geometry
circumcircle
trigonometry
geometric transformation
homothety
reflection
incenter

We restate the problem as follow

[b]Proposition.[/b] We are given a triangle $\triangle ABC$ $(AC>AB).$ Circle $\omega$ with center $X$ and radius $\rho$ is  tangent to the circumcircle $(O)$ and $\overline{BC}$ through $A,W,$ respectively. Perpendicular to $BC$ through $C$ cuts the external bisector of $ \angle BAC$ at $P.$ Then $B,X,P$ are colllinear.

Obviously, $ AW$ bisects $ \angle BAC$ and $X \in AO,$ hence $ \angle XAW = \frac {_1}{^2}(\angle B -\angle C)$

$\Longrightarrow \ \rho \equal{} \frac {AW}{2 \cos(\frac {\angle B-\angle C}{2})} \ (1)$

Since $APCW$ is cyclic $\Longrightarrow$ $ \angle CPW= \angle CAW = \frac {_1}{^2}\angle A$ $\Longrightarrow$ $PC= CW \cdot \cot\frac {A}{2}$

On the other hand, $ CW = AW \cdot \frac {\sin\frac {A}{2}}{\sin C} \ (2).$ Then, combining $(1)$ and $(2)$ yields

$\frac {PC}{\rho} = \frac {2\cos\frac {(\angle B -\angle C)}{2} \cdot \cos\frac {A}{2}}{\sin C}$

Combining this latter expression with the identities

$\frac{\cos \frac {\angle B - \angle C}{2}}{\sin \frac {A}{2}}= \frac {b + c}{a}$ and $\sin A = 2\sin \frac {A}{2} \cdot \cos \frac {A}{2}$ $\Longrightarrow \  \frac {PC}{\rho}= \frac {b \plus{} c}{c}.$

But $ \frac {b + c}{c} = \frac {CB}{WB} \Longrightarrow \ \frac {PC}{\rho} \equal{} \frac {CB}{WB}$ $\Longrightarrow P,X,B$ are collinear.

Let $ K$ be, the center of the circle $ (K),$ instead of $ \omega$ and let $ O$ be, the center of the circle $ (O),$ instead of $ \Gamma$ and we denote as $ X,$ the midpoint of the arc $ AB,$ not containing the point $ S.$

From $ KT\parallel OX$ and $ \frac {SK}{SO} \equal{} \frac {KT}{OX},$ we conclude that the points $ S,\ T,\ X,$ are collinear.

So, the line segment $ ST,$ is the angle bisector of the angle $ \angle S,$ of the triangle $ \bigtriangleup SAB.$

We denote the point $ P,$ as the intersection point of the line segments through vertices $ S,\ B$ and perpendicular to $ ST,\ AB$ respectively $ ($ $ SP$ is the external bisector of $ \angle S$ $ )$ and it is enough to prove that the line segment $ BP,$ passes through the point $ K.$

[b][size=100][color=DarkBlue]EQUIVALENT PROBLEM. – A triangle $ \bigtriangleup ABC$ is given with circumcircle $ (O)$ and let $ T,\ X$ be, the points of intersections of $ BC,\ (O)$ respectively, from the angle bisector of the angle $ \angle A.$ Through the vertex $ C,$ we draw the line perpendicular to $ BC,$ which intersects the external bisector of $ \angle A,$ at point so be it $ P$ and let be the point $ K\equiv OA\cap BP.$ Prove that $ KT\perp BC.$[/color][/size][/b]

[b][size=100]PROOF.[/size][/b] – [ [b][size=100]t=265876[/size][/b] ] - Through the vertex $ B,$ we draw the line perpendicular to $ BC,$ which intersects the line segment $ AP,$ as the external bisector of $ \angle A,$ at point so be it $ Q$ and let be the point $ A'\equiv (O)\cap AO.$

It is easy to show that the line segments $ BQ,\ BA',$ are isogonal conjugates with respect to the angle $ \angle B,$ from $ QB\perp BC$ and $ AB\perp BA'.$

Similarly, the line segments $ CP,\ CA',$ are isogonal conjugates with respect to the angle $ \angle C.$

Because of now, the external angle bisector of $ \angle A,$ is a particular case of two lines through the vertex $ A,$ isogonal conjugates with respect to the angle $ \angle A,$ based on the [b][size=100]Jacobi theorem [/size][/b] ( I like better to say  the [b][size=100]Isogonic theorem[/size][/b] ), we conclude that the line segments $ AA'\equiv AO,\ BP,\ CQ,$ are concurrent at one point so be it $ K$ and we will prove that $ KT\perp BC.$

$ \bullet$ From $ BQ\parallel CP$ $ \Longrightarrow$ $ \frac {KB}{KP} \equal{} \frac {BQ}{CP}$ $ ,(1)$

Because of the triangles $ \bigtriangleup AQB,\ \bigtriangleup APC$ have $ \angle BAQ \equal{} \angle CAP$ and $ \angle AQB \plus{} \angle APC \equal{} 180^{o},$ we conclude that $ \frac {BQ}{CP} \equal{} \frac {AB}{AC}$ $ ,(2)$

Because of $ AT$ is the angle bisector of $ \angle A,$ we have that $ \frac {AB}{AC} \equal{} \frac {TB}{TC}$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac {KB}{KP} \equal{} \frac {TB}{TC}$ $ ,(4)$

From $ (4)$ we conclude that $ KT\parallel PC$ $ \Longrightarrow$ $ KT\perp BC$ and the proof of the equivalent problem is completed.

[b][size=100]REMARK.[/size][/b] - It is easy to show that  the circle $ (K)$ centered at $ K$ with radius $ KA \equal{} KT$ ( easy to prove ), tangents internaly to $ (O)$ at point $ A$ and also tangents to $ BC,$ at point $ T$ and then, we have the configuration as the proposed problem states.

[b][size=100]REFERENCE.[/size][/b] - http://www.mathlinks.ro/Forum/viewtopic.php?t=154396

Kostas Vittas.

Let the tangents at $ S,T$ of $ \omega$ meet at $ Q$

For any point $ X$ denote by $ h_X$ the distance of the point $ X$ from line $ QO$

We have $ QT^2 \equal{} QA\cdot QB\implies \frac {QB}{QT} \equal{} \frac {QT}{QA}\implies \frac {QB}{QT} \equal{} \frac {QT}{QA} \equal{} \frac {QB \minus{} QT}{QT \minus{} QA} \equal{} \frac {TB}{TA}$

And since $ \frac {h_T}{h_A} \equal{} \frac {QT}{QA}$ we have $ \frac {h_T}{h_A} \equal{} \frac {TB}{TA}$

Since $ QO\perp TS$

$ PS\perp TS\iff PS\parallel{}QO\iff h_P \equal{} h_S\iff h_A\cdot \frac {PO}{AO} \equal{} h_T\iff$
$ \frac {PO}{AO} \equal{} \frac {h_T}{h_A}\iff \frac {PO}{AO} \equal{} \frac {TB}{TA}\iff\Delta PAB\sim \Delta OAT\iff PB\perp AB$

$ QED$

Dear Mathlinkers,
1. (0) the circle going through S, A and B
2. (1) the circle tangent to AB and internally to 0
2. X the midpoint of the arc AB, not containing the point S
3. Tx the tangent to 0 at X
4. According to Reim's theorem applied to (0) and (1), Tx // ATB
5. 2 the circle with diameter PT
6. According to a converse of Reim's theorem aplied to (0) and (2), X, T and the second intersection of 0 and 2à (other than B) are collinear.
7. We are done...
Sincerely
Jean-Louis

Dear Jean-Louis,

Can you please state the Reim's theorem?
(I searched in google and almost all the hits refer to ML-AOPS)

I know and can prove (without Reim's theorem) that $ X,T,S$ are collinear. But could not use it to solve this problem when I tried. So I want to understand your solution.

Thank you very much
Akashnil

Dear Mathlinkers,
for the Reim's theorem (general case and particular case), for the two converses, see
http://perso.orange.fr/jl.ayme   then:    à propos
Sincerely
Jean-Louis

I do not understand the language ...  :blush:  :maybe:  :(

[quote="vittasko"][color=darkred]Let $ ABC$ be a triangle with the circumcircle $ w = C(O,R)$  and let $ T$ , $ X$ be the intersections of $ BC$ , $ w$

respectively with the bisector of the angle $ \angle A$ . Through the vertex $ C$ we draw the perpendicular line to $ BC$

which intersects the external bisector of $ \angle A$ in the point $ P$ . Denote $ K\in OA\cap BP$ . Prove that $ KT\perp BC$ .[/color][/quote]

[color=darkblue][b][u]Proof[/u].[/b] Suppose w.l.o.g. $ b > c$ . Observe that $ \|\begin{array}{c} m(\angle KAB) = 90^{\circ} - C \\
 \\
m(\angle KAP) = 90^{\circ} - \frac {B - C}{2} \\
 \\
m(\angle APC) = 90^{\circ} - \frac {B - C}{2}\end{array}\|$ . Denote $ \|\begin{array}{c} D\in BC \\
\ AD\perp BC\end{array}\|$ . Therefore,

$ \frac {KB}{KP} = \frac {AB}{AP}\cdot\frac {\sin\widehat {KAB}}{\sin\widehat {KAP}} =$ ${ |\frac {c\cdot\sin (90^{\circ} - C)}{AP\cdot \cos\frac {B - C}{2}}}| =$ $ \frac {c\cdot |\cos C|}{DC} = \frac cb$ . In conclusion, $ \frac {KB}{KP} = \frac {TB}{TC}$ $ \implies$ $ KT\parallel PC$ $ \implies$ $ KT\perp BC$ .[/color]

[quote="vittasko"][color=darkred]Let $ ABC$ be a triangle with the circumcircle $ w \equal{} C(O,R)$  and let $ T$ , $ X$ be the intersections of $ BC$ , $ w$

respectively with the bisector of the angle $ \angle A$ . Through the vertex $ C$ we draw the perpendicular line to $ BC$

which intersects the external bisector of $ \angle A$ in the point $ P$ . Denote $ K\in OA\cap BP$ . Prove that $ KT\perp BC$ .[/color][/quote]

[color=darkblue][b][u]Proof (proiectively)[/u].[/b] Denote $ L\in BC\cap AP$ , the diameter $ [XY]$ ,  the line $ d$ for which $ K\in d$ , $ d\parallel PC$, i.e. $ d\perp BC$ and the points $ U\in AP\cap d$ , $ V\in PT\cap LK$ , $ W\in LK\cap PC$ , $ T_1\in d\cap AX$ , $ T_2\in d\cap BC$ . Observe that $ KU\equal{}KT_1$ because $ K\in AO$ - the $ A$-median in $ \triangle AXY$ and $ \overline {UKT_1}\parallel \overline {XOY}$ . Thus, the division $ \{B,C;L,T\}$ is harmonically $ \Longleftrightarrow$ the pencil $ P\{B,C;L,T\}$ is harmonically $ \Longleftrightarrow$ the divison $ \{K,W;L,V\}$ is harmonically. Therefore, $ KT_2\parallel WC$  $ \Longrightarrow$ $ KT_2\equal{}KU$ $ \Longrightarrow$ $ KT_1\equal{}KT_2$ $ \Longrightarrow$ $ T_1\equiv T_2\equiv T$ $ \Longrightarrow$ $ KT\perp BC$ .[/color]

Akashnil, Reim's theorem is stated here:[url]http://alt1.mathlinks.ro/viewtopic.php?t=35337[/url]

Thank you dgreenb801.
But I still can't understand the way it is used here. Can anyone help me?

This problem have many good geometry method，[url]http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=15&Id=15330[/url]
see one of then：

It is sufficient to show that the perpendicular to $ ST$ at $ S$, the perpendicular to $ AB$ at $ B$, and $ AO$ concur. We show this by trig Ceva on triangle $ SAB$:

Let the perpendicular to $ ST$ at $ S$ meet $ AO$ at $ P$, and the perpendicular to $ AB$ at $ B$ meet $ AO$ at $ P'$. Then it suffices to show that

\[ \frac{\sin{\angle BSP}}{\sin{\angle ASP}} \frac{\sin{\angle SAP}}{\sin{\angle BAP}} \frac{\sin{\angle ABP}}{\sin{\angle SBP}} = 1\]

Let $ \angle SAB = \alpha, \angle SBA = \beta$. We prove a lemma:

[b]Lemma.[/b] $ \angle TSA = \angle TSB$.

This is fairly well-known; extend $ ST$ to hit the outer circle at $ R$, and let $ AS$ intersect $ \omega$ at $ A'$. Then, $ \angle ATA' = \angle TSA'$ because $ AT$ is tangent to $ \omega$. However, by the homothety centered at $ S$ taking $ \omega$ to the outer circle, $ A'T  \parallel{} AR$. Hence, $ \angle ATA' = \angle TAR$, which equals $ \angle RSB = \angle TSB$ because they intercept the same arc. $ \blacksquare$

So, we may let $ \angle TSB = \angle TSA = \gamma$.

Now, $ \angle BSP = 90+\gamma, \angle aSP = 90-\gamma$ because $ SP \perp TS$. Hence, $ \displaystyle\frac{\sin{\angle BSP}}{\sin{\angle ASP}} = 1$. Also, $ \angle SAP = 90-\alpha$, $ \angle BAP = 90$.

It suffices to show that $ \displaystyle \frac{\sin{\angle ABP}}{\sin{\angle SBP}} = \frac{1}{\sin{90-\alpha}$. But this is evident from considering triangles $ BTO, BSO$: $ \sin \angle ABP = \frac{TO}{BO} = \frac{SO}{BO} = \frac{\sin{\angle SBP}}{\sin \angle OSB}$, but $ \angle OSP$ = $ 90-\angle OSM = 90-\angle SAB = 90-\alpha$, where $ M$ is an arbitrary point on the common tangent to the circles at $ S$ on the same side of $ OS$ as $ B$. So, we are done.

First, I will show that if $PB \perp PA$ implies that $SP \perp ST$, then $SP \perp ST$ implies $PB \perp PA$. If $SP \perp ST$, define $Q$ on $AO$ such that $P'B \perp BA$. By assumption, we must have $SQ \perp ST$. Hence, both $P$ and $Q$ lie on the intersection of $OS$ and the perpendicular to $ST$ through $S$, yielding $P = Q$. The result follows.

Consider the homothety centered at $S$ that maps $\omega$ to $\Gamma$. Let the images of $A$, $B$, $O_1$, $P_1$, and $T$ under this homothety be $A_1$, $B_1$, $O_1$, $P_1$, and $T_1$, respectively. The tangent to $\Gamma$ at $T_1$ is parallel to $AB$, so $T$ is the midpoint of arc $AB$, whence $\angle B_1 S T_1 = \angle T _1 S A_2$.

Let $P_1'$ be the intersection of the perpendicular to $ST_1$ through $S$ and the perpendicular to $A_1 B_1$ through $B_1$. It is sufficient to show that $P_1'$, $O_1$, and $A_1$ are collinear, for then both $P_1$ and $P_1'$ must lie on $AO$ and the perpendicular to $B_1 A_1$ through $B$, which implies they are equal.

Perform an inversion centered at $T_1$ with arbitrary radius; let the images of $P_1'$, $S$, $A_1$, $O_1$, and $B_1$ be $P_2$, $S_2$, $A_2$, $O_2$ and $B_2$, respectively. $\angle T_1 S_1 B_1 = \angle T_1 S_1 A_1$, yielding $\angle T_1 B_2 S_2 = \angle T_1 A_2 S_2$, whence $S_2 A_2 = S_2 B_2$. $O_1$ is sent to the reflection of $T_1$ across the line through $S_2$ parallel to $A_2 B_2$. $P_2$ lies on $B_2 S_2$ and $P_2 T_1 \perp B_2 S_2$ (since $T _1 B_1 P_1 S$ is cyclic, and $T_1 B_1 \perp P_1 B_1$.) We wish to show that $T_1 P_2 O_2 A_2$ is cyclic.

Let $X$ be the intersection of $B_2 S_2$ and the line through $A_2$ perpendicular to $A_2 B_2$. Note that $XO_2 || A_2 T_1$. It follows that $XP_2 T_1 A_2$ and $T_1 P_2 O_2 X$ are cyclic, so $O_2 P_2 T_1 A_2$ is cyclic, as desired.

my solution: consider $M$ and $M'$ are midpoint of arc $AB$ and $ASB$, respectively. as we know $S,T,M$ are colliner(a useful lemma!). i want to prove that if $P$ is a point on $AO$ such that $PS \bot ST$ then $ PB \bot BT$ .
 $PS\bot ST$ so $P,S,M'$ are colliner. so it's sufficient to prove that$ STBP$  is cyclic. for this i will prove that $\angle BSP=\angle BTP$ but this is equivalence to $TP \| AM'$. so i define $L$ as a line through $T$ parall to $AM'$. so i will  prove that $L,AO,SM'$ are concurrent. prove this by using ceva(sinus form) in triangle $STA$.

I have proved this problem on this site yet , but i don't remember where
Let TH is diameter of omega
Let point P such that angle PBT = TSP = 90
let's prove that P is on line AO
Let line AS intersect w at points J and S
Let line AS intersect (BTS) at points S and I
Well known that TS is inner angle bissector of BSA , so angle BSP = PSI , so BP = IP
After Reim's Theorem easy to see that TJ || BI
TO || BP and TO = JO , so A is Homotety center of triangles TJO and BIP . done

Let $O_1,r_1,r$ be the center of $\Gamma$, radius of $\Gamma$, and radius of $\omega$, respectively. Let $M=ST\cap \Gamma$; it's well-known that $M$ is the midpoint of arc $AB$ not containing $S$.

Let $Q$ be the intersection of tangent $BB$ and $(STB)$; then $QB\perp AB$ and $QS\perp TS$, and we want to show that $A,O,Q$ are collinear. Now $\angle{BQT}=\angle{BST}=\alpha$, so
\[ \frac{AT}{OT}\frac{AB}{QB} = \frac{AT}{r}\frac{TB\cot\alpha}{2r_1\sin{2\alpha}} = \frac{TS\cdot TM}{4rr_1\sin^2\alpha}. \]But $\triangle{MTA}\sim\triangle{MAS}$ by simple angle chasing, so $MT\cdot MS = MA^2 = 4r_1^2\sin^2\alpha$. We thus obtain
\[ \frac{AT}{OT}\frac{AB}{QB} = \frac{TS}{MS} \frac{MT\cdot MS}{4rr_1\sin^2\alpha} = \frac{OS}{O_1S} \frac{r_1}{r} = 1, \]whence $A,O,Q$ are collinear, as desired.

My solution:

Let another tangent of $ \omega $ through $ A $ cut $ \Gamma $ at $ C $ and tangent to $ \omega $ at $ R $ .

From [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=407366]incenter of triangle[/url] we get the midpoint $ H $ of $ TR $ lie on $ (TBS) $ ,
so $ AB \perp BP \Longleftrightarrow S, H, T, B, P $ are concyclic $ \Longleftrightarrow TS \perp SP $ .

Q.E.D

Dear Mathlinkers,
with harmonic division we have also a proof...
Sincerely
Jean-Louis

Take an inversion with center T and any circle. Then $\omega$ circle transforms into line $\ell$ and the $\Gamma$ circle tranforms into a circle tangent to line $\ell$.And the line $\ell$ bisects $OT$ (which is under inversion) .The line AO tranforms into the circle passing through points $A,O,T$(which are under inversion). The circle that passes through points $S,T,B$ tranforms into the line that passes through $S$ and $B$(which are under inversion) . Point $P$ under inversion transforms into the intersection between $SB$ and the circle that passes through points $A,O,T$.
Our condition is equvalent to showing $ TP\perp SB.$ Let line through $A,S$ intersect the circle that passes through points $A,O,T$ the second time at point K.Let $OP$ intersect $\ell$ at point $R$.Then since $ \ell\parallel AB$ and $ \triangle ABS$ is isosceles(which is obvious) we find that $ \angle OAK = \angle PAT=\angle TOP$.
$\angle SPO = \angle SRO-\angle RSB = 90^{\circ}-\angle TOP -\angle SAT= 90^{\circ}-\angle OAT=\angle AOT=\angle APT $ .And because $\angle ATO = \angle APO=90^{\circ}$, we find $\angle TPS=90^{\circ} $. So $ TP\perp SB$ and we are done.

$Z$ is the antipode of $T$ at $\omega$ and $SZ \cap AB \equiv X \Rightarrow (A,B,X,T)=-1$
$PS \perp TS \Longleftrightarrow  P \in SZ \Longleftrightarrow P(Z,T,A,B)=-1 \Longleftrightarrow PB \parallel ZT \Longleftrightarrow PB \perp AB$

Let $ST$ cut $\Gamma$ for a second time at $M$, and let the perpendicular to $AB$ through $B$ cut $AO$ at $Q.$ Let the tangent to $\omega$ at $S$ meet $AB$ at $Z$, and denote $O_1 \equiv OZ \cap QT.$ We will prove that $QS \perp TS.$

First, note that the homothety with center $S$ that takes $\omega \mapsto \Gamma$ also takes $T \mapsto M.$ It follows that this homothety takes $AB$ to the line tangent to $\Gamma$ at $M.$ Therefore, the tangent to $\Gamma$ at $M$ is parallel to $AB$, which implies that $M$ is the midpoint of arc $\widehat{AB}.$ Hence, $ST$ bisects $\angle ASB.$ Now, from the Menelaus Theorem for the transversal $ZOO_1$ w.r.t. $\triangle AQT$, we obtain \[1 = \frac{ZT}{ZA} \cdot \frac{OA}{OQ} \cdot \frac{O_1Q}{O_1T}.\] Observe that $\triangle AOT \sim \triangle AQB$ and $\triangle ZAS \sim \triangle ZSB.$ It follows from these two relations and the Angle Bisector Theorem that $OA : OQ = TA : TB = SA : SB = ZA : ZS.$ Because $ZS = ZT$ by equal tangents, we obtain $O_1Q = O_1T.$ Hence, $O_1$ is the midpoint of $\overline{QT}$ in right triangle $\triangle BQT.$ It follows that $O_1B = O_1Q = O_1T.$ Moreover, since $O_1$ lies on $OZ$, the perpendicular bisector of $\overline{ST}$, we have $O_1S = O_1T.$ Thus, $B, Q, T, S$ are concyclic, and $QT$ is a diameter of their circumcircle. Hence, $QS \perp TS$ as desired. $\square$

My solution :
Considering the equivalent problem :set intersection of $BP$ and circumcircle of $ABC$   $X$ , and set $\angle XAC=x$. Now we can calculate that angles $\angle KAO=(B-C)/2$$\angle OAP=90-(b-c)/2$$\angle AKn=(B-C)/2$
$\angle APB=x-(B-C)/2$$\angle BP K=90-x-a/2$$\angle nKP=a/2$( $n$ is perpendicular to $BC$ at $K$ ) which now implies that Cevas theorem hold for triangle $AKP$. So we are done.

My solution, take $P$ such that $\angle TSP=\angle PBA=90$. Then, we will prove that $A$, $O$ and $P$ are collinear,. Set $SP \cap AB=X$ and $SP \cap \omega=Y$. Considering homotethy centred at $S$ which brings $\omega$ to $\Gamma$ gives us that $ST$ bissects $\angle ASB$. Notice that $\angle XST$ is adjacent to $\angle TSP$, so $\angle XST=90$ meaning that $(X,T,A,B)$ are harmonic quadruple. Also, $\angle YST \equiv \angle TSP=90$. So $O$ is the midpoint of $YT$. we will prove that $PA$ bissects $YT$, which completes solution. Indeed, consider harmonic pencil $(PX,PT,PA,PB)$ intersecting line $YT$. $PB$ is perpendicular to $AB$ from our choice of $P$, $YT$ is perpendicular to $AB$ from is diametr of $\omega$ and $AB$ is tangent to $\omega$ at $T$, so $PB$ and $YT$ are parallel, $PX \cap YT=Y$, $PT \cap YT=T$, so $PA$ intersects $YT$ at the midpoint, which is $O$, $QED$.

[hide=Solution][quote=China TST 2009 Quiz 1 P1]Given that circle $ \omega$ is tangent internally to circle $ \Gamma$ at $ S.$  $ \omega$ touches the chord $ AB$ of $ \Gamma$ at $ T$. Let $ O$ be the center of $ \omega.$ Point $ P$ lies on the line $ AO.$ Show that $ PB\perp AB$ if and only if $ PS\perp TS.$[/quote]
[b][color=#000]Solution:[/color][/b] Invert around $S$ with arbitrary radius followed by Inversion at $T$ with radius $\sqrt{TA \cdot TB}$ and we get the following problem
[quote=Inverted Problem]Let $\Delta ABC$ be a triangle and $M$ be midpoint of arc $BC$ not containing $A$. Let $AM$ $\cap$ $BC$ $=$ $D$. Let $D'$ be the reflection of $D$ over tangent at $M$ to $\odot (ABC)$. If $\odot (DCD')$ $\cap$ $BM$ $=$ $X$. Prove, $\angle BXD=90^{\circ}$[/quote]
[b][color=#f00]Proof:[/color][/b] Let $E$ $\in$ $BM$ such $\angle BCE=90^{\circ}$. Thus, $DCED'$ is rectangle $\implies$ $\angle BXD$ $=$ $90^{\circ}$ $\qquad \blacksquare$[/hide]


Support calculus derivative invariant ratio Putnam Columbia

Ever since I was little, I had amazing language acquisition skills, even though I absolutely hated mathematics, and it was only until recently that I started feeling envy towards mathematicians. However, I have discovered that my lack of commitment to problem solving isn't a result of "laziness", but of prenatal brain development.

My mother has had an estrogen dominance problem when she was pregnant with me for during the entire course, including the pivotal brain developmental time-frame that occurs during the first trimester. It's also known among neurologists that testosterone exposure is key for developing the spatial ability required for mathematics (van Anders & Hampson 2005), and estrogen exposure tends to lead itself to more verbal skills and intuitive feelings. In essence, due to this estrogen exposure, I have only average spatial and mathematical ability, and above average verbal ability.

There is some good to this. Namely, I know my chances of developing Schizophrenia are nonexistent, and I'm also extremely fluid when it comes to social interaction and writing. However, I also have anxiety and panic attack problems (to the point where it's been causing insomnia for quite some time), in addition to being bisexual (which has caused me some grief growing up) and having an extremely bad time with sports of any kind.

However, as a result of this, my neurological wiring is not limited to not being adept at math. I also have the problem of not being adept at either music, art, or programming. I feel pathetic that my world is limited to just writing and perhaps language acquisition. The fact that I can't really become extremely good at art, music, nor math really upsets me.

But perhaps worst of all, I feel as I'm an evolutionary leftover. Society seems to be consistently favoring those gifted in spatial visualization, which I'm not extremely good at. As society moves more and more forward, exalting those skills, I feel more and more alienated.

I guess some people will write this off as attention-drawing or me being histrionic, but I feel that based on neurology, I'm extremely suffocated. Should I just spend my time being suffocated but working on my natural abilities, or should I spend my time trying to improve my weaker skills, knowing that I'll never really go far with it?

You've come to the conclusion that most other people come to -- that you're not the best in the world at mathematics. This shouldn't come as a [i]surprise[/i] -- how many people are the best in the world?

The problem, as I perceive it, is your apparent need to justify it. Apart from being a turn-off, it's also highly pessimistic and won't serve you well. It seems like you're taking one fact, applying it incorrectly, and using it to justify everything that you perceive to be wrong with your life (which isn't really wrong with your life).

[b]Conclusion:[/b] This is an issue of attitude/outlook on life, rather than one of neurology. My best advice is to [i]forget[/i] neurology, do your own, un-biased self-assessment of your abilities, decide which ones you want to invest in (an easier task when you're not biased by unnecessary thoughts of neurology), and then DO IT.

If it makes you feel any better, keep in mind that very few people are remembered for brilliance of mind. Most great people are remembered for what they [i]do[/i]. You are certainly intelligent and able to do something influential. And if you really think your social and communication skills are good, then that's excellent, because it's more likely that you can get the necessary support from others.

[b]Read on if you're not convinced:[/b] I'll take up a different tone for this part.
[hide][b]A.[/b] I'm skeptical about the premise of your post. You're probably pretty good at mathematics anyway. Probably better than most people. I noticed you have posts on implicit differentiation and stuff like that over the forum. Think about where that automatically places you on the bell curve. Now, take your age into consideration. This is why I think it's an attitude/outlook issue.

[b]B.[/b] You previously made a thread on how you wanted to be immortal so that you could achieve something "great" in mathematics. This supports the conclusion of [A].

[b]C.[/b] Inaccuracies in your post

- Estrogen dominance does not imply testosterone deficiency
- Estrogen and testosterone levels have been shown to be invariant of sexual orientation
- Who says estrogen significantly improves social interaction?

This supports that you're probably looking to justify the misconception addressed in [A], rather than trying to look objectively at the facts.[/hide]

I actually feel somewhat embarrassed for creating this topic, since I seemed to have done it out of impulse.

But yes, I think that you're correct. Most of my information was acquired out of a need to consistently justify my perceived failure.

[quote]
You've come to the conclusion that most other people come to -- that you're not the best in the world at mathematics. This shouldn't come as a surprise -- how many people are the best in the world?
[/quote]

Nobody wants to be dispensable, and this might sound extremely odd, but I did think that I was the best person in the world at not just mathematics, but basically everything (intellectual, that is). It's almost like I felt a need to constantly demonstrate in some way to the world that I am the best, which sounds extremely egotistical (and probably is), but nonetheless was how I felt. It then usually follows through the conclusion that I'm not exactly all that remarkable.

Usually, it gets much worse, sense I end up using pseudo-philosophy to find a justification for [i]why[/i] I must be the best at everything, and therefore end up digging myself into a deeper hole when I come to the realization that I'm not.

[quote]If it makes you feel any better, keep in mind that very few people are remembered for brilliance of mind. Most great people are remembered for what they do. You are certainly intelligent and able to do something influential. And if you really think your social and communication skills are good, then that's excellent, because it's more likely that you can get the necessary support from others.[/quote]

Probably, and arguably, that's what matters the most in the grand scheme of life. Otherwise, assuming that only brilliance of mind determines worth in society, then any genius sociopath would be determined as being a valuable contributer to humanity. :wink:

Although I think that the advice that you give is probably better; as it's better to look at it from the perspective of production and ability as opposed to a skill comparison.

[hide]
I think the general message behind your section here is correct, but nonetheless I want to correct what I found to be scientific errors. If there's anything that I put here that's incorrect, please tell me.

[quote]
- Estrogen dominance does not imply testosterone deficiency
- Estrogen and testosterone levels have been shown to be invariant of sexual orientation
- Who says estrogen significantly improves social interaction?
[/quote]

1. Probably correct here.
2. Not True, to quote:

[quote="Wikipedia paraphrasing of Rahman Q. The neurodevelopment of human sexual orientation. Neurosci Biobehav Rev. 2005;29(7):1057-1066."]
Independent studies indicate that homosexual women have masculinized (lower) [2D:4D digit] ratios and homosexual men show either hyper-masculinized or feminized ratios. These findings reinforce the prenatal androgen model - abnormal prenatal hormone exposure is related to the development of human homosexuality.
[/quote]

3. Androgen exposure (to be more specific, Testosterone) during the first Trimester has been shown to non-trivially increase the chances of Autism and Schizophrenia, both of which have prominent symptoms in the form of social withdraw and difficulty in communication. Likewise, Estrogen has been shown to increase symptoms in anxiety disorders and depression. (Bailey & Hurd, 2005b). Plus, the fact that there are reliable sex differences in communication ability (McGlone 1996; Halpern 1992), which leads me to think that estrogen influences communication to a large extent.

Also, the reason why the ration between the second and fourth digits is brought up so much is due to the fact that both are significantly influenced by exposure to androgens (McIntyre 2006).
[/hide]

How good are you in math? Can you solve IMO or Putnam problems?

You don't need to be embarrassed about anything. A few responses:

[hide][2] I realized now that you meant prenatal exposure to estrogen, rather than actual hormone levels. Oops! I was looking at [hide="this"]In 1984 Heino Meyer-Bahlburg, a neurobiologist at Columbia University, analyzed the results of twenty-seven studies undertaken to test the theory. According to Meyer-Bahlburg, a score of the studies in fact showed no difference between the testosterone or estrogen levels of homosexual and heterosexual men. Three studies did show that homosexuals had significantly lower levels of testosterone, but Meyer-Bahlburg believed that two of them were methodologically unsound and that the third was tainted by psychotropic drug use on the part of its subjects. Two studies actually reported higher levels of testosterone in homosexual men than in heterosexual men, and one unhelpfully showed the levels to be higher in bisexuals than in either heterosexuals or homosexuals.[/hide]

[3] I agree that estrogen makes it less likely to develop several conditions that reduce social ability. But if you look at how much of the population these people comprise, you'll see that this alone has very little overall effect on social interaction. It's not like half of men are autistic, or something. So I don't think it's accurate to conclude that prenatal estrogen exposure improved your social abilities.[/hide]


Of the following statements, the one that is incorrect is:

$ \textbf{(A)}\ \text{Doubling the base of a given rectangle doubles the area.}$
$ \textbf{(B)}\ \text{Doubling the altitude of a triangle doubles the area.}$
$ \textbf{(C)}\ \text{Doubling the radius of a given circle doubles the area.}$
$ \textbf{(D)}\ \text{Doubling the divisor of a fraction and dividing its numerator by 2 changes the quotient.}$
$ \textbf{(E)}\ \text{Doubling a given quantity may make it less than it originally was.}$

[hide]
For $ A$, we can let $ b$ be the base and $ h$ be the height. $ A=bh$, so $ (2b)h=2A$.
For $ B$, we can let $ b$ be the base and $ h$ be the altitude. $ A=\frac{1}{2}bh$, so $ b(2h)=2A$.
For $ C$, we can let $ r$ denote the radius. $ A=\pi r^2$, so $ \pi (2r)^2=\pi 4r^2%Error. "nequal" is a bad command.
2A$, so $ \boxed{C}$ is our answer.
[/hide]


geometry
3D geometry
algebra
factorization
sum of cubes
number theory unsolved
number theory

A natural number $ n$ is called good if and only if it can be represented as $ n =x^{3}+y^{3}+z^{3}$ where $ x,y,z\in Z_{+}$
For all $ i =0,1,2,3$ there exist infinitely many natural numbers $ n$ such that there excactly $ i$ good numbers among
$ {n,n+2,n+28}$

Please note that $ x,y,z\in Z_{+}$.I think your hint is not true

[quote="Nbach"]A natural number $ n$ is called good if and only if it can be represented as $ n =x^{3}+y^{3}+z^{3}$ where $ x,y,z\in Z_{+}$. [Prove] there exist infinitely many natural numbers $ n$ such that there are exactly $ 3$ good numbers among $ {n,n+2,n+28}$[/quote]
Actually, it's enough to prove there exists just [i]one[/i] $ 4$-tuple $ (a,b,c,d) \in \mathbb{N}^{\;\!4}$ such that $ a^{3}+b^{3}+c^{3}= d^{3}$. Then taking $ n = (\alpha d)^{3}$, for arbitrary $ \alpha \in \mathbb{N}$, yields $ n+2 = (\alpha d)^{3}+1^{3}+1^{3}$ and $ n+28 = (\alpha d)^{3}+3^{3}+1^{3}$. Now note $ 3^{3}+4^{3}+5^{3}= 6^{3}$. Here as in fashion, $ \mathbb{N}= \{1, 2, \ldots\}$.

[quote="Nbach"]A natural number $ n$ is called good if and only if it can be represented as $ n =x^{3}+y^{3}+z^{3}$ where $ x,y,z\in Z_{+}$. [Prove] there exist infinitely many natural numbers $ n$ such that there is exactly [i]one[/i] good numbers among $ {n,n+2,n+28}$[/quote]
Let $ \mathcal{R}= \{x^{3}\bmod 9: x \in \mathbb{Z}\}$ and $ \mathcal{S}= \{(x^{3}+y^{3}+z^{3}) \bmod 9: (x,y,z) \in \mathbb{Z}^{\;\!3}\}$, where obviously we think of both $ \mathcal{R}$ and $ \mathcal{S}$ as subsets of $ \mathbb{Z}/9\mathbb{Z}$. Then $ \mathcal{R}= \{0, \pm 1\}$ and consequently $ \mathcal{S}= \{0, \pm 1, \pm 2, \pm 3\}$. Then take $ n = 3a^{3}$, for arbitrary $ a \in \mathbb{N}$ such that $ a \equiv 1 \bmod 9$. Since $ n+2\equiv 5 \bmod 9$ and $ 5 \not\in \mathcal{S}$, thus clearly $ n \not\equiv x^{3}+y^{3}+z^{3}\bmod 9$, for any $ (x,y,z) \in \mathbb{Z}^{\;\!3}$. In a similar way, $ n+28 \equiv 4 \bmod 9$ and $ 4 \not\in \mathcal{S}$, so $ n \not\equiv x^{3}+y^{3}+z^{3}\bmod 9$, no matter what is $ (x,y,z) \in \mathbb{Z}^{\;\!3}$. Hence there is exactly [i]one[/i] element in the set $ \{n, n+2, n+28\}$ which is representable as a sum of cubes of positive integers (namely, $ n$). Finally recall that $ a \in \mathbb{N}$ is any arbitrary integer $ \equiv 1 \bmod 9$ and conclude.


I proved them by explaining the reasons with a lot of words, but I want some more mathematical proofs.

(1) Prove that if $ A$ is lower triangular and $ B_{ij}$ is the matrix that results when the $ i$-th row and $ j$-th column of $ A$ are deleted, then $ B_{ij}$ is lower triangular if $ i > j$.

(2) Prove that if $ A$ is an invertible lower triangular matrix, then $ A^{ \minus{} 1}$ is lower triangular.

(2) since $ A$ is lower triangular, we may reduce $ A$ to the identity matrix by multiplying $ A$ by the product of some lower triangular elementary matrices. $ A^{\minus{}1}$ is the product of $ I$ with these same elementary matrices, and since the product of lower triangular matrices is lower triangular, $ A^{\minus{}1}$ is then lower triangular.


Find all numbers "n" such that the equation $ x^{n}\plus{}y^{n}\equal{}z^{n}$ doesn't have any solutions.

I don't know if this is a joke, but I will answer your question anyway.

This equation and the statment related to it is called "Fermat's Last Theorem".
Fermat stated that he had proved that for any n>2 there are no integer solutions to this equation. However, he never posted his own proof, so it remained an open question. For over 300 years mahematicians from all around the world tried to solve it, but no one succeded. Finally in September of the year 1994, Andrew Wiles presented a proof to this theorem. It was a very long, and complicated proof, but it was correct.

So the answer to your question is: for every n>2 there are no integer solutions to this equation. For n=1,2 we have an infinite number of integer solutions.

You never said integer solutions, in that case, I think n=0 is the only such number.

I'm not sure what you mean.  There are clearly no solutions for $ n \equal{} 0$ (we have $ 1 \plus{} 1 \equal{} 1$), any $ x, y$ determines $ z$ when $ n \equal{} 1$, the Pythagorean triples are $ n \equal{} 2$, and elementary proofs exist that there are no solutions for $ n \equal{} 3, 4$ if Wiles' proof doesn't convince you.

[b]dgreenb801[/b] is right! I said [color=red]any[/color] solutions. The values for "n" must be only      n=0.


function
combinatorics solved
combinatorics

Let n \in N , n  \geq 3 ,Z={0,1,2,..,n-1}
A_n={(a,b,c)/a,b,c \in Z ,a<b<c ,a+b+c is divisible by n}
B_n={(a,b,c)/a,b,c \in Z , a \leq b \leq c,a+b+c is divisible by n}
Prove that : |A_(n+3)|=|B_n| , |B_n|=|A_n|+n

First note that if (a,b,c) is in B_n then 0  \leq  a+b+c  \leq  3(n-1) < 3n. Then (a,b,c) = (0,0,0) or a+b+c = n or a+b+c = 2n.
Thus the sets B'_n = {(a,b,c) in B_n / a+b+c = n or (a,b,c) = (0,0,0)} and B''_n = {(a,b,c) in B_n / a+b+c = 2n} form a partition of B_n.

In a similar way, if (a,b,c) is in A_(n+3) then 3  \leq  a+b+c  \leq (n+2)+(n+1)+n = 3n+3 < 3(n+3). And since n  \geq 3, we have 3n+3  \geq  2(n+3). Thus a+b+c = n+3 or 2(n+3).
It follows that the sets A'_(n+3) = {(a,b,c) in A_(n+3) / a+b+c = n+3} and A''_(n+3) = {(a,b,c) in A_(n+3) / a+b+c = 2n+6} form a partition of A_(n+3).

Let f be the function defined on B_n by :
f(0,0,0) = (0,1,n+2)
f(a,b,c) = (a,b+1,c+2) if (a,b,c) is in B'_n-{(0,0,0)}
f(a,b,c) = (a+1,b+2,c+3) if (a,b,c) is in B''_n.
We prove that f is a bijection from B'_n onto A'_(n+3) and from B''_n onto A''_(n+3), which will prove that |B_n| = |A_(n+3)|.

a) If (a,b,c) is in B'_n :
f(0,0,0) = (0,1,n+2) is in A'_(n+3)
If (a,b,c) =/= (0,0,0) then a \leq b \leq c leads to a < b+1 < c+2 and a+(b+1)+(c+2) = n+3. Moreover c  \leq  n-1 then c+2 \leq  n+2. It follows that f(a,b,c) is in A'_(n+3) and that f(a,b,c) =/= f(0,0,0).
Clearly f is injective from B'_n.

Now let (x,y,z) in A'_(n+3).
If (x,y,z) = (0,1,n+2) then (x,y,z) = f(0,0,0). Thus we assume that (x,y,z) =/= (0,1,n+2). Thus x+y \geq  2, from which we deduce that z  \leq  n+1.
Thus a = x, b = y-1 and c = z-2 are positive integers, such that a \leq b \leq c \leq n-1. And a+b+c = n. Thus (a,b,c) is in B'_n, and clearly f(a,b,c) = (x,y,z).
Then f is surjective from B'_n onto A'_(n+3).
Thus f is a bijection from B'_n onto A'_(n+3).

b) If (a,b,c) is in B''_n :
From c  \leq  n-1, we deduce that c+3  \leq  n+2, then 0  \leq  a+1 < b+2 < c+3  \leq  n+2, and (a+1)+(b+2)+(c+3) = 2n+6, thus f(a,b,c) is in A''_(n+3).
Clearly f is injective on B''_n.

Now let (x,y,z) in A''_(n+3) :
Then 2n+6 = x+y+z  \leq  x + n+2 + n+1 that gives x  \geq  3.
Moreover, as above we have z \leq n+2 then z-3 \leq n-1.
Thus a = x-1, b = y-2 and c = z-3 are integers such that 0 \leq a \leq b \leq c \leq  n-1, and a+b+c = 2n. Thus (a,b,c) is in B''_n and f(a,b,c) = (x,y,z).
Thus f is surjective from B''_n onto A''_(n+3), from which that f is bijective from B''_n onto A''_(n+3).
And we are done.

Now, we prove that |B_n| = |A_n| + n :
Clearly A_n is a subset of B_n. Thus it suffices to prove that C_n = B_n\A_n has exactly n elements.
But C_n = {(a,b,c) in B_n / a=b=c or a=b < c or a < b =c}.
- Note that (0,0,0) is in C_n
- (a,b,c) is in C_n \cap B'_n iff (a,b,c) = (x,x,y) or (y,x,x) where x,y are two non-negative integers such that 2x+y = n (note that we may have x=y) and x,y \leq  n-1.
If x= 0 then y = n thus no solution in that case.
For each x = 1,...,[n/2] (where [.] denotes the integer part), we have exactly one value of y satisfying the desired conditions. And for x  \geq  [n/2]+1 the value of y is negative so as to be rejected.
Thus there are exactly [n/2] elements in C_n \cap B'_n not counting (0,0,0).

- (a,b,c) is in C_n \cap B''_n iff (a,b,c) = (x,x,y) or (y,x,x) where x,y are two non-negative integers such that 2x+y = 2n (as above, we may have x=y) and x,y \leq  n-1.
This time, we easily prove as above that each of the values of x = [n/2]+1,...,n-1 gives exactly one value of y and no other value of x gives a solution.
Thus there are exactly n - 1 - [n/2] elements in C_n \cap B''_n.

Since B'_n and B''_n form a partition of B_n, we deduce that |C_n| = 1 + [n/2] + n - 1 - [n/2] = n, and we are done.

Pierre.

Congrats, you just made the longest post on the site :D

I hope for the students of those old books there is a shorter way :?

[quote="pbornsztein"]Well, I think I can do longer  :D

Pierre.[/quote]

Lol, you got a lifetime to type those things? :D
It's almost like longer than IMO answers :P


function
real analysis
integration
limit
real analysis unsolved

Let $ E$ be a measurable subset of $ \mathbb{R}$ with $ |E|\equal{}1$. Let $ f: \mathbb{R} \to \mathbb{R}$ be given by

\[ f(x)\equal{}|E\plus{}x \cap  E|\]
where $ E\plus{}x\equal{}\{e\plus{}x, e \in E\}$.

I've been trying to prove that $ f$ is continuos.

\[ |E\plus{}x \cap E| \minus{} |E\plus{}x_0 \cap E| \leq |E\plus{}x \cap E| \minus{} |E\plus{}x_0 \cap E \cap E\plus{}x| \equal{} |E\plus{}x \cap E \cap (E\plus{}x_0)^c|\]

It think that the last member of the equality may go to 0 when $ x \to x_0$ but couldn't manage to prove it... I've tried integrating characteristic funcions... Can someone give me any hint?

Let $ x_n \to x_0$.

\[ |E\plus{}x_n \cap E \cap (E\plus{}x_0)^c| \equal{} \int \chi_{E\plus{}x_n}(t) \chi_E(t) \chi_{E\plus{}x_0}(t) dt\]
Using that $ \chi_{E\plus{}x_n}(t) \equal{} \chi_E (t\minus{}x_n)$ and $ \chi_{(E\plus{}x_0)^c}(t) \equal{} \chi_{E^c}(t\minus{}x_0)$ we have

\[ |E\plus{}x_n \cap E \cap (E\plus{}x_0)^c| \equal{} \int_E \chi_E (t\minus{}x_n) \chi_{E^c}(t\minus{}x_0) dt\]
I'd like to prove that $ \chi_E (t\minus{}x_n) \chi_{E^c}(t\minus{}x_0)$ goes to 0 when $ n\to\infty$ and then use dominated convergence, but as the characteristic function is not continuous I dont see the way to do it... Maybe I'm missing some important detail...  :maybe:

Use that if $ f \in L^{p}(\mathbb{R})$, then $ \lim_{x \to x_0}\int_{\mathbb{R}} |f(t \plus{} x) \minus{} f(t \plus{} x_0)|^p \, dt \equal{} 0$. It's obviously true for compactly supported continuous functions, and $ f$ can be approximated in norm by such functions.


Cred ca pentru obisnuitii forumului e floare la ureche:

Sunt date a,b,c >0 sa se demonstreza ca:

a/b + b/c + c/a >3

Rog cine stie demonstratia sa o posteze, am nevoie si nu-i dau de capat...

Salut !

Aplici inegalitatea mediilor pentru $a/b, b/c, c/a$ si gata ! :roll: