Here is a barycentric coordinate solution.

Observe that $A$ is the miquel point of $PBQC$. This means that we must have $(APBR)$ and $(QARC)$ are cyclic. Since
\[\angle ARP = \angle ABP = \angle ACQ = \angle ARQ\]
this means $A$ lies on the angle bisector of $\angle BRC$.

We now proceed with barycentric coordinates. Let our reference triangle be $\triangle BRC$, so $R = (1:0:0), B = (0:1:0), C = (0:0:1)$. For convenience, let $BC = a, CR = b, BR = c$. Since $A$ lies on the angle bisector, let $A = (t:b:c)$, for some number $t$. Now, the equation for $(ARB)$ is
\[-a^{2}yz - b^{2}xz - c^{2}xy + (wz)(x+y+z)\]
because $(ARB)$ passes through $B,R$. Plugging in $A$ gives
\[-a^{2}bc - b^{2}tc - c^{2}tb + wc(t+b+c) = 0\Rightarrow w = \frac{a^{2}b + b^{2}t + cbt}{t+b+c}\]
Now, since $P = (ARB)\cap RC$, if $P = (p:0:1-p)$, then we can plug this into our equation for $(ARB)$ to get
\[-b^{2}p(1-p) + \frac{a^{2}b + b^{2}t + cbt}{t+b+c}\cdot (1-p) = 0 \Rightarrow p = \frac{a^{2} + bt + ct}{bt + b^{2} + bc}\]
Therefore, by symmetry,
\[P = \left(\frac{a^{2} + bt + ct}{b^{2}+bt + bc} : 0 : \frac{b^{2} + bc - ct - a^{2}}{bt + b^{2} + bc}\right), Q = \left(\frac{a^{2} + bt + ct}{c^{2} + ct + bc} : \frac{c^{2} + bc -bt - a^{2}}{c^{2} + ct + bc} : 0 \right)\]
To show that $AO\perp PQ$, we can use generalized perpendicularity (Theorem 7.25 of EGMO). Assume that $O$ was the origin. We have
\[\overrightarrow{PQ} = (a^{2} + bt + ct)\cdot \frac{c-b}{bc(b+c+t)}\vec{R} + \frac{bt + a^{2} - bc - c^{2}}{c(b+c+t)}\vec{B} + \frac{b^{2} + bc - ct - a^{2}}{c(b+c+t)}\vec{C}\]
Since $O$ is the origin, we have $A = \frac{t}{t+b+c}\vec{R} + \frac{b}{b+c+t}\vec{B} + \frac{c}{b+c+t}\vec{C}$. By Theorem 7.25 of EGMO, we must show the following expression is $0$:
\[0 = a^{2}\left(\frac{bt+a^{2}-bc-c^{2}}{b+c+t} + \frac{b^{2} + bc -ct - a^{2}}{b+t+c}\right)+ b^{2}\left(\frac{c-b}{b(b+c+t)}(a^{2} + bt + ct) + \frac{t(b^{2} + bc - ct - a^{2})}{b(b+c+t)}\right)+ c^{2}\left(\frac{c-b}{c(b+c+t)}(a^{2} + bt + ct) + \frac{t(bt + a^{2}-bc-c^{2})}{c(b+c+t)}\right)\]
\[\Rightarrow 0 = a^{2}(b^{2}-c^{2} + bt - ct) + b((c-b)(a^{2} + bt + ct) + b^{2}t + bct - ct^{2} - a^{2}t)+ c((c-b)(a^{2}+bt+ct) + bt^{2} + bct - ct^{2} - at^{2})\]
\[\Rightarrow 0 = a^{2}(b^{2} - c^{2} + bt - ct) + b[a^{2}c + bct + c^{2}t - a^{2}b - ct^{2} - a^{2}t] + c[a^{2}c - a^{2}b - b^{2}t + bt^{2} + a^{2}t - bct]\]
\[= a^{2}b^{2} - a^{2}c^{2} + a^{2}bt - a^{2}ct + a^{2}bc + b^{2}ct + bc^{2}t - a^{2}b^{2} - bct^{2}-a^{2}tb + a^{2}c^{2} - a^{2}bc - b^{2}ct + bct^{2} + a^{2}ct - bc^{2}t = 0\]
Therefore, this expression is $0$, which means $AO\perp PQ$.

[quote=k12byda5h]Triple Jacobi's theorem :-D
Let $S$ be a point outside $\triangle ABC$ and $\angle SBC = \angle ABP=\angle SCB$. By Jacobi, $AS,BQ,CP$ are concurrent at $R$ and $S$ lies on $\odot(BCR)$. Hence, $\angle OBC = \angle OCB=90^{\circ}-\angle BAP$. Let $B',C'$ be the foot of perpendicular from $B,C$ to $AP,AQ$ respectively.  By Jacobi, $AO,BC',CB'$ are concurrent.Then Jacobi $\triangle AB'C'$, $AO \perp B'C'$ but $B'C' \parallel PQ$ ($\frac{AB'}{AP} = \frac{AC'}{AQ} \implies \triangle AB'C' \sim \triangle APQ$). Therefore, $AO \bot PQ$.[/quote]

"Then Jacobi $\triangle AB'C'$, $AO \perp B'C'$" Why'd have this perpendicularity?

[quote]"Then Jacobi $\triangle AB'C'$, $AO \perp B'C'$" Why'd have this perpendicularity?[/quote]

Let $BC' \cap CB' = X$.By Jacobi, $AO,BC',CB'$ are concurrent $\implies A,O,X$ are collinear. Then Jacobi $\triangle AB'C'$ with point $\infty_{\perp B'C'},C,B$. We get $AOX \perp B'C'$

Let $R'=\overline{PB}\cap\overline{QC}$. By spiral sim $A,B,P,R$ and $A,C,R,Q$ are cyclic. Under an inversion at $A$ with radius $\sqrt{AP\cdot AQ}$ and a reflection about the angle bisector of $\angle PAQ$, $R$ goes to $R'$. Since $\angle APR'=\angle AQR'$, if we let $R''$ be the point such that $R'PR''Q$ is a parallelogram, $\overline{AR''},\overline{AR'}$ are isogonal in $\angle PAQ$, so $R-A-R''$. Let $O_1$ be the center of $(APR)$ and let $O_2$ be the center of $(AQR)$. Then the perpendicular from $P$ to $\overline{AO_2}$, the perpendicular from $Q$ to $\overline{AO_1}$, and the perpendicular from $R$ to $\overline{O_1O_2}$ meet at $R''$. Thus $\triangle AO_1O_2,\triangle RPQ$ are orthologic, as desired.

A very, very funny solution.

Let $\angle BAP = \angle CAQ = \theta$. Since $\triangle APB\sim \triangle ACQ$, we have $\triangle APC\sim \triangle ABQ$, so $\angle (PC, BQ) = \theta$ (the spiral similarity at $A$ sending $APC$ to $ABQ$ has a rotation of angle $\theta$). Thus, $\angle BRC = 180^\circ - \theta$, so $\angle BOC = 2\theta$.

It suffices to show that $\overrightarrow{AO}\cdot \overrightarrow{PQ} = 0$. Note that $\angle(PA, OC) = \measuredangle (PA, AB) + \measuredangle (AB, BC) + \measuredangle (BC, CO) = \theta - \angle B +90^\circ - \theta = 90^\circ - \angle B$. Similarly, $\angle(QA, OB) = 90^\circ - \angle C$.

Now, we compute
\begin{align*}
\overrightarrow{AO}\cdot \overrightarrow{AP} &= \left(\overrightarrow{AC} + \overrightarrow{CO}\right)\cdot \overrightarrow{AP}\\
& = AB\cdot AC \cdot \cos (\angle A + \theta) + AB\cdot CO \cdot \cos(90^\circ - \angle B)\\
& = AB\cdot AC \cdot \cos(\angle A + \theta) + AB\sin \angle B\cdot CO\cdot \\
\end{align*}
which is clearly symmetric. Thus, $\overrightarrow{AO}\cdot \overrightarrow{AP} = \overrightarrow{AO}\cdot \overrightarrow{AQ}$, and we are done.

Note that $\triangle AQC$ and $\triangle ABP$ are rotations about $A$. In particular, $APBR$ and $AQCR$ are cyclic. It follows that $\angle ARP = \angle ABP = \angle ACQ = \angle ARQ$. Call this common angle $\theta$. Now let $\triangle XYZ$ be congruent to $\triangle AQC$ and $\triangle ABP$, and let $K$ and $L$ be on $YZ$ so that $\angle XKY = \angle XLZ = \theta$. Let $M$ be the foot from $X$ to $YZ$.  For points $T$ in the plane, define $f(T) = \text{pow}_{(BRC)}(T) - TA^2$. We have

$$f(P) - f(Q) = (PR\cdot PC - PA^2) - (QR\cdot QB - QA^2) = $$

$$= (YK\cdot YZ - YX^2) - (ZL\cdot ZY - ZX^2)$$

$$= YZ(YK - ZL) - YX^2 + ZX^2$$

$$=YZ\cdot YM - ZY\cdot ZM - YX^2 + ZX^2$$

$$=\text{pow}_{(XZ)}(Y) - \text{pow}_{(XY)}(Z) - YX^2 + ZX^2$$

If $D$ and $E$ are the midpoints of $XY$ and $XZ$, this is

$$(YE^2 - EX^2) - (ZD^2 - DX^2) - YX^2 + ZX^2$$

$$=(\frac{1}{2}YX^2 + \frac{1}{2}YZ^2 - \frac{1}{4}XZ^2 - \frac{1}{4}XZ^2) - (\frac{1}{2}ZX^2 + \frac{1}{2}ZY^2 - \frac{1}{4}XY^2 - \frac{1}{4}XY^2) - YX^2  + ZX^2$$

$$=0$$

so $PQ$ is parallel to the radical axis of $(BRC)$ and the point circle at $A$, which is perpendicular to $AO$, as desired.

Clearly $\triangle APB \sim \triangle ACQ$, so by spiral similarity facts $ABPR$ and $ACQR$ are cyclic. Obviously $\overline{AR}$ bisects $\angle BRC$ as well. Thus we may view the problem with reference triangle $RBC$, restating it as follows.
[quote=Restated]
Let $ABC$ be a triangle with circumcenter $O$ and $D$ be a point on its internal $\angle A$-bisector. Let $(ABD)$ and $(ACD)$ intersect $\overline{AC}$ and $\overline{AB}$ at $E$ and $F$ respectively. Then $\overline{OD} \perp \overline{EF}$.
[/quote]
Let $D=(t:b:c)$. If the equation of $(ABD)$ is $-a^2yz-b^2zx-c^2xy+(ux+vy+wz)(x+y+z)=0$, then plugging in $A$ and $B$ implies $u=v=0$, and then plugging in $D$ implies $w=\tfrac{a^2b+b^2t+bct}{b+c+t}$. To compute $E$, we solve for $r_1$ given that $(r_1,0,1-r_1)$ lies on the circle: this will yield (after dividing out $1-r_1$, which would correspond to $A$ itself) $r_1=\tfrac{a^2+bt+ct}{b(b+c+t)}$. Likewise, if $F=(r_2,1-r_2,0)$ then $r_2=\tfrac{a^2+bt+ct}{c(b+c+t)}$.

Now, by "strong EFFT" we just have to prove that
\begin{align*}
0&=a^2\left(\frac{a^2+bt+ct}{b+c+t}-c+b-\frac{a^2+bt+ct}{b+c+t}\right)\\
&+b^2\left(\left(\frac{c}{b}-1\right)\left(\frac{a^2+bt+ct}{b+c+t}\right)+t-\frac{t}{b}\left(\frac{a^2+bt+ct}{b+c+t}\right)\right)\\
&+c^2\left(\left(1-\frac{b}{c}\right)\left(\frac{a^2+bt+ct}{b+c+t}\right)+\frac{t}{c}\left(\frac{a^2+bt+ct}{b+c+t}\right)-t\right)\\
&=a^2(b-c)(b+c+t)+b(c-b-t)(a^2+bt+ct)+b^2t(b+c+t)+c(c-b+t)(a^2+bt+ct)-c^2t(b+c+t)\\
&=(b-c)(a^2+bt+ct)(b+c+t)-(b+c)(b-c)(a^2+bt+ct)-t(b-c)(a^2+bt+ct))\\
&=(b-c)(a^2+bt+ct)(b+c+t-b-c-t)\\
&=0,
\end{align*}
so we're done. $\blacksquare$

Started around 8:25 CST (since Hazard is pro god at math I need to compete with him)

Let $U\in PC$ satisfy $UP=UB$, define $V$ similarly. Consider ellipse $\Gamma$ with foci at $B,C$ through $U,V$ such that $AU$ and $AV$ are tangents (not sure if it'll come in handy later).

Also note that there is a circle $\Omega$ centered at $A$ which is tangent to $BU,CV,RU,RV$. This entire ellipse configuration comes from djmathman's handout, yay!

Hence showing $PQ\perp AO$ is basically showing $PQ$ is parallel to some radical axis. Let $K$ be the foot of the altitude from $A$ to $PC$, and define $L$ similarly.

Notice that it suffices to show
\[PK^2-PR\cdot PC=QL^2-QR\cdot QB.\]
We're basically almost done I think. Notice that this equates to
\[(PK+QL)(PK-QL)=PR\cdot PC-QR\cdot QB\implies PC(PK-QL)=PR\cdot PC-QR\cdot PC\implies PK-QL=PR-QR\]
or just
\[PK-PR=QL-QR\implies RK=RL\]
which is obviously true. Ta-da! 22 minutes! WOOOOOOOOOOOOOO

Solved with hints

First, notice that since the angle of rotation is the same, there is a spiral similarity centered at $A$ taking $BP$ to $QC$. This means that $A$ is the Miquel point of $BPCQ$, so $ARBP$ is cyclic. This means that $$\angle BOC=2\angle BRC=2\angle BAP.$$ Thus, if we let $M$ denote the arc midpoint of $BC$ not containing $R$, we have that $\angle BOM=\angle BAP$. Since they are both isosceles, we have $\triangle BOM\sim\triangle BAP$ as well.\\

Use complex numbers with unit circle $(BRC)$. Let $B=b^2,C=c^2,R=r^2$. Furthermore, we let $A=a$ as another free variable as well (the problem has four degrees of freedom, so this decision is ok).\\

We can use $\triangle BOM\sim\triangle BAP$ to compute $p$. We have $$\frac{a-p}{a-b^2}=\frac{0-(-bc)}{0-b^2},$$ which we can solve for $p$ to get $$p=\frac{ab+ac-b^2c}{b}.$$ Similarly, $$q=\frac{ab+ac-bc^2}{c}.$$ Thus, we have $$p-q=\frac{abc+ac^2-b^2c^2-ab^2-abc+b^2c^2}{bc}=\frac{ac^2-ab^2}{bc}.$$ Finally, dividing this by $a$ gives $\frac{c^2-b^2}{bc}$, which is clearly imaginary after conjugating and multiplying top and bottom by $b^2c^2$, so we are done.

[color=#f0f][b]Remark:[/b][/color]

There are many things to take away from this problem. First of all, two rotations of the same angle around the same point should motivate using a spiral similarity, and spiral similiarity often works well with complex numbers.

Next, there is a very "tempting" setup, in fact the one I originally used and failed with, where we set $(ABC)$ as the unit circle as usual, then note that there is an "angle of rotation" of some $\theta$ to get $P$ and $Q$, and setting $a$, $b$, $c$, and $d=e^{i\theta}$ as the four "free" variables to represent the four degrees of freedom. While this quickly gives $p$ and $q$, the computation for $r$ is now very messy as it requires the general intersection formula, much to the detriment of this setup, as $R$ is required to compute the circumcenter.

The motivation for setting $(BRC)$ as the unit circle is that, rather than having to go through the tedious process of computing the circumcenter, this gives us the circumcenter "for free". Then, we can make $a$ our remaining free variable.

Next, the introduction of the arc midpoint allows us to turn the condition $\angle BOC=2\angle BAP$ into a similarity condition, which is actually a much stronger condition than just the angle equivalence, as it effectively encodes two degrees of freedom rather than just one. If we were to use $\angle BOC=2\angle BAP$ directly, we would actually need two equations to compute $P$ (the other one probably being $PRC$ collinear), which will be much less elegant.

Finally, this problem illustrates another important point of computing points "indirectly". In this solution, rather than plugging into some formula to compute $p$ and $q$, we used the fact that $\triangle BAP$ is similar to $\triangle BOM$ to indirectly compute $P$.

We will employ complex numbers, setting $A$ as the origin, $P$ at $(0, 1)$, $B$ at $|b| = 1$, $C$ at $c$, and $Q$ at $bc$.

Obviously, $APRB$ and $AQCR$ are cyclic, say by considering the pair of congruent triangles. Letting $\angle PAB = \theta$, it follows that $\angle BOC = 2\theta$. In other words, we have $\frac{o-b}{o-c} = b^2$, so $o = \frac{b(bc-1)}{b^2-1}$, and thus $$\frac o{p-q} = \frac b{b^2-1} \in i\mathbb R.$$ Hence $\overline{AO} \perp \overline{PQ}$ as required.


easy one: find all complex polynomials $P$ such that $P''$ divides $P$

i think this an X or Centrale Oral
am i wrong?


Is $Q^{4}=\langle \begin{bmatrix}1\\1\\1\\1 \end{bmatrix}, \begin{bmatrix}1\\1\\0\\1 \end{bmatrix}, \begin{bmatrix}0\\1\\0\\1 \end{bmatrix}\rangle ?$

A fourdimensional space can never be spanned by a set consisting of less than four vectors.


induction
number theory unsolved
number theory

Solve in positive integers n.

$n | 2^n + 1$

There are infinitely many such $n$. Did you find them all?

BTW, why do you use "South Korea" flag?

I doubt they can all be found, but who knows?

Anyway, we can construct an infinite sequence like this: take $a_1=3$, then take $a_{n+1}=2^{a_n}+1$. We easily get $a_n|2^{a_n}+1,\ \forall n\in\mathbb N^*$.

You correctly understood my idea, grobber!  ;)  :D

I had changed it to USA flag a few weeks ago, but with the server crash it reverted.

I also noticed that any power of 3 suffices. Any proof?

Heh. What I meant was a proof covering all solutions.


Putnam
inequalities
geometric series
college contests

Let $ A$ be a positive real number. What are the possible values of $ \displaystyle\sum_{j\equal{}0}^\infty{x_j^2}$, given that $ x_0,x_1,\ldots$ are positive numbers for which $ \displaystyle\sum_{j\equal{}0}^\infty{x_j}\equal{}A$?

[hide]My instinct tells me the range is the open interval $ (0,A^2)$. To approach the upper edge just take $ x_0 \equal{} (A \minus{} \epsilon)$ and $ x_1,x_2,\ldots$ as a convergent series adding up to $ \epsilon$. To approach the bottom edge, take any convergent series summing to $ A$, and using the fact that for positive $ a,b$, $ (a \plus{} b)^2 > a^2 \plus{} b^2$, just split terms as many times as you need to get as low as you want.

I'm pretty sure I could come up with a rigorous proof that both $ 0$ and $ A^2$ are achievable (as limits). But I'm not sure how to show that everything in between can also be hit.[/hide]

[hide="My writeup for my class that year."]The set of possible values of the sum  $ \sum_{j\equal{}0}^{\infty}x_j^2$ is the open interval $ (0, A^2).$

We have two things to prove: first, that the values are confined to this interval, and second, that all numbers in this interval are possible. As to the first: the values are obviously positive, so we must show that the sum of the squares is strictly less than $ A^2.$ We note that $ \sum_{j\equal{}0}^{\infty}\left(\frac{x_j}A\right)   \equal{} 1.$ Since each $ x_j > 0,$ we must have that $ 0 <  \frac{x_j}A   < 1.$ Then,   $ \left(\frac{x_j}A\right)^2   <  \frac{x_j}A,$ with the inequality being strict. Hence,  $ \sum_{j\equal{}0}^{\infty}\left(\frac{x_j}A\right)^2   <  \sum_{j\equal{}0}^{\infty}\left(\frac{x_j}A\right)   \equal{} 1.$ Multiply through by $ A^2$ to get that $ \sum_{j\equal{}0}^{\infty}x_j^2  <  A^2.$

As for the second part, let $ 0 < r < 1$ and let $ x_j \equal{} A(1\minus{} r)r^j.$ Then  $ \sum_{j\equal{}0}^{\infty}x_j  \equal{}  A$ as a geometric series and  $ \sum_{j\equal{}0}^{\infty} x_j^2 \equal{} \sum_{j\equal{}0}^{\infty}A^2(1\minus{} r)^2r^{2j}  \equal{}  \frac{A^2(1\minus{} r)^2}{1\minus{} r^2}   \equal{}  \frac{A^2(1\minus{}r)}{1 \plus{} r}  .$ Clearly, we can make  $ \frac{1\minus{} r}{1 \plus{} r}$   take on any value in $ (0, 1)$ by choosing an appropriate value of $ r$ in $ (0, 1).$ (In particular,  $ x  \equal{}  \frac{1 \minus{} r}{1 \plus{} r}$   if $ r \equal{}  \frac{1\minus{} x}{1 \plus{} x}$  .)[/hide]


linear algebra
matrix
real analysis
real analysis unsolved

Let $ \Omega  \subset \mathbb{R}^{n}$ be bounded subset and $ u \in C^{2}(\Omega) \cap C^{2}(\overline{\Omega})$ and $ u$ satisfies:
\[ \Delta u\minus{}c(x)u\geq 0 \ \ \ in \ \ \  \Omega\]
where $ c(x) \in C(\Omega)$  and $ c\geq 0$ in $ \Omega$. Show that if $ \sup_{\Omega}u(x)>0$ then :
\[ \sup_{\Omega}u(x)\equal{}\max_{\partial \Omega}u(x)\]

[quote="zibi"]Let $ \Omega \subset \mathbb{R}^{n}$ be bounded subset and $ u \in C^{2}(\Omega) \cap C^{2}(\overline{\Omega})$ and $ u$ satisfies:
\[ \Delta u \minus{} c(x)u\geq 0 \ \ \ in \ \ \ \Omega\]
where $ c(x) \in C(\Omega)$  and $ c\geq 0$ in $ \Omega$. Show that if $ \sup_{\Omega}u(x) > 0$ then :
\[ \sup_{\Omega}u(x) \equal{} \max_{\partial \Omega}u(x)\]
[/quote]

Yes, this is nothing but the Weak Maximum Principle.

Can you explain it little more ? I still don't see solution...

For any $ \varepsilon >0$ consider $ v(x)\equal{}u(x)\plus{}\varepsilon e^{\alpha x_1}$ with $ \alpha$ to be determined. Then

$ \Delta v \minus{} c(x) v \equal{} \Delta u \minus{} c(x) u \plus{} \varepsilon (\alpha^2 \plus{} c(x))e^{\alpha x_1} .$

By chosing $ \alpha$ large enough, we get $ \alpha^2 \plus{} c(x) >0$ for any $ x \in \Omega$. This implies $ \Delta v \minus{} c(x) v>0$.

If $ v$ attains its non-negative maximum value of $ \bar \Omega$ in $ x_0 \in \Omega$, then the matrix $ \big(v_{ij}(x_0)\big)_{i,j}$ is semi-negative definite. Since the identity matrix is positive definite, then $ \Delta v(x_0) \leq 0$. This is a contradiction.

Then $ v$ attains its non-negative maximum only on $ \partial \Omega$. By letting $ \varepsilon \to 0$ we finish the proof.

There should be
\[ \Delta v \minus{} c(x) v \equal{} \Delta u \minus{} c(x) u \plus{} \varepsilon (\alpha^{2} \minus{} c(x))e^{\alpha x_{1}}.\]
But we don't know whether c(x) is bounded ? so we can't find such $ \alpha$ that $ \alpha^{2}\plus{}c(x) >0$ ?

[quote="zibi"]
But we don't know whether c(x) is bounded ? so we can't find such $ \alpha$ that $ \alpha^{2} \plus{} c(x) > 0$ ?[/quote]

Oh you are right, normally $ c$ must be continuous up to the boundary $ c \in C(\bar \Omega)$.

[quote="zibi"]There should be
\[ \Delta v \minus{} c(x) v \equal{} \Delta u \minus{} c(x) u \plus{} \varepsilon (\alpha^{2} \minus{} c(x))e^{\alpha x_{1}}.\]
But we don't know whether c(x) is bounded ? so we can't find such $ \alpha$ that $ \alpha^{2} \plus{} c(x) > 0$ ?[/quote]

In order to avoid this difficulty, you can assume $ \sup_{\overline \Omega} u(x)$ is archived (since $ u$ is continuous on a compact set $ \overline \Omega$) by an $ x_0 \in \Omega$. Then you shall work on a small neighborhood of $ x_0$. Having the contradiction shows that $ \sup_{\overline \Omega} u(x)$ is archived only on the boundary $ \partial \Omega$.


trigonometry
search
geometry
similar triangles
Comc
Canada

A line segment BC has length 6. Point A is chosen such that angle BAC is right. For any position of A, a point D is chosen in BC such that AD is perpendicular to BC. A circle with AD as diameter has tangents drawn from C and B to touch the circle at M and N, respectively, with these tangents intersecting at Z.
Prove ZB + ZC is constant.

The given solution on the internet is kind of trig bashy :(. Does anyone have a nicer solution?

This isn't so nice either but works.

First $ A$ is on the circunferenc of diameter $ BC$.
Let $ E$ and $ F$ be the other points of tangency for $ B$ and $ C$ respectively.

$ ZB \plus{} ZC \equal{} ZE \plus{} ZF \plus{} BE \plus{} CF \equal{} BC \plus{} 2ZE$, we search to prove that $ ZE$ is constant.

It is easy to see that $ \frac {AD}{2} \equal{} \sqrt {\frac {BD*CD*ZE}{BD \plus{} CD \plus{} ZE}} \equal{} \frac {\sqrt {BD*CD}}{2}$
wich implies:
$ 4ZE \equal{} BD \plus{} CD \plus{} ZE$ or $ ZE \equal{} \frac {BC}{3}$

 :)

[quote="ElChapin"]
It is easy to see that $ \frac {AD}{2} \equal{} \sqrt {\frac {BD*CD*ZE}{BD \plus{} CD \plus{} ZE}} \equal{} \frac {\sqrt {BD*CD}}{2}$
[/quote]

Sorry, could you explain that statement some more?

A triangle with sides $ a$, $ b$, $ c$, semiperimeter $ s$ and inradi $ r$ has area

        $ \sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}\equal{}sr$(if you didn't know it, is a good exercise to prove this relation)

From this:  $ r\equal{}\sqrt{(s\minus{}a)(s\minus{}b)(s\minus{}c)}{s}$

Also the lenghts of the tangents from each vertex to incircle are $ s\minus{}a$, $ s\minus{}b$ and $ s\minus{}c$, and note that $ s\equal{}(s\minus{}a)\plus{}(s\minus{}b)\plus{}(s\minus{}c)$

In a triangle with a right angle: the altitud $ h$ from the right angle divides the hypotenuse in two segments of lenghts $ d$ and $ e$, it can be prove (only using similar triangles) that $ h^2\equal{}d*e$

This should be enough to solve the problem, if you didnt know or don't understand something try it by yourself, this helps a lot to get experience and to solve more of this kind of problems.

 :)


geometry
3D geometry
number theory unsolved
number theory

Find all perfect powers whose last $ 4$ digits are $ 2,0,0,8$, in that order.

$ 10^4|a^b \minus{} 2008\implies 2^4|a^b \minus{} 251\cdot 2^3$
The largest power of $ 2$ dividing $ a^b$ must be precisely $ 2^3$. So $ a$ even but not divisible by $ 4$ and $ b \equal{} 3$
Let $ a \equal{} 4k \plus{} 2$
$ 10^4|(4k \plus{} 2)^3 \minus{} 251\cdot 2^3\implies 5^4|k(4k^2 \plus{} 6k \plus{} 3) \minus{} 125$
But $ 4k^2 \plus{} 6k \plus{} 3$ is never divisible by $ 5$ (checking $ a \equal{} 1,2,3,4$ suffice). So $ k$ must be pf the form $ l\cdot 5^3$
So $ 5|l(4k^2 \plus{} 6k \plus{} 3) \minus{} 1\implies 5|3l \minus{} 1$. So $ l$ must be of the form $ 5t \plus{} 2$
Going back, we see that all such powers are represented by $ \boxed{(2500t\plus{}1002)^3}$

Since such a number is divisible by 8 and not by 16, we conclude that the number is a cube.

let $ n^{3} \equal{} 10000x \plus{} 2008$ which implies $ m^{3} \equal{} 1250x \plus{} 251$ implying $ m^{3} \minus{} 1 \equal{} 250(5x \plus{} 1)$ ,(n=2m)

hence, $ m^{3} \equal{} 1(mod250)$
By Euler's theorem , $ m^{100} \equal{} 1(mod250)$
since (3,100)=1, we get that m=1(mod250), so m=250t+1 for some integer t

$ (250t \plus{} 1)^{3} \minus{} 1 \equal{} 250(5x \plus{} 1)$ . Dividing both sides by 250 and reducing modulo 5, we get t=2(mod5)

Hence, all such cubes are of the form $ (2(250(5u \plus{} 2) \plus{} 1))^{3}$, that is $ (2500u \plus{} 1002)^{3}$


geometry
circumcircle
parallelogram
geometric transformation
reflection
geometry unsolved

[quote="mr.danh"][quote="Doctor A"](A.Zaslavsky) The circumradius of triangle $ ABC$ is equal to $ R$. Another circle with the same radius passes through the orthocenter $ H$ of this triangle and intersect its circumcirle in points $ X$, $ Y$. Point $ Z$ is the fourth vertex of parallelogram $ CXZY$. Find the circumradius of triangle $ ABZ$.[/quote]
It should be "parallelogram CXYZ". Then, the circumcircle (ABZ) is equal to R.[/quote]
dear mr.danh,I think Doctor A is right
X,C,Y,Z lie clockwisely

[quote="mr.danh"][quote="Doctor A"](A.Zaslavsky) The circumradius of triangle $ ABC$ is equal to $ R$. Another circle with the same radius passes through the orthocenter $ H$ of this triangle and intersect its circumcirle in points $ X$, $ Y$. Point $ Z$ is the fourth vertex of parallelogram $ CXZY$. Find the circumradius of triangle $ ABZ$.[/quote]
It should be "parallelogram CXYZ". Then, the circumcircle (ABZ) is equal to R.[/quote]
how to do it?could you post your idea?
thank you in advance!

[b]NO! In the original language of Russian, is CXZY paralelogram![/b]

Denote the intersection of XY and OO1 is N
1.We make the reflection of H w.r.t N is H’
We know ON=O1N  so HOH’O1 is a parallelogram  => HO1=OH’=R => H’lies on ⊙O
=> ZHCH’a parallelogram
2. Extend AO meet ⊙O at M
HCMB is a parallelogram  =>  MB∥CH∥ZH’ MB=CH=ZH’ => ZBMH’ is a parallelogram
we have ZB∥H’M
3.∠MH’C=∠BZH =∠MAC=∠H1AB  =>  B,Z,A,H are concyclic
∠BZA=∠ACB=180-∠AHB
4.using sin law
sin∠AZB=AB/2R1    sin∠ACB=AB/2R   => R=R1
DONE!

[b]Remark:[/b] the circle HAB is reflection of  ABC  relatively the line AB ---> (circumcircle radius of $ \triangle {HAB})\equal{}$ (circumcircle radius of $ \triangle {ABC})\equal{}R.$

Yes. Sorry, I think X,Z are intersections of the circle which passes through H and the circumcircle (ABC). And this problem is not hard.
We should denote the reflections of O wrt AB and XY, then make a use of properties of parallelogram.
Best regards, mr.danh

This problem can be easily solved by complex number. :P


AMC
AMC 12
AIME
USA(J)MO
USAMO
Harvard
college

I've been doing quite a bit of thinking the past few days about many, many things, and I'd just like to share a piece of one of my thoughts with you...

I still remember how I was a year ago. I was a sophomore back then. I remember how I was still a little boy, running around during math club meetings studying cool problems and puzzles with my friends. Mathematics was so fascinating and elegant. I loved math. I really did.

When I heard my math coach announce that we would be taking the AMC12, I became interested in what exactly the competition was. I had taken it the year before too, but no one at my school really studied for it. It wasn't a very big deal, and I had completely forgotten that it had existed. So I googled up the exam and found the official AMC website. I learned about the AMC, AIME, USAMO, MOSP. and IMO. I thought: "Wow, wouldn't it be cool to attend a math summer camp with the top math students in the country?" So from that point onwards, I studied really really hard in the hope of at least qualifying for the USAMO exam next year. I had a lot of fun too. I loved both the math and the competition. Life was looking good.

However, soon, the evitable college process started looming over everyone, and I was thrown into the chaos and tension of college admissions. Everyone told me: "Hey, you know, many of the USAMO qualifiers go to really competitive schools like Harvard and MIT. If you qualify, you could probably get admitted at one of those universities." But I said: "Oh...but you know, I really only study this stuff for fun, sort of as an leisurely activity, you might say." Yet of course, everyone began fretting over grades, over SAT scores, over extracurricular activities, etc... And of course, my parents were obsessing over the ordeal as well: almost everything that they walked to me about was related to college admissions. So I looked at myself and thought: "You know, my grades and SAT's aren't really spectacular, and I don't really have many extracurricular activities or achievements. But...you know...if I make USAMO, maybe that could be a huge hook for me in terms of college admissions."

So I began studying even harder, with an even greater determination to do well on the competitions. However, I found myself questioning: Am I studying so hard because I love math, or because I want to get admitted to the highly-selective colleges? Why am I doing what am doing? What is the driving force of my life right now? I felt so much pressure to make the USAMO. I go to a very good private school, many of whose seniors go on to Ivy League schools, so I felt even greater pressure to get accepted into the most selective colleges. I thought that the USAMO would have been my ticket. I was up against it the wall, and it was my only way out. I would either make it or break it--and the second choice wasn't really an option.

Yet, I sit here now having failed to qualify for the USAMO. I'm rambling now, but you know, as I look back upon the past year, I can't help thinking that the above story contains some pretty messed up stuff.

That was some pretty messed up stuff.

Hi, Fiery, I want to post a longer answer to you on that other thread you started, but meanwhile let me just say that my philosophy as a parent is that college admission should never be a basis for NOT DOING something, but rather a basis for DOING something. (This will become my slogan: this is the second time tonight I've had occasion to type this.) Too many people these days are WAY too obsessed about college admission, and way too willing to give up personally fulfilling, interesting activities because they feel they should only do what is on the formula for getting into the top college. But as JBL has been trying to point out in several threads, there is no sure-fire formula anyway. Dare to be yourself. Last year you did math for fun, and you can do math for fun for the entirety of your life.

My oldest son is younger than you are, and I think people reading between the lines of my posts can sense that I am a little disappointed that when I was a kid I heard about the top schools (and even had one adult close to me advise me to completely drop my then preferred major subject unless I could get into M.I.T., because studying that subject at State U. would be a waste of my time) but didn't hear about sensible things to do to prepare to get into top schools. I don't believe in letting the college admissions system grind someone down. Yes, do what is interesting and satisfying to you, learn about how to do it as well as you can, and then just don't worry about the mistakes some admission officer might make. People with superimpressive on-paper credentials don't get into the top schools sometimes, for mystifying reasons. People who were dying to get into the top schools have to make do with State U., and then discover a job, a cause, a fiance, or great scenery that makes them glad they ended up there.

I had a funny conversation about a year ago with a guy in my homeschooling parent support group who confessed that he had always wanted to be in a different occupation: an occupation that I used to have and didn't like. That was funny because I thought I had always wanted to be in HIS occupation--until that conversation. Our wives each got a good laugh out of hearing about our conversation. Do what's fun, do what pays the bills, but most of all learn how to have fun while doing what pays the bills. 99.9 percent of the time it doesn't matter where, or even whether, you went to college. The other 0.1 percent of the time is subject to other variables anyhow, so don't worry about it.

Fiery - are you Asian? :)

Anyway, the first thing I'd say is: calm down.  Do math because you love it, not because it looks good on a college application. Having some contest on an application is only a fringe benefit that is nothing compared to your genuine appreciation for the subject. I do math because I like doing it, not to garner awards and boost my application.  I'm sure that, deep down, you feel the same.

The second thing is, wherever you go to school, you will do well. Why? Because I believe your own drive and self-motivation for math will be your guide. Don't get too hung up on where. It's far more important to appreciate the four years that you stay in college as opposed to a piece of paper that either begins with either "Congratulations!" or "We regret to inform you."

With that said, I completely empathize with your situation. I am Asian. I have Asian parents. I go to a predominantly Asian school (around 65%). And yes, if you're not Asian, let me be the first to tell you that the stereotypes are all true, and my parents drive me insane sometimes with college admissions. Lately I've also been feeling a lot of pressure to get into the big schools: HYPMS. I sometimes catch myself forgetting to take my own advice--that admissions aren't everything. Yes, parents and friends would like to have you think that a silly little piece of paper matters, but in the end, it's what you do later in college that matters.

I know that I'm just an internet stranger to you, but trust me, I feel you. I'm pretty disappointed I didn't qualify for USAMO either, but then again, I can't say I share your grief because I really wasn't very qualified to begin with. But don't read these words and scoff, "He doesn't understand," because I do. You let yourself down, but that happens to everyone. The best thing to do is move on and try harder next year.

And let me reiterate that the entire admissions process is overrated. Just do your best, and whatever happens happens.

Hope that helped.

rlee: Yes, I'm Asian, Chinese in fact.

Hmm, ok, let me comment on something that you folks mentioned in this thread (and something that has been mentioned in other threads on this forum as well, I believe). Many here have commented that college admissions is not that important, and that what a person does during college is much more important than what college a person goes to. To a very great extent, I agree with this assertion. I don't know how much better of an undergraduate education in mathematics a student could get at MIT as compared to, say, UChicago (as a random school).

However, the part about the assertion that I disagree with is based upon this: that going to a highly selective college like HYPMS will give a student quite a leg up after the student graduates. That is to say, a student who graduates from a highly selective college will be seen by job employers, graduate schools, and other organizations as more "qualified" than a student with the same intelligence who graduates from a less selective school. This student will thus have an easier time after he/she graduates simply because of the name of the university that he/she attended. I think this is the reason why many people want to get into HYPMS as opposed to supposed "second-tier" colleges -- for the prestige and name, not necessarily because of the education.

I don't really like this *thing* that society has created that I described above, but I think its the reality.

I don't think that's true at all. For one thing, it is quite well-known that graduate schools do not like to accept their own undergraduates, so if you go to one of the five schools that you consider to be the absolute best, then you will only have four top graduate school choices rather than five most likely. (That is one reason why people advised me not to go to Berkeley, incidentally.) Also, I think if you do a very good job at a ``second-tier'' school, then grad schools will be more likely to accept you than if you do fairly well at a ``first-tier'' school.

By the way, University of Chicago is not exactly a random school. I would have gone to UChicago over MIT for sure. (It helped that I applied to UChicago and not MIT.) And where did Yale come from? I don't believe people consider Yale one of the great math schools. But I'm just being picky now, so I should stop.

[quote="Rep123max"]I thought Yale was liberal arts or something.[/quote]

Mathematics is a liberal art. I myself am puzzled that Yale isn't mentioned more often as a math school. Roger Howe and Serge Lang are both quite famous math professors, and I think that there are others there. I have been to the campuses of Yale, Harvard, M.I.T., Princeton, Stanford, U. Chicago, and various other famous schools several times on business trips, and I think the one great disadvantage Yale has in attracting students is being located in New Haven, Connecticut, which is just not as attractive a town as Boston or Princeton or Chicago or Palo Alto. It's hard to be everyone's second choice, but that is not the same as saying Yale is a bad school. (Yale is arguably at the very top in law school rankings, for example, so it has its strong programs. I would love to do a graduate degree in psychology with Robert Sternberg there.)

That's enough of this digression. The point is that Fiery or any of the rest of us can have a successful life without having to go to any of the above-mentioned schools.

Well, I thought I was going to have something productive to add here, but rlee and tokenadult and Simon have roughly covered all the territory I might have wanted to, and quite elegantly.  (By the way -- I also applied to U. Chicago and not MIT -- they have a very good math program there.  The reason it's not as popular as others has nothing to do with the quality of its academics, and far more to do with the structure of its curriculum and the particular type of student it tends to attract.)

Anyhow, the thing I wanted to comment on was the "How come Yale isn't a math school?" question.  The answer, I think, doesn't lie so much in their not trying to attract math people -- I believe they are rather hoping to attract math people, and may even have tried to draw a similar group of students (Olympiadists, etc.) to waht Harvard and MIT get in recent years.  (That was a terrible sentence, but I hope it's meaning is clear.)  Anyhow, I think they fail for two reasons:  (1) they don't already have any olympiadists, so it's harder to attract them, and (2) they don't have a good freshman math class.  Harvard has Math 55; Chicago has Honors Analysis; MIT lets people take whatever the heck math class they want, give or take; but the most theoretical freshman math class at Yale is roughly as theoretical and tough as the third-most such class at Harvard.  (I have this via a very talented friend who is currently at Yale and in that class and thinks that it's silly, rather.)  I don't know how they expect to attract people without offering a class those people would want to take -- it seems very silly.  So, there was my (badly written) two cents.

Oh, yeah -- Fiery, relax! :)

Hey Fiery.
I think I can understand how you feel, even though my problem is somewhat different.
I love learning. Just learning in general, just getting to explore anything and everything. Biology's my favorite subject, but I adore lit, chem, languages, math, drama, phys, history... pretty much anything I can get my hands on. There's always something interesting you can find in any subject, regardless of what it may be. And I love all of it.
My problem? I can't focus. I'm one of those "all-around" kids who's never outstanding in anything because she's "good" in everything. I cant devote myself wholly to one particular pursuit because there are so many others that are so unbearably [i]beautiful[/i] and impossible to let go of.
Of course my other problem is that Im not smart enough excel at everything, but thats a separate issue. Heh.
So this year Im taking a few APs. I dont really mind APs because learning the material is interesting enough and fun in its own way. Yet sometimes, when Im exhausted from studying or otherwise worn out, I ask myself - whats all this in aid of? Am I working because I want to embellish my transcript or because I truly love the material? I would like to sincerely say that its the former, and more often than not I believe that thats the case; but there are points when Im flopped senseless on my bed wondering [i]what in the world am I trying to achieve here[/i]? Is there any meaning to all of this?
As for the SAT, I had a little episode with it that some of you know about and others can PM me if they care to learn. Suffice it to say that just added to my general sense of disillusionment  a state I dont really care for. Id rather be hyper and spunky anytime. =)
*laughs and shakes head* I dont think this letter/note/reply is at all coherent. Excuse me. Im sick this week. My mum thinks its from over-exhaustion. No comment.

You are the reason that liberal-arts colleges were invented :).  Try and relax a little bit -- being good at many things should not be complained about :)


find all points(a,b) such that b^3 -a^3=2007ab+2009


MATHCOUNTS
probability
geometry
number theory
AMC 8

7th in Massachusetts

[color=darkblue][size=75]<You may list your rank.  No remarks about the competition.  Period.  Edited by moderator rcv>[/size][/color]

State: Utah
Place: 8th individual and countdown

[color=darkblue][size=75]<You may list your rank.  No remarks about the competition.  Period.  Edited by moderator rcv>[/size][/color]

Let me reemphasize what moderator rcv is repeatedly pointing out: we don't want any discussion of the state meets until after all of the state meets are over. That's simply more fair to all the mathletes, and fair play is part of what you should be learning through MATHCOUNTS participation.

As before, I wish all the AoPSers the best success in their math competition participation.

NO, LOL, worse than 24th in Indiana, though I have two more years to try to win :)

ours is next Saturday.... trying to study right now!

Same here... just hoping to get in... :crosses his fingers:

same i really want to go to nationals too

Arrggh, 3rd countdown but only 12th individual. Sadly countdown isn't official in Maryland.

well, i'm just hoping to get a good score, i'm pretty sure i won't be able to make it.... only four people goes, and i'm only fourth in my chapter.... but i'm hoping maybe i'll go next year as an eighth grader :D

Lucy,

I've been coaching for 6 years, and twice I've had a kid finish 4th at chapter and then make it to nationals(both times they were 3rd at state)  So don't give up hope.  Here is a list of how my kids who did well at chapter faired at states

Chapter --->  State
2nd                4th
4th                 10th
3rd                 3rd
1st                 3rd
2nd                8th
3rd                 2nd
4th                 10th
4th                 3rd(one point out of 1st)
3rd                 2nd
6th                 6th
1st                 7th
2nd                4th
3rd                 11th
4th                 3rd
9th                 6th
Those last 5 were all last year.  You can see that you don't always do worse at state than you did at chapter, just depends on how hard you work!!

Good Luck

State: Utah
Rank:  3rd.

ps.  For anyone who has already made it to nationals, or knows what to do...do I have to list a "preferred" roomate?

my state is on march 12 i hope i get to top 4

[quote="Secret Asian"]State: Utah
ps.  For anyone who has already made it to nationals, or knows what to do...do I have to list a "preferred" roomate?[/quote]

Usually at the meeting right after states, we have the kids pair off and put down a preferred roommate.  If the 4 kids are 3 boys and a girl, or 3 girls and a boy, then they obviously can't pair off and are pretty much placed at random.  If you don't put a preferred roommate and your state is sending all boys or all girls, you will be paired with a kid from your state.

i dunno but some how i doubt it. my mc coach relies more on the team to make it to nationals. oh well. i doubt it too so i cant argue. (haha) but our competitions this satuday on march 12. i hope i hope i hope! :?  :?  :?

i just watch my clock and if im halfway and its not 15 min yet then i do it in the order they come but if it is 15 min and im not halfway i look for easy questions

I never check my watch, I just go from 1-30. I've never had trouble doing it in time.  This years problems weren't that hard and didn't take that long

sometimes i almost run out of time sometimes i finsih very early that i can check it 2 times

Mark, you always finish the test.
You sit at the same table as me and you always turn the page in like 1 minute...

dont you think that the 6 minute for the pair of problems for the target round is too much? I finish the problems in 2 minutes and then fiddle around with my calculator.

For the target problems it really depends on the test. Some tests I fininish the targets in 30 seconds others I'm struggling to finish within the time period.

Yeah, I've had times where I want to just get out of my seat and stretch until the 5 remaining minutes are up, and then there's the times when I don't see how the problem can be done.

I always work to the last second, I check, double check, triple check, etc until time is up if I'm done.

[quote="09husbri"]...If that applies to you, then... :lol:[/quote]

hey, be quiet  :blush: that one was hard!

My bad, didn't mean to offend any fellow Mathlete :)

Greg (1st place one) told me he played tetris on his calculator after each set of target round problems at state :D.  But at nationals you can never check too many times.

[quote="09husbri"]...If that applies to you, then... :lol:[/quote]Hey, I missed number one!  For some reason I thought 40+8=52.

Are you going to nationals? If you are, at least I'm not getting 228th...JK :lol:


inequalities function calculus derivative

Let $ a,b,c\in\mathbb{R}$ and $ a\plus{}b\plus{}c\equal{}1$. Prove that:
$ \frac{a}{1\plus{}a^2}\plus{}\frac{b}{1\plus{}b^2}\plus{}\frac{c}{1\plus{}c^2}\le \frac{9}{10}$

It has one more condition $ a,b,c\ge\frac{\minus{}3}{4}$
$ 28^{th}$ problem in Mildroff's inequalities.
[hide]Even i was not able to prove it[/hide]

[quote="Kyoshiro"]Let $ a,b,c\in\mathbb{R}$ and $ a \plus{} b \plus{} c \equal{} 1$. Prove that:
$ \frac {a}{1 \plus{} a^2} \plus{} \frac {b}{1 \plus{} b^2} \plus{} \frac {c}{1 \plus{} c^2}\le \frac {9}{10}$[/quote]
this problem is not difficult, we can solve it by a formal solution :lol:
I suggest that we can find k which satisfy this ineq.:
$ \frac {a}{1 \plus{} a^2} \le \frac{3}{10}\plus{} k(\frac{1}{3}\minus{}a)$

WLOG $ a\geq b \geq c$ ,and use Chebyshev's inequality

We get

$ LHS \leq \frac{1}{3} \sum_{cyc}{\frac{1}{1\plus{}a^2}}$

Let $ f(x)\equal{}\frac{1}{x^2 \plus{}1}$ ,now using Jensen's inequality we have

$ \frac{1}{3} \left(\frac{1}{a^2 \plus{}1} \plus{}\frac{1}{b^2 \plus{}1} \plus{}\frac{1}{c^2 \plus{}1} \right) \leq f\left(\frac{a\plus{}b\plus{}c}{3} \right)\equal{}\frac{9}{10}$

Are sequences $ (a,b,c)$ and $ \left( \frac{1}{1\plus{}a^2}, \frac{1}{1\plus{}b^2} , \frac{1}{1\plus{}c^2} \right)$ similarly sorted? I think they aren't.

If $ a\geq b \geq c$ then $ \frac{1}{1\plus{}a^2 } \leq \frac{1}{1\plus{}b^2 } \leq \frac{1}{1\plus{}c^2}$

Then using Chebyshev's inequality

$ LHS \leq \frac{1}{3}\sum_{cyc}{\frac{1}{1\plus{}a^2}}$

Yes of course it is correct.

Nice solution. :)

[quote="enndb0x"]Let $ f(x) \equal{} \frac {1}{x^2 \plus{} 1}$ ,now using Jensen's inequality we have

$ \frac {1}{3} \left(\frac {1}{a^2 \plus{} 1} \plus{} \frac {1}{b^2 \plus{} 1} \plus{} \frac {1}{c^2 \plus{} 1} \right) \leq f\left(\frac {a \plus{} b \plus{} c}{3} \right) \equal{} \frac {9}{10}$[/quote]

In order to use Jensen's Inequality, you must have a function that is strictly convex, or strictly concave over the interval containing all possible values of the variables (in this case, the interval must cover all possible values of $ a,b,c$).

However: $ f''(x) \le 0$ for $ x \in \left[\minus{}\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right]$, and $ f''(x) \ge 0$ for $ x \in \left(\minus{}\infty,\minus{}\dfrac{1}{\sqrt{3}}\right] \bigcup \left[\dfrac{1}{\sqrt{3}},\infty\right)$.

[url]http://www.wolframalpha.com/input/?i=second+derivative+1%2F(x^2%2B1)[/url]

So, unless you have either $ a,b,c \in \left[\minus{}\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right]$ OR $ a,b,c \in \left(\minus{}\infty,\minus{}\dfrac{1}{\sqrt{3}}\right] \bigcup \left[\dfrac{1}{\sqrt{3}},\infty\right)$, you can't use Jensen's.

[quote="enndb0x"]If $ a\geq b \geq c$ then $ \frac {1}{1 \plus{} a^2 } \leq \frac {1}{1 \plus{} b^2 } \leq \frac {1}{1 \plus{} c^2}$ [/quote]

So you proved that sequences are sorted in opposite way. Apart from that you are not sure about sign of $ a$ and $ b$.

@micf

Ah..you're right , all the time I was looking for a solution in positive real numbers , do you think the solution is OK in positive reals.

@Xantos
Are you saying that $ f(a) \equal{} \frac {1}{1 \plus{} a^2}$ is not concave in interval $ (0,\infty)$ ?

That is exactly what I'm saying.

The link that I tried to post shows that $ \dfrac{1}{1 \plus{} x^2}$ is concave down on $ \left( \minus{} \frac {1}{\sqrt {3}},\frac {1}{\sqrt {3}}\right)$ and concave up on $ \left( \minus{} \infty, \minus{} \frac {1}{\sqrt {3}}\right)\bigcup\left(\frac {1}{\sqrt {3}},\infty\right)$.

If we are given that $ a,b,c$ are non-negative:

Then $ a,b,c \in [0,1]$ and we can use Jensen's Inequality on $ f(x) \equal{} \dfrac{x}{1 \plus{} x^2}$ which is concave down on $ [0,1]$.

Therefore, $ f(a) \plus{} f(b) \plus{} f(c) \le 3f\left(\dfrac{a \plus{} b \plus{} c}{3}\right) \equal{} \dfrac{9}{10}$ as desired.

thus,the problem is solved just for positive real numbers ... :(

From Enndbox's post (chebyshev's ineq) we have:
$ LHS\leq \frac{1}{3}\sum \frac{1}{a^2\plus{}1}$
Now we prove that $ \sum\frac{1}{1\plus{}a^2}\leq \frac{27}{10}$
this is an interesting inequality and has been discussed here many times before.
Now, $ (4\minus{}3t)(3t\minus{}1)^2\geq 0 \forall t\leq 1$
Hence  $ \frac{1}{1\plus{}x^2}\leq \frac{27}{50}(2\minus{}x)$
So that $ \sum \frac{1}{1\plus{}x^2}\leq \frac{27}{10}$

It is a hard solution to find.....  :rotfl:

[quote="Potla"]From Enndbox's post (chebyshev's ineq) we have:
$ LHS\leq \frac {1}{3}\sum \frac {1}{a^2 \plus{} 1}$
Now we prove that $ \sum\frac {1}{1 \plus{} a^2}\leq \frac {27}{10}$
this is an interesting inequality and has been discussed here many times before.
Now, $ (4 \minus{} 3t)(3t \minus{} 1)^2\geq 0 \forall t\leq 1$
Hence  $ \frac {1}{1 \plus{} x^2}\leq \frac {27}{50}(2 \minus{} x)$
So that $ \sum \frac {1}{1 \plus{} x^2}\leq \frac {27}{10}$

It is a hard solution to find.....  :rotfl:[/quote]
Not really, if we take the derivative at the place where we guess the extreme value occurs, and find the equation of the tangent line.

a^2/a+a^3+b^2/b+b^3+c^2/c+c^3.
We use titu's lemma.
>=(a+b+c)^2/a^3+b^3+c^3+a+b+c=1/1+a^3+b^3+c^3
a^3+b^3+c^3>=3abc
a+b+c>=3sqrtabc=(maximized when this equals 1/3)
abc=1/9 at maximum. so the inequality is true.1/1+1/9=9/20


Anyone got a nice short solution for this problem?

The points $z_1, z_2, z_3$ form an equilateral triangle.
Prove that: $z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1$.

It reminds me of the well known and easy to prove inequality $x^2+y^2+z^2\geq xy+yz+xz$, though i can't see any link with it =/

Suppose we have an equilateral triangle with center $z_c$.  We substitute $(z_i-z_c)$ for $z_i$ in the equations and note that the the condition is equivalent (just multiply it out).  Thus, we need only show it holds when the triangle is centered at the origin.  Note that now $|z_1|=|z_2|=|z_3|$, since they form an equilateral triangle at the origin.  We replace each $z_i$ by $\frac{z_i}{z_1}$ (multiplying out shows that this condition is equivalent).  Now, $z_1=\text{cis}\left( 0\right)$, $z_2=\text{cis}\left( \frac{2\pi}{3}\right)$, $z_3=\text{cis}\left( \frac{4\pi}{3}\right)$, and we only have to plug them in and check that it works.

Thus, an equilateral triangle centered at the origin with $z_1=1$ works, and so any equilateral triangle centered at the origin works, and so any equilateral triangle works, and we are done.

think of the roots of the equation
$(z-w)^3=m$, w will be the center of triangle, the roots will be the points of the triangle, (since i ran out of letters, $m\in \mathbb{C}$ also
so we have
$z^3-3z^2w+3zw^2-w^3-m=0$
then by viete's sums, we have $3w^2=z_1z_2+z_1z+z_2z_3$
and using newton's sums (we want $s_2$), we have
$1*s_1+(-3w)=0$
$s_1=3w$
$1(s_2)+(-3w)(3w)+2(3w^2)=0$
$s_2=3w^2$
using that, we have
$s_2=3w^2=z_1z_2+z_1z+z_2z_3=z_1^2+z_2^2+z_3^2$


[quote="koredim1989"]ean i sinechis sinartisi g(x) ine gnisios fthinousa konta sto +apeiro kai to orio tis sto +apiro ine 0,na ipologisete to orio kathos to x tini sto +apiro tou orismenou olokliromatos apo x eos x+1 tis g(x).

sorry,alla den echo to programma gia ta mathimatika simvola...[/quote]

Γράψε

 $\int_{x}^{x+1}g(x)dx = g(k) , x<k<x+1$΄, από το ΘΜΤ για τα ολοκλ.

Αλλά $\lim_{x \to \infty }g(k) = 0$ , διότι .....

 Μπάμπης


the answer is $ \frac{\minus{}1}{3}$.... solved it by using by parts....

put u as the log function and put parts and use limits to get the value when x tends to zero...

i hope this is right...

[quote="valeriummaximum"]the answer is $ \frac { \minus{} 1}{3}$.... solved it by using by parts....

put u as the log function and put parts and use limits to get the value when x tends to zero...

i hope this is right...[/quote]

Why dont you post your steps using Latex than just the algorithm. This is pretty irritating. And I'm sorry I forgot a small part of a question. Question has been changed.

[hide="is the answer"]
$ 8!$
?
i will post solution after u confirm the answer :)
[/hide]
[hide="hint"]
consider $ I_n \equal{} \int_{0}^{1} (log(\frac {1}{x}))^ndx$ :wink: [/hide]

Maybe. But you are going to use reduction formula? :rotfl:

Pretty Simple using Gamma function :D
Make the substitution 1/x=t
u get the integral to be gamma (9)=8!

my method is obviously longer than elastiboy's, but thats coz i derive the gamma function :D
anyway, consider
$ I_n\equal{} \int_{0}^{1} (log(\frac {1}{x}))^ndx$
doing parts, we get
$ I_n \equal{} n \int_{0}^{1} (log(\frac {1}{x}))^{n\minus{}1}dx$
(the UV term after doing parts = 0)
$ \equal{}> I_n \equal{} n. I_{n\minus{}1}$
$ I_0 \equal{} 1$
hence,
$ I_8 \equal{} 8!$
:)

[quote="hash_include"]my method is obviously longer than elastiboy's, but thats coz i derive the gamma function :D
anyway, consider
$ I_n \equal{} \int_{0}^{1} (log(\frac {1}{x}))^ndx$
doing parts, we get
$ I_n \equal{} n \int_{0}^{1} (log(\frac {1}{x}))^{n \minus{} 1}dx$
(the UV term after doing parts = 0)
$ \equal{} > I_n \equal{} n. I_{n \minus{} 1}$
$ I_0 \equal{} 1$
hence,
$ I_8 \equal{} 8!$
:)[/quote]

Its more elegant this way but stating the obvious doesn't make it more obvious. :D


In TeXNiC i cant space my lines so that it appears well on a page. For example:
When I type :
"PROMYS app Problem: 1
blah blah blah"

It comes out
"PROMYS app Problem: 1 blah blah blah

You have to tell LaTeX you want a new line with \\. This allows you to lay out the code in any way you want without affectng the final text. So in your example type:
[code]PROMYS app Problem: 1 \\
blah blah blah[/code]


Hi......
@skand  which all topics have u prepared for the nsea(senior).... i am planning to give it this year...... i am in class 10 so i know i dont stand much chance but just to have an outing.......
  :)  :)

And can u give me some links for the papers.....
i have nsea senior paper for 2007

Can anyone provide me the papers for the previous years....????  :?:

welcome to AoPS :)

[url]http://www.iapt.org.in[/url]
you can have question papers there

about me,i am not planning to do anything about NSEA.Will just give it as timepass :rotfl:

contact Arnav,he was NSEA topper previous year :)

hey seems like nsea's attendance is increasing day by day

you need not take any tension..
wat match am i for u???
anyways
where is the topper..... cant he provide me some help :)

@skand i have 2007 paper as i told u.....
do u have other years papers??

prev yr papers r usually not available on internet as long as some1 doesn't hack them or scan them :D  :roll:

hackers at their work!
ritu u know hacking?

I remember ritu telling that he hacked somebody's account in aops itself

admin's account :rotfl:  :rotfl:  :rotfl: (adiarya)


number theory
relatively prime
number theory proposed

Let p,q,r,a natural numbers with $pq=ra^2 $ r is a prime number and (p,q)=1
There exist prime number $ p $ such that the number $ p\left(2^{p+1}-1\right) $ is a perfect square?

What is that first sentence doing there?

For the question "Find all primes $p$ such that $p(2^{p + 1} - 1)$ is a perfect square..."

For $p = 2$ we get 14 which is not a perfect square.

Now suppose that $p$ is odd, $p = 2q - 1$, say.

Suppose that $p(2^{p + 1} - 1)$ is a perfect square. Since it is divisible by $p$, it must be divisible by $p^2$, so that $p$ divides $2^{p + 1} - 1$. It is easy to see that $2^{p + 1} - 1$ is in fact equal to $pa^2$ where $a$ is a positive integer. Hence $(2^q - 1)(2^q + 1) = pa^2$, and since the two factors on the left hand side are relatively prime, one of them must be a perfect square. However the first one cannot be a perfect square since it is equal to 3 (mod 4). So the second one must be a square. Suppose that $2^q + 1 = t^2$, then $(t - 1)(t + 1) = 2^q$. The only powers of 2 differing by 2 are 2 and 4. Hence $t = 3$ and $q = 3$. Hence $p = 5$. However for $p =5$, we have that $p(2^{p + 1} - 1) = 3^2 \cdot 5 \cdot 7$ which is not a perfect square. Hence there are no solutions.

Arne is right.
You can also solve this one by using fermat's little theorem.
Because since $p$ divides $ 2^{p+1} - 1$ and $2^p -2$ we get that $p$ divides $2^{p+1} - 1 -2(2^p -2) = 3$.
But $p$ can't equal $3$. So there are no solutions.

This problem is from the Greek M.O "Archimedes" 2001 . It coresponds the level of the problems posted in greek competitions (It was the last problem then!!!!  :( ).

It is a quite simple question without the first condition.
But I am interested in the first condition.
Are there any relation between two questions?

I think there are some mistakes.


induction
combinatorics proposed
combinatorics

Let $ n$ be an integer, $ n\geq 2$, and the integers $ a_1,a_2,\ldots,a_n$, such that $ 0 < a_k\leq k$, for all $ k \equal{} 1,2,\ldots,n$. Knowing that the number $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n$ is even, prove that there exists a choosing of the signs $ \plus{}$, respectively $ \minus{}$, such that
\[ a_1 \pm a_2 \pm \cdots \pm a_n\equal{} 0.
\]

It is very similar to Chvatal Theorem about Hamiltonian paths.
Consider a graph with vertices of degree $ a_i$. Then they satisfy $ \sum a_i \equal{}2E$.
Then the for a Hamiltonian path pass we add degrees for even vertices on the path and subtract odd.
It is not the same problem, but remind me about graph theory.

[quote=mto]By induction on n

Suppose that we could choose the sign +, - when n=k and we have to show that we could find an arrane when n=k+1

 If a(k+1) does not equal ak, then apply induction hypothes for k number

 a1,a2,..., a(k-1), |a(k+1) -ak|

 If a(k+1)=ak, then apply induction hypothes for k-1 number   a1,a2,.., a(k-1) and put a - between ak, a(k+1)[/quote]

VERY SMOOTH  :coool:


On the circle given by 16x^2 + 16y^2 + 48x - 8y - 43 = 0 find the point M which is closest to the straight line given by 8x- 4y + 73 = 0 and calculate the distance d from M to this Line

The equation of the circle is $(x+\frac{3}{2})^{2}+(y-\frac{1}{4})^{2}=(\sqrt{5})^{2}$.

[hide]  We can find the point closest to $8x-4y+73=0$ by drawing the tangent to the circle with the same slope of 2.  Then the radius from M to the center of the circle is perpindicular to this line, so it has a slope of $-\frac{1}{2}$.  Therefore $M$ is at $(-\frac{3}{2}+\frac{2}{\sqrt{5}}\times\sqrt{5},\frac{1}{4}-\frac{1}{\sqrt{5}}\times \sqrt{5})=(\frac{1}{2},-\frac{3}{4})$ or
$(-\frac{3}{2}-\frac{2}{\sqrt{5}}\times\sqrt{5},\frac{1}{4}+\frac{1}{\sqrt{5}}\times \sqrt{5})=(-\frac{7}{2},\frac{5}{4})$.  The line $8x-4y+73=0$ passes above the circle, so $M=(-\frac{7}{2},\frac{5}{4})$.

Then the projection of $M$ to the line is at the intersection of $8x-4y+73=0$ and $x+2y+1=0$, which is $(-\frac{15}{2},\frac{13}{4})$.  This distance is equal to $\sqrt{4^{2}+2^{2}}=2\sqrt{5}$.[/hide]


probability
Pascal\u0027s Triangle
conditional probability

Mrs. Witts has two children. One of them is a female. If having a male or female child is equally likely, what is the probability that the other one is a male? Express your answer as a common fraction.

The answer is given as 2/3, and everyone says it is. But I think its 1/2 because the probabilities are independent right?

Why don't you read the posts in the other thread in FTWPD?

I did, and I fail to see why MF is more likely than FF once you know there is 1 female.

Sigh...last try.

Say Mrs. Witts has two children, A and B, who I hope you realize are distinct. :wink:

We are only given that at least one of them is female.

Assume that A is female. Then B can either be female or male, giving us two possibilities.
Assume that A is male. Then B can either be female or male--but wait, B can only be female, since at least one of them must be female (do you see why???)!!!

This gives us three different possibilities:

A=female, B=female
A=male, B=female
A=female, B=male

Notice how choice two and three are different (can you tell the difference?--good job.).

wait whoa. A is 50% chance of being male right?

So case 1 where a is female occurs only 50% of the time.

If B is male then FM (50% chance) or female FF (50%) so combining 25% FF 25% FM.

Case 2, a is male (50%) b is femal so FM 100%.

Thus its 25% of FF and 75% FM?

That's not how I would recommend you think about it...just notice that the three cases I posted are equally likely--another way to go about it is $ \frac{2\cdot0.5^2}{1\minus{}0.25}\equal{}\frac23$ (do you see why it works? What do the numerator and denominators represent?)

NUmerator=proabbility of MF so shouldnt denominator be 1?

And why are the cases equally likely? I just shwoed why they [b]werent[/b]

Which cases are you referring to? My cases (in my posts) are equally likely. There is a $ 1$ probability that there are $ 2$ children here, but we must exclude MM, which has probability $ 0.25^2$.

Oh ok i get it now! Thank you so much everyone! It's because you only count the 2 right parts of pascals triangle so its $ \frac{2}{2\plus{}1}\equal{}\frac{2}{3}$

I fail to see how Pascal's Triangle fits in.

What math154 said is as good of an explanation as you're going to get.  There are $ 4$ possibilities without restrictions, but with the restriction, we cannot have $ MM$.  Then the possibilities left are $ MF$, $ FM$, and $ FF$, so the probability is clearly $ \boxed{\frac{2}{3}}$.

Again, I don't see how Pascal fits in.

[quote="AIME15"]I fail to see how Pascal's Triangle fits in.

What math154 said is as good of an explanation as you're going to get.  There are $ 4$ possibilities without restrictions, but with the restriction, we cannot have $ MM$.  Then the possibilities left are $ MF$, $ FM$, and $ FF$, so the probability is clearly $ \boxed{\frac {2}{3}}$.

Again, I don't see how Pascal fits in.[/quote]

The Pascal's Triangle fits in like this:

                                                                      1
                                                                    1  1
                                                                   1 2 1

In the 121 place, the left 1 stands for All girls, middle 2 stands for 1 girl and one boy, and the right 1 stands for 2 boys.  mewto i think got 1/2 because they want 1 girl and 1 boy, so it would be $ \frac12$, but since 2 boys is not possible, you keep those three and get $ \boxed{\frac23}$

[quote="Dojo"]I first interpreted the question like mew. For example:

Maria flipped a coin 9 times, and was surprised to see that they were all heads. What is the probability that the next flip will be heads?[/quote]

I remember doing a problem like that on some test.  It took me for ever, as I made Pascal's Triangle all the way to 10 (I wasn't very smart back then, and I'm still not), and finally got the answer.  Then I found out the answer was $ \frac12$.  It wasted a lot of time...

:rotfl:

BTW dojo, don't put problems from your test here :D

The confusion here is a distinction between this question and the following question:

Mrs. Witts has two children.  [b]The older child[/b] is female.  What is the probability that the younger child is male?

Here the two events are, in fact, independent, and the answer is $ \frac{1}{2}$ as intuition suggests.  Can you see why this question is not the question being asked?  There are two ways that one of the children can be female - it can be the younger or the older child - and that influences the [url=http://en.wikipedia.org/wiki/Conditional_probability]conditional probability[/url] involved.  As an exercise, answer the corresponding question:

Given that one of the children is male, what is the probability that the other child is also male?

[hide="Click for solution"]
Without restrictions, the cases are:

FF
FM
MF
MM

However, we do have restrictions!  FF is now an invalid case!  The probability that the other child is male is thus $ \frac{1}{3}$.
[/hide]

i think the answer is 1/2 becaouse the chances r equal so....

Okay I will try to explain this again.
There are 2 kids, let's call them A and B. And there are 3 different ways they can, well, be arranged (or something like that)
1.  A=female, B=female
2.  A=male, B=female
3.  A=female, B=male
So there's a $ \frac{2}{3}$ chance, see?

I saw a problem like this in a math book I had. I was something involving that game show where you can take what's behind door 1 or take what's behind door 2 or 3.

Like the [url=http://en.wikipedia.org/wiki/Monty_hall_problem]Monty Hall problem[/url]?

Exactly. Except it gave a name for the talke show. I think it was iin the book [i]"Hexaflexagons and other mathematical diversions[/i]

[quote="joyofpi"]Oh that one!  :D
I prefer to put it like this...

There are 3 doors, one contains 1 million dollars and the others contain nothing but air. You open a door and it contains nothing but air. You are allowed choose another door, or stay with what you have. What is the probability that you'll get the money?

[hide=" This is probably the wrong answer but..."]The probability would 1/3, the same as before, not 1/2 because you have the option to keep the air, so then it's a 1/3 probability. Right? I think it's something like that.[/hide][/hide][/quote]

$ \frac{1}{3}$ should be correct.  Why:

Right now you have air.  these are the following possibilities that can happen:

1. Re choose: get money.
2. Re Choose: get door 3 which contains some anonymous thing.
3. Stay with air.

Thus the probability is $ \frac13$

I'm sorry, but I still seriously don't get it.

Aren't the children independent?

I mean, even for that coin flipping problem you brought up - "If you flip a coin 9 times and they are all heads, find the probability the next one is head", it could just mean that you're really lucky. Each flip is independent and shouldn't affect the other flips. It's not like the coin one day just randomly decided to become a heads, or something.  :P

If we are given that the [b]first[/b] child is a female, then you would be right in saying that the probability the other is a male is $ \frac{1}{2}$

However, in this problem, we are just given that [b]one[/b] of the children is a female, so we have 3 equally likely cases, not two
MF
FM
FF
so $ \frac{2}{3}$

[hide="How 2 solve this..."]
The births of Mrs. Witts two children are two successive Bernoulli trials.

Let f = the probability that a birth is female, and likewise let m = the probability that a birth is male. Then, notwithstanding that in this case male and female births are equally likely, $ f \plus{} m \equal{} 1$.

For two children, i.e. two Bernoulli trials, the probabilities are distributed like this: $ (f \plus{} m)^2 \equal{} 1$.

Now, instead of invoking Pascal's Triangle, I'm going to expand the expression on the left hand side using FOIL because doing so allows us to see each of the terms clearly.

\[ (f \plus{} m)^2 \equal{} (f \plus{} m)(f \plus{} m) \equal{} f^2 \plus{} fm \plus{} mf \plus{} m^2 \equal{} 1 \]

The probability that we are being asked to find is the conditional probability $ P(C_2 \equal{} \text{male}|C_1 \equal{}\text{female})$.

(Any time what you're being asked for is subject to initial conditions like, "What's the probability that the second is a boy [b]given[/b] that the first is a girl?" think [u]conditional probability[/u]. That is you have some bit of prior knowledge; knowledge affects your assessment of how likely something is, i.e. its probability.)

\[ P(C_2 \equal{} \text{male}|C_1 \equal{}\text{female})\stackrel{\triangle} \equal{} \frac{P(C_1 \equal{}\text{female},C_2 \equal{}\text{male})}{P(C_1 \equal{}\text{female})}\]

According to Bayes Theorem, $ P(C_2 \equal{}\text{male}|C_1\equal{}\text{female}) \equal{}\frac{P(C_2\equal{}\text{male})\cdot P(C_1 \equal{}\text{female}|C_2 \equal{}\text{male})}{P(C_1 \equal{}\text{female})}$

$ P(C_2\equal{}\text{male}) \equal{} \frac{1}{3}$,  of the three possible outcomes involving females only one has a second born male.

$ P(C_1\equal{}\text{female}|C_2 \equal{}\text{male}) \equal{} 1$,  you were given that the first born was female, so the first born is female with probability 1; the condition that the second is male is superfluous.

Finally, the [i]a priori[/i] probability of a first born female is one-half, $ P(C_1\equal{}\text{female}) \equal{}\frac{1}{2}$.

Putting these all together with Bayes Theorem, $ P(C_2\equal{}\text{male}|C_1\equal{}\text{female}) \equal{}\frac{\frac{1}{3}\cdot (1)}{\frac{1}{2}} \equal{} \boxed{\frac{2}{3}}$.[/hide]

MF
FM
FF

But the older child CAN NOT be male.

edit: nvm, Was following t0ra's question.


If n=72, then what does 13(n-8)+n/9 equal?
 :play_ball:  :starwars:

We have
\begin{align*}
13(72-8) + \frac{72}{9} &= 13(64) + 8 \\
&= \boxed{840}.
\end{align*}
Should this be in the classroom math forum as it's only computation/order of operations? :maybe:


floor function
irrational number
number theory
number theory proposed

Let $ a,b$ be real nonzero numbers, such that number $ \lfloor an \plus{} b \rfloor$  is an even integer for every $ n \in \mathbb{N}$. Prove that $ a$ is an even integer.

[quote="Ahiles"][color=brown]Let $ a,b$ be real nonzero numbers, such that number $ [an + b]$  is an even integer for every $ n \in \mathbb{N}$. Prove that $ a$ is an even integer. [/color][/quote]
A very good problem .
To solve this problem ,we need a lemma .
[color=blue]Lemma [/color]
If $ a$ is a irrational number then $ \{na\}$ is dense in $ (0,1)$
Proof of this lemma quite hard ,it can be found in a number theory book .
Now consider problem .
$ x_n = an + b$
Suppose $ [a_n]\equiv 0 (\mod 2 ),\forall n\in N$ then $ [x_{n + 1}]\geq [x_n] + 2$
We only need to consider $ a\in [0,2)$
First step we will prove that a is a rational number.
Suppose a is a irrational number then
$ x_{n + 1} - x_n = a$
Exist an integer n_0 such that $ \{an_0 + b\} < 2 - a$
So $ [x_{n_0 + 1}] = [an_0 + b + a] = [x_{n_0}] + [\{an_0 + b\} + a] < [x_{n_0}] + 2$
It gives contradiction .
So a must be a rational number.
We also only to consider $ b\in [0,2)$
Let $ a = \frac {p}{q},\gcd (p,q) = 1$ and $ q > 1$

${ [x_n] = [\frac {np}{q}] + [\frac {np}{q}} + b]$

Take $ n = kq$ then easy to check that $ p\equiv 0 (\mod 2)$
[u]Case 1[/u] $ b\in [0,1)$
Exist an integer n such that $ np\equiv 1 (\mod q )$
Because $ p\equiv 1 (\mod 2)$ so $ [\frac {np}{q}]\equiv 1 (\mod 2)$
$ \Rightarrow [\frac {1}{q} + b] = 1,3$
Because $ 0 < \frac {1}{p} + b < 2$ so $ [\frac {1}{p} + b] = 1$
So $ 1 > b > 1 - \frac {1}{p}$
It follows that $ [b + \frac {np}{q}] = 1$
$ \Rightarrow [\frac {np}{q}]\equiv 0 (\mod 2 ),\forall n\in N$
It gives $ q = 1$
Case 2 :$ 1 < b < 2$
Consider $ b_1 = b - 1$ and
$ y_n = x_n - 1$ then $ [y_n]\equiv 1 (\mod 2)$
Prove as case 1 ,it gives $ q = 1$
So $ a = p = 2k$ and $ [b]\equiv 0(\mod 2)$
Problem claim  .

[quote="TTsphn"][color=blue]Lemma [/color]
If $ a$ is a irrational number then $ \{na\}$ is dense in $ (0,1)$
Proof of this lemma quite hard [...][/quote]
This is not hard! See http://www.mathlinks.ro/Forum/viewtopic.php?t=27475 (scroll down for grobbers proof, post #16).

The hard one is http://www.mathlinks.ro/viewtopic.php?t=172978 .

The lemma $ \{na\}$ is dense in $ (0,1)$ with $ a$ is a irrational is true.

I tihnk I have reached a solution.Hope it is OK. Let a=2k+r (k integer and   with r non negative and r  smaller than 2). Then if we substitue in the original question n=1 we have that [2k+r+b]=2m  (m is an integer)  [r+b]=2(m-k) =2t ....by doing the same thing we obtain that [nr+b]=2P  P is an integer.  Suppose that b=2h+q q  q can be in [0,1) or in [1,2)  if it is in [0,1) we say that it "adds 0 to the equation" . Then we have   [nr]=2P' then if n=1 we have [r]=0 so there will be an n such
that [nr]=1 if r>0  so we have that r=0 so a=2k as we wish to prove. Then if q is in [1,2) we say that "it adds 1 to the equation" so  [nr+1]=2P  then repating the procedure we have [r]=1 then we can se that there will be an n such that [nr+1]=2L+1 ( you can start doing things like these: the numbersin [1,1.5) whem we set n=2 will contradict the condition of the problems, if you continue doing that it is easily checked that they can't be an answer to the problem)
So we conclude that a=2k
Excellent problem!
I hope you understand the idea and I wish peple can improve the parts where are not completly proven
 :) Daniel[/code]

[quote="TTsphn"][quote="Ahiles"][color=brown]

${ [x_n] = [\frac {np}{q}] + [\frac {np}{q}} + b]$

Take $ n = kq$ then easy to check that $ p\equiv 0 (\mod 2)$
[u]Case 1[/u] $ b\in [0,1)$
Exist an integer n such that $ np\equiv 1 (\mod q )$
Because $ p\equiv 1 (\mod 2)$ so $ [\frac {np}{q}]\equiv 1 (\mod 2)$
$ \Rightarrow [\frac {1}{q} + b] = 1,3$
Because $ 0 < \frac {1}{p} + b < 2$ so $ [\frac {1}{p} + b] = 1$
So $ 1 > b > 1 - \frac {1}{p}$
It follows that $ [b + \frac {np}{q}] = 1$
$ \Rightarrow [\frac {np}{q}]\equiv 0 (\mod 2 ),\forall n\in N$
It gives $ q = 1$
[/quote][/quote]

why ?  show me please